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BayArea/md/en-handouts/en-BAMO-original.md

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+# 1st Bay Area Mathematical Olympiad 
+
+February 23, 1999
+
+## Problems and Official Solutions
+
+The time limit for this exam is 4 hours. Your solutions should be clearly written arguments. Merely stating an answer without any justification will receive little credit. Conversely, a good argument which has a few minor errors may receive much credit.
+
+The five problems below are arranged in roughly increasing order of difficulty. In particular, problems 4 and 5 are quite difficult. We don't expect many students to solve all the problems; indeed, solving just one problem completely is a fine achievement. We do hope, however, that you find the experience of thinking deeply about mathematics for 4 hours to be a fun and rewarding challenge. We hope that you find BAMO interesting, and that you continue to think about the problems after the exam is over.
+
+1. Prove that among any 12 consecutive positive integers there is at least one which is smaller than the sum of its proper divisors. (The proper divisors of a positive integer $n$ are all positive integers other than 1 and $n$ which divide $n$. For example, the proper divisors of 14 are 2 and 7.) ${ }^{1}$
+
+Solution. One of the twelve numbers must be a multiple of twelve; call it $a=12 n$. Among the proper divisors of $a$ are the integers $2 n, 3 n, 4 n, 6 n$. These sum to $15 n>a$.
+
+2. Let $C$ be a circle in the $x y$-plane with center on the $y$-axis and passing through $A=(0, a)$ and $B=(0, b)$ with $0<a<b$. Let $P$ be any other point on the circle, let $Q$ be the intersection of the line through $P$ and $A$ with the $x$-axis, and let $O=(0,0)$. Prove that $\angle B Q P=\angle B O P$.
+
+Solution. We make use of the fact that an angle inscribed in a circle has measure equal to one-half of the arc subtended. Since the $x$ - and $y$-axes meet in a right angle, the circle $C_{1}$ through $B, O$, and $Q$ has $Q B$ as a diameter. Also, $\angle A P B$ is a right angle, since $A B$ is the diameter of $C$. But this means that $\angle Q P B$ and $\angle Q O B$ are both right angles, so that $P, O, Q, B$ all lie on circle $C_{1}$. Thus the two angles in question, $\angle B Q P$ and $\angle B O P$, are inscribed in $C_{1}$, subtend the same arc, and are therefore equal.
+
+3. A lock has 16 keys arranged in a $4 \times 4$ array, each key oriented either horizontally or vertically. In order to open it, all the keys must be vertically oriented. When a key is switched to another position, all the other keys in the same row and column automatically switch their positions too (see diagram). Show that no matter what the starting positions are, it is always possible to open this lock. (Only one key at a time can be switched.)[^0]
+
+Solution. The problem is solved if there is a way to change the orientation of any specified single key, without changing any of the others. This is equivalent to finding a way to switch the chosen key an odd number of times, while switching all other keys a even number of times.
+
+This can be done by switching all keys on the same row and column of the chosen key (including the chosen key). To see why, choose a key $K$. If we switch it and all of its "sisters" that share the same row and column, $K$ will be switched 7 times. Now, examine the other 15 keys in the lock. There are two cases: either the key is a sister of $K$, or not. Suppose that $L$ is a sister of $K$, say, sharing a row with $K$. Then $L$ will be switched 4 times. For the other case, suppose $M$ is not a sister of $K$. Then $M$ will be switched twice,
+
+because among the 6 sisters of $K$ which are turned, exactly two of them share a row or column with $M$. Consequently, of all the keys in the lock, only $K$ is switched an odd number of times. All other keys are switched either 2 or 4 times, leaving their orientation unchanged. Thus we will be able to open the lock by selecting each horizontal key one-by-one, and turning it and all of its sister keys.
+
+4. Finitely many cards are placed in two stacks, with more cards in the left stack than the right. Each card has one or more distinct names written on it, although different cards may share some names. For each name, we define a "shuffle" by moving every card that has this name written on it to the opposite stack. Prove that it is always possible to end up with more cards in the right stack by picking several distinct names, and doing in turn the shuffle corresponding to each name. $^{2}$
+
+Solution. Let the number of cards be $c$ and let the number of distinct names be $n$. Each card contains a set of names; denote these sets by $S_{1}, S_{2}, \ldots, S_{c}$ (some of these sets may share elements). Now let $E$ be a subset of the set of $n$ names, and denote by $D(E)$ the difference of the number of cards in the left stack and the number of cards in the right stack, after the cards are shuffled by the names in $E$. We are given that $D(\emptyset)>0$; we must show that there exists a subset $E$ such that $D(E)<0$.
+
+Consider the sum of all $D(E)$ as $E$ ranges through all $2^{n}$ possible subsets of names. If we can show that this sum is equal to zero, we will be done, since $D(\emptyset)>0$ is just one term in this sum, forcing at least one other term to be negative. For $i=1,2, \ldots, c$, let $v_{i}=+1$ or -1 if the $i$ th card is initially in the left or right stack, respectively. Then
+
+$$
+D(\emptyset)=v_{1}+v_{2}+\cdots+v_{c}=\sum_{i=1}^{c} v_{i}
+$$
+
+Now consider what happens when we shuffle the cards corresponding to a subset $E$ of names. The $i$ th card will move back and forth from one stack to the other a total of $\left|E \cap S_{i}\right|$ times $(|A|$ means the number of elements in the set $A$ ). The only thing that matters is whether this value is even or odd. Thus we have
+
+$$
+D(E)=\sum_{i=1}^{c}(-1)^{\left|E \cap S_{i}\right|} v_{i}
+$$[^1]
+
+It remains to sum this expression over all subsets $E$. Let us examine what happens just for one card; i.e., let us compute
+
+$$
+\sum_{E}(-1)^{\left|E \cap S_{i}\right|} v_{i}
+$$
+
+for a fixed $i$ as $E$ ranges over all $2^{n}$ subsets of names. If we can show that this equals zero, then the entire sum will equal zero and we are done. Let $\left|S_{i}\right|=k$. Then $\left|E \cap S_{i}\right|$ will range from 0 to $k$ inclusive. For $0<r<k$, how many subsets $E$ are there such that $\left|E \cap S_{i}\right|=r$ ? There are $\left(\begin{array}{l}k \\ r\end{array}\right)$ subsets of $S_{i}$ with $r$ elements. Fix one of them, call it $T$. Then $E$ must contain all the elements of $T$, plus any subset of the names that are not contained in $S_{i}$. In other words, there are $2^{n-k}$ subsets containing $T$, and thus there are $\left(\begin{array}{l}k \\ r\end{array}\right) 2^{n-k}$ subsets $E$ altogether satisfying $\left|E \cap S_{i}\right|=r$. Therefore
+
+$$
+\sum_{E}(-1)^{\left|E \cap S_{i}\right|} v_{i}=v_{i} \sum_{r=0}^{k}(-1)^{r}\left(\begin{array}{l}
+k \\
+r
+\end{array}\right) 2^{n-k}=2^{n-k} v_{i} \sum_{r=0}^{k}(-1)^{r}\left(\begin{array}{l}
+k \\
+r
+\end{array}\right)
+$$
+
+By the binomial theorem, we have
+
+$$
+2^{n-k} v_{i} \sum_{r=0}^{k}(-1)^{r}\left(\begin{array}{l}
+k \\
+r
+\end{array}\right)=2^{n-k} v_{i}(1-1)^{k}=0
+$$
+
+5. Let $A B C D$ be a cyclic quadrilateral (a quadrilateral which can be inscribed in a circle). Let $E$ and $F$ be variable points on the sides $A B$ and $C D$, respectively, such that $A E / E B=C F / F D$. Let $P$ be the point on the segment $E F$ such that $P E / P F=A B / C D$. Prove that the ratio between the areas of triangle $A P D$ and $B P C$ does not depend on the choice of $E$ and $F .^{3}$
+
+Solution. There are two cases to consider. First, assume that the lines $A D$ and $B C$ are not parallel and meet at $S$. Since $A B C D$ is cyclic, $\triangle A S B$ and $\triangle C S D$ are similar. Since $A E / A B=$ $C F / C D$, then $A E / C F=A B / C D=A S / C S$, and $\triangle A S E$ and $\triangle C S F$ are also similar $(\angle S A E=$ $\angle S C D)$. Therefore, $\angle D S E=\angle C S F$.
+
+By similarity, we have
+
+$$
+\frac{S E}{S F}=\frac{S A}{S C}=\frac{A B}{C D}=\frac{P E}{P F},
+$$
+
+which means that $S P$ is the bisector of angle $S$ in $\triangle F S E$. This implies that $\angle E S P=\angle F S P$ and hence $\angle A S P=\angle B S P$, so $S P$ is also the bisector of angle $S$ in $\triangle A S B$. This means that $P$ is equidistant from the lines $A D$ and $B C$. Thus
+
+$$
+[A P D] /[B P C]=A D / B C,
+$$
+
+which is a constant (we use the notation $[A B C]$ for the area of $\triangle A B C$ ).[^2]
+
+For the second case, assume that $A D$ and $B C$ are parallel. Then $A B C D$ is an isosceles trapezoid with $A B=C D$, and we have $B E=D F$. Let $M$ and $N$ be the midpoints of $A B$ and $C D$, respectively. Then $M E=N F$ and $E$ and $F$ are equidistant from the line $M N$. Thus $P$, the midpoint of $E F$, lies on $M N$. This implies that $P$ is equidistant from $A D$ and $B C$, and hence
+
+$$
+[A P D] /[B P C]=A D / B C .
+$$
+
+## Brilliancy Award Solution to Problem 4
+
+by Oaz Nir, Monta Vista High School, grade 10
+
+We will prove the statement by induction on $n$, the number of distinct names present. Call the left stack $L$ and the right stack $R$. For the case $n=1$, let $L$ have $x$ cards and $R$ have $y$ cards with $x>y$; all cards have the same single name written on them. We perform the shuffle corresponding to the single name present, and we are done (now we have $x$ in $R$ and $y$ in $L$ ).
+
+Next, assume that we have proven the statement for $n$ names, and consider the case with $n+1$ names. Call the first $n$ names $a_{1}, a_{2} \ldots, a_{n}$, and let the new $(n+1)$-st name be $a$. There are two cases:
+
+Case 1. The number of $L$ cards containing only name $a$ is less than or equal to the number of $R$ cards containing only name $a$. In this case, we need only look at the remaining cards (the ones that do not contain only $a$.) We can ignore the name $a$, and use the inductive hypothesis to perform the required shuffles using some subset of the $n$ names $a_{1}, a_{2} \ldots, a_{n}$, and we will be done: there are now more of these remaining cards in stack $R$ than in stack $L$, and since there were at least as many "only $a$ " cards in $R$ than in $L$, the final configuration has more cards in $R$ than in $L$. (Note that if the original configuration contains as many "remaining" cards in $R$ as in $L$, we don't have to perform any shuffles at all, given that the "only $a$ " cards in $R$ are more than those in $L$.)
+
+Case 2: The number of $L$ cards containing only name $a$ is greater the number of $R$ cards containing only $a$. In this case, we first perform the shuffle corresponding to the name $a$. This will switch the "only $a$ " cards, so that there are more of them in $R$ than in $L$. Unfortunately, some of the remaining cards may also have moved around. But now there are two possibilities: if there are at least as many remaining cards in $R$ than in $L$, we are done. If not, then we can again employ the inductive hypothesis, ignoring the name $a$, and just applying the shuffles to some subset of the $n$ names $a_{1}, a_{2} \ldots, a_{n}$. The end result will be that there will be as many remaining cards in $R$ than in $L$, and we are done.
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_ef0e15aebc386b0201feg-4.jpg?height=414&width=1021&top_left_y=1887&top_left_x=536)
+
+You may freely disseminate this exam, but please do attribute its source (Bay Area Mathematical Olympiad, 1999, created by the BAMO organizing committee, bamo@msri.org).
+
+
+[^0]:    ${ }^{1}$ Adapted from Moscow Math Olympiad, 1972
+
+[^1]:    ${ }^{2}$ Adapted from USSR Math Olympiad, 1968
+
+[^2]:    ${ }^{3}$ Shortlisted for the International Math Olympiad, 1998.
+

BayArea/md/en-handouts/en-sol-sample99.md

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+# BAMO 1999: SOLUTIONS TO PRACTICE PROBLEMS 
+
+BAMO PROBLEMS COMMITTEE
+
+Problem 1. Let $A$ and $B$ be two different hospitals that treat exactly the same number of patients during a year. Each patient suffers from one of two diseases, $X$ or $Y$. Hospital $A$ cures a greater percentage of its patients than hospital $B$. Is it possible that hospital $B$ cures both a greater percentage of $X$-patients than $A$, and a greater percentage of $Y$-patients than $A$ ?
+
+Solution. This is the well-known Simpson's Paradox: just make $B$ specialize in a riskier disease. For example, let $B$ treat 90 cancer patients and 10 acne patients, with respective cure rates of $50 \%$ and $100 \%$. Let $A$ treat 10 cancer and 90 acne patients, with cure rates of $0 \%$ and $70 \%$, respectively.
+
+Problem 2. Bildert works in a cubicle in an office which consists of 27 cubicles arranged in a $3 \times 3 \times 3$ cube. Any two cubicles sharing a wall have a connecting door on this wall; for example, the corner cubicles have exactly 3 doors, while the center cubicle has 6 doors: one on each wall, one on the floor, and one on the ceiling. If Bildert starts at the central cubicle, can he visit each of the other 26 cubicles exactly once (i.e. without revisiting any cubicles)?
+
+Solution I. The answer is no. Denote the central cubicle by $C$, and denote the vertex, edge and face cubicles by $V, E$ and $F$, respectively. The trip must start with $C$ and include every one of the $8 V$ 's, $6 F$ 's, and $12 E$ 's. The sequence must begin with $C F E$. Each cubicle $V$ is adjacent only to $E$ cubicles, and each $F$ cubicle except for the very first one, is adjacent only to $E$ cubicles. This means that for the remaining $13 \mathrm{~V}$ 's and $F$ 's that follow the initial $C F E$, at least 12 new $E$ 's are needed. This means that we need at least $13 E$ 's, impossible.
+
+Solution II. Divide the cubicles into two subsets depending on the parity of the sums of coordinates of each cubicle. (Thus, in the above notation, one subset consists of cubicles $C$ and $E$ 's, and the other subset consists of $F^{\prime}$ 's and $V^{\prime}$ 's.) Each move alternates between the two subsets. The starting subset has 13 cubicles, while the other subset has 14 cubicles - obviously we cannot keep alternating, because we'll run short of cubicles in the starting subset. Hence, Bildert cannot visit all cubicles without repetitions.
+
+Problem 3. Let $a, b, c, d, e, f$ be positive integers, each at least 2 , whose sum is $S$. Prove that
+
+$$
+a(a-1)+b(b-1)+c(c-1)+d(d-1)+e(e-1)+f(f-1) \leq(S-10)(S-11)+10 .
+$$
+
+When is equality achieved?
+
+Solution I. Adding $-3 S+24$ to both sides, makes the inequality equivalent to
+
+$$
+(a-2)^{2}+(b-2)^{2}+(c-2)^{2}+(d-2)^{2}+(e-2)^{2}+(f-2)^{2} \leq(S-12)^{2} .
+$$
+
+Substituting $A=a-2, B=b-2$, etc., this is the same as
+
+$$
+A^{2}+B^{2}+C^{2}+D^{2}+E^{2}+F_{1}^{2} \leq(A+B+C+D+E+F)^{2}
+$$
+
+On the left side we have the sum of squares of nonnegative numbers and on the right side we have the square of the sum. The latter is always larger except when all pairwise products $A B, A C, D F \ldots$ are zeros. This happens when all but one of $A, B, C, D, E, F$ are zero, correspondingly, when all but one of $a, b, c, d, e, f$ are 2 .
+
+Solution II. The given inequality can be written as
+
+$$
+a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2} \leq(S-10)^{2}+20 .
+$$
+
+Now, for any two numbers $x, y \geq z$, we have
+
+$$
+x^{2}+y^{2} \leq z^{2}+(x+y-z)^{2} .
+$$
+
+Indeed, this is equivalent to
+
+$$
+x^{2}-z^{2} \leq(x+y-z)^{2}-y^{2} \Leftrightarrow(x-z)(x+z) \leq(x-z)(x+2 y-z) \Leftrightarrow 0 \leq 2(x-z)(y-z),
+$$
+
+and the last is true because $x \geq z$ and $y \geq z$. Note that (1) replaces the numbers $(x, y)$ by $(z, x+y-z)$ without changing the sum of the two numbers, but increases the sum of their squares. In the original problem, we do this for $a, b \geq 2$ : we replace $(a, b)$ by $(2, a+b-2)$ :
+
+$$
+a^{2}+b^{2} \leq 2^{2}+(a+b-2)^{2} .
+$$
+
+We then do the same for $(a+b-2, c)$ : replace them by $(2, a+b+c-4)$, and so on. In the end, we will have replaced five of the original numbers by 2's, and the last by $a+b+c+d+e+f-10=S-10$.
+
+$$
+\Rightarrow a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2} \leq 2^{2}+2^{2}+2^{2}+2^{2}+2^{2}+(S-10)^{2}=20+(S-10)^{2} .
+$$
+
+Equality is achieved if and only if there are equalities each time we apply (1), i.e. five of the given numbers are 2's, and the remaining number is therefore $S-10$.
+
+Solution III. We first show the following inequality:
+
+$$
+x(x-1)+y(y-1) \leq(x+y-2)(x+y-3)+2, x, y \geq 2
+$$
+
+Consider two complete graphs, with $x$ and $y$ vertices, respectively. (A complete graph has all possible edges drawn.) Thus we have $\left(\begin{array}{l}x \\ 2\end{array}\right)$ and $\left(\begin{array}{l}y \\ 2\end{array}\right)$ edges in the two graphs. If we glue the graphs together on an edge, we produce a new graph with $x+y-2$ vertices. Count edges: the original configuration had $\left(\begin{array}{l}x \\ 2\end{array}\right)+\left(\begin{array}{l}y \\ 2\end{array}\right)$ edges, while the new configuration has at most $\left(\begin{array}{c}x+y-2 \\ 2\end{array}\right)$ edges. Since we lost an edge when we glued the two graphs together, we conclude that
+
+$$
+\left(\begin{array}{l}
+x \\
+2
+\end{array}\right)+\left(\begin{array}{l}
+y \\
+2
+\end{array}\right) \leq\left(\begin{array}{c}
+x+y-2 \\
+2
+\end{array}\right)+1
+$$
+
+This is equivalent to (2) after multiplying by 2 . Equality is attained if and only if the new graph is also complete, i.e. one of the original graphs must have been just an edge $(x=2$ or $y=2)$. From here, apply consecutively (2) to the desired inequality. Again, maximum is attained if and only if five of the given numbers are 2 's.
+
+Problem 4. In the $O-E$ game, a round starts with player $A$ paying $c$ cents to player $B$. Then $A$ secretly arranges the numbers $1,3,5,7,9,11,13$ in some order as a sequence $a_{1}, a_{2}, \ldots, a_{7}$, and $B$ secretly arranges $2,4,6,8,10,12,14$ as a sequence $b_{1}, b_{2}, \ldots, b_{7}$. Finally, the players show their sequences and $B$ pays $A$ one cent for each $i$ in $X=\{1,2,3,4,5,6,7\}$ such that $a_{i}<b_{i}$. This finishes the round. What number $c$ would make the game fair? (The game is fair if the total payments by $A$ to $B$ equals the total payments by $B$ to $A$ after all possible distinct rounds are played exactly once.)
+
+Solution I. Let $k$ be in $X$. There are $7 ! 6 !(8-k)$ choices of the sequences $a_{1}, a_{2}, \ldots, a_{7}$ and $b_{1}, b_{2}, \ldots, b_{7}$ for which $2 k-1$ is an $a_{j}$ with $a_{j}<b_{j}$. Indeed, the $a$ 's can be any of the 7 ! permutations of the 7 odd integers; then $j$ is the subscript such that $a_{j}=2 k-1$, and $b_{j}$ must be one of the $8-k$ numbers in $\{2 k, 2 k+2, \ldots, 14\}$; the remaining 6 even integers can be arranged in 6 ! ways.
+
+The total of the payments by $B$ to $A$ for the $(7 !)^{2}$ possible rounds is then
+
+$$
+\sum_{k=1}^{7} 7 ! 6 !(8-k)=7 ! 6 !(7+6+\cdots+1)=7 ! 6 ! 28=(7 !)^{2} 4
+$$
+
+$A$ pays to $B$ a total of $(7 !)^{2} c$; so $c=4$ makes the game fair.
+
+Solution II. For those who know that the sum of expected values (averages) is the expected value of the sum, we note that in the first spot, 1, 3, 5, 7, 9, 11, 13 are equally likely, and hence the average payment is the average of the payments for each of these numbers, $7 / 7,6 / 7,5 / 7,4 / 7$, $3 / 7,2 / 7,1 / 7$. Hence in the first spot the average payment is
+
+$$
+\frac{(7 / 7+6 / 7+5 / 7+4 / 7+3 / 7+2 / 7+1 / 7)}{7}=4 / 7 \text { of a cent. }
+$$
+
+Since there are 7 spots, with an average payment of $4 / 7$ cent each, the total payment averages 4 cents; so $c=4$ makes the game fair.
+
+Problem 5. $\triangle A B C$ is inscribed in a circle $k$ with center $O$ so that $\angle A C B=120^{\circ}$.
+
+(a) If $H$ is the orthocenter of $\triangle A B C$, prove that $A, B, O, H$ lie on a circle with center the midpoint of the arc $A C B$. (The orthocenter of $\triangle A B C$ is the intersection point of its three altitudes.)
+
+(b) If $G$ is the centroid of $\triangle A B C$, and $I$ is the incenter of $\triangle A B H$, prove that the points $O, G, I, H$ lie on a line. (The centroid of $\triangle A B C$ is the intersection point of its three medians: a median connects a vertex of $\triangle A B C$ with the midpoint of the opposite side; the incenter of $\triangle A B C$ is the intersection of its three angle bisectors.)
+
+Solution. ${ }^{1}$ Let $O_{1}$ be the midpoint of the arc $A C B$ and let $R$ be the radius of $k$.
+
+$\triangle A O O_{1}$ and $\triangle B O O_{1}$ are equilateral $\left(\angle A C B=120^{\circ} \Rightarrow \angle A O B=120^{\circ}\right.$.) The segments $A B$ and $O O_{1}$ intersect each other in their midpoint, $D$. If line $A O$ intersects $k$ in point $P$, then $\angle A B P=\angle A C P=90^{\circ}$, i.e. $P B \| C H$ and $P C \| B H$, and thus $P B H C$ is a parallelogram. From here, $C H=P B$. Since $O D$ is a midsegment in $\triangle A B P$, then $P B=2 O D=O O_{1}$ and $C H=O O_{1}=R$. Again, $O C H O_{1}$ is a parallelogram; moreover, it is a rhombus for $O C=R$. Thus, $O_{1} H=R$ and the points $A, B, O, H$ lie on a circle $k_{1}$ with center $O_{1}$ and radius $R$.
+
+Look at the quadrilateral $C M H N$ : it contains two right angles (at $M$ and $N$ ), and the angle at $C$ is $120^{\circ}$, so the angle at $H$ is $60^{\circ}: \angle A H B=60^{\circ}$. But $\angle A O B=120^{\circ}$ (as above), so $A H B O$ do lie on the same circle. Since $A, O, B$ lie on a circle with center $O_{1}$ (as above), $H$ is forced to lie on the same circle.[^0](b) Solution I. In $k_{1}, \angle B O_{1} H$ is central, and $\angle B A H$ is inscribed, so that $\angle B O_{1} H=2 \angle B A H$ and $\angle B H O_{1}=90^{\circ}-\angle B A H=\angle A H C$. In the rhombus $O C H O_{1}$ the diagonal $O H$ is the bisector of $\angle C H O_{1}$. Hence, $\angle A H O=\angle A H C+\angle C H O=\angle B H O_{1}+\angle O H O_{1}=\angle B H O \Rightarrow O H$ is the angle bisector also of $\angle A H B$. This means that point $I$ lies on $O H$.
+
+Let $O H$ intersect the median $C D$ in point $G$. We will show that $G$ is the medicenter of $\triangle A B C$ by showing first that $C G=2 G D$. Indeed, if $E$ and $F$ are the midpoints of $C G$ and $H G$, then $E F$ is a midsegment in $\triangle C H G$ with $E F=C H / 2=O D$ and $E F \| O D$ (from (a)). Then $\triangle E F G \cong \triangle D O G$, and $E G=G D, C G=2 E G=2 G D$. Thus, $G$ is the medicenter of $\triangle A B C$, and the points $I$ and $G$ lie on the line $O H$.
+
+For those who know about the Euler line: points $O, G, H$ lie on the Euler line of $\triangle A B C$ (we even know the ratio $O G: G H=1: 2$.) Thus, it remains to show that $I$ lies on the line $O H$, or equivalently, that $O H$ is the angle bisector of $\angle A H B$. Recall that $k_{1}$ was the circle described around $A H B O$ from part (a). Since $O A=O B$, the corresponding $\operatorname{arcs} O A$ and $O B$ on $k_{1}$ are equal, and hence the inscribed angles are equal: $\angle A H O=\angle B H O$.
+
+
+[^0]:    ${ }^{1}$ This problem was given at a national contest in Bulgaria in 1994.
+

BayArea/md/en-handouts/en-solution99-2.md

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+# BAMO 1999: SOLUTIONS TO PRACTICE PROBLEMS II 
+
+BAMO PROBLEMS COMMITTEE<br>COMPILED BY ZVEZDELINA STANKOVA-FRENKEL
+
+Participation in BAMO is already an achievement - four hours thinking over hard math problems is what counts, not who solved how many problems! The problems are arranged in approximately increasing difficulty; in particular, problems 4 and 5 are quite hard. We expect no more than 1 or 2 students (if any) to solve all problems, and few to solve completely 3 problems. Thus, solving even one problem is already a FINE performance! Remember that you are competing against students in your age group - if the problems seem hard (simple) to you, they are probably hard (simple) for the other contestants too. The idea of BAMO is that students find it a fun experience and give it all they have during the contest; and after the contest is over, students continue thinking over the proposed problems.
+
+Problem 1. Prove that $1993^{1993}+1994^{1994}+1995^{1995}+1996^{1996}$ is divisible by 10 .
+
+Solution. Since $5^{2}=25$ ends in 5 again, and $6^{2}=36$ ends in 6 again, no matter to what power we raise 5 or 6 , the resulting numbers will end in 5 or 6 , respectively. Thus, $1995^{1995}$ ends in 5 and $1996^{1996}$ ends in 6 . Further, $4^{1}=1,4^{2}=16,4^{3}=\ldots 4,4^{4}=\ldots 6$, etc, i.e. an odd power of 4 ends in 4, and an even power of 4 ends in 6 (why?) Hence, $1994^{1994}$ ends in 6 . So, the last digit of $1994^{1994}+1995^{1995}+1996^{1996}$ is the same as the last digit of $6+5+6=17$, i.e. it ends in 7 . It remains to show that $1993^{1993}$ ends in 3 (so that $3+7=10$.)
+
+Indeed, $3^{1}=3,3^{2}=9,3^{3}=\ldots 7,3^{4}=\ldots 1,3^{5}=\ldots 3$, and the pattern will be $3,9,7,1,3,9,7,1, \ldots$, etc. So we need to figure out into which of these slots will the 1993-rd power of 3 fall. Equivalently, we need to find the remainder of 1993 when divided by 4 (the length of the pattern above is 4). Now, $1993=4 \cdot 498+1: 3^{1993}$ will complete 498 cycles of the pattern, and will end in 3 . This shows that $1993^{1993}$ ends in 3 . Therefore, the last digit of the total sum is the last digit of $3+6+5+6=20$, 0 , which makes the sum divisible by 10 .
+
+Problem 2. King Arthur and his 100 knights are having a feast at a round table. Each person has a glass with white or red wine in front of him. Exactly at midnight, every person moves his glass in front his left neighbor if the glass has white wine, or in front of his right neighbor if the glass has red wine. It is known that there is at least one glass with red and one glass with white wine.
+
+(a) Prove that after midnight there will be at least one person without a glass of wine in front of him.
+
+(b) If Lady Guinivera, the Queen, calls the King off the table before midnight, will the conclusion of part (a) still going to be correct?
+
+Solution (a). Suppose not. Label all seats as $1,2, \ldots, 101$ by going around the table anticlockwise. If each person still has a glass after midnight, then no one has received glasses from both of his
+neighbors, i.e. any two people with exactly one person between must have the same color wine. Thus, the following seats must correspond to the same color wine: $1,3,5, \ldots, 99,101,2,4, \ldots, 98,100$. But these are all seats - this contradicts the hypothesis that there are glasses with red and glasses with white wine! Therefore, the supposition is wrong, and at least one person will be left without a glass after midnight.
+
+Solution (b). The conclusion won't be correct. For example, give white wine to all odd seated people $1,3,5, \ldots, 99$, and red wine to all even seated people $2,4,6, \ldots, 100$. After midnight the two groups will switch the colors of their wines.
+
+Problem 3. On the bases $A B$ and $C D$ of a trapezoid $A B C D$ draw two squares externally to $A B C D$. Let $O$ be the intersection point of the diagonals $A C$ and $B D$, and let $O_{1}$ and $O_{2}$ be the centers of the two squares. Prove that $O_{1}, O$ and $O_{2}$ lie on a line (i.e. they are collinear; see Figure.)
+
+Solution. The idea is to show that $\triangle O C O_{2} \sim \triangle O A O_{1}$. Indeed, first notice that $\triangle A O_{1} B \sim$ $\triangle C O_{2} D$ - both are right isosceles triangles. Therefore, $A O_{1}: C O_{2}=A B: C D$. But $\triangle A O B \sim$ $\triangle C O D(A B \| C D \Rightarrow$ all three angles are the same), so $A B: C D=A O: C O$. This implies $A O_{1}: C O_{2}=A O: C O$. Further, $\angle O_{1} A O=\angle O_{1} A B+\angle B A O=45^{\circ}+\angle D C O=\angle O C O_{2}$, and we finally conclude that $\triangle O C O_{2} \sim \triangle O A O_{1}$.
+
+Hence, $\angle A O O_{1}=\angle C O O_{2}$. Since $A O C$ is a line, then $O_{2} O O_{1}$ is also a line.
+
+Note: Alternatively, apply dilation with center $O$ and ratio $A O: O C$ - it will take one square to the other, and correspondingly, one center to the other.
+
+Problem 4. The real positive numbers $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ satisfy the relation $a_{n+1}^{2}=a_{n}+1$ for all $n=1,2, \ldots$ Prove that at least one of the $a_{i}$ 's must be an irrational number.
+
+Solution. Argue by contradiction. Suppose that for all $i, a_{i}=p_{i} / q_{i}$ for some relatively prime positive integers $p_{i}, q_{i}$. Substitute in the given relation:
+
+$$
+\frac{p_{n+1}^{2}}{q_{n+1}^{2}}=\frac{p_{n}+q_{n}}{q_{n}} \Rightarrow p_{n+1}^{2} q_{n}=q_{n+1}^{2}\left(p_{n}+q_{n}\right)
+$$
+
+But $p_{n+1}^{2}$ and $q_{n+1}^{2}$ are relatively prime, and $q_{n}$ and $q_{n}+p_{n}$ are also relatively prime (why?). This means that $p_{n+1}^{2}=p_{n}+q_{n}$ and $q_{n}=q_{n+1}^{2}$. In particular, $q_{n+1}=\sqrt{q_{n}}=\sqrt[4]{q_{n-1}}=\cdots=\sqrt[2 n]{q_{1}}$. Unless $q_{1}=1$, we will eventually run out of perfect squares in $q_{1}$ and $q_{n}$ will not be an integer, a contradiction. Thus, the only possibility is $q_{1}=1$, so that all $q_{n}=1$, and our numbers $a_{n}$ are after all integers.
+
+But this is absurd! Indeed, $a_{n+1}=\sqrt{a_{n}+1}<a_{n}$ (because $a_{n}+1<a_{n}^{2}$ for all integers $a_{n}>1$, and $a_{1}=1$ implies an immediate contradiction for $a_{2}=\sqrt{2}$ is irrational,) so that we obtain an infinite strictly descreasing sequence of positive integers, and there isn't simply such a thing! This shows that our supposition was wrong and at least one of the $a_{n}$ 's will be irrational.
+
+Problem 5. [Simpson's Line] Let $\triangle A B C$ be inscribed in a circle $k$, and let $M$ be an arbitrary point on $k$ different from $A, B, C$. Prove that the feet of the three perpendiculars from $M$ to the sides of $\triangle A B C$ are collinear. (Note: You may have to extend some sides to find these feet: see Figure.)
+
+Solution. It will suffice to show that $\angle A Q R=\angle C Q P$ (compare with Problem 3.) Note that both quadrilaterals $M Q A R$ and $M Q P C$ are cyclic: $\angle M R A=90^{\circ}=\angle M Q A$, and $\angle M Q C=$ $90^{\circ}=\angle M P C$. Using inscribed angles in these quadrilaterals, we obtain that the two angles in questions equal correspondingly to:
+
+$$
+\angle A Q R=\angle A M R, \angle C Q P=\angle C M R \text {. }
+$$
+
+Look carefully at $\triangle M A R$ and $\triangle M C P$ : both have one right angle, and we hope to show that two more angles are the same. Equivalently, we want to show that these two triangles are similar. Can we do that? Yes. Their third angles are the same: $\angle M C P=\angle M A R$ because $\angle M A R=$ $180^{\circ}-\angle M A B=\angle M C B$ from the inscribed quadrilateral $M C B A$.
+
+Thus, we conclude that $\triangle M A R \sim \triangle M C P$ (two same angles), so their third angles are also the same: $\angle C M P=\angle R M A$. Combining this with (1), we finally obtain $\angle A Q R=\angle C Q P$, and therefore $P, Q, R$ are collinear.
+

BayArea/md/en-monthly/en-0001-mc1sol.md

@@ -0,0 +1,80 @@
+# Berkeley Math Circle 2000-2001 Monthly Contest \#1 — Solutions 
+
+1. Two sets of points in the coordinate plane are given: $\{(-1,1),(-1,2), \ldots,(-1,2000)\}$ and $\{(1,1),(1,2)$, $\ldots,(1,2000)\} .2000$ line segments are drawn connecting these points so that each point in the first set is connected to exactly one point in the second set, and vice versa. Find, with proof, the sum of the $y$-intercepts of the segments.
+
+Solution: Note that, for any real numbers $a$ and $b$, the segment connecting $(-1, a)$ and $(1, b)$ has midpoint $\left(0, \frac{a+b}{2}\right)$, so its $y$-intercept is $\frac{a+b}{2}$. Now suppose that our given segments connect $(-1,1)$ to $\left(1, y_{1}\right),(-1,2)$ to $\left(1, y_{2}\right),(-1,3)$ to $\left(1, y_{3}\right), \ldots,(-1,2000)$ to $\left(1, y_{2000}\right)$. The sum of the $y$-intercepts is then
+
+$\frac{1+y_{1}}{2}+\frac{2+y_{2}}{2}+\frac{3+y_{3}}{2}+\cdots+\frac{2000+y_{2000}}{2}=\frac{1}{2}\left[(1+2+3+\cdots+2000)+\left(y_{1}+y_{2}+y_{3}+\cdots+y_{2000}\right)\right]$.
+
+But it follows from the given that $y_{1}, y_{2}, \ldots, y_{2000}$ are just some reordering of $1,2, \ldots, 2000$, so our sum simplifies to
+
+$$
+\frac{1}{2}[(1+2+3+\cdots+2000)+(1+2+3+\cdots+2000)]=1+2+3+\cdots+2000
+$$
+
+By the formula for the sum of an arithmetic progression, this equals $2000 \cdot 2001 / 2=2,001,000$.
+
+Remark: One way to "check" this answer is to assume that $(-1, y)$ is connected to $(1, y)$ for each $y=1,2,3, \ldots, 2000$; then the $y$-intercepts of the segments are simply $1,2, \ldots, 2000$, giving the desired sum.
+
+2. Find all functions $f$ from the set $\mathbf{R}$ of real numbers to itself such that $f(x y+1)=x f(y)+2$ for all $x, y \in \mathbf{R}$.
+
+Solution: The only such function is $f(x)=2 x$. First, we note that this function really is a solution of the equation, since $2(x y+1)=2 x y+2=x(2 y)+2$ for all $x, y$.
+
+Now let $f$ be any function satisfying the equation. First put $x=y=0$; the equation gives $f(0 \cdot 0+1)=$ $0 f(0)+2$ or $f(1)=2$. Now suppose $y=1$ (and $x$ may be anything); we have $f(1 x+1)=x f(1)+2$ or $f(x+1)=2 x+2$ for all $x \in \mathbf{R}$. Equivalently, $f(x)=f((x-1)+1)=2(x-1)+2=2 x$ for all $x$, so we have verified that this is the only function satisfying the equation.
+
+3. Let $A B C$ be a triangle and $D, E, F, G, H, I, J$ points on sides $B C, C A, A B, B C, C A, A B, B C$ respectively, such that $E F$ and $H I$ are parallel to $B C$; $I J$ and $F G$ are parallel to $C A$; and $D E, G H$ are parallel to $A B$. Prove that $D=J$.
+
+Solution: First we prove the following lemma: If $P Q R$ is a triangle with points $S, T$ on sides $P Q, Q R$ such that $S T$ is parallel to $R P$, then (a) $S Q / P Q=T Q / R Q$ and (b) $P S / P Q=R T / R Q$. Proof: First, observe that triangles $P Q R, S Q T$ are similar (because parallel lines imply equal angles). Equation (a) then follows because the ratios of sides in similar triangles are equal. Manipulating this equation gives
+
+$$
+\frac{S Q}{P Q}=\frac{T Q}{R Q} \Rightarrow 1-\frac{S Q}{P Q}=1-\frac{T Q}{R Q} \Rightarrow \frac{P Q-S Q}{P Q}=\frac{R Q-T Q}{R Q} \Rightarrow \frac{P S}{P Q}=\frac{R T}{R Q}
+$$
+
+which is equation (b).
+
+Now we simply apply this lemma repeatedly. We obtain
+
+$$
+\frac{B D}{B C}=\frac{A E}{A C}(\text { by part (b), using line } D E)=\frac{A F}{A B}(\text { part (a), line } E F)=\frac{C G}{C B}((\mathrm{~b}), \text { line } F G)
+$$
+
+$$
+=\frac{C H}{C A}((\mathrm{a}), \text { line } G H)=\frac{B I}{B A}((\mathrm{~b}) \text {, line } H I)=\frac{B J}{B C}((\mathrm{a}), \text { line } I J) \text {. }
+$$
+
+Thus $B D / B C=B J / B C$, or $B D=B J$. But there is only one point on segment $B C$ at distance $B D$ from $B$; thus we conclude $D=J$.
+
+4. There are 30 members of the math faculty at UC Berkeley. The evil Gastropod places a red or green stamp on each member's forehead, then places them all in separate rooms. Each day, the Gastropod convenes all the faculty in one room and asks any member who has conclusively determined that his/her stamp is red to raise his/her hand. No other communication between them is allowed. As it turns out, everyone has a red stamp (but since each can only see the others' stamps, none of them realize this at first). The Gastropod remarks on the first day: "At least one of you has a red stamp." Given that all Berkeley faculty members are perfect reasoners, and each knows all the others are perfect reasoners, prove that everyone will eventually determine that his/her stamp is red.
+
+Solution: We prove the following generalization: if exactly $n$ faculty members were given red stamps and the rest have green stamps (and the procedure is as before), no hands will be raised for the first $n-1$ days, and all $n$ faculty members will raise their hands on the $n$th day. This will be shown by induction. If $n=1$, the one faculty member sees on the first day that everyone else has green stamps and realizes that he/she must be red.
+
+If $n>1$, then each of the red faculty sees $n-1$ red stamps. (The people with green stamps can be disregarded since they cannot offer any useful information: they can never raise their hands, and all the red people already know they are green.) Suppose $A$ has a red stamp. Using the induction hypothesis, $A$ reasons as follows: "If I am green, then there are $n-1$ red stamps, so nothing will happen for the first $n-2$ days, and all those with red stamps will raise their hands on the $n-1$ th day." Also, by symmetry, all the red faculty reason (and act) identically. Consider the $i$ th day for $i=1,2, \ldots, n-1$; we claim no hands are raised on any of these days. This holds by strong induction on $i$ : For each such value of $i$, on the $i$ th day, $A$ remembers that no hands were raised on any previous day (by the induction hypothesis), which is consistent (from $A$ 's point of view) with the hypothesis that $A$ is green. So $A$ will not raise his hand on the $i$ th day, because he still believes he could be green; by symmetry, none of the other faculty raise their hands. On the $n$th day, however, $A$ notices that no hands were raised on the $n-1$ th day, which contradicts the behavior expected if $A$ were green. So $A$ realizes he is red and raises his hand. Likewise, all the other red faculty raise their hands. This completes the induction step.
+
+In the given case $n=30$, we see that all the faculty will be silent for 29 days, and they will all realize on the 30th day that their stamps are red.
+
+5. Let $a_{1}, a_{2}, \ldots, a_{2000}$ be real numbers in the interval $[0,1]$. Find the maximum possible value of
+
+$$
+\sum_{1 \leq i<j \leq 2000}(j-i)\left|a_{j}-a_{i}\right|
+$$
+
+Solution: The answer is $1,000,000,000$. First, note that the desired sum $S$ is convex as a function of each $a_{i}$. (Indeed, $\left|a_{i}-a\right|$ is convex for any real number $a$, and a sum of convex functions is itself convex.) Consequently, it attains a maximum at some point where each variable lies at an endpoint of its interval, i.e. $S$ has a maximum with each $a_{i}$ equal to either 0 or 1 . So we may restrict our attention to these cases.
+
+Then $\left|a_{j}-a_{i}\right|=0$ if $a_{i}=a_{j}$ and 1 otherwise, so $S$ is really the sum of all values of $j-i$ where $a_{i} \neq a_{j}$. Equivalently, it equals $\sum_{1 \leq i<j \leq 2000}(j-i)-$ (sum of $j-i$ over pairs for which $a_{i}=a_{j}$ ). So our goal is equivalent to minimizing the latter sum. Fix a positive integer $k$, and suppose there are $k$ values of $i$ with $a_{i}=0$ (and $2000-k$ values with $a_{i}=1$ ). If we consider the sum of $j-i$ over all pairs $i<j$ with $a_{i}=a_{j}=0$, this sum is clearly minimized when all the $i$ 's with $a_{i}=0$ are consecutive; likewise the sum over all pairs with $a_{i}=a_{j}=1$ is minimized when all the $1 \mathrm{~s}$ are consecutive. So we may restrict our attention to the case where $a_{1}=a_{2}=\cdots=a_{k}=0$ and $a_{k+1}=\cdots=a_{2000}=1$. (The case where the block of 1 s precedes the block of 0 s is symmetrically equivalent.)
+
+In this case, the sum $S$ comes out to be
+
+$$
+\begin{aligned}
+& \sum_{\substack{1 \leq i<j \leq 2000 \\
+a_{i} \neq a_{j}}}(j-i)=\sum_{i=1}^{k} \sum_{j=k+1}^{2000}(j-i)=\sum_{j=k+1}^{2000}\left(\sum_{i=1}^{k} j\right)-\sum_{i=1}^{k}\left(\sum_{j=k+1}^{2000} i\right)=\sum_{j=k+1}^{2000} k j-\sum_{i=1}^{k}(2000-k) i \\
+= & \frac{(2000-k)(2000+[k+1])}{2}-(2000-k) \frac{k(k+1)}{2}=1000 k(2000-k)=1000\left(1000^{2}-[k-1000]^{2}\right)
+\end{aligned}
+$$
+
+which is at most $1000^{3}$, with equality holding iff $k=1000$. So $1000^{3}$ is our desired maximum, attained when $a_{i}=0$ for $i \leq 1000$ and $a_{i}=1$ for $i>1000$.
+
+Of course, essentially the same proof shows that, if 2000 is replaced by any positive integer $n$, the maximum of $S$ is $n^{3} / 8$ when $n$ is even and $\left(n^{3}-n\right) / 8$ when $n$ is odd.
+
+Solutions (C) 2000, Berkeley Math Circle.
+

BayArea/md/en-monthly/en-0001-mc2sol.md

@@ -0,0 +1,82 @@
+# Berkeley Math Circle 2000-2001 
+
+## Monthly Contest \#2 - Solutions
+
+1. Given $n+1$ distinct integers from the set $\{1,2, \ldots, 2 n\}(n \geq 1)$, prove that some two of them are relatively prime.
+
+Solution: As is well known, any two consecutive integers are relatively prime. (Proof: let $d$ be the greatest common divisor of $k$ and $k+1$ for some integer $k$; then, since $k+1$ and $k$ are both multiples of $d$, their difference, 1 , is also a multiple of $d$; so $d=1$.) Now partition the set $\{1,2, \ldots, 2 n\}$ into $n$ pairs $\{1,2\},\{3,4\},\{5,6\}, \ldots,\{2 n-1,2 n\}$. Since we have $n+1$ integers but only $n$ pairs, the Pigeonhole Principle tells us that some two integers must belong to the same pair. But that means that they are consecutive, so they are relatively prime.
+
+2. Three grasshoppers are on a straight line. Every second one grasshopper jumps. It jumps across one (but not across two) of the other grasshoppers. Prove that after 1999 seconds the grasshoppers cannot be in the initial position.
+
+Solution: Assume that the grasshoppers are initially in the order $(A, B, C)$ on the line. Call $(A, B, C),(B, C, A),(C, A, B)$ "even" orders and $(A, C, B),(B, A, C),(C, B, A)$ "odd" orders. We claim that if the grasshoppers are in an even order, they will be in an odd order the next second, and vice versa. This can be proved by simply checking all the possible situations. For example, if we start from the even order $(A, B, C)$, then either $A$ jumps across $B, B$ jumps across $A, B$ jumps across $C$, or $C$ jumps across $B$. These yield the respective orders $(B, A, C),(B, A, C),(A, C, B),(A, C, B)$, which are odd. A similar analysis starting from each of the other five positions confirms that we switch parity with every jump.
+
+Now we claim that after an even number of seconds, the grasshoppers are in an even order, and after an odd number of seconds, they are in an odd order. This is a simple induction: after $n=0$ seconds, they are in an order defined to be even. If $n>0$, suppose $n$ is even. Then $n-1$ is odd, so the grasshoppers are in an odd order after $n-1$ seconds (by induction) and will be in an even order after $n$ seconds. Likewise, if $n$ is odd, then the grasshoppers are in an even order after $n-1$ seconds and therefore in an odd order after $n$ seconds. This completes the induction step.
+
+But this means that after 1999 seconds, the grasshoppers are in an odd order; since they initially were in an even order, they cannot be in the same position.
+
+Remark: "Even" and "odd" permutations are defined — and of considerable importance — for all numbers of grasshoppers, not just 3. In general, a permutation can be obtained by either an even number or an odd number of transpositions (switches of two grasshoppers), but not by both.
+
+3. Do there exist three different prime numbers such that the sum of any two of them is a square?
+
+Solution: The answer is no. We work modulo 4 (i.e. we look at remainders on division by 4). Note that any square is $\equiv 0$ or $1 \bmod 4$. Indeed, if $n=2 k$ is even, $n^{2}=4 k^{2} \equiv 0$, while if $n=2 k+1$ is odd, $n^{2}=4 k^{2}+4 k+1 \equiv 1$.
+
+Now suppose some three such primes exist. If they are all odd, then each of them is $\equiv 1$ or $3(\bmod$ 4); some two must then have the same value modulo 4 (both 1 or both 3 ), and then their sum is $\equiv 2$ $(\bmod 4)$, so it cannot be a square.
+
+Thus, the primes are not all odd, so one of them is 2 . Since they are all different, the other two are odd. If these two are both $\equiv 3$, then again their sum is $\equiv 2$ and cannot be a square. So one of them is $\equiv 1$, but then this prime plus 2 is congruent to 3 , so that sum cannot be a square. So, once again, we can find two of the three primes whose sum is not a square.
+
+4. Let $S=\{1,2, \ldots, n\}$, and let $T$ be the set consisting of all nonempty subsets of $S$. The function $f: T \rightarrow S$ is "garish" if there do not exist sets $A, B \in T$ such that $A$ is a proper subset of $B$ and $f(A)=f(B)$. Determine, with proof, how many garish functions exist.
+
+Solution: There are $n$ ! such functions. If $g$ is any bijective map from $S$ to itself (i.e. a permutation of $S$ ), then the function $f: T \rightarrow S$ defined by $f(A)=g(|A|)$ (here $|A|$ is the cardinality of set $A$ ) is garish. To see this, just note that if $A$ is a proper subset of $B$, then $|A|<|B|$, so $g(|A|) \neq g(|B|)$. There are $n$ ! possible choices of the map $g$, and all of them give different maps $f$ (this follows from the fact that every element of $S$ is the cardinality of some set in $T$ ), so we get $n$ ! garish functions this way. We now wish to show that every garish function is of this form.
+
+Let $f$ be garish; our crucial observation is the following: If $a_{1}, a_{2}, \ldots, a_{n}$ are the elements of $S$ in some order, we shall say that this ordering "produces" a sequence of $n$ values $f\left(\left\{a_{1}\right\}\right), f\left(\left\{a_{1}, a_{2}\right\}\right), \ldots$, $f\left(\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}\right)$. These values are all different, since, of any two of these sets, one is properly contained in the other. But there are exactly $n$ possible values for $f$, namely $1,2, \ldots, n$, so any produced sequence consists of exactly these in some order.
+
+Now choose any integer $k, 1 \leq k \leq n$. We will show that $f$ has the same value on all $k$-element sets. We can assume $k<n$ (otherwise there is only one $k$-element set). First consider any two sets that differ by at most one element; let them be $\left\{a_{1}, a_{2}, \ldots, a_{k}\right\}$ and $\left\{a_{1}, a_{2}, \ldots, a_{k-1}, b\right\}$. Assume $a_{k} \neq b$. Let the remaining elements of $S$ (if there are any) be $a_{k+2}, a_{k+3}, \ldots, a_{n}$ in any order. Consider the sequences produced from the two orderings $a_{1}, a_{2}, \ldots, a_{k-1}, a_{k}, b, a_{k+2}, \ldots, a_{n}$ and $a_{1}, a_{2}, \ldots, a_{k-1}, b, a_{k}, a_{k+2}, \ldots, a_{n}$. From the above, each produced sequence contains every element of $S$ exactly once. But these two sequences are identical except that the first has $f\left(\left\{a_{1}, \ldots, a_{k}\right\}\right)$ where the second has $f\left(\left\{a_{1}, \ldots, a_{k-1}, b\right\}\right)$, so we conclude that these two values of $f$ are identical. Also, if $a_{k}=b$, then the two values are equal (trivially).
+
+This shows that $f$ has the same value on two $k$-element sets differing by $\leq 1$ element. Now if $\left\{a_{1}, \ldots, a_{k}\right\}$ and $\left\{b_{1}, \ldots, b_{k}\right\}$ are any two sets in $T$, we have (applying this repeatedly) that
+
+$$
+f\left(\left\{a_{1}, a_{2}, \ldots, a_{k}\right\}\right)=f\left(\left\{b_{1}, a_{2}, \ldots, a_{k}\right\}\right)=f\left(\left\{b_{1}, b_{2}, a_{3}, \ldots, a_{k}\right\}\right)=\cdots=f\left(\left\{b_{1}, b_{2}, \ldots, b_{k}\right\}\right) .
+$$
+
+So $f$ is constant over all $k$-element sets. Thus we can define the function $g: S \rightarrow S$ by letting $g(k)$ be the value of $f$ on any $k$-element set, and $g$ uniquely determines $f$ since $f(A)=g(|A|)$ for all $A \in T$. Moreover, if $g(k)=g(l)$ for some $k<l$, then letting $A=\{1,2, \ldots, k\}, B=\{1,2, \ldots, l\}$, we have $f(A)=f(B)$, a contradiction since $A$ is a proper subset of $B$. Thus, $g$ is one-one; but since it maps the finite set $S$ to itself, it is actually bijective. Thus all garish functions are indeed in the form claimed above.
+
+Alternate Solution: (Submitted by Inna Zakharevich) We observe that $n$ ! garish functions can be constructed, precisely as described in the first solution, and we wish to show that all garish functions are of this form. We use induction on $n$. In the base case $n=1$, there is only one possible function, given by $f(\{1\})=1$, and it is garish. Now consider any $n$, and suppose the statement is true for $n-1$. If $f$ is a garish function, let $a_{k}=f(\{1,2, \ldots, k\})$ for each $k=1,2, \ldots, n$. Then all $a_{k}$ are different: if $a_{k}=a_{l}$ for some $k<l$, then $f$ would not be garish, since $\{1,2, \ldots, k\}$ is a proper subset of $\{1,2, \ldots, l\}$. Thus, the elements $a_{1}, \ldots, a_{n}$ equal $1, \ldots, n$ in some order. Partition $T$ into three subsets: let $T_{1}$ consist of the nonempty subsets of $\{1,2, \ldots, n-1\}$, let $T_{2}=\{S\}$, and let $T_{3}$ be the rest of $T$ (the proper subsets of $S$ containing $n$ ). We claim that, if $A \in T_{1}$, then $f(A)=a_{|A|}$. This follows from the induction hypothesis. Formally, we first observe that $f(A) \neq a_{n}$ for $A \in T_{1}$ since $A \subset S$. Then, if we define $h:\left\{a_{1}, a_{2}, \ldots, a_{n-1}\right\} \rightarrow\{1,2, \ldots, n-1\}$ by $h\left(a_{k}\right)=k$ (and this is well-defined since $a_{1}, \ldots, a_{n-1}$ are distinct), then the composite $h \circ f$ is a garish function from $T_{1}$ to $\{1,2, \ldots, n-1\}$, so, by the induction hypothesis, it takes the same value on all $k$-element sets; since $h$ is injective, $f$ takes the same value on all $k$-element sets, as claimed. Also, if $A \in T_{2}$ (so $A=S$ ) then $f(A)=a_{n}$, by assumption.
+
+Now consider any set $A \in T_{3}$; we have $|A|<n$. We claim that $f(A)=a_{|A|}$; this will be shown by strong downward induction on the cardinality $|A|$. If $|A|=n-1$, then $A \subset S$ and $f(S)=a_{n}$. On the other hand, the set $A-\{n\}$ has $n-2$ elements and is contained in $T_{1}$; it certainly has subsets $A_{n-2}, A_{n-3}, \ldots, A_{1}$ of respective cardinalities $n-2, n-3, \ldots, 1$, all of which lie in $T_{1}$; thus, $f\left(A_{i}\right)=a_{i}$ from the previous paragraph. From this, garishness gives $f(A) \neq a_{n}$ and $f(A) \neq a_{1}, a_{2}, \ldots, a_{n-2}$, so $f(A)=a_{n-1}$.
+
+The induction step is similar: suppose we have proven that, for all $A \in T_{3}$ with $|A|>m, f(A)=a_{|A|}$, and we wish to move to the case $|A|=m$. By successively adding elements to $A$, we can construct sets $A_{m+1}, A_{m+2}, \ldots, A_{n}=S$ of respective cardinalities $m+1, m+2, \ldots, n$, which all lie in $T_{3}$, except for $A_{n} \in T_{2}$. Also, if $m>1$, then $A-\{n\}$ has $m-1$ elements and lies in $T_{1}$; repeatedly removing elements, we get sets $A_{m-1}, A_{m-2}, \ldots, A_{1} \in T_{1}$ of respective sizes $m-1, m-2, \ldots, 1$. So we have $f\left(A_{k}\right)=a_{k}$ for $k>m$ by the induction hypothesis, and $f\left(A_{k}\right)=a_{k}$ for $k<m$ because these sets lie in $T_{1}$. But each of the $A_{k}$ either contains $A$ or is contained in $A$, so $f(A)$ cannot equal any of these $a_{k}$; hence, it must equal $a_{m}$, completing the induction step.
+
+At this point, we have shown that $f(A)=a_{|A|}$ for each $A \in T$, and one sees as in the previous solution that $f(A)=g(|A|)$ for some bijective function $g: S \rightarrow S$, as needed.
+
+Remark: Another clever solution, submitted by Neil Herriot, involved reducing the $n$-element case to the $n$-1-element case for each of the $n$ possible subsets obtainable by removing one element from $S$, rather than just the subset $\{1,2, \ldots, n-1\}$.
+
+5. Let $A B C D$ be a quadrilateral, and let $O$ be the intersection of $A C$ and $B D$. Quadrilateral $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is obtained by rotating $A B C D$ about $O$ by some angle. Let $A_{1}, B_{1}, C_{1}, D_{1}$ be the intersection ponts of the lines $A^{\prime} B^{\prime}$ and $A B, B^{\prime} C^{\prime}$ and $B C, C^{\prime} D^{\prime}$ and $C D, D^{\prime} A^{\prime}$ and $D A$, respectively. Prove that quadrilateral $A_{1} B_{1} C_{1} D_{1}$ is cyclic if and only if $A C$ is perpendicular to $B D$.
+
+Solution: First we prove a lemma: Suppose some line $S T$ is rotated by angle $\theta$ about $O$ to obtain line $S^{\prime} T^{\prime}$. Let $S T$ and $S^{\prime} T^{\prime}$ intersect at $S_{1}$, and let $S_{2}$ be the foot of the perpendicular from $O$ to $S T$. Then a rotation about $O$ of angle $\theta / 2$, together with a homothety about $O$ of ratio $\sec \theta / 2$, sends $S_{2}$ to $S_{1}$. Proof: If $P$ is the image of $S_{2}$ under this transformation, then $\triangle P S_{2} O$ is a right triangle (right angle at $S_{2}$ ), so $P$ lies on $S T$. Let $Q$ be the reflection of $P$ across line $S_{2} O$; then $Q$ also lies on $S T, O Q=O P$, and $\angle Q O P=\theta$. Thus, a rotation about $O$ through angle $\theta$ sends $Q$ to $P$. But, since $Q \in S T$, we have $P \in S^{\prime} T^{\prime}$, so that $P=S_{1}$ as desired.
+
+Now, in the given situation, let $A_{2}, B_{2}, C_{2}, D_{2}$ denote the feet of the perpendiculars from $O$ to $A B, B C$, $C D, D A$ respectively; we have shown that a rotation and homothety about $O$ sends $A_{2}, B_{2}, C_{2}, D_{2}$ to $A_{1}, B_{1}, C_{1}, D_{1}$. Thus, quadrilaterals $A_{1} B_{1} C_{1} D_{1}$ and $A_{2} B_{2} C_{2} D_{2}$ are similar, so it suffices to show that the latter is cyclic iff $A C$ is perpendicular to $B D$. For this, note that $\angle O A_{2} B=\pi / 2=\angle O B_{2} B$, so quadrilateral $O A_{2} B B_{2}$ is cyclic; similarly, quadrilaterals $O B_{2} C C_{2}, O C_{2} D D_{2}, O D_{2} A A_{2}$ are cyclic. From these cyclicities we get
+
+$$
+\begin{gathered}
+\angle A_{2} B_{2} O=\angle A_{2} B O ; \quad \angle O B_{2} C_{2}=\angle O C C_{2} ; \quad \angle C_{2} D_{2} O=\angle C_{2} D O ; \quad \angle O D_{2} A_{2}=\angle O A A_{2} \\
+\Rightarrow \angle A_{2} B_{2} C_{2}+\angle C_{2} D_{2} A_{2}=\angle A_{2} B O+\angle O C C_{2}+\angle C_{2} D O+\angle O A A_{2}=(\angle A B O+\angle O A B)+(\angle O C D+\angle C D O) \\
+=(\pi-\angle B O A)+(\pi-\angle D O C)=2 \pi-2 \angle D O C .
+\end{gathered}
+$$
+
+From this, we see that $\angle A_{2} B_{2} C_{2}+\angle C_{2} D_{2} A_{2}=\pi$ iff $\angle D O C=\pi / 2$; i.e. quadrilateral $A_{2} B_{2} C_{2} D_{2}$ is cyclic iff $A C$ and $B D$ are perpendicular.
+
+Alternate Solution: (Submitted by several contestants) Observe that $\angle O A A_{1}=\angle O A B=$ $\angle O A^{\prime} B^{\prime}=\angle O A^{\prime} A_{1}$, so that quadrilateral $O A A^{\prime} A_{1}$ is cyclic. Likewise, $\angle O A D_{1}=\angle O A D=\angle O A^{\prime} D^{\prime}=$ $\angle O A^{\prime} D_{1}$, so quadrilateral $O A^{\prime} A D_{1}$ is cyclic. Thus, $A_{1}, D_{1}$ both lie on the circumcircle of $\triangle O A^{\prime} A$, so the pentagon $O D_{1} A A^{\prime} A_{1}$ is cyclic. Similar reasoning gives that pentagons $O A_{1} B B^{\prime} B_{1}, O B_{1} C C^{\prime} C_{1}$, and $O C_{1} D D^{\prime} D_{1}$ are cyclic. But now we simply compute:
+
+$$
+\begin{gathered}
+\angle A_{1} B_{1} C_{1}+\angle C_{1} D_{1} A_{1}=\angle A_{1} B_{1} O+\angle O B_{1} C_{1}+\angle C_{1} D_{1} O+\angle O D_{1} A_{1} \\
+=\angle A_{1} B O+\angle O C C_{1}+\angle C_{1} D O+\angle O A A_{1} \quad \text { (because of the previously cited cyclicities) } \\
+=\angle A B O+\angle O C D+\angle C D O+\angle O A B=(\pi-\angle B O A)+(\pi-\angle D O C)=2 \pi-2 \angle D O C,
+\end{gathered}
+$$
+
+and now the result follows exactly as in the previous solution.
+
+Solutions (c) 2000, Berkeley Math Circle.
+

BayArea/md/en-monthly/en-0001-mc3sol.md

@@ -0,0 +1,59 @@
+# Berkeley Math Circle 2000-2001 
+
+## Monthly Contest \#3 - Solutions
+
+1. I have three opaque jars filled with coins. One contains only nickels; one contains only dimes; one contains some nickels and some dimes. Also, the three jars are labeled "Nickels," "Dimes," and "Nickels and Dimes," but it is known that all of the jars are labeled incorrectly. You are allowed to choose one jar and look at one coin randomly drawn from it. How can you determine which jar is which?
+
+Solution: Since the jar labeled "Nickels and Dimes" is mislabeled, it contains either only nickels or only dimes. This suggests drawing a coin from this jar.
+
+If the coin drawn is a nickel, then the "Nickels and Dimes" jar actually contains only nickels. Then, the "Dimes" jar (because it is also mislabeled) must contain nickels and dimes, and the "Nickels" jar contains only dimes. If, on the other hand, the coin drawn is a dime, then the "Nickels and Dimes" jar contains only dimes; this forces the "Nickels" jar to contain coins of both types, and the "Dimes" jar then contains nickels. Thus, either way, it is possible to identify all three jars correctly.
+
+2. An $n \times n$ matrix of integers is called "golden" if, for every row and every column, their union contains all of the numbers $1,2,3, \ldots, 2 n-1$. Find all golden matrices (of all sizes).
+
+Solution: The $1 \times 1$ matrix whose only entry is 1 is certainly golden; we claim there are no others. To see this, note that each of the numbers $1,2, \ldots, 2 n-1$ must appear at least $n$ times in the matrix. (Proof: Suppose some number $k$ appears less than $n$ times. Because there are $n$ columns, some column does not contain $k$. Likewise, some row does not contain $k$; then the union of this row and this column fails to contain $k$, contrary to the given). So we have at least $n(2 n-1)$ entries in the matrix. But an $n \times n$ matrix has $n^{2}$ entries, so $n^{2} \geq n(2 n-1) \Rightarrow n \geq 2 n-1 \Rightarrow 1 \geq n$, and only the $1 \times 1$ matrix can be golden.
+
+Alternate Solution: As before, the $1 \times 1$ matrix consisting only of a 1 is golden. Now, note that the union of any row and any column in an $n \times n$ golden matrix contains exactly $2 n-1$ entries, so each of the numbers $1,2, \ldots, 2 n-1$ must appear exactly once in this union. Therefore, there is no room in the matrix for any numbers other than $1,2, \ldots, 2 n-1$. Moreover, if any of these entries appeared twice, say once at the intersection of the $i$ th row and $j$ th column and once at the $i^{\prime}$ th row and $j^{\prime}$ th column, then the union of the $i$ th row and $j^{\prime}$ th column would contain this number twice, not leaving enough room for the remaining $2 n-2$ numbers. So each number can appear only once in the matrix, meaning it can have only $2 n-1$ entries. Thus $n^{2}=2 n-1 \Rightarrow n^{2}-2 n+1=0 \Rightarrow n=1$, and only the $1 \times 1$ matrix is golden.
+
+3. An equilateral triangle in the coordinate plane has vertices $(a, b),(c, d),(e, f)$. Prove that $a, b, c, d, e, f$ cannot all be integers.
+
+Solution: Suppose otherwise. If the triangle has side length $s$, then $s^{2}=(a-c)^{2}+(b-d)^{2}$ is a positive integer, so we may choose the triangle for which $s^{2}$ is as small as possible; this makes $s$ as small as possible. We may further assume that $e=f=0$, since otherwise this assumption can be made true by translating the triangle by the vector $(-e,-f)$, an operation which does not change the triangle's side length. Then we have
+
+$$
+a^{2}+b^{2}=c^{2}+d^{2}=(a-c)^{2}+(b-d)^{2}=s^{2} .
+$$
+
+We consider this chain of equalities modulo 4 , i.e. we look at remainders on dividing by 4 . Note that $n^{2} \equiv 0$ modulo 4 if $n$ is even (since $n=2 k \Rightarrow n^{2}=4 k^{2}$ ), and $n^{2} \equiv 1$ when $n$ is odd $(n=2 k+1 \Rightarrow$ $n^{2}=4\left(k^{2}+k\right)+1$ ). So a sum of two squares is either 0,1 , or 2 modulo 4 , according as the two squares are both even, one even and one odd, or both odd, respectively.
+
+If $s^{2} \equiv 0 \bmod 4$, then the above equation forces $a, b, c, d$ all to be even. Then scaling our triangle by a factor of $1 / 2$ gives us an equilateral triangle with vertices $(a / 2, b / 2),(c / 2, d / 2),(0,0)$. These coordinates
+are still integers, and the new triangle is smaller than the previous one, contradicting our assumption of minimality.
+
+If $s^{2} \equiv 1 \bmod 4$, then each of the pairs $\{a, b\},\{c, d\},\{a-c, b-d\}$ contains one even and one odd number. Adding up these six numbers, we get that the even number $2 a+2 b$ is a sum of three even and three odd numbers, so it should be odd. So we have a contradiction here as well.
+
+If $s^{2} \equiv 2 \bmod 4$, then $a, b, c, d, a-c, b-d$ are all odd. But if $a$ and $c$ are both odd, then $a-c$ should be even, another contradiction. So in every possible case we have a contradiction; hence, our initial assumption - that all coordinates were integers - must be wrong.
+
+4. Let $f, g$ be two functions from the set $\mathbf{R}$ of real numbers to itself, such that $f(x)<g(x)$ for all $x \in \mathbf{R}$. Prove that there exists an infinite subset $S \subseteq \mathbf{R}$ such that $f(x)<g(y)$ for all $x, y \in S$.
+
+Solution: Note that, for every $x$, we can choose a rational number $h(x)$ such that $f(x)<h(x)<$ $g(x)$. Proof: since $g(x)-f(x)>0$, we can choose an integer $n$ larger than $1 /(g(x)-f(x))$. Then the open interval $(n f(x), n g(x))$ has width $n(g(x)-f(x))>1$, so it contains some integer $m$. So we have $f(x)<m / n<g(x)$, and we can take $h(x)=m / n$.
+
+Now $h$ is a function mapping an uncountably infinite set, $\mathbf{R}$, to a countably infinite set, $\mathbf{Q}$ (the set of rational numbers). For any $z \in \mathbf{Q}$, consider $h^{-1}(z)$, the set of elements of $\mathbf{R}$ whose image under $h$ is $z$. Certainly $\mathbf{R}$ is the union of all such sets, since, for any $x \in \mathbf{R}, x$ lies in $h^{-1}(h(x))$. If the set $h^{-1}(z)$ is finite (or even countably infinite) for all $z \in \mathbf{Q}$, then $\mathbf{R}$ consists of a union of countably many countable sets, so it is countable. But this is false, so some $h^{-1}(z)$ must be uncountably infinite. Let $S=h^{-1}(z)$ accordingly. Then, for all $x, y \in S$, we have $f(x)<h(x)=z=h(y)<g(y)$, and $S$ is infinite, as needed.
+
+5. A permutation of the numbers $1,2, \ldots, n$ is called "bad" if it contains a subsequence of 10 numbers in decreasing order, and "good" otherwise. For example, for $n=15$,
+
+$$
+15,13,1,12,7,11,9,8,10,6,5,4,3,2,14
+$$
+
+is a bad permutation, because it contains the subsequence
+
+$$
+15,13,12,11,10,6,5,4,3,2
+$$
+
+Prove that, for each $n$, the number of good permutations is at most $81^{n}$.
+
+Solution: Consider any permutation of $1,2, \ldots, n$. Let the "height" of a number $i$ in the permutation be the length of the longest decreasing subsequence ending in $i$. Then, the permutation is bad if and only if some number has height at least 10. Also note that all numbers with the same height must be arranged in increasing order. To see that this is so, suppose that two numbers $i, j$ both have height $h$, where $i<j$, but $j$ precedes $i$ in the permutation Then there exists a decreasing subsequence of the permutation having length $h$, ending in $j$. By attaching $i$ to the end of this, we get a decreasing sequence of length $h+1$ ending in $i$; this contradicts the assumption that $i$ had height $h$.
+
+Now we know that, in any good permutation, every number has a height in the range $1,2, \ldots, 9$. We also know that, for each height $h$, the numbers with height $h$ occur in the permutation in increasing order. This means that, if we know which of the numbers $1,2, \ldots, n$ have height $h$ and which positions in the permutation are occupied by numbers of height $h$ for each $h=1,2, \ldots, 9$, we can uniquely reconstruct the permutation from this information. Since every possible good permutation can be represented in this form, the number of good permutations is at most equal to the number of possible assignments of heights $1,2, \ldots, 9$ to the $n$ numbers and the $n$ positions in the permutation. There are $9^{n}$ ways of assigning heights from this range to $1,2, \ldots, n$, and another $9^{n}$ ways of assigning heights to the $n$ positions; altogther this makes for at most $\left(9^{n}\right)^{2}=81^{n}$ possible good permutations.
+
+Solutions (c) 2000, Berkeley Math Circle.
+

BayArea/md/en-monthly/en-0001-mc4sol.md

@@ -0,0 +1,144 @@
+# Berkeley Math Circle 2000-2001 
+
+## Monthly Contest \#4 - Solutions
+
+1. Two players play a game with pennies, which are circles of radius 1 , on an $m \times n$ rectangular table. Each player takes turns putting a penny on the table so that it touches no other penny. The first player who is unable to do so loses. The table starts with no pennies on it. Assuming that there is an infinite supply of pennies, for what values of $m$ and $n$ does the first player have a winning strategy?
+
+Solution: If $m<2$ or $n<2$, then it is not possible to fit even one penny on the table, so the first player loses immediately. On the other hand, if $m \geq 2$ and $n \geq 2$, we will show that the first player can win. Let $O$ denote the center of the table, and let the first player place his first penny so that it is centered at $O$. The resulting configuration is centrally symmetric (i.e. it remains the same under a rotation of $180^{\circ}$ about $O$ ). Now, whenever the second player places a penny with some center $A$, the first player should respond with a penny centered at $B$, the $180^{\circ}$ rotation of $A$ about $O$. It is evident that the first player always leaves a centrally symmetric configuration; we must show that this move is actually legal.
+
+Since placing the penny with center $A$ was legal, the symmetry of the previous configuration means that, before this penny was placed, a move centered at $B$ was also legal. So, after the second player's move at $A$, the first player can still legally move at $B$ unless the penny centered at $B$ intersects the penny centered at $A$. If $P$ is an intersection point, then $A P \leq 1$ and $P B \leq 1$, so $A B \leq 2$. But $O$ is the midpoint of $A B$, so $A O=A B / 2 \leq 1$. This means that $A$ lies inside the penny placed with center $O$; this is a contradiction. Hence, the move with center $B$ is, in fact, legal.
+
+Thus, the first player can always legally move so as to preserve central symmetry. This move's availability means that the first player never loses. But since only a finite number of pennies can be placed without overlap, the game eventually ends, and it is the second player who loses.
+
+2. Suppose that $k$ and $n$ are integers with $0 \leq k \leq n-3$. Prove that
+
+$$
+\left(\begin{array}{c}
+n \\
+k
+\end{array}\right),\left(\begin{array}{c}
+n \\
+k+1
+\end{array}\right),\left(\begin{array}{c}
+n \\
+k+2
+\end{array}\right),\left(\begin{array}{c}
+n \\
+k+3
+\end{array}\right)
+$$
+
+cannot form an arithmetic progression.
+
+Solution: Suppose otherwise. The common difference $d$ satisfies $\left(\begin{array}{c}n \\ k+1\end{array}\right)-\left(\begin{array}{l}n \\ k\end{array}\right)=d=\left(\begin{array}{c}n \\ k+2\end{array}\right)-\left(\begin{array}{c}n \\ k+1\end{array}\right)$, giving $2\left(\begin{array}{c}n \\ k+1\end{array}\right)=\left(\begin{array}{l}n \\ k\end{array}\right)+\left(\begin{array}{c}n \\ k+2\end{array}\right)$. Expanding the binomial coefficients in terms of factorials gives
+
+$$
+2 \frac{n !}{(k+1) !(n-k-1) !}=\frac{n !}{k !(n-k) !}+\frac{n !}{(k+2) !(n-k-2) !}
+$$
+
+or, multiplying through by $(k+2) !(n-k) ! / n !$,
+
+$$
+2(k+2)(n-k)=(k+2)(k+1)+(n-k)(n-k-1) .
+$$
+
+A completely analogous analysis gives $2\left(\begin{array}{c}n \\ k+2\end{array}\right)=\left(\begin{array}{c}n \\ k+1\end{array}\right)+\left(\begin{array}{c}n \\ k+3\end{array}\right)$ and
+
+$$
+2(k+3)(n-k-1)=(k+2)(k+3)+(n-k-2)(n-k-1) .
+$$
+
+Subtracting (1) from (2) gives
+
+$$
+2(k+3)(n-k-1)-2(k+2)(n-k)=2(k+2)-2(n-k-1) .
+$$
+
+If we divide by 2 , this becomes $-n+2 k+3=(k+3)(n-k-1)-(k+2)(n-k)=n-2 k-3$, which yields $n=2 k+3$. Plugging this back into (1) gives us $2(k+2)(k+3)=(k+2)(k+1)+(k+3)(k+2)$; since $k \geq 0$, we can divide by $k+2$ to get $2 k+6=2 k+4$, which yields $6=4$. This is nonsense, so our assumption - that the numbers form an arithmetic progression - must be false.
+
+3. A point $M$ and a circle $k$ are given in the plane. If $A B C D$ is an arbitrary square inscribed in $k$, prove that the sum $M A^{4}+M B^{4}+M C^{4}+M D^{4}$ is independent of the positioning of the square.
+
+Solution: By scaling, assume $k$ has radius 1. Position the circle in the coordinate plane with its center $O$ at the origin, and points $A, B, C, D$ at $(1,0),(0,1),(-1,0),(0,-1)$, respectively. Let $M$ have coordinates $(x, y)$. Then, using the Pythagorean theorem to expand $M A^{2}, M B^{2}, M C^{2}, M D^{2}$, we get
+
+$$
+\begin{aligned}
+M A^{4} & +M B^{4}+M C^{4}+M D^{4}=\left[(x-1)^{2}+y^{2}\right]^{2}+\left[x^{2}+(y-1)^{2}\right]^{2}+\left[(x+1)^{2}+y^{2}\right]^{2}+\left[x^{2}+(y+1)^{2}\right]^{2} \\
+& =\left(x^{2}-2 x+1+y^{2}\right)^{2}+\left(x^{2}+y^{2}-2 y+1\right)^{2}+\left(x^{2}+2 x+1+y^{2}\right)^{2}+\left(x^{2}+y^{2}+2 y+1\right)^{2}
+\end{aligned}
+$$
+
+If we set $s=x^{2}+y^{2}+1=O M^{2}+1$, this becomes
+
+$$
+\begin{gathered}
+(s-2 x)^{2}+(s-2 y)^{2}+(s+2 x)^{2}+(s+2 y)^{2} \\
+=\left(s^{2}-4 s x+4 x^{2}\right)+\left(s^{2}-4 s y+4 y^{2}\right)+\left(s^{2}+4 s x+4 x^{2}\right)+\left(s^{2}+4 s y+4 y^{2}\right) \\
+=4 s^{2}+8\left(x^{2}+y^{2}\right)=4 O M^{4}+16 O M^{2}+4 .
+\end{gathered}
+$$
+
+Thus, our sum depends only on the circle's radius and the distance $O M$, which is what we need.
+
+4. Find all pairs of integers $x, y$ such that $2 x^{2}-6 x y+3 y^{2}=-1$.
+
+Solution: We first characterize all solutions to the equation $a^{2}-3 b^{2}=1$ in nonnegative integers. This is a Pell equation, and the method used is standard. Suppose we have $(a+b \sqrt{3})(a-b \sqrt{3})=$ $a^{2}-3 b^{2}=1$ with $a>1$; then we can set $a^{\prime}=2 a-3 b, b^{\prime}=2 b-a$. Then $(a+b \sqrt{3})=(2+\sqrt{3})\left(a^{\prime}+b^{\prime} \sqrt{3}\right)$ and $a^{\prime 2}-3 b^{2}=\left(4 a^{2}-12 a b+9 b^{2}\right)-3\left(4 b^{2}-4 a b+a^{2}\right)=a^{2}-3 b^{2}=1$. Moreover, we claim $a^{\prime}$ and $b^{\prime}$ are also nonnegative, with $a^{\prime}<a$. To see this, note that $9 b^{2}=3\left(a^{2}-1\right)<4 a^{2} \Rightarrow 3 b<2 a \Rightarrow$ $a^{\prime}=2 a-3 b>0 ; 3 b^{2}=a^{2}-1>0 \Rightarrow b>0 \Rightarrow a^{2}=3 b^{2}+1 \leq 4 b^{2} \Rightarrow b^{\prime}=2 b-a \geq 0$; and $b>0 \Rightarrow 9 b^{2}>3 b^{2}+1=a^{2} \Rightarrow a-a^{\prime}=3 b-a>0$.
+
+Thus, a solution $(a, b)$ in nonnegative integers with $a>1$ yields a new solution $\left(a^{\prime}, b^{\prime}\right)$ such that $a^{\prime}<a$ and $(a+b \sqrt{3})=(2+\sqrt{3})\left(a^{\prime}+b^{\prime} \sqrt{3}\right)$. If $a^{\prime}>1$, we can repeat this process to obtain nonnegative integers $a^{\prime \prime}, b^{\prime \prime}$ such that $a^{\prime \prime}<a^{\prime}$ and $\left(a^{\prime}+b^{\prime} \sqrt{3}\right)=(2+\sqrt{3})\left(a^{\prime \prime}+b^{\prime \prime} \sqrt{3}\right)$. Continuing in this manner, we obtain a sequence of solutions with $a>a^{\prime}>a^{\prime \prime}>\cdots$. But this sequence of nonnegative integers cannot decrease forever, so eventually we get to some $n$ for which $a^{(n)}=1$. The equation then implies $b^{(n)}=0$, so $a+b \sqrt{3}=(2+\sqrt{3})^{n}\left(a^{(n)}+b^{(n)} \sqrt{3}\right)=(2+\sqrt{3})^{n}$. Then, it follows that $(2-\sqrt{3})^{n}=a-b \sqrt{3}$. (This can be proven by comparing the binomial expansions of $(2+\sqrt{3})^{n}$ and $(2-\sqrt{3})^{n}$, or by a straightforward induction on $n$; note that the equation $a+b \sqrt{3}=(2+\sqrt{3})^{n}$ uniquely determines the integers $a$ and b.) Thus, we conclude that
+
+$$
+a=\frac{(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}}{2}, \quad b=\frac{(2+\sqrt{3})^{n}-(2-\sqrt{3})^{n}}{2 \sqrt{3}} .
+$$
+
+If we had wanted all integer solutions to the equation $a^{2}-3 b^{2}=1$, rather than all nonnegative integer solutions, we simply note that the equation is invariant under switching signs of $a$ and/or $b$, so the set of all solutions is simply the set of solutions given above with possible sign changes thrown in.
+
+Now we consider the given equation, $2 x^{2}-6 x y+3 y^{2}=-1$. This can be rewritten as $x^{2}-3(y-x)^{2}=1$. Thus, by letting $a=x, b=y-x$, we have an equivalence between our given equation and the Pell equation, so we can express our solutions to the former in terms of those of the latter: $x=a$ and $y=a+b$ gives
+
+$$
+x= \pm \frac{(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}}{2}, \quad y=x \pm \frac{(2+\sqrt{3})^{n}-(2-\sqrt{3})^{n}}{2 \sqrt{3}}
+$$
+
+where $n$ is any nonnegative integer. It can be checked by computation that any such pair is a valid, integral solution, and the foregoing argument shows that all solutions are of this form.
+
+5. Suppose that $S_{1}, S_{2}, S_{3}, \ldots$ are sets of integers such that no integer is contained in more than one $S_{n}$; every $S_{n}$ has exactly two elements; and the sum of the elements of $S_{n}$ is $n$. Prove that there exist infinitely many values of $n$ with the following property: one of the elements of $S_{n}$ is greater than $13 n / 7$.
+
+Solution: Consider any set $S$ representable as the union of $n$ disjoint pairs of integers such that, if any negative number $-m$ is in $S$, then the other element of its pair (henceforth called its "partner") is $\geq 13 m / 6$. Any such $S$ will be called "plausible" of order $n$. We will work with plausible sets (and their partner decompositions) in general; later, we will show how these are relevant to the original problem.
+
+Let $S$ be plausible of order $n$. Then it consists of $n$ disjoint pairs, each of which has a nonnegative sum. Hence, the sum of the elements of $S$ is a nonnegative integer. Now, if we consider the set of all possible values for the sum of $S$, this set consists of nonnegative integers, and it is nonempty (since at least one plausible set exists). Hence, it has a minimal element. So we can meaningfully attempt to characterize a plausible set $S$ (of order $n$ ) with minimal sum.
+
+This set consists of some negative numbers $-m_{1},-m_{2}, \ldots,-m_{k}$, their (positive) partners, and $2(n-k)$ other nonnegative numbers, all distinct. Clearly, for any fixed choice of the $m_{i}$ and their partners, the sum is minimized by choosing the extra numbers to be the $2(n-k)$ smallest nonnegative integers not equal to the partner of any $-m_{i}$, so we may assume the set to be of this form. Now define $f(m)=\lceil 13 m / 6\rceil$ for any integer $m$. We claim it is no loss to assume the partner of $-m_{i}$ is $f\left(m_{i}\right)$ for each $i$. Indeed, if not, then this partner is $p>f\left(m_{i}\right)$. If $f\left(m_{i}\right) \notin S$, then replacing $p$ by $f\left(m_{i}\right)$ gives us a new plausible set with smaller sum than before, contradicting minimality. So $f\left(m_{i}\right) \in S$. But then $f\left(m_{i}\right)$ has some partner $q$. We claim we can switch partnerships, matching $-m_{i}$ with $f\left(m_{i}\right)$ and $p$ with $q$, so that our partnership decomposition remains plausible. Indeed, if $q \geq 0$, this is clear. If $q<0$ then $q=$ some $-m_{j} \Rightarrow f\left(m_{i}\right) \geq f\left(m_{j}\right)$ since $f\left(m_{i}\right)$ is the partner of $-m_{j}$, but $p>f\left(m_{i}\right) \geq f\left(m_{j}\right)$, so we are again safe. Moreover, in either case, neither of the pairs involved is initially of the form $-m_{l}, f\left(m_{l}\right)$ (in the case $q=-m_{j}$, we cannot have $f\left(m_{i}\right)=f\left(m_{j}\right)$ because $f$ is injective), and after the switch, one of them is of this form, so the total number of such pairs strictly increases. So we can perform this switch as many times as necessary, and eventually, we have each $-m_{i}$ partnered with $f\left(m_{i}\right)$.
+
+Next, we claim that, in the context of the partnering result obtained in the previous paragraph, $-j \in S$ implies $-i \in S$ for $0<i<j$. Indeed, suppose a counterexample exists. If $f(i) \notin S$, then we can replace $-j$ and its partner by $-i$ and $f(i)$ to obtain a new plausible set $S$ whose sum has been changed by $\leq-i+f(i)+j-f(j)=\lceil 7 i / 6\rceil-\lceil 7 j / 6\rceil<0$, contradicting minimality. So we do have $f(i) \in S$. It cannot be partnered with a negative number, since this number would have to be $-i$ (by injectivity of $f)$ but $-i \notin S$; so it is partnered with some positive number. Then, changing this number to $-i$ would again give us a new plausible set with strictly smaller sum, contradiction. This proves the claim.
+
+But this claim shows that the negative elements of $S$ must be exactly $-1,-2, \ldots,-k$, and now we are in a position to estimate the sum of $S$. $S$ consists of these negative elements, their partners $f(1), \ldots, f(k)$, and $2(n-k)$ other, distinct "leftover" nonnegatives which are as small as possible. Let $a$ be the largest leftover, so $0, \ldots, a$ are all in $S$. We claim $b=\lfloor 6 a / 13\rfloor \leq k$; if not, then $-b \notin S$ but $f(b) \leq a$ is in $S$, so it is a leftover (because if its partner is $-c$, injectivity implies $c=b$, impossible), and replacing its nonnegative partner by $-b$ gives a new plausible set with smaller sum, contradiction. So $k \geq b$, and it follows that fully $b$ of the numbers $0, \ldots, a$ are partners of negative numbers. Hence, the number of leftovers can be counted in two ways: $a+1-b=2(n-k)$. But $b \geq 6 a / 13-1 \Rightarrow 2(n-k) \leq a-6 a / 13+2=7 a / 13+2 \Rightarrow a \geq 26(n-k-1) / 7$. Now, the elements of $S$ consist of $-k, \ldots,-1,0, \ldots, a$, and the partners of $-k, \ldots,-(b+1)$. So the sum of these numbers is
+
+$$
+\frac{a(a+1)}{2}-\frac{k(k+1)}{2}+\sum_{i=b+1}^{k}\lceil 13 i / 6\rceil \geq \frac{a(a+1)}{2}-\frac{k(k+1)}{2}+\frac{13}{6}\left(\frac{k(k+1)}{2}-\frac{b(b+1)}{2}\right) ;
+$$
+
+using the (rather crude) estimates $0 \leq a, b, k \leq 2 n$ and $b \leq 6 a / 13$ to simplify our calculations, this is
+
+$$
+\begin{gathered}
+\geq \frac{a^{2}}{2}-\frac{k^{2}}{2}+\frac{13 k^{2}}{12}-\frac{13 b^{2}}{12}-4 n \geq \frac{a^{2}}{2}+\frac{7 k^{2}}{12}-\frac{13(6 a / 13)^{2}}{12}-4 n=\frac{7 a^{2}}{26}+\frac{7 k^{2}}{12}-4 n \\
+\geq \frac{26(n-k-1)^{2}}{7}+\frac{7 k^{2}}{12}-4 n \geq \frac{26(n-k)^{2}}{7}-8 n+\frac{7 k^{2}}{12}-4 n=\frac{361 k^{2}}{84}-\frac{52 n k}{7}+\frac{26 n^{2}}{7}-12 n
+\end{gathered}
+$$
+
+$$
+=\frac{361}{84}\left(k-\frac{312 n}{361}\right)^{2}+\frac{1274}{2527} n^{2}-12 n \geq \alpha n^{2}-12 n
+$$
+
+where $\alpha=1274 / 2527>1 / 2$. So we now know that any plausible set of order $n$ has sum $\geq \alpha n^{2}-12 n$.
+
+Now let us return to the original problem statement. Suppose that the statement is false; we seek a contradiction. Let $i_{1}<i_{2}<\cdots<i_{c}$ be the (finitely many) values of $i$ for which $S_{i}$ has an element $>13 i / 7$, and let $d=i_{1}+i_{2}+\cdots+i_{c}$. Now suppose $i$ is not equal to any of $i_{1}, \ldots, i_{c}$, and suppose $-m \in S_{i}(m>0)$. Then the other element of $S_{i}$ is $i+m$, and we have $i+m \leq$ $13 i / 7 \Rightarrow m \leq 6 i / 7 \Rightarrow i+m \geq 13 m / 6$. Consequently, if we take any $n>i_{c}-c$, we see that the set $S=\left(S_{1} \cup S_{2} \cup \cdots \cup S_{n+c}\right)-\left(S_{i_{1}} \cup S_{i_{2}} \cup \cdots \cup S_{i_{c}}\right)$ is plausible of order $n$. So the sum of its elements is $\geq \alpha n^{2}-12 n$. However, this sum should also equal
+
+$$
+(1+2+\cdots+(n+c))-\left(i_{1}+i_{2}+\cdots+i_{c}\right)=(n+c)(n+c+1) / 2-d
+$$
+
+This is a quadratic function of $n$ with leading coefficient $1 / 2$, so it will actually be less than $\alpha n^{2}-12 n$ when $n$ is sufficiently large. This is our contradiction.
+
+Remark: The messy numbers appearing above suggest that $13 / 7$ is not the optimal bound. In fact, completely analogous logic can be used when $13 / 7$ is replaced by any positive number less than $1+\sqrt{3} / 2$. On the other hand, one can show that, if the sets $S_{n}$ are constructed via the greedy algorithm (i.e. for each $n$, choose the two smallest nonnegative numbers not previously used if their sum is $n$; otherwise, choose the largest unused negative number and an appropriate partner), then both elements of $S_{n}$ are $<(1+\sqrt{3} / 2) n$ for all $n$.
+
+Solutions (c) 2001, Berkeley Math Circle.
+

BayArea/md/en-monthly/en-0001-mc5sol.md

@@ -0,0 +1,78 @@
+# Berkeley Math Circle 2000-2001 
+
+## Monthly Contest \#5 — Solutions
+
+1. 2000 distinct positive integers are written down, and it so happens that the product of any 3 different numbers from this list is a square. Prove that each one of them is a square.
+
+Solution: Let $a$ be any number in the list; we wish to prove that $a$ is a square. Let $b, c, d$ be any three other elements of the list, all different from each other. Then, the given implies that there are integers $p, q, r$ with $a b c=p^{2}, a b d=q^{2}, a c d=r^{2}$. Multiplying these together gives $a^{3} b^{2} c^{2} d^{2}=p^{2} q^{2} r^{2}$; equivalently, $a=\left(p^{2} q^{2} r^{2}\right) /\left(a^{2} b^{2} c^{2} d^{2}\right)=(p q r / a b c d)^{2}$. So $a$ is the square of a rational number, but since it is an integer, we conclude that it is the square of an integer, as needed.
+
+2. Martha is a chicken who lives in a coop whose finitely many residents have a definite "pecking order": for every pair of distinct chickens, exactly one pecks the other. (However, it is not necessarily true that if $X$ pecks $Y$ and $Y$ pecks $Z$, then $X$ pecks $Z$.) A chicken $X$ is called a "leader" of the coop if every other chicken is pecked by $X$ or pecked by a chicken who is pecked by $X$. Prove that if no one in the coop pecks more chickens than Martha, then Martha is a leader.
+
+Solution: Suppose that no one pecks more chickens than $M$, yet $M$ is not a leader; we will seek a contradiction. Let $X_{1}, X_{2}, \ldots, X_{n}$ be the chickens pecked by $M$. Since $M$ is not a leader, there is a chicken $N$ who is neither pecked by $M$ nor by any of the $X_{i}$. But it is given that of any two distinct chickens, one pecks the other; hence, $N$ pecks all of $M, X_{1}, X_{2}, \ldots, X_{n}$. Thus, we have found at least $n+1$ chickens pecked by $N$. But $M$ only pecks $n$ chickens. Since we assumed nobody pecked more chickens than $M$, we have a contradiction, as needed.
+
+3. If $a, b, c, d, e, f$ are real numbers such that $a+b+c+d+e+f=0$ and $a^{3}+b^{3}+c^{3}+d^{3}+e^{3}+f^{3}=0$, prove that $(a+c)(a+d)(a+e)(a+f)=(b+c)(b+d)(b+e)(b+f)$.
+
+Solution: First, let $S$ be the sum of all possible products of three distinct variables from the set $\{a, b, c, d, e, f\}$, i.e. $S=a b c+a b d+a b e+\cdots+d e f$. Now, we assert that
+
+$$
+\begin{aligned}
+& (a+b+c+d+e+f)^{3}+2\left(a^{3}+b^{3}+c^{3}+d^{3}+e^{3}+f^{3}\right)- \\
+& 3(a+b+c+d+e+f)\left(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2}\right)=6 S
+\end{aligned}
+$$
+
+To see this, note that, when the left-hand side is expanded, the terms obtained come in three forms: $a^{3}$ (i.e. cubes of one variable), $a^{2} b$ (squares of one variable times another), and $a b c$ (products of three different variables). Each term of the form $a^{3}$ has coefficient 1 in $(a+\cdots+f)^{3}, 2$ in $2\left(a^{3}+\cdots+f^{3}\right)$, and -3 in $-3(a+\cdots+f)\left(a^{2}+\cdots+f^{2}\right)$, for a total coefficient of 0 ; the terms of the form $a^{2} b$ have respective coefficients $3,0,-3$, for a total of 0 ; those of the form $a b c$ have coefficients $6,0,0$, for a total of 6 . Since the right side of (1) consists of the $a b c$-type terms with coefficient 6 , the assertion is proven.
+
+However, the given tells us that the left-hand side of (1) is zero, so $S=0$. Now, let $p=a b-(c d+c e+$ $c f+d e+d f+e f)$. We claim that the following holds for all numbers $x$ :
+
+$$
+(x+c)(x+d)(x+e)(x+f)=(x-a)(x-b)\left(x^{2}-p\right)+a b p+c d e f .
+$$
+
+We can verify this by comparing coefficients for corresponding powers of $x$.
+
+The coefficient of $x^{4}$ on both sides is 1 ; the coefficient of $x^{3}$ is $c+d+e+f$ on the left and $-(a+b)$ on the right, but the given implies that these are equal. The coefficient of $x^{2}$ is $c d+c e+c f+d e+d f+e f$ on the left and $a b-p$ on the right; these are equal by the definition of $p$. The coefficient of $x$ is $c d e+c d f+c e f+d e f$ on the left and $a p+b p$ on the right. To see that these are equal, observe that
+
+$$
+\begin{gathered}
+c d e+c d f+c e f+d e f-(a+b) p=c d e+c d f+c e f+d e f-(a+b) a b+(a+b)(c d+c e+c f+d e+d f+e f) \\
+=(S-[a b c+a b d+a b e+a b f])-(a+b) a b=-a b(c+d+e+f+a+b)=0
+\end{gathered}
+$$
+
+where we have used the definition of $S$ for the second equality and the fact that $S=0$ for the third. Finally, the constant term equals $c d e f$ on both sides. Thus, equation (2) holds. Plugging in $x=a$ and $x=b$, we find
+
+$$
+\begin{aligned}
+& (a+c)(a+d)(a+e)(a+f)=(a-a)(a-b)\left(a^{2}-p\right)+a b p+c d e f=a b p+c d e f \\
+& (b+c)(b+d)(b+e)(b+f)=(b-a)(b-b)\left(b^{2}-p\right)+a b p+c d e f=a b p+c d e f
+\end{aligned}
+$$
+
+Since the right-hand sides match, we are done.
+
+4. Let $A B C D$ be a cyclic quadrilateral (i.e. inscribed in a circle) such that $D C=A D+B C$. Prove that the intersection point of the angle bisectors of angle $A$ and angle $B$ lies on $C D$.
+
+Solution: Let $E$ lie on segment $C D$ so that $C E=B C$. Then, $D E=D C-C E=A D$. So, $\triangle B C E$ and $\triangle A D E$ are both isosceles. Let $\alpha=\angle D A E=\angle A E D$, and let $\beta=\angle C E B=\angle E B C$. Then $\angle C D A=\angle E D A=\pi-2 \alpha$, but since the quadrilateral is cyclic, this gives $\angle A B C=2 \alpha$. Likewise, $\angle B C D=\pi-2 \beta$ and $\angle D A B=2 \beta$.
+
+Now, let the circumcircle of triangle $A B E$ intersect line $C D$ again at $F$. We will assume that $F$ lies on the same side of $E$ as $D$ (the other case is completely analogous). Since quadrilateral $A B E F$ is cyclic, we have $\angle A B F=\angle A E F=\alpha=\angle A B C / 2$, and $\angle F A B=\pi-\angle B E F=\angle C E B=\beta=\angle D A B / 2$. So $F$ lies on the bisectors of angles $D A B$ and $A B C$, which means it is precisely the intersection point specified in the problem statement. Since $F$ was constructed to lie on $C D$, we are done.
+
+5. In the UC Berkeley bureaucratic hierarchy, certain administrators report to certain other administrators. It so happens that if $A$ reports to $B$ and $B$ reports to $C$, then $C$ reports to $A$. Also, administrators do not report to themselves. Prove that all the administrators can be divided into three disjoint groups $X, Y, Z$ so that, if $A$ reports to $B$, then either $A$ is in $X$ and $B$ is in $Y$, or $A$ is in $Y$ and $B$ is in $Z$, or $A$ is in $Z$ and $B$ is in $X$.
+
+Solution: Form a graph whose vertices represent the administrators, with an edge between $A$ and $B$ if one of $A, B$ reports to the other. We will temporarily assume that the graph is connected. Consider any walk on this graph (i.e. a finite sequence of vertices, any two consecutive members of which are connected by an edge). Whenever this walk traverses an edge from $A$ to $B$, assign this edge a weight of 1 if $A$ reports to $B$ and -1 if $B$ reports to $A$; then let the "value" of the walk be the sum of the weights of its edges. This definition is unambiguous, i.e. we cannot have both $A$ reporting to $B$ and $B$ reporting to $A$, since then we would also have to have $A$ reporting to $A$, which is prohibited.
+
+We claim that if a walk is a cycle (it starts and ends at the same vertex), its value is divisible by 3 . If not, choose a counterexample that uses as few edges as possible. Let its value be $v$. We consider the possible weights of successive edges in our cycle. First, suppose there are two consecutive 1's, representing some movements from $A$ to $B$ and from $B$ to $C$. So $A$ reports to $B$, and $B$ reports to $C$; hence, from the given, $C$ reports to $A$, and we can replace these two edges $A B C$ by one edge $A C$, whose weight will be -1 . So the new cycle we have obtained has one edge less than the old one, and its value is $v-3$, which is not divisible by 3 (since $v$ wasn't). This contradicts the minimality of the original cycle, so this situation is impossible. Similarly, if there were two consecutive -1 's, corresponding to edges $C B A$, we could replace them with an edge $C A$ of weight 1 , and we would have a new cycle, shorter than the previous one, of value $v+3$, which again is not divisible by 3 . So this, too, is impossible.
+
+We conclude that the edge weights of our cycle must alternate between 1 and -1 . The initial and final edges must also have opposite weights, or else we can replace them with one edge to obtain a shorter counterexample, precisely as in the previous paragraph. Actually, we avoid this problem if the initial and final edges are the same, so that our minimal cycle consists of only one edge, but then this edge runs from $A$ to $A$, which is illegal. So, the only remaining possibility is that our cycle weights are of the form $1,-1,1,-1, \ldots,-1$, or $-1,1,-1,1, \ldots, 1$, but either way, the sum is 0 , which is in fact a multiple of 3. So we have ruled out all the cases and found that a counterexample to our claim is impossible. Thus, the claim holds.
+
+Now choose any vertex $P$. For $Q$ any other vertex, we can define the value of $Q$ to be the element of the set $\{0,1,2\}$ which is congruent mod 3 to the value of a walk from $P$ to $Q$. (Such a walk exists, since
+the graph is connected). This is well-defined as long as we know that the value of $Q$ is independent of the walk chosen. But if there exist two walks from $P$ to $Q$ with respective values $v, w$, then traversing the first walk from $P$ to $Q$ and traversing the second walk backwards from $Q$ to $P$ gives a cycle with value $v-w$. The above claim then shows that $v-w$ is divisible by 3 , so both walks do give the same value, as required.
+
+Now just let $X$ consist of all vertices of value 0 (including $P$ ), $Y$ all vertices of value 1 , and $Z$ those of value 2. If $A B$ is an edge (with $A$ reporting to $B$ ), then it is clear that a walk from $P$ to $A$, followed by the edge $A B$, gives a walk from $P$ to $B$. So, the value of $B$ equals the value of $A$ plus $1(\bmod 3)$. Consequently: if $A \in X$ then $B \in Y$; if $A \in Y$ then $B \in Z$; if $A \in Z$ then $B \in X$.
+
+Thus, the problem is solved in the connected case. But if our administrator graph was disconnected, let the components be $C_{i}$ for some set of indices $i$. For each $C_{i}$, partition it into sets $X_{i}, Y_{i}, Z_{i}$ by the connected case; then, let $X=\cup X_{i}, Y=\cup Y_{i}, Z=\cup Z_{i}$. Since any two administrators connected by an edge must lie in the same $C_{i}$, it easily follows that these sets $X, Y, Z$ do the trick.
+
+Remark: Work submitted by Neil Herriot indicated an alternate solution worth mentioning, though difficult to present in full rigor: the administrators can be successively numbered and assigned to sets $X, Y, Z$ in such a manner that, if there is an "inconsistency" (the requirements of the problem statement are not fulfilled) between two administrators, there must also be an inconsistency between two administrators with smaller numbers, leading to an infinite descent and contradiction. Although the wording of the problem suggests that there are finitely many administrators, the solution presented here works if there are infinitely (even uncountably) many, an advantage over the inductive solution.
+
+Solutions (C) 2001, Berkeley Math Circle.
+

BayArea/md/en-monthly/en-0001-mc6sol.md

@@ -0,0 +1,59 @@
+# Berkeley Math Circle 2000-2001 
+
+## Monthly Contest \#6 - Solutions
+
+1. In triangle $A B C$, let $D$ be the midpoint of side $B C$. Let $E$ and $F$ be the feet of the perpendiculars to $A D$ from $B$ and $C$, respectively. Prove that $B E=C F$.
+
+Solution: We have $\angle D F C=\pi / 2=\angle D E B$. Also, $\angle C D F=\angle B D E$ since they are vertical angles. (It seems possible that $E$ and $F$ could lie on the same side of $D$, so that $\angle C D F$ and $\angle B D E$ would be supplementary rather than equal; however, if they were supplementary and unequal, then one of them, say $\angle B D E$, would be $>\pi / 2$, so the sum of the angles of $\triangle B D E$ would be $>\pi$, a contradiction.) These equal angles imply $\triangle D F C \sim \triangle D E B$. But $C D=B C / 2=B D$, so in fact $\triangle D F C \cong \triangle D E B$, giving $C F=B E$.
+
+Alternate Solution: (Thanks to Inna Zakharevich) Let $H$ be the foot of the perpendicular from $A$ to $B C$. Then, using the $b h / 2$ formula and the fact that $D$ is the midpoint of $B C$, we get
+
+$$
+\frac{A D \cdot B E}{2}=\operatorname{Area}(\triangle A B D)=\frac{B D \cdot A H}{2}=\frac{C D \cdot A H}{2}=\operatorname{Area}(\triangle A C D)=\frac{A D \cdot C F}{2} .
+$$
+
+Multiplying by $2 / A D$ now gives $B E=C F$.
+
+2. Let $A B C$ be an equilateral triangle, and let $P$ be a point on minor $\operatorname{arc} B C$ of the circumcircle of $A B C$. Prove that $P A=P B+P C$.
+
+Solution: Extend line $P C$ through $C$ to point $D$ such that $C D=B P$. Note that $\angle A C D=$ $\pi-\angle P C A=\angle A B P$ (since quadrilateral $A B P C$ is cyclic), and $A C=A B$ since $\triangle A B C$ is equilateral. Consequently, $\triangle A C D \cong \triangle A B P$. In particular, we have $\angle P D A=\angle C D A=\angle B P A=\angle B C A$ (again by cyclicity) $=\pi / 3$. But also $\angle A P D=\angle A P C=\angle A B C$ (cyclicity) $=\pi / 3$. We conclude that triangle $A P D$ is equilateral. So, $P A=P D=P C+C D=P C+B P$.
+
+Remark: This is a special case of Ptolemy's Theorem: if $R S T U$ is any convex cyclic quadrilateral, then $R S \cdot T U+S T \cdot R U=R T \cdot S U$. The proof of the theorem is similar to the above.
+
+3. Determine all triples $(x, y, n)$ of integers such that $x^{2}+2 y^{2}=2^{n}$.
+
+Solution: It is easy to check that $\left( \pm 2^{r}, 0,2 r\right)$ and $\left(0, \pm 2^{r}, 2 r+1\right)$ satisfy this equation for any nonnegative integer $r$. We will show that these are all the solutions by an infinite descent method.
+
+So suppose we have some solution $\left(x_{0}, y_{0}, n_{0}\right)$. If $x_{0}$ is odd, then $2^{n_{0}}$ is odd, which forces $n_{0}=0$ and then $x_{0}^{2}+2 y_{0}^{2}=1$, so $y_{0}=0$ (or else $2 y_{0}^{2}>1$ already) and then $x_{0}= \pm 1$. On the other hand, if $x_{0}$ is even, we can let $x_{0}=2 x_{0}^{\prime}$ and then $4 x_{0}^{\prime 2}+2 y_{0}^{2}=2^{n_{0}} \Rightarrow y_{0}^{2}+2 x_{0}^{\prime 2}=2^{n_{0}-1}$, so $\left(x_{1}, y_{1}, n_{1}\right)=\left(y_{0}, x_{0} / 2, n_{0}-1\right)$ is another solution to our equation, where $n_{0}$ has been replaced by $n_{0}-1$. Now if $x_{1}$ is even, we can repeat this construction to get another new solution $\left(x_{2}, y_{2}, n_{2}\right)$ with $n_{0}-1$ replaced by $n_{0}-2$, and so on. These integers $n$ cannot go on decreasing forever, since there does not exist an integral solution where $n<0$. Thus, eventually our process terminates, which means we get to a solution $\left(x_{k}, y_{k}, n_{k}\right)$ with $x_{k}$ odd. By the above, this is possible only if $x_{k}= \pm 1, y_{k}=0, n_{k}=0$.
+
+On the other hand, the above construction can be performed in reverse: we have $x_{i}=2 y_{i+1}, y_{i}=$ $x_{i+1}, n_{i}=n_{i+1}+1$ for each value of $i \geq 0$. Now we claim that $\left(x_{i}, y_{i}, n_{i}\right)=\left( \pm 2^{(k-i) / 2}, 0, k-i\right)$ when $k-i$ is even, and $\left(0, \pm 2^{(k-i-1) / 2}, k-i\right)$ when $k-i$ is odd. The proof is by downward induction: the base case $i=k$ is certainly true; given that the statement holds for some $i>0$, it is simple algebra to check that it holds for $i-1$ by applying our reverse construction. Thus, the claim is true for each $i \geq 0$. In particular, $\left(x_{0}, y_{0}, n_{0}\right)=\left( \pm 2^{k / 2}, 0, k\right)$ or $\left(0, \pm 2^{(k-1) / 2}, k\right)$, which fits the form above. So, every solution is of this form.
+
+Remark: For those who like heavy machinery, this problem can also be solved quite rapidly by using unique factorization in the ring $\mathbb{Z}[\sqrt{-2}]$, factoring $x^{2}+2 y^{2}$ as $(x+\sqrt{-2} y)(x-\sqrt{-2} y)$.
+
+4. Suppose that $S$ is a set of 2001 positive integers, and $n$ different subsets of $S$ are chosen so that their sums are pairwise relatively prime. Find the maximum possible value of $n$. (Here the "sum" of a finite set of numbers means the sum of its elements; the empty set has sum 0.)
+
+Solution: The answer is $2^{2000}+1$. To see that we cannot do better than this, note that at least half of the $2^{2001}$ possible subsets of $S$ have even sums. Indeed, if all elements of $S$ are even, then all subsets have even sums; on the other hand, if there exists some odd $s \in S$, we can divide the subsets of $S$ into pairs of the form $\{T, T \cup\{s\}\}$ for each subset $T$ not containing $s$. Since the sum of $T$ and that of $T \cup\{s\}$ are of opposite parity, each pair contains exactly 1 subset with an even sum. So, in this case, half the subsets of $S$ have even sums. The upshot is that, in either case, there are at most $2^{2000}$ subsets of $S$ with odd sums. Since our chosen subsets can include at most one subset whose sum is even (because no two sums can have a common factor of 2), we cannot choose more than $2^{2000}+1$ subsets altogether.
+
+Now, we must construct an example to show that we can have $n=2^{2000}+1$. To do this, let $k=\left(2^{2000}\right)$ !, and let $S=\left\{k, 2 k, 4 k, 8 k, \ldots, 2^{1999} k, 1\right\}$. We consider the $2^{2000}$ subsets containing the element 1 , plus the one subset $\{k\}$. It is evident that $k$, the sum of the last subset, is relatively prime to the sum of any subset containing 1 , since this latter sum is of the form $a k+1$ for some $a$. So now we just need to prove that any two distinct subsets containing 1 have relatively prime sums. Well, any such set consists of several distinct powers of 2 , multiplied by $k$, plus 1 . The sum of these powers of 2 is some number $a, 0 \leq a<2^{2000}$. Thus the subset's sum is $a k+1$. However, it follows from the uniqueness of binary representation that, for each possible value of $a$, there is only one subset whose sum is $a k+1$. Consequently, if we choose another, different subset (also containing 1), its sum is $b k+1$ for some $b, 0 \leq b<2^{2000}$ with $a \neq b$. Now suppose $a k+1$ and $b k+1$ are not relatively prime; then they have some common prime factor $p$. So $p \mid a k+1$ and $p \mid b k+1$, hence $p \mid(a k+1)-(b k+1)=(a-b) k$. Then, $p \mid a-b$ or $p \mid k$. But $a-b$ is nonzero and has absolute value $<2^{2000}$, so $a-b$ is one of the factors in the product $1 \cdot 2 \cdot 3 \cdots 2^{2000}=k$, and we get $a-b \mid k$. Thus, we are guaranteed that $p$ divides $k$. But then $p$ cannot divide $a k+1$, so we have a contradiction. We conclude that our subset sums are, in fact, pairwise relatively prime, completing the proof.
+
+5. Let $x_{1}, x_{2}, \ldots, x_{1000}, y_{1}, y_{2}, \ldots, y_{1000}$ be 2000 different real numbers, and form the $1000 \times 1000$ matrix whose $(i, j)$-entry is $x_{i}+y_{j}$. If the product of the numbers in each row is 1 , show that the product of the numbers in each column is -1 .
+
+Solution: The given says that $\left(x_{i}+y_{1}\right)\left(x_{i}+y_{2}\right) \cdots\left(x_{i}+y_{1000}\right)=1$ for each $i=1,2, \ldots, 1000$. So, if we let $P(x)$ be the polynomial $\left(x+y_{1}\right)\left(x+y_{2}\right) \cdots\left(x+y_{1000}\right)-1$, the numbers $x_{i}$ are all roots of $P$. These numbers are all distinct, and there are 1000 of them. But $P$, being of degree 1000, can only have 1000 roots, so these are all the roots of $P$ and the polynomial factors as $P(x)=c\left(x-x_{1}\right)\left(x-x_{2}\right) \cdots\left(x-x_{1000}\right)$ for some constant $c$. Since the leading coefficient of $P$ is 1 , we conclude that $c=1$. Thus,
+
+$$
+\left(x+y_{1}\right)\left(x+y_{2}\right) \cdots\left(x+y_{1000}\right)-1=\left(x-x_{1}\right)\left(x-x_{2}\right) \cdots\left(x-x_{1000}\right)
+$$
+
+is a polynomial identity, valid for all $x$.
+
+Now choose any $j(1 \leq j \leq 1000)$; we wish to show that the product of the numbers in the $j$ th column of the matrix is -1 . Letting $x=-y_{j}$ in the above equation, we get
+
+$$
+\begin{aligned}
+\left(-y_{j}+y_{1}\right)\left(-y_{j}+y_{2}\right) \cdots\left(-y_{j}+y_{1000}\right)-1 & =\left(-y_{j}-x_{1}\right)\left(-y_{j}-x_{2}\right) \cdots\left(-y_{j}-x_{1000}\right) \\
+& =(-1)^{1000}\left(x_{1}+y_{j}\right)\left(x_{2}+y_{j}\right) \cdots\left(x_{1000}+y_{j}\right)
+\end{aligned}
+$$
+
+However, the product on the left-hand side is 0 , since the $j$ th factor is $-y_{j}+y_{j}=0$; also, $(-1)^{1000}=1$. Thus, we obtain $-1=\left(x_{1}+y_{j}\right)\left(x_{2}+y_{j}\right) \cdots\left(x_{1000}+y_{j}\right)$, which is what we wanted to prove.
+

BayArea/md/en-monthly/en-0001-mc7sol.md

@@ -0,0 +1,50 @@
+# Berkeley Math Circle 2000-2001 
+
+## Monthly Contest \#7 - Solutions
+
+1. Show that there exist infinitely many natural numbers $n$ with the following property: the sum of all the positive divisors of $n$, excluding $n$ itself, equals $n+12$.
+
+Solution: Let $p$ be any prime number greater than 3 ; we show that $n=6 p$ has the desired property. The positive divisors of $6 p=2 \cdot 3 \cdot p$ are $1,2,3, p, 2 \cdot 3,2 \cdot p, 3 \cdot p$, and $2 \cdot 3 \cdot p$. The sum of all the factors other than $6 p$ is equal to $6 p+12$, as needed.
+
+It is well known that there are infinitely many prime numbers. Since each value of $p$ gives a different value for $n=6 p$, we obtain infinitely many values for $n$.
+
+2. 5 married couples gather at a party. As they come in and greet each other, various people exchange handshakes - but, of course, people never shake hands with themselves or with their own respective spouses. At the end of the party, one woman goes around asking people how many hands they shook, and she gets nine different answers. How many hands did she herself shake?
+
+Solution: Suppose that there were $n$ couples, and the woman asked all $2 n-1$ other attendees how many hands they shook and received $2 n-1$ different answers. We will show that she herself shook $n-1$ hands; hence, in our particular case, the answer is 4 .
+
+We work by induction. When $n=1$, there is one couple, and no handshakes can occur, proving the base case. Now suppose the result holds for $n$ couples; we will prove it is valid for $n+1$ couples. With $n+1$ couples present, the woman receives $2 n+1$ different answers to her question. But no person $P$ can shake more than $2 n$ hands (for $2 n+2$ people, minus $P$ and $P$ 's spouse); hence, these $2 n+1$ numbers must be exactly $0,1,2, \ldots, 2 n$ in some order. In particular, one of these people, $A$, shook everyone else's hand except $A$ 's own spouse (that accounts for the " $2 n$ " answer), and another, $B$, shook no hands (the " 0 " answer). Because $B$ did not shake $A$ 's hand, $A$ and $B$ must be married to each other. The remaining $2 n$ people include the woman who asked the question, together with those who answered $1,2, \ldots, 2 n-1$ to her question. Now pretend that $A$ and $B$ had not attended the party, so we are left with $n$ couples. Each of these people shook hands with $A$ and not with $B$; therefore, when $A$ and $B$ are removed, their handshake counts become $0,1,2, \ldots, 2 n-2$. Hence, by the induction hypothesis, the questioner shook $n-1$ hands. But now, if we put $A$ and $B$ back in, we note that the woman shook $A$ 's hand as well (and not $B$ 's). So, altogether, she shook $n$ hands. This completes the induction step, and now the proof is done.
+
+3. Let $A B C D$ be a square and $E$ a point on side $C D$. The circle inscribed in triangle $A D E$ touches $D E$ at $F$, and the circle inside quadrilateral $A B C E$, tangent to sides $A B, B C, E A$, touches $A B$ at $G$. Prove that lines $A E, B D$, and $F G$ meet in a point.
+
+Solution: Extend lines $B C$ and $A E$ to intersect at $H$. Then the circle inside quadrilateral $A B C E$, tangent to $A B, B C$, and $E A$, is really the inscribed circle of $\triangle H B A$. (Actually, this is only true if the circle lies inside $\triangle H B A$ rather than outside it. However, the fact that $C E$ is parallel to $A B$ with $C E<C D=A B$ readily implies that $E$ lies between $H$ and $A$, and $C$ lies between $H$ and $B$, so that the whole quadrilateral $A B C E$ lies within $\triangle H B A$, so the circle drawn inside it does too.) Now let $P$ be the intersection point of lines $B D$ and $A E$. Consider the homothety (scaling) about $P$ that sends point $D$ to point $B$. Since homotheties preserve directions of lines, this map takes line $A D$ to the line through $B$ and parallel to $A D$, namely line $H B$. Similarly, it takes line $D E$ to line $B A$. And line $E A$ passes through $P$, the center of the homothety, so it goes to itself.
+
+Thus, our homothety takes lines $A D, D E, E A$ to lines $H B, B A, A H(=A E)$, respectively, so it takes $\triangle A D E$ to $\triangle H B A$. Consequently, the incircle of $\triangle A D E$ is mapped to the incircle of $\triangle H B A$, and the map also matches corresponding tangency points: $F$ goes to $G$. But if a homothety about $P$ takes $F$ to $G$, then $P, F, G$ must be collinear. We now know that $P$ lies on lines $A E, B D$, and $F G$, which is what we need.
+
+4. There are 3, 999,999 cities in Antarctica, and some pairs of them are connected by roads. It is known that, given any two cities, there is a sequence of roads leading from one to the other. Prove that the cities can be divided into 1999 groups (of 2001 cities each) such that, given any two cities in the same group, it is possible to get from one to the other using at most 4000 roads.
+
+Solution: First, we provide some relevant graph-theoretic background. Any finite, connected graph can be turned into a tree (a connected graph without cycles) by removing some edges. Proof: If our graph has a cycle, any edge of that cycle can be removed without disconnecting the graph. So remove this edge, leaving a new graph. If it has a cycle, we can again remove an edge; continuing in this manner, we must eventually stop, since there are only finitely many edges to remove. We then have a graph with no cycles; since no edge removal ever disconnected the graph, it must still be connected.
+
+Also, given a tree, we can choose a root vertex $r$. Then, for any vertex $v$, there is a unique path from $v$ to $r$, never repeating a vertex (uniqueness follows from the absence of cycles). We call $v$ a descendant of $w$ if this path goes through $w$. Every vertex is considered to be a descendant of itself and of $r$. Suppose $v$ is a descendant of $w$; then the path from $v$ to $r$ consists of the path from $v$ to $w$ followed by the path from $w$ to $r$. It follows that descent is transitive: if $w$ in turn is a descendant of $u$, then $v$ is a descendant of $u$. It also follows that $d(v, r)=d(v, w)+d(w, r)$, where $d(x, y)$ denotes the distance (i.e. number of edges in the path) from $x$ to $y$. Finally, a vertex $v$ with no descendants can be removed and the graph will remain connected. Proof: every other vertex is connected to $r$ by a path that does not pass through $v$, so these vertices will remain connected to $r$ - and hence to each other - when $v$ is removed.
+
+Now we can solve our original problem. We state the graph-theoretic translation: given a connected graph $G$ on $k n$ vertices $(k \geq 0, n \geq 1)$, these vertices can be partitioned into $k$ sets of size $n$ such that $d(v, w) \leq 2 n-2$ whenever $v, w$ are in the same set. (In our case, $k=1999, n=2001$.) We prove this by induction on $k$. If $k=0$, we form no vertex sets, and the statement is vacuously true. Now suppose the statement holds for $k-1$, where $k \geq 1$, and we have a graph $G$ on $k n$ vertices. It suffices to prove the result when $G$ is a tree, since otherwise we can make it a tree by removing some edges and partition the vertices of this tree appropriately. The same partition will then work for the original graph $G$, since the distance between two vertices cannot increase when we put the deleted edges back.
+
+So suppose $G$ is a tree, and arbitrarily choose a root $r$. Now let $a$ be a vertex for which $d(a, r)$ is maximal. Let $a=v_{1}, v_{2}, \ldots, v_{q}=r$ be the path from $a$ to $r$, and choose the smallest $i$ such that $v_{i}$ has at least $n$ descendants. (Some such $i$ surely exists, since $r$ has $k n$ descendants.) Let $S$ be the set of descendants of $v_{i}$; note that if $v \in S$, then every descendant of $v$ is in $S$, by transitivity. Notice that $v_{1}, v_{2}, \ldots, v_{i-1}$ are all descendants of $v_{i-1}$, so the minimality of $i$ implies $i-1<n$. Thus, $d\left(a, v_{i}\right)=i-1 \leq n-1$. Now we claim the distance between any two elements of $S$ is at most $2 n-2$. Indeed, suppose $b, c \in S$. We have $d\left(b, v_{i}\right)=d(b, r)-d\left(v_{i}, r\right) \leq d(a, r)-d\left(v_{i}, r\right)$ (by choice of $a$ ) $=d\left(a, v_{i}\right) \leq n-1$. Similarly, $d\left(c, v_{i}\right) \leq n-1$, and so $d(b, c) \leq d\left(b, v_{i}\right)+d\left(v_{i}, c\right) \leq 2(n-1)$, as claimed.
+
+Now let $a_{1}$ be an element of $S$ at maximal distance from $r$. (For example, take $a_{1}=a$.) Then $a_{1}$ can have no descendants (except itself) in $G$, since if $b$ were a descendant of $a_{1}$, we would have $d(b, r)=d\left(a_{1}, b\right)+d\left(a_{1}, r\right)$, contradicting maximality. Thus, we can remove $a_{1}$ from $G$ to leave a graph $G_{1}$, which is still connected - in fact, still a tree with root $r$. Now a similar argument shows that, if $a_{2} \in S \cap G_{1}$ is chosen to have maximal distance from $r$, then $a_{2}$ can have no descendants in $G_{1}$ : any descendant would lie in $S$, by transitivity, and it would also be farther from $r$ than $a_{2}$, violating maximality. So, deleting $a_{2}$ from $G_{1}$ gives another rooted tree, $G_{2}$. Then, we can choose $a_{3} \in S \cap G_{2}$ to be maximally distant from $r$, and so forth. We continue removing vertices in this manner; since $S$ has at least $n$ elements, we can remove $n$ vertices. Thus, we choose distinct vertices $a_{1}, a_{2}, \ldots, a_{n}$, all of which lie in $S$; this means that any two of these vertices are at distance $\leq 2 n-2$ from each other, and the remaining graph, $G-\left\{a_{1}, \ldots, a_{n}\right\}$, is still a tree. Now using the induction hypothesis, this remaining graph can be partitioned to form the remaining $k-1$ sets of vertices, and the desired partition of $G$ is accomplished.
+
+Remark: In fact, given an arbitrary connected graph with a vertex $r$ selected, we can define $v$ to be a descendant of $w$ whenever $d(v, r)=d(v, w)+d(w, r)$, and the same solution works, without having to assume the graph is a tree. However, the case of a tree helps to motivate the definition.
+
+5. Let $a_{1}, a_{2}, a_{3}, \ldots$ be a sequence of positive integers with the following property: if $S$ is any nonempty set of positive integers, there exists $s \in S$ with $a_{s} \leq \operatorname{gcd}(S)$. Prove that $n$ ! is divisible by $a_{1} a_{2} \cdots a_{n}$ for every positive integer $n$.
+
+Solution: Fix $n$. Arrange the integers $a_{1}, \ldots, a_{n}$ in nonincreasing order, $a_{i_{1}} \geq a_{i_{2}} \geq \cdots \geq a_{i_{n}}$. We claim there exists a bijective function $f:\{1,2, \ldots, n\} \rightarrow\{1,2, \ldots, n\}$ such that $a_{i_{k}} \mid f(k)$ for each $k=1,2, \ldots, n$. To demonstrate this, we construct $f$ inductively. Suppose that $f(k)$ has been defined for all values of $k$ less than some $j$, and we wish to define $f(j)$. Let $d$ be the greatest common divisor of $i_{1}, i_{2}, \ldots, i_{j}$. By the hypothesis, there exists some $i \in\left\{i_{1}, i_{2}, \ldots, i_{j}\right\}$ such that $a_{i} \leq d$; since $a_{i_{j}}=\min \left\{a_{i_{1}}, a_{i_{2}}, \ldots, a_{i_{j}}\right\}$, we have $a_{i_{j}} \leq a_{i} \leq d$. Now, we know of $j$ distinct multiples of $d$ lying in the set $\{1,2, \ldots, n\}$ (namely, $i_{1}, i_{2}, \ldots, i_{j}$ ); this many multiples can only exist if $j d \leq n$. Then, the numbers $a_{i_{j}}, 2 a_{i_{j}}, \ldots, j a_{i_{j}}$ are also all in $\{1,2, \ldots, n\}$, since $j a_{i_{j}} \leq j d \leq n$. At most $j-1$ of these can have been used up by the previously defined values $f(1), f(2), \ldots, f(j-1)$, so some value is left over; we define $f(j)$ to be such a value. Then $a_{i_{j}} \mid f(j)$, as required.
+
+So we can recursively define $f(1), f(2), \ldots, f(n)$ by the above method, and our construction ensures that $f$ is injective. Since it maps the finite set $\{1,2, \ldots, n\}$ to itself, it must actually be bijective. Since $a_{i_{k}} \mid f(k)$ for each $k$, we have
+
+$$
+a_{1} a_{2} \cdots a_{n}=a_{i_{1}} a_{i_{2}} \cdots a_{i_{n}} \mid f(1) f(2) \cdots f(n)=1 \cdot 2 \cdots n=n !
+$$
+
+Remark: In fact, $n$ ! is the smallest positive integer that necessarily satisfies this condition. Indeed, if $p$ is any prime, then we can define $a_{n}$ to be the largest power of $p$ dividing $n$, for each $n$, and this produces a sequence meeting the condition of the problem statement. Then $a_{1} a_{2} \cdots a_{n}$ is the highest power of $p$ dividing $n$ !, so taking the least common multiple of these values over all choices of $p$ gives us $n !$.
+
+Solutions (c) 2001, Berkeley Math Circle.
+

BayArea/md/en-monthly/en-0001-mc8sol.md

@@ -0,0 +1,52 @@
+# Berkeley Math Circle 2000-2001 
+
+## Monthly Contest \#8 — Solutions
+
+1. Every point of the plane is colored either red or blue. Prove that there exists an equilateral triangle all of whose vertices are the same color.
+
+Solution: Suppose that no such triangle exists; we will obtain a contradiction. Let $A B C$ be any equilateral triangle of side 1; then, by assumption, two vertices of $\triangle A B C$ are one color and the other vertex is the second color. Without loss of generality, we may suppose $A$ and $B$ are red, and $C$ is blue. Construct equilateral $\triangle A B D(D \neq C)$; then $D$ must also be blue. Extend ray $A B$ to point $E$ such that $B E=1$, and note that $\triangle C D E$ is equilateral, with $C D=C E=D E=\sqrt{3}$. Since $C$ and $D$ are both blue, $E$ must be red.
+
+But if we draw equilateral $\triangle B E F$, with $F$ and $C$ on the same side of $A B$, then $F$ must be blue (since $B, E$ are red). Now complete the equilateral triangle $C F G(G \neq B)$. We see that $C$ and $F$ are blue, so $G$ is red. However, $\triangle A E G$ is also equilateral with $A$ and $E$ red, so $G$ should be blue. This is our contradiction, so our assumption - that no monochrome equilateral triangles existed - must be false.
+
+2. The UC Berkeley math department is about to move into a new, one-story building consisting of a $2001 \times 2001$ square grid of rooms. They would like to install doors between adjacent rooms so that each room has exactly two doors. Prove that this cannot be done.
+
+Solution: Suppose that it can be done. Color the building in checkerboard fashion, and suppose that we obtain $a$ white rooms and $b$ black rooms. Each door connects a white room with a black room. So, if we consider, for each white room, the number of doors adjoining it, we will count each door exactly once. Since every room is to have 2 doors, the total number of doors will be $2 a$ by this count. But similarly, if we consider, for each black room, the number of adjoining doors, each door will be counted once, so that the total number of doors is also equal to $2 b$. So, $2 a=2 b$, or $a=b$. It follows that the total number of rooms is $a+b=2 a$, an even number. But we also know the number of rooms is $2001^{2}$, an odd number - contradiction. Hence, the desired condition cannot be met.
+
+3. The manager of Chez Gastropod wants to write a menu consisting of 15 dishes. A "meal" is defined to be a subset of this menu (possibly empty), but some meals are legal and others are not. The manager may choose which meals are legal, but there is a requirement that the intersection of any two legal meals should still be legal. He wants there to be exactly 2001 legal meals. Can he do it?
+
+Solution: The answer is yes. Take any arbitrary 15-element set (menu), and call a collection of subsets (meals) "valid" if the intersection of any two sets in the collection is again in the collection. Thus, the objective is to show that there exists a valid collection containing exactly 2001 sets. We will show, by downward induction, that there exists a valid collection with exactly $n$ sets for each $n, 0 \leq n \leq 2^{15}=32768$.
+
+The base case $n=2^{15}$ is clear: the collection of all subsets of the menu is certainly valid. For the induction step, suppose that there is a valid collection $C$ of $n$ subsets $(1 \leq n \leq 32768)$; we will show that there is a valid collection of $n-1$ subsets. Choose a subset in $C$ of maximum possible size, and remove it; let $C^{\prime}$ denote the remaining collection, so that it consists of $n-1$ subsets. We claim $C^{\prime}$ is still valid. Indeed, if $S, T \in C^{\prime}$, then the removed subset cannot be contained within either $S$ or $T$ (because of maximality), hence it certainly cannot equal their intersection. But $S \cap T$ was in $C$; hence, it is also in $C^{\prime}$, as needed. Thus, the claim holds. Now, let $n=2001$, and the problem is solved.
+
+4. Given a line segment $A B$, construct a segment half as long as $A B$ using only a compass. Construct a segment one-third as long as $A B$ using only a compass.
+
+Solution: First, we provide (part of) an algorithm for circular inversion. Suppose we are given point $O$ and a circle centered at $O$ of some radius $r$. If $P$ is a point outside the circle, we wish to construct point $Q$ on ray $O P$, satisfying $O P \cdot O Q=r^{2}$. Let the circle centered at $P$, with radius $O P$, intersect the given circle centered at $O$ at points $X$ and $Y$. Then let the circle with center $X$,
+radius $r$ (which is constructible since $r=O X$ ), and the circle with center $Y$, radius $r$, intersect at $O$ and $Q$. We claim this point $Q$ is what we want. Indeed, it is clear from symmetry that $O P$ is the perpendicular bisector of $X Y$. Since $X Q=r=Y Q, Q$ lies on this bisector as well - that is, on line $O P$. Moreover, let $H$ denote the intersection of $X Y$ with $O P$; then $P H<P X=P O \Rightarrow H$ lies on ray $O P$, and, since $Q$ is the reflection of $O$ across $X Y, Q$ will lie on the opposite side of $H$ from $O$. This shows that $Q$ lies on ray $O P$, not just on line $O P$. Finally, observe that $P X=P O$ and $X O=X Q \Rightarrow \angle O X P=\angle P O X=\angle Q O X=\angle X Q O$, so, by equal angles, $\triangle P O X \sim \triangle X Q O$. Thus, $O P / O X=Q X / Q O \Rightarrow O P \cdot O Q=O X \cdot Q X=r^{2}$, as needed.
+
+Now that this is done, we return to the original problem. By scaling, assume $A B=1$. By drawing circles centered at $A$ and $B$ of radius 1 , and letting $C$ be one of their intersection points, we obtain an equilateral $\triangle A B C$. Similarly, we successively construct equilateral triangles $B C D, B D E$ (with $D \neq A, E \neq C)$. We have $\angle A B E=\angle A B C+\angle C B D+\angle D B E=3(\pi / 3)=\pi$, so $A, B, E$ are collinear, and $A E=A B+B E=2$. Then, since we have drawn the circle with center $A$ and radius 1 , we can invert $E$ across it according to the paragraph above, obtaining $F$ such that $A F=1 / 2$. In fact, $F$ lies on the given segment $A B$, so the segment $A F$ is fully drawn.
+
+Similarly, to construct a segment of length $1 / 3$, let $A B$ be given; extend $A B$ to $E$ as above so that $B E=1$, and then repeat the process, extending $B E$ to $G$ so that $E G=1$. Then $A G=3$, and inverting $G$ across our circle (center $A$, radius 1 ) will give what we need.
+
+5. Let $a_{1}=3$ and define $a_{n+1}=\left(3 a_{n}^{2}+1\right) / 2-a_{n}$ for $n \geq 1$. If $n$ is a power of 3 , prove that $a_{n}$ is divisible by $n$.
+
+Solution: The main trick is finding a closed-form expression for $a_{n}$, which requires some experimentation. We will show that $a_{n}=\left(2^{2^{n}+1}+1\right) / 3$ for all $n$ by induction. It is easy to check that the formula holds for $n=1$. And if it holds for some $n$, then
+
+$$
+\begin{aligned}
+a_{n+1} & =\frac{3 a_{n}^{2}+1}{2}-a_{n} \\
+& =\frac{1}{2}\left(3\left[\frac{2^{2^{n}+1}+1}{3}\right]^{2}+1\right)-\frac{2^{2^{n}+1}+1}{3} \\
+& =\frac{1}{2}\left(\frac{\left(2^{2^{n}+1}\right)^{2}+2\left(2^{2^{n}+1}\right)+1}{3}+1\right)-\frac{2^{2^{n}+1}+1}{3} \\
+& =\frac{1}{2}\left(\frac{2^{2^{n+1}+2}+2\left(2^{2^{n}+1}\right)+4}{3}\right)-\frac{2^{2^{n}+1}+1}{3} \\
+& =\frac{2^{2^{n+1}+1}+2^{2^{n}+1}+2}{3}-\frac{2^{2^{n}+1}+1}{3} \\
+& =\frac{2^{2^{n+1}+1}+1}{3},
+\end{aligned}
+$$
+
+giving the induction step.
+
+Now, by Euler's theorem, $3^{k}$ divides $2^{2 \cdot 3^{k-1}}-1$, for any nonnegative integer $k$, since $\phi\left(3^{k}\right)$ (i.e. the number of integers in $\left\{1,2, \ldots, 3^{k}\right\}$ relatively prime to $3^{k}$ ) equals $2 \cdot 3^{k-1}$. But notice that $2^{2 \cdot 3^{k-1}}-1=$ $\left(2^{3^{k-1}}-1\right)\left(2^{3^{k-1}}+1\right)$, and $3^{k-1}$ is odd $\Rightarrow 2^{3^{k-1}} \equiv 2(\bmod 3) \Rightarrow 2^{3^{k-1}}-1$ is relatively prime to $3^{k}$, so, in fact, $3^{k}$ divides $2^{3^{k-1}}+1$. Also, for any integers $c, d$ with $d$ odd, $2^{c}+1 \mid 2^{c d}+1$. We conclude that $3^{k} \mid 2^{a}+1$ whenever $3^{k-1} \mid a$ and $a$ is odd.
+
+Applying this result twice in succession, we find that $3^{k} \mid 2^{3^{k}}+1$ and then that $3^{k+1} \mid 2^{2^{3^{k}}+1}+1$, so that $3^{k} \mid\left(2^{2^{3^{k}}+1}+1\right) / 3=a_{3^{k}}$ for any integer $k \geq 0$, and this is what we wanted to prove.
+
+Solutions (C) 2001, Berkeley Math Circle.
+

BayArea/md/en-monthly/en-0304-mc1sol.md

@@ -0,0 +1,63 @@
+![](https://cdn.mathpix.com/cropped/2024_04_17_6ca0f9eb2283e780b632g-1.jpg?height=358&width=1524&top_left_y=282&top_left_x=276)
+
+# Berkeley Math Circle Monthly Contest 1 <br> Due October 19, 2003 
+
+1. a) One Sunday, Zvezda wrote 14 numbers in a circle, so that each number is equal to the sum of its two neighbors. Prove that the sum of all 14 numbers is 0 .
+
+b) On the next Sunday, Zvezda wrote 21 numbers in a circle, and this time each number was equal to half the sum of its two neighbors. What is the sum of all 21 numbers, if one of the numbers is 3 ?
+
+Solution.
+
+a) Denoting the numbers $a_{1}, a_{2}, \ldots, a_{14}$, and their sum as $S$ we have $a_{i}=$ $a_{i-1}+a_{i+1}$ for $i=1, \ldots, 14$ (we take $a_{15}=a_{1}, a_{0}=a_{14}$ ). Summing all these equalities we get $S=2 S$ (since each $a_{i}$ appears exactly once on the left and exactly twice on the right). Therefor $S=0$.
+
+b) Again, let the numbers be $a_{1}, \ldots, a_{21}$. Let $a_{i}$ be maximal among the numbers. Then we have $a_{i} \geq a_{i-1}, a_{i} \geq a_{i+1}$ and so $a_{i} \geq \frac{a_{i-1}+a_{i+1}}{2}$. But $a_{i}=\frac{a_{i-1}+a_{i+1}}{2}$, so both inequalities are actually equalities. Next, considering now $a_{i-1}$ we conclude $a_{i+1}=a_{i+2}$. Proceeding in this way, we conclude that all the numbers are equal. As one of them is 3 , the sum of all numbers is $3 * 21=63$.
+
+2. A grasshopper lives on a coordinate line. It starts off at 1. It can jump either 1 unit or 5 units either to the right or to the left. However, the coordinate line has holes at all points with coordinates divisible by 4 (e.g. there are holes at $-4,0,4,8$ etc.), so the grasshopper can not jump to any of those points. Can it reach point 3 after 2003 jumps?
+
+Solution.
+
+Each jump changes the parity of grasshopper's coordinate. After 2003 jumps the grasshopper will be at an even point on the coordinate line, and therefore can not be at 3 .
+
+3. The sets $A$ and $B$ and be form a partition of positive integers if $A \cap B=\emptyset$ and $A \cup B=N$. The set $S$ is called prohibited for the partition, if $k+l \neq s$ for any $k, l \in A, s \in S$ and any $k, l \in B, s \in S$.
+
+a) Define Fibonacci numbers $f_{i}$ by letting $f_{1}=1, f_{2}=2$ and $f_{i+1}=f_{i}+f_{i-1}$, so that $f_{3}=3, f_{4}=5$ etc. How many partitions for with the set $F$ of all Fibonacci numbers is prohibited are there? (We count $A, B$ and $B, A$ as the same partition.)
+
+b) How many partitions for which the set $P$ of all powers of 2 is prohibited are there? What if we require in addition that $P \subseteq A$ ?
+
+Solution.
+
+b) We prove the following: Given a partition of the set of all powers of 2 (i.e. two sets $Q$ and $R$ such that each $2^{k}$ is in exactly one of $Q, R$ ) there exists unique partition $A, B$ of positive integers with all powers of 2 prohibited and with $Q \subseteq A, R \subseteq B$. Note that this implies that there are infinitely many partitions of integers with powers of 2 prohibited, and exactly one such partition with $P \subseteq A$.
+
+First we show that required partition is unique if it exists. Without loss of generality 1 is in $A$. Suppose now that we have been able to place all integers up to some $k$ unambiguously, i.e. for any $l<k$ we know whether $l$ is in $A$ or in $B$. If $k$ is a power of 2 we know where to put it. Otherwise let $2^{i}$ be the smallest power of 2 strictly greater than $k$. Then $d=2^{i}-k$ is positive and less than $k$ Therefore we know to which of two sets $(A$ or $B$ )the number $d$ belongs. Then we have no choice but to place $k$ in the other set (otherwise $k$ and $d$ would be in the same set, and since $k+d=2^{i}$ this would contradict $A, B$ being acceptable). So there exists no more than one acceptable partition.
+
+We now show that desired partition exists. Since we have already defined above a recursive construction producing $A, B$, we just need to check that the resulting partition is in fact acceptable. Suppose not. Then there exist $n<m$ such that $n+m=2^{i}$ for some $i$ and $m, n$ are in the same set of the partition. Then $m$ is not a power of 2 (if it were, $n$ would be greater than $m$ ). Note that $n<m$ implies that $2^{i}$ is the smallest power of 2 strictly greater than $m$. Then by construction $m$ is in the set different from $n$. Contradiction. So there is no such pair $m, n$. The partition constructed above works. This completes the proof.
+
+a) The proof is similar to that of part b. We will call a partition acceptable if $F$ is prohibited for it. First, we show that there exists no more than one acceptable partition. Since we do not distinguish between $A, B$ we may assume without loss of generality that 1 is in $A$. Suppose now that we have been able to place all integers up to some $k$ unambiguously, i.e. for any $l<k$ we know whether $l$ is in $A$ or in $B$. Let $f_{i}$ be the smallest Fibonacci number strictly greater than $k$. Then $d=f_{i}-k$ is positive and less than $k$ (to see that, note that if $f_{i}-k>k$ then $2 k<f_{i}, k<f_{i} / 2 \leq f_{i-1}$, contradicting the choice of $f_{i}$ ).
+
+Therefore we know to which of two sets $(A$ or $B$ )the number $d$ belongs. Then we have no choice but to place $k$ in the other set (otherwise $k$ and $d$ would be in the same set, and since $k+d=f_{i}$ this would contradict $A, B$ being acceptable). So there exists no more than one acceptable partition.
+
+We now show that there actually exists an acceptable partition. Since we have already defined above a recursive construction producing $A, B$, we just need to check that the resulting partition is in fact acceptable. Suppose not. Then there exist a smallest $m$ such that $n+m=f_{i}$ for some $n<m$, and $m, n$ are in the same set (by renaming $A$ and $B$ if necessary we may assume $m, n$ are in $A$ ), i.e. the first $m$ for which there is a problem with the above recursive construction. Then if $f_{j}$ is as before the smallest Fibonacci number bigger than $m$ we have $f_{j-1} \leq m<f_{j}$ which together with $n<m$ gives $f_{j-1}<m+n<2 f_{j}<f_{j+2}$. But $m+n=f_{j}$ is excluded by construction (recall that $m$ is assigned to the set other than that of $f_{j}-m$ ). So it must be that $m+n=f_{j+1}$. On the other hand for $\hat{m}=f_{j}-m, \hat{n}=f_{j}-n$ we have:
+
+1. $\hat{n}>\hat{m}>0$ and $\hat{n}=f_{j}-n<m$, so the bigger of $\hat{n}, \hat{m}$ is less than the bigger of $m, n$.
+2. $\hat{m}+\hat{n}=f_{j}+f_{j}-(m+n)=f_{j}+f_{j}-f_{j+1}=f_{j}-f_{j-1}=f_{j-2}$.
+3. $\hat{m}$ is in $B$ by construction.
+4. $\hat{n}$ is in $B$, because otherwise $\hat{n}, n$ will be a pair of elements of $A$ adding up to a Fibonacci number with the maximal element in the pair less than $m$, contradicting our choice of $m$.
+
+These observations together mean that $\hat{n}, \hat{m}$ is a pair of elements of $B$ adding up to a Fibonacci number with the maximal element in the pair less than $m$, contradicting our choice of $m$. This contradiction shows that $A, B$ constructed above is indeed an acceptable partition.
+
+4. The circle $\omega$ is drawn through the vertices $A$ and $B$ of the triangle $A B C$. If $\omega$ intersects $A C$ at point $M$ and $B C$ at point $P$. The segment $M P$ contains the center of the circle inscribed in $A B C$. Given that $A B=c, B C=a$ and $C A=b$, find $M P$.
+
+Solution 1: Since $A M P B$ is cyclic, $\angle C M P=\angle C B A$ and $\angle C P M=\angle C A B$. Therefore the triangle $C M P$ is similar to the triangle $C B A$. Let $C M=x C B=$ $x a$, where $x$ is the coefficient of similarity. Then $C P=x C A=x b, M P=$ $x A B=x c$. If $I$ is the center of the circle inscribed in $A B C$ and $r$ - its radius, we have the following equality for the areas: $S_{C M P}=S_{C M I}+S_{C P I}$. But $S_{C M P}=\frac{1}{2} x a x b \sin C$, and $S_{C P I}=\frac{1}{2} x b r, S_{C P M}=\frac{1}{2} x a r$. Plugging these in and solving for $x$ we get $x=f r a c(b+a) r b a \sin C$. But $\frac{1}{2} a b \sin C=S_{A B C}=$ $(a+b+c) r$. Hence $x=\frac{a+b}{a+b+c}, M P=\frac{c(a+b)}{a+b+c}$.
+
+Solution 2: Keeping the notation from the first solution, we have
+
+$M P=M I+I P=r\left(\frac{1}{\sin B}+\frac{1}{\sin A}\right)=r\left(\frac{a c}{2 S_{A B C}}+\frac{b c}{2 S_{A B C}}\right)=\frac{c(a+b)}{a+b+c}$.
+
+5. For which $n$ is it possible to fill the $n$ by $n$ table with 0 's, 1 's and 2 's so that the sums of numbers in rows and columns take all different values from 1 to $2 n$ ?
+
+Solution.
+
+For odd $n$ it is impossible to create a table like that. Indeed, in such a table the sum of all column sums and row sums would be $1+2+\ldots+2 n=n(2 n+1)$, and so would be odd. But it would also be twice the sum of all the numbers in the table (each number counted twice - once in the column sum, once in the row sum), and so would have to be even. This contradiction proves that no such table exists.
+
+For any even $n$ such a table exists. If one takes $2 k$ by $2 k$ table and fills in all the elements above the main diagonal (running from upper left to lower right corner of the table) with 2's, all the elements below the main diagonal with 0 's, first $k$ elements on the diagonal with $1 \mathrm{~s}$ and the last $k$ elements on the main diagonal with 2's one gets a table that satisfies the conditions. Indeed, the sums of elements in the first $k$ rows (top to bottom) are $4 k-1,4 k-3, \ldots, 2 k+1$, the next $k$ rows $-2 k, 2 k-2, \ldots, 2$. The sums of elements in the first $k$ columns (left to right) are $1,3, \ldots, 2 k-1$, the next $k$ columns $-2 k+2, \ldots, 4 k$. We see that all the numbers from 1 to $4 k$ appear exactly once, as wanted.
+

BayArea/md/en-monthly/en-0304-mc2sol.md

@@ -0,0 +1,70 @@
+![](https://cdn.mathpix.com/cropped/2024_04_17_352bf5696dff36172414g-1.jpg?height=355&width=1524&top_left_y=283&top_left_x=276)
+
+# Berkeley Math Circle <br> Monthly Contest 2 <br> Solutions 
+
+1. Solve
+
+$$
+2 \sqrt{1+x \sqrt{1+(x+1) \sqrt{1+(x+2) \sqrt{1+(x+3)(x+5)}}}}=x
+$$
+
+Solution.
+
+As the left hand side is nonnegative, we see that any solution will have $x \geq 0$. For such $x$ we have $\sqrt{1+(x+3)(x+5)}=\sqrt{x^{2}+8 x+16}=\sqrt{(x+4)^{2}}=\mid x+$ $4 \mid=x+4$. Proceeding similarly we get
+
+$$
+\begin{aligned}
+& 2 \sqrt{1+x \sqrt{1+(x+1) \sqrt{1+(x+2) \sqrt{1+(x+3)(x+5)}}}} \\
+= & 2 \sqrt{1+x \sqrt{1+(x+1) \sqrt{1+(x+2)(x+4)}}} \\
+= & 2 \sqrt{1+x \sqrt{1+(x+1)(x+3)}}=2 \sqrt{1+x(x+2)}=2(x+1)
+\end{aligned}
+$$
+
+Solving $2(x+1)=x$ gives $x=-2$, which is negative. Therefore the equation has no solutions.
+
+2. The circle $\omega$ passes through the vertices $A$ and $B$ of a unit square $A B C D$. It intersects $A D$ and $A C$ at $K$ and $M$ respectively. Find the length of the projection of $K M$ onto $A C$.
+
+Solution.
+
+Let $T$ be the point of intersection of $\omega$ with $B C$. Then, as $\angle A B T$ a right angle, $A T$ is a diameter, and $\angle A M T$ is also a right angle. Therefore the projections of $K M$ and $K T$ on $A C$ coincide. But the length of the projection of $K T$ is $\frac{\sqrt{2}}{2}$ because the length of $K T$ is one, and the angle between $K T$ and $A C$ is 45 .
+
+3. A king is placed in the left bottom corner of the 6 by 6 chessboard. At each step it can either move one square up, or one square to the right, or diagonally - one up and one to the right. How many ways are there for the king to reach the top right corner of the board?
+
+Solution.
+
+We shall make a $6 \times 6$ table. In each cell of the table we will write a number of ways in which the king can reach that cell. We will fill it out gradually starting with a row of ones at the bottom and a column of ones at the left. To fill out the rest we use the following rule: the number in each cell is equal to the sum of the numbers immediately below, to the left, and diagonally (to the left and below). The result is:
+
+| 1 | 11 | 61 | 231 | 681 | 1683 |
+| :---: | :---: | :---: | :---: | :---: | :---: |
+| 1 | 9 | 41 | 129 | 321 | 681 |
+| 1 | 7 | 25 | 63 | 129 | 231 |
+| 1 | 5 | 13 | 25 | 41 | 61 |
+| 1 | 3 | 5 | 7 | 9 | 11 |
+| 1 | 1 | 1 | 1 | 1 | 1 |
+
+4. In the triangle $A B C$ the angle $B$ is not a right angle, and $A B: B C=k$. Let $M$ be the midpoint of $A C$. The lines symmetric to $B M$ with respect to $A B$ and $B C$ intersect $A C$ at $D$ and $E$. Find $B D: B E$.
+
+Solution.
+
+As $B C$ is the angle bisector in the triangle $M B E$, we have $\frac{C E}{B E}=\frac{C M}{B M}$ (by a well-known property of the angle bisector). Similarly, $\frac{A D}{B D}=\frac{A M}{B M}$. Draw a line $B M^{\prime}$ symmetric to $B M$ with respect to the angle bisector of $A B C$ (point $M^{\prime}$ is on the line $A C$ ). $B M^{\prime}$ bisects the angle $D B E$. Using the same property of the angle bisector, we get $\frac{E M^{\prime}}{B E}=\frac{D M^{\prime}}{B D}$. Subtracting from this $\frac{C E}{B E}=\frac{C M}{B M}$ we get $\frac{C M^{\prime}}{B E}=\frac{A M^{\prime}}{B D}$ or $\frac{B D}{B E}=\frac{A M^{\prime}}{C M^{\prime}}$.
+
+Now it remains only to find the ratio in which $M^{\prime}$ divides $A C$. To do that, note that $M B C$ and $M B A$ have equal areas: $\frac{1}{2} B M \cdot B C \cdot \sin \angle M B C=$ $\frac{1}{2} B M \cdot B A \cdot \sin \angle M B A$. Therefore $\frac{\sin \angle M B C}{\sin \angle M B A}=\frac{A B}{B C}=k$. Hence
+
+$$
+\begin{aligned}
+\frac{A M^{\prime}}{C M^{\prime}}=\frac{S_{A B M^{\prime}}}{S_{B C M^{\prime}}} & =\frac{\frac{1}{2} B M^{\prime} \cdot B A \cdot \sin \angle M^{\prime} B A}{\frac{1}{2} B M^{\prime} \cdot B C \cdot \sin \angle M^{\prime} B C} \\
+& =k \cdot \frac{\sin \angle M^{\prime} B A}{\sin \angle M^{\prime} B C}=\frac{\sin \angle M B C}{\sin \angle M B A}=k \cdot k=k^{2}
+\end{aligned}
+$$
+
+5. One marks 16 points on a circle. What is the maximum number of acute triangles with vertices in these points?
+
+Solution.
+
+Consider the set of all angles $M_{1} M_{2} M_{3}$, where $M_{1}, M_{2}$ and $M_{3}$ is an arbitrary triple of selected points. There are $\frac{16 \cdot 15 \cdot 14}{2}=1680$ different angles in this
+set. Suppose $n$ of them are not acute. We shall prove $n \geq 392$. For each integer $m$ between 1 and 7 , take a chord with endpoints among the selected points such that there are exactly $m$ selected points to one side of the chord (not including the endpoints). We will call such a chord an m-chord. Each $m$-chord subtends not less than $m$ nonacute angles with vertices among the marked points. For each $m \leq 6$ there are exactly $16 m$-chords, and for $m=7$ there are exactly 8 of them. So the total number of nonacute angles is at least $16(1+2+\ldots+6)+8 * 7=392$.
+
+There are $\frac{16 \cdot 15 \cdot 14}{6}=560$ triangles with vertices among the marked points. Each nonacute angle will "spoil" exactly one triangle, so the number of acute triangles is not greater than $560-392=168$.
+
+It is only left to construct an example with exactly 168 acute triangles. Mark eight consecutive vertices of a right 16 -gon $V_{1}, \ldots V_{8}$. Draw a line through the center of the 16-gon not parallel to $V_{1} V_{8}$, such that all $V$ 's lie on the same side of that line (and not on the line). Reflecting the $V^{\prime} s$ with respect to that line we get points $V_{1}{ }^{\prime}, \ldots V_{8}{ }^{\prime}$. We claim that the set $V_{1}, \ldots V_{8}, V_{1}{ }^{\prime} \ldots V_{8}{ }^{\prime}$ is as wanted. Indeed, there are no diametrically opposite points in this set (otherwise we would get $V_{1} V_{1}{ }^{\prime}=0$ or $V_{8} V_{8}{ }^{\prime}=0$ ), and so for $m=7$ each $m$-chord subtends exactly 8 nonacute angles. Moreover, for each $m \leq 6$ each $m$-chord subtends exactly $m$ nonacute angles, and so $n=392$, and the number of acute triangles is 168 .
+

BayArea/md/en-monthly/en-0304-mc3sol.md

@@ -0,0 +1,99 @@
+# Berkeley Math Circle Monthly Contest 3 Solutions 
+
+1. Find all integer solutions to $x y=2003(x+y)$
+
+Solution.
+
+We get
+
+$$
+\begin{gathered}
+2003^{2}-2003 x-2003 y-x y=2003^{2} \\
+(2003-x)(2003-y)=2003^{2}
+\end{gathered}
+$$
+
+So $(2003-x)$ is a divisor of $2003^{2}$. As 2003 is a prime number, we conclude that $2003-x$ is $-2003^{2},-2003,-1,1,2003$, or $2003^{2}$, and $x$. Then for $x, y$ we obtain
+
+$$
+\begin{array}{ll}
+x=2003^{2}+2003=4014012, & y=2004, \\
+x=2003+2003=4006, & y=4006, \\
+x=1+2003=2004, & y=4014012, \\
+x=-1+2003=2002, & y=-401006, \\
+x=-2003+2003=0, & y=0, \\
+x=-2003^{2}+2002=-401006, & y=2002 .
+\end{array}
+$$
+
+2. Show that for each $n \geq 17$ one can cut a square into $n$ smaller squares.
+
+Solution.
+
+One can cut a square into $n$ smaller squares for all $n \geq 6$. Here is how.
+
+For $n=4$
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_40dfa36454fa0dbb3b15g-1.jpg?height=109&width=114&top_left_y=1693&top_left_x=503)
+
+For $n=6$
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_40dfa36454fa0dbb3b15g-1.jpg?height=140&width=159&top_left_y=1889&top_left_x=503)
+
+For $n=8$
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_40dfa36454fa0dbb3b15g-1.jpg?height=186&width=186&top_left_y=2117&top_left_x=501)
+
+Suppose we know how to cut a square into $n$ smaller squares. If we take one of the square "pieces" and cut it onto four smaller squares (as in the picture above for $n=4$. Now we have cut our original square into $n+3$ pieces. Since we know how to cut a square into 4 pieces, we can cut it into $7,10, \ldots, 3 k+1, \ldots$ pieces. Similarly, starting with 6 pieces we can get $6,9, \ldots, 3 k, \ldots$ pieces, and from 8 pieces $-8,11, \ldots, 3 k+2, \ldots$ pieces. Hence we can get any number of pieces bigger or equal to 6 .
+
+3. In each cell of $n \times n$ table there is an arrow pointing in one of eight principal directions (i.e. one of the following: $\rightarrow, \leftarrow, \uparrow, \downarrow, \nearrow, \searrow, \nwarrow, \swarrow$ ), in such a way that the arrows in any two adjacent cells form an angle of no more than 45 degrees, and the arrows in any two cells adjacent diagonally (i.e. sharing a vertex) form an angle of no more than 90 degrees. We start at some cell and follow the arrows (e.g. if the arrow in our cell is $\uparrow$ we move up by one, if it is $\searrow$ we move diagonally down and to the right, etc.). Prove that we will eventually escape from the table (i.e. will reach a cell on the boundary of the table where the arrow will point "out of the table").
+
+Solution.
+
+Suppose that there exists such a table of arrows and a cell in it so that we can follow the arrows without ever escaping the table. Then, since the number of cells in the table is finite, we must at some point return to a cell we have already visited, and then continue in a loop. Let's rotate all the arrows in our original $n \times n$ table by 45 degrees, so that the arrows in the cells along our loop point into the loop after the rotation. Note that because the adjacent arrows of the loop are at no more than 90 degrees, the rotated arrows will indeed point to the cells inside of the loop. The new table of (rotated) arrows still has the property that any two adjacent cells form an angle of no more than 45 degrees, and the arrows in any two cells adjacent diagonally form an angle of no more than 90 degrees. If we now start at some point on the loop and follow the (rotated) arrows, then since all the arrows on the boundary of the loop point inward, we will never escape from inside of the loop, and so there is going to be a yet smaller loop inside of our first loop. We can now repeat the process, obtaining smaller and smaller loops along the way. But the number of cells inside the loop (counting those on the loop itself) is an integer, and so it can not decrease indefinitely. This contradiction shows that one will always escape from the table.
+
+4. Let $\mathbf{R}^{+}$be the set of all positive real numbers. Find all functions $f$ : $\mathbf{R}^{+} \rightarrow \mathbf{R}^{+}$such that
+
+$$
+f(x) f(y f(x))=f(x+y)
+$$
+
+for all $x, y \in \mathbf{R}^{+}$.
+
+Solution.
+
+If $f(x)>1$ for some $x$, then for $y=\frac{x}{f(x)-1}$ we get
+
+$f(x) f\left(\frac{f(x) x}{f(x)-1}\right)=f(x) f(y f(x))=f(x+y)=f\left(x+\frac{x}{f(x)-1}\right)=f\left(\frac{f(x) x}{f(x)-1}\right)$,
+which leads to $f(x)=1$, a contradiction. Hence $f(x) \leq 1$ for all $x$. From this we get $f(x+y)=f(x) f(y f(x)) \leq f(x)$ for all $x, y$, so $f$ is non-increasing.
+
+Suppose there exists an $x_{0}$ with $f\left(x_{0}\right)=1$. Then $f\left(x_{0}+y\right)=f(y)$ for all $y$, and then $f\left(k x_{0}+y\right)=f(y)$ by induction on $k$. Also, by monotonicity $f(x)=1$ for all $x \leq x_{0}$, and since any $y$ can be written as $k x_{0}+y_{0}$ with $y_{0} \leq x_{0}$, we conclude $f(y)=f\left(y_{0}\right)=1$ for all $y$.
+
+If there is no $x$ with $f(x)=1$. Then $f$ is monotone decreasing. Now
+
+$$
+\begin{aligned}
+f(x) f(y f(x)) & =f(x+y)=f(y f(x)+x+(1-f(x)) y) \\
+& =f(y(f(x))) f((x+(1-f(x)) y) f(y(f(x))),
+\end{aligned}
+$$
+
+so $f(x)=f((x+(1-f(x)) y) f(y(f(x))$. As $f$ is monotone decreasing, this implies $x=x+(1-f(x)) y) f(y(f x)$. Setting $x=1, z=x f(1)$ and $a=f(1)$ we get $f(z)=\frac{1}{1+a z}$. Combining this with the case $f(z)=1$, we conclude that $f(x)=\frac{1}{1+a x}$ for each $x$ with $a \geq 0$. Conversely, a direct verification shows that the functions of this form satisfy the initial equality.
+
+5. A circle is tangent to the continuations of sides $C A$ and $C B$ of the triangle $A B C$, and is also tangent to the side $A B$ at point $P$. Prove that the radius of the circle tangent to $A P, C P$ and the circle circumscribed around $A B C$ is equal to the radius of the circle inscribed in $A B C$.
+
+Solution.
+
+Let $K$ and $M$ be the tangency points of the circle with $A P$ and $C P$ respectively, $L$ - the point where it touches the circumscribed circle of $A B C, T$ - the middle of the arc $A B$ of the circumscribed circle, $I$ - the center of the inscribed circle of $A B C$.
+
+The tangent to the circle $A B C$ at $T$ is parallel to the line $A B$. So the similarity transformation ("stretching") centered at $K$ taking the circle $A B C$ to the circle $K L M$ takes $A B$ to this tangent, and hence $T$ to $L$. So $K, L$ and $T$ are collinear. We now prove that points $K, M$ and $I$ are collinear. Let $M^{\prime}$ be the point of intersection of the line $K I$ with the circle $K L M$. We want to show $M=M^{\prime}$.
+
+First, let's note that the quadrilateral $L C I M^{\prime}$ can be inscribed into a circle. In fact, $\angle L C T$ is equal is equal to the half-sum of the $\operatorname{arcs} L A$ and $A T$, which is equal to $\frac{1}{2}(\breve{L A}+T B)=\angle L K A$. But $L K A$ is the angle formed by the tangent $A K$ and the chord $L K$ of the circle $L K M$, and is therefore equal to $L M^{\prime} K$, subtended by this chord. Hence $\angle L C T=\angle L M^{\prime} K$ and the quadrilateral $L C M I$ can be inscribed in a circle. Therefore $\angle L M^{\prime} C=\angle L I C$.
+
+Further, $\angle A K T=\frac{1}{2}(\breve{A T}+\breve{B L})=\frac{1}{2}(\check{T B}+\check{B L})=\angle L A T$. Hence the triangles $T A K$ and $T L A$ are similar, and so $T K \cdot T L=T A^{2}$. Also $\angle A I T=$ $\angle C A I+\angle A C I=\angle C A I+\angle T A B=\angle B A I+\angle T A B=\angle T A I$. Therefore $T A I$ is an equilateral triangle, and so $T I^{2}=T A^{2}=T K \cdot T L$, which implies
+that the triangles $T K I$ and $T I L$ are similar, and so $\angle L I C=\angle L K I$. Hence $\angle L M^{\prime} C=\angle L I C=\angle L K M^{\prime}$, so the angle subtended by the chord $L M^{\prime}$ is equal to the angle between this chord and $M^{\prime} C$, so $M^{\prime} C$ is a tangent, and so $M^{\prime}=M$.
+
+It is time to use the definition of the point $P$. Draw a tangent to the inscribed circle of $A B C$ parallel to $A B$. Let it be tangent to this circle at point $F$. The circle tangent to the continuations of $A C, B C$ and to $A B$ at $P$ can be obtained from the inscribed circle of $A B C$ by a similarity transformation centered at $C$. This transformation would take $F$ to $P$. Hence $C, F$ and $P$ are collinear. Let $V$ be the midpoint of $P E$, where $E$ is the tangency point of $A B$ with the inscribed circle of $A B C$. As $I V \| F P$ ( $I V$ connects the midpoints of the sides of $F E P$ ), $K I V$ is similar to $K M P$, and so $V I=V K$. Hence $I E^{2}=V I^{2}-V E^{2}=V K^{2}-V E^{2}=(V K+V E)(V K-V E)=E K \cdot P K$.
+
+Now let $D$ be the center of the circle $K L M$. As $\angle K D P=\angle I K E$, the triangles $K D P$ and $E K I$ are similar. Hence $D K \cdot I E=E K \cdot P K$, and so $I E^{2}=E K \cdot P K=D K \cdot I E$, and so $I E=D K$, as wanted.
+

BayArea/md/en-monthly/en-0506-mc1sol.md

@@ -0,0 +1,21 @@
+# Berkeley Math Circle <br> Monthly Contest 1 - Solutions 
+
+1. Whenever the child takes two apples from the tree, the number of green apples will decrease by 2 or remain unchanged - the parity of green apples remains the same. Since that number was odd at the beginning, it must be odd at the end. Hence, the last apple must be green.
+2. It is easy to verify that $x^{3}$ has only three remainders upon division by 9 , namely 0,1 and 8 . Writing down all possible combinations, we see that $x^{3}+y^{3}+z^{3} \not \equiv 5(\bmod 9)$, while $500 \equiv 5(\bmod 9)$. This proves the statement.
+3. $F$ is the orthocenter of the triangle $A B E$ (or $E$ is the orthocenter of $\triangle A B F$ ) hence $E F \perp A B$. From $\varangle C S D=90^{\circ}$ we conclude that $\varangle C A D=45^{\circ}$ and $\triangle A C F$ is right-angled with $A C=C F$. Since $\varangle E C F=\varangle B C A=90^{\circ}$ and $\varangle E F C=$ $\varangle B A C$ (both of them are equal to $90^{\circ}-\varangle C E F$, as can be easily seen from $\triangle C F E$ and triangle formed by lines $A E, E F$ and $A B$ ), triangles $E C F$ and $B C A$ are congruent, hence $E F=A B$.
+4. Recall that $x^{2}+y^{2} \geq 2 x y$, where equality holds if and only if $x=y$. Applying this inequality to the pairs of numbers $(a / 2, b)$, $(a / 2, c)$, and $(a / 2, d)$ yields:
+
+$$
+\begin{aligned}
+& \frac{a^{2}}{4}+b^{2} \geq a b, \\
+& \frac{a^{2}}{4}+c^{2} \geq a c, \\
+& \frac{a^{2}}{4}+d^{2} \geq a d .
+\end{aligned}
+$$
+
+Note also that $a^{2} / 4>0$. Adding these four inequalities gives us $a^{2}+b^{2}+c^{2}+d^{2} \geq a(b+c+d)$. Equality can hold only if all the inequalities were equalities, i.e. $a^{2}=0, a / 2=b, a / 2=c, a / 2=d$. Hence $a=b=c=d=0$ is the only solution of the given equation.
+
+5. Yes, see the following example:
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_2fffa64495ba2d15d9b4g-1.jpg?height=411&width=414&top_left_y=1439&top_left_x=844)
+

BayArea/md/en-monthly/en-0506-mc2sol.md

@@ -0,0 +1,32 @@
+# Berkeley Math Circle Monthly Contest 2 - Solutions 
+
+1. The frog can't manage the described trip. We will take advantage of the coloring of the chessboard. Notice that at each step frog is jumping to the square that is of different color than the square from which frog has jumped. Since the frog will make a total of 63 jumps, it must finish the trip on a cell that is of different color than the cell from which the trip has started. It remains to notice that the lower-left and the upper-right corners are of the same color.
+2. Since $A C$ is the diameter of $k$, we conclude that $\angle A E C=90^{\circ}$. Since $A C \perp B C, B C$ is a tangent to the circle $k$. Hence, $D E$ and $D C$ are two segments that are tangent to $k$ which implies that $D E=D C$. We now have that $D$ is the point on hypothenuse of $\triangle E C B$ such that $D E=D C$ which in turn gives that $D$ is the midpoint of $B C$ and $D E=D B$. The statement is proved.
+3. If we apply the inequality $x^{2}+y^{2} \geq 2 x y$ to the numbers $x=\frac{a b}{c}$ and $y=\frac{b c}{a}$ we get
+
+$$
+\frac{a^{2} b^{2}}{c^{2}}+\frac{b^{2} c^{2}}{a^{2}} \geq 2 b^{2}
+$$
+
+Similarly we get
+
+$$
+\begin{gathered}
+\frac{b^{2} c^{2}}{a^{2}}+\frac{c^{2} a^{2}}{b^{2}} \geq 2 c^{2}, \text { and } \\
+\frac{c^{2} a^{2}}{b^{2}}+\frac{a^{2} b^{2}}{c^{2}} \geq 2 a^{2} .
+\end{gathered}
+$$
+
+Summing up (1), (2) and (3) gives
+
+$$
+2\left(\frac{a^{2} b^{2}}{c^{2}}+\frac{b^{2} c^{2}}{a^{2}}+\frac{c^{2} a^{2}}{b^{2}}\right) \geq 2\left(a^{2}+b^{2}+c^{2}\right)=2,
+$$
+
+hence $S \geq 1$. The equality holds if and only if $\frac{a b}{c}=\frac{b c}{a}=\frac{c a}{b}$, i.e. $a=b=c=\frac{1}{\sqrt{3}}$.
+
+4. Divide the numbers into 1002 pairs in the following way: $\{1,2004\},\{2,2003\},\{3,2002\}, \ldots,\{1002,1003\}$. Since $1002+k$ numbers are blue, there are at least $k$ pairs whose both elements are blue. If we choose $2 k$ numbers from these pairs their sum will be $2005 k$ (since the sum of numbers in each pair is equal to 2005). The number $2005 k$ is, obviously, divisible by 2005 .
+5. Suppose that there are no four colinear red points with the stated property. Let $P$ be any blue point (there must be at least one blue point). Consider the circle $k$ with center $P$ and radius 1 . If some point on the circle is blue, then the claim will follow immediately. Thus suppose that all points on the circle are red. Let $A B C D E F$ be an arbitrary regular hexagon whose vertices belong to $k$. Let $X$ be intersection of $A B$ with $C D, Y$ the intersection of $C D$ with $E F$, and $Z$ the intersection point of $E F$ with $A B$. If at least two of the points $X, Y, Z$ are red (say $X$ and $Y$ ) then we can find four colinear red points (in our case, $X, C, D, Y$ ) with the stated property, and that would be a contradiction. Hence at least two of the points $X, Y, Z$ are blue.
+
+Now, consider the rotation with the center $P$ by which the points $A, B, C, D, E, F, X, Y, Z$ are mapped to the points $A^{\prime}, B^{\prime}$, $C^{\prime}, D^{\prime}, E^{\prime}, F^{\prime}, X^{\prime}, Y^{\prime}, Z^{\prime}$ such that $X X^{\prime}=Y Y^{\prime}=Z Z^{\prime}=1$. By the same argument we get that two of the points $X^{\prime}, Y^{\prime}, Z^{\prime}$ are blue. The problem is now solved, since at least one of the segments $X X^{\prime}, Y Y^{\prime}, Z Z^{\prime}$ has both ends painted in blue.
+

BayArea/md/en-monthly/en-0506-mc3sol.md

@@ -0,0 +1,37 @@
+# Berkeley Math Circle <br> Monthly Contest 3 - Solutions 
+
+1. Given $n$ positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$ such that $a_{1} a_{2} \cdots a_{n}=1$, prove that
+
+$$
+\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right) \geq 2^{n}
+$$
+
+When does the equality hold?
+
+Solution. By the inequality $a+b \geq 2 \sqrt{a b}$ which holds for positive numbers $a, b$ (and equality is if and only if $a=b$ ), we see that $1+a_{1} \geq 2 \sqrt{a_{1}}, 1+a_{2} \geq 2 \sqrt{a_{2}}, \ldots, 1+a_{n} \geq 2 \sqrt{a_{n}}$. Multiplying these inequalities we get $\left(1+a_{1}\right) \cdots\left(1+a_{n}\right) \geq$ $2^{n} \sqrt{a_{1} \cdots a_{n}}=2^{n}$. The equality holds if and only if $a_{1}=a_{2}=\cdots=a_{n}=1$.
+
+2. There are three prisoners in a prison. A warden has 2 red and 3 green hats and he has decided to play the following game: He puts the prisoners in a row one behind the other and on the had of each prisoner he puts a hat. The first prisoner in the row can't see any of the hats, the second prisoner can see only the hat at the had of the first one, and the third prisoner can see the hats of the first two prisoners. If some of the prisoners tells the color of his own hat, he is free; but if he is wrong, the warden will kill him. If a prisoner remain silent for sufficiently long, he is returned to his cell. Of course, each of them would like to be rescued from the prison, but if he isn't sure about the color of his hat, he won't guess.
+
+After noticing that second and third prisoner are silent for a long time, first prisoner (the one who doesn't see any hat) has concluded the color of his hat and told that to the warden. What is the color of the hat of the first prisoner? Explain your answer! (All prisoners know that there are 2 red and 3 green hats in total and all of them are good at mathematics.)
+
+Solution. If the first two prisoners had red hats, the third one won't be silent (he would conclude that his hat is green). Hence, at least one of the first two prisoners has a green hat, and everybody knows that (because the third prisoner is silent). Thus if the first prisoner had red hat, the second one would conclude that he must be the one with the green hat. However, the second prisoner was silent and the hat at the hat of the first prisoner must be green.
+
+3. Given three squares of dimensions $2 \times 2,3 \times 3$, and $6 \times 6$, choose two of them and cut each into 2 figures, such that it is possible to make another square from the obtained 5 figures.
+
+Solution. The solution is shown in the picture below
+![](https://cdn.mathpix.com/cropped/2024_04_17_dfe4584ca362f653de74g-1.jpg?height=770&width=616&top_left_y=1402&top_left_x=708)
+
+4. Let $A B C D$ be a rectangle. Let $E$ be the foot of perpendicular from $A$ to $B D$. Let $F$ be an arbitrary point of the diagonal $B D$ between $D$ and $E$. Let $G$ be the intersection of the line $C F$ with the perpendicular from $B$ to $A F$. Let $H$ be the intersection of the line $B C$ with the perpendicular from $G$ to $B D$. Prove that $\varangle E G B=\varangle E H B$.
+
+Solution. Let $X$ be the intersection of lines $B G$ and $A E$. Since $B X \perp A F$ and $A X \perp B F, X$ is the orthocenter of $\triangle A B F$ hence $F X \perp A B \Rightarrow F X\|B C\| A D$. It follows:
+
+$$
+\frac{E F}{E D}=\frac{F X}{D A}=\frac{F X}{B C}=\frac{G F}{G C}
+$$
+
+(the first equality follows from $\triangle E F X \sim \triangle E D A$, and the last from $\triangle G F X \sim \triangle G C B$ ). We conclude that $G E \| C D \perp$ $B H$. Since $B E \perp G H$ it follows that $E$ is the orthocenter of $\triangle G H B$ and $\varangle E G B=\varangle E H B$ follows easily (if we denote by $H^{\prime}$ and $G^{\prime}$ the intersections of $G E$ and $H E$, the triangles $G G^{\prime} B$ and $H H^{\prime} B$ are similar).
+
+5. Does there exist an integer such that its cube is equal to $3 n^{2}+3 n+7$, where $n$ is an integer?
+
+Solution. Suppose that there exist integers $n$ and $m$ such that $m^{3}=3 n^{2}+3 n+7$. Then from $m^{3} \equiv 1(\bmod 3)$ it follows that $m=3 k+1$ for some $k \in \mathbb{Z}$. Substituting into the initial equation we obtain $3 k\left(3 k^{2}+3 k+1\right)=n^{2}+n+2$. It is easy to check that $n^{2}+n+2$ cannot be divisible by 3 , and so this equality cannot be true. Therefore our equation has no solutions in integers.
+

BayArea/md/en-monthly/en-0506-mc4sol.md

@@ -0,0 +1,65 @@
+# Berkeley Math Circle Monthly Contest 4 - Solutions 
+
+1. A cube $3 \times 3 \times 3$ is made of cheese and consists of 27 small cubical cheese pieces arranged in the $3 \times 3 \times 3$ pattern. A mouse is eating the cheese in such a way that it starts at one of the corners and eats smaller pieces one by one. After he finishes one piece, he moves to the adjacent piece (pieces are adjacent if they share a face). Is it possible that the last piece mouse has eaten is the central one?
+
+Remark: Pieces don't fall down if a piece underneath is eaten first.
+
+Solution. Color the pieces of chiese alternatively in red and green such that corners are green and any two adjacent cubes are of different colors. We easily see that the mouse is moving always from the cube of one color to the cube of the other color. There are 14 green and 13 red cubes, the central cube being red. Since mouse has started from green piece, it will finish at the green piece (after 27 moves), hence it can't finish at the central cube.
+
+2. Let $M$ be the midpoint of the side $A C$ of triangle $A B C$. If $N$ is the point on the side $A B, O$ intersection of the lines $B M$ and $C N$, and if the areas of triangles $B O N$ and $C O M$ are equal, prove that $N$ is the midpoint of $A B$.
+
+Solution. Since the areas of $\triangle B O N$ and $\triangle C O M$ are equal we see that the areas of triangles $\triangle B C N$ and $\triangle C B M$ are also equal. Since these two triangles share the side, they must have the corresponding altitudes equal. Hence the length of perpendiculars from $M$ and $N$ to $B C$ are equal, implying that $N M \| B C$. Thus $M N$ is the midsegment of $\triangle A B C$ and consequently $N$ is the midpoint of $A B$.
+
+3. Determine whether the number
+
+$$
+\frac{1}{2 \sqrt{1}+1 \sqrt{2}}+\frac{1}{3 \sqrt{2}+2 \sqrt{3}}+\frac{1}{4 \sqrt{3}+3 \sqrt{4}}+\cdots+\frac{1}{100 \sqrt{99}+99 \sqrt{100}}
+$$
+
+is rational or irrational. Explain your answer.
+
+Solution. Notice that
+
+$$
+\begin{aligned}
+\frac{1}{(n+1) \sqrt{n}+n \sqrt{n+1}} & =\frac{1}{\sqrt{n(n+1)}} \cdot \frac{1}{\sqrt{n+1}+\sqrt{n}} \\
+& =\frac{1}{\sqrt{n(n+1)}} \cdot \frac{\sqrt{n+1}-\sqrt{n}}{(\sqrt{n+1}+\sqrt{n})(\sqrt{n+1}-\sqrt{n})} \\
+& =\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}} .
+\end{aligned}
+$$
+
+Now we see that the given sum is equal to $1-\frac{1}{\sqrt{100}}=\frac{9}{10}$.
+
+4. Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers whose sum is equal to 1 . If
+
+$$
+\begin{aligned}
+S= & \frac{a_{1}^{2}}{2 a_{1}}+\frac{a_{1} a_{2}}{a_{1}+a_{2}}+\frac{a_{1} a_{3}}{a_{1}+a_{3}}+\cdots+\frac{a_{1} a_{n}}{a_{1}+a_{n}} \\
+& +\frac{a_{2} a_{1}}{a_{2}+a_{1}}+\frac{a_{2}^{2}}{2 a_{2}}+\frac{a_{2} a_{3}}{a_{2}+a_{3}}+\cdots+\frac{a_{2} a_{n}}{a_{2}+a_{n}} \\
+& \vdots \\
+& +\frac{a_{n} a_{1}}{a_{n}+a_{1}}+\frac{a_{n} a_{2}}{a_{n}+a_{2}}+\frac{a_{n} a_{3}}{a_{n}+a_{3}}+\cdots+\frac{a_{n}^{2}}{2 a_{n}}
+\end{aligned}
+$$
+
+prove that $S \leq \frac{n}{2}$.
+
+Solution. It is easy to show that $\left(\frac{a+b}{2}\right)^{2} \geq a b$. Now applying this inequality to $a_{1}, a_{2}$ we get $\frac{a_{1} a_{2}}{a_{1}+a_{2}} \leq \frac{\left[\left(a_{1}+a_{2}\right) / 2\right]^{2}}{a_{1}+a_{2}}=$ $\frac{a_{1}+a_{2}}{4}$. We get similar inequalities for all terms of the given sum. Now, using that $a_{1}+a_{2}+a_{3}+\cdots+a_{n}=1$ we get:
+
+$$
+\frac{a_{1}^{2}}{2 a_{1}}+\frac{a_{1} a_{2}}{a_{1}+a_{2}}+\frac{a_{1} a_{3}}{a_{1}+a_{3}}+\cdots+\frac{a_{1} a_{n}}{a_{1}+a_{n}} \leq \frac{a_{1}+a_{1}}{4}+\frac{a_{1}+a_{2}}{4}+\frac{a_{1}+a_{3}}{4}+\cdots+\frac{a_{1}+a_{n}}{4}=\frac{n a_{1}+1}{4} .
+$$
+
+Similarly
+
+$$
+\frac{a_{2} a_{1}}{a_{2}+a_{1}}+\frac{a_{2}^{2}}{2 a_{2}}+\frac{a_{2} a_{3}}{a_{2}+a_{3}}+\cdots+\frac{a_{2} a_{n}}{a_{2}+a_{n}} \leq \frac{n a_{2}+1}{4}, \text { etc. }
+$$
+
+Adding these inequalities and using again that $a_{1}+\cdots+a_{n}=1$ we get the desired inequality. Equality holds if and only if $a_{1}=a_{2}=\cdots=a_{n}=1 / n$.
+
+5. Each of three schools contain $n$ students. Each student has at least $n+1$ friends among students of the other two schools. Prove that there are three students, all from different schools who are friends to each other. (Friendship is symmetric: If $A$ is a friend to $B$, then $B$ is a friend to $A$.)
+
+Solution. Suppose to the contrary that there doesn't exist three students from different schools who know each other. Denote schools by $A, B$ and $C$. Consider the student (or one of the students if there are more of them) who has the maximal number of friends in one of the schools. Let $a$ be the name of that student, suppose he is from the school $A$ and suppose that he has $k$ friends denoted by $b_{1}, b_{2}, \ldots, b_{k}$ from the school $B$. Then he must have at least $n+1-k$ friends in the school $C$, say $c_{1}, c_{2}, \ldots, c_{n+1-k}$. Then student $c_{1}$ doesn't know any of $b_{1}, b_{2}, \ldots, b_{k}$ (otherwise, if he knows $b_{i}$, then $a, b_{i}$ and $c_{1}$ would be the desired triple), so he can have at most $n-k$ friends in the school $B$. Since $c_{1}$ has at least $n+1$ friends in the schools $B$ and $A$, he must have at least $n+1-(n-k)=k+1$ friends in $A$. However, this is impossible since we assumed that $k$ is the maximal number of friends a student can have in one of the schools.
+
+The obtained contradiction proves the result.
+

BayArea/md/en-monthly/en-0506-mc6sol.md

@@ -0,0 +1,53 @@
+# Berkeley Math Circle Monthly Contest 6 - Solutions 
+
+1. Find all pairs $(m, n)$ of natural numbers such that $200 m+6 n=2006$.
+
+Solution. First we divide both sides of the equation by 2 and get: $100 m+3 n=1003$. Since $m$ and $n$ are natural numbers we immediately get that $m \leq 10$. Since $3 n$ is divisible by 3 and 1003 gives remainder 1 upon division by 3 , we conclude that $100 m$ must also give the remainder 1 upon division by 3 . Since $100 m=99 m+m$ and $99 m$ is divisible by 3 we see that $m$ must give remainder 1 when divided by 3 . Thus $m$ has to be one of the numbers $1,4,7,10$. Corresponding $n$ s are 301, 201,101 and 1 , respectively.
+
+2. Circles $k_{1}$ and $k_{2}$ intersect at points $A$ and $B$. Line $l$ is the common tangent to these circles and it touches $k_{1}$ at $C$ and $k_{2}$ at $D$ such that $B$ belongs to the interior of the tirangle $A C D$. Prove that $\angle C A D+\angle C B D=180^{\circ}$.
+
+Solution. Here we use the fact that the angle between the tangent and the chord is equal to the peripheral angle corresponding to the chord (this fact is easy to verify and can be found in any book on geometry). We have $\angle C A B=\angle D C B$ and $\angle D A B=$ $\angle D C B$. Hence $\angle C A D=\angle C A B+\angle D A B=\angle D C B+\angle D C B=180^{\circ}-\angle D B C$. The statement now follows directly.
+
+3. If $x$ and $y$ are two positive numbers less than 1 , prove that
+
+$$
+\frac{1}{1-x^{2}}+\frac{1}{1-y^{2}} \geq \frac{2}{1-x y}
+$$
+
+Solution. First we use the inequality $a+b \geq 2 \sqrt{a b}$ and get $\frac{1}{1-x^{2}}+\frac{1}{1-y^{2}} \geq \frac{2}{\sqrt{\left(1-x^{2}\right)\left(1-y^{2}\right)}}$. Now we notice that $\left(1-x^{2}\right)\left(1-y^{2}\right)=1+x^{2} y^{2}-x^{2}-y^{2} \leq 1+x^{2} y^{2}-2 x y=(1-x y)^{2}$ which implies that $\frac{2}{\sqrt{\left(1-x^{2}\right)\left(1-y^{2}\right)}} \geq \frac{2}{1-x y}$ and this completes the proof.
+
+4. If $n$ is natural number such that $2 n+1$ and $3 n+1$ are perfect squares, prove that $5 n+1$ can't be a prime number.
+
+Solution. This problem contains a typo in its original formulation. Aaron Wilkowski, one of the BMC contestants, found a counter-example, e.g. $n=3960$. Indeed, $2 n+1=89^{2}, 3 n+1=109^{2}$, and $5 n+1=19801$ which is a prime.
+
+The correct formulation of the problem is: If $n$ is natural number such that $2 n+1$ and $3 n+1$ are perfect squares, prove that $5 n+3$ can't be a prime number.
+
+The solution is:
+
+Suppose that $5 n+3$ is a prime number. Let $x$ and $y$ be natural numbers such that $x^{2}=2 n+1$ and $y^{2}=3 n+1$. Then $5 n+3=4(2 n+1)-(3 n+1)=4 x^{2}-y^{2}=(2 x-y)(2 x+y)$. Since $5 n+1$ is prime number and $x$ and $y$ are natural numbers we must have $2 x-y=1$ implying that $y=2 x-1$. Since $n=y^{2}-x^{2}$ we have that $x^{2}=2 n+1=2\left(y^{2}-x^{2}\right)+1=$ $2\left((2 x-1)^{2}-x^{2}\right)+1=6 x^{2}-8 x+3$. Thus $5 x^{2}-8 x+3=0$ and the solutions are $x=1$ or $x=3 / 5$. Since $x^{2}=2 n+1$ for a natural number $n$ neither of these $x$ would work and we have a contradiction.
+
+5. Every two members of a certain society are either friends or enemies. Suppose that there are $n$ members of the society, that there are exactly $q$ pairs of friends, and that in every set of three persons there are two who are enemies to each other. Prove that there is at least one member of the society among whose enemies there are at most $q \cdot\left(1-\frac{4 q}{n^{2}}\right)$ pairs of friends.
+
+Solution. Denote by $S$ the set of all members of the society, by $A$ the set of all pairs of friends, and by $N$ the set of all pairs of enemies. For every $x \in S$, denote by $f(x)$ number of friends of $x$ and by $F(x)$ number of pairs of friends among enemies of $x$. It is easy to prove:
+
+$$
+\begin{gathered}
+q=|A|=\frac{1}{2} \sum_{x \in S} f(x) ; \\
+\sum_{\{a, b\} \in A}(f(a)+f(b))=\sum_{x \in S} f^{2}(x) .
+\end{gathered}
+$$
+
+If $a$ and $b$ are friends, then the number of their common enemies is equal to $(n-2)-(f(a)-1)-(f(b)-1)=n-f(a)-f(b)$. Thus
+
+$$
+\frac{1}{n} \sum_{x \in S} F(x)=\frac{1}{n} \sum_{\{a, b\} \in A}(n-f(a)-f(b))=q-\frac{1}{n} \sum_{x \in S} f^{2}(x)
+$$
+
+Using the inequality between arithmetic and quadratic mean on the last expression, we get
+
+$$
+\frac{1}{n} \sum_{x \in S} F(x) \leq q-\frac{4 q^{2}}{n^{2}}
+$$
+
+and the statement of the problem follows immediately.
+

BayArea/md/en-monthly/en-0506-mc7sol.md

@@ -0,0 +1,57 @@
+# Berkeley Math Circle Monthly Contest 7 - Solutions 
+
+1. If $k$ is an integer, prove that the number $k^{2}+k+1$ is not divisible by 2006 .
+
+Solution. The number $k^{2}+k+1=k(k+1)+1$ is odd, because $k(k+1)$ is even. Hence it can't be divisible by 2006 .
+
+2. Given an $n \times n$ matrix whose entries $a_{i j}$ satisfy $a_{i j}=\frac{1}{i+j-1}, n$ numbers are chosen from the matrix no two of which are from the same row or the same column. Prove that the sum of these $n$ numbers is at least 1 .
+
+Solution. Suppose that $a_{i j}$ and $a_{k l}$ are among the chosen numbers and suppose that $i<k$ and $j<l$. It is straightforward to show that $a_{i j}+a_{k l} \geqslant a_{i l}+a_{k j}$. Hence, whenever $a_{i j}$ and $a_{k l}$ with $i<k$ and $j<l$ are among chosen numbers, we can lower the sum by replacing these two numbers with $a_{i l}$ and $a_{k j}$. Hence the smallest possible sum is when we choose $a_{1 n}, a_{2, n-1}, \ldots, a_{n 1}$ - and in that case the sum is 1 .
+
+3. Given a triangle $A B C$ such that $\angle B=90^{\circ}$, denote by $k$ the circle with center on $B C$ that is tangent to $A C$. Denote by $T$ a point of tangency of $k$ and the tangent from $A$ to $k$ (different from $A C$ ). If $B^{\prime}$ is the midpoint of $A C$ and $M$ the intersection of $B B^{\prime}$ and $A T$, prove that $M B=M T$.
+
+Solution. Let $O$ be the center of $k$. The quadrilateral $A B T O$ can be inscribed in a circle hence $\angle C B B^{\prime}=\angle B^{\prime} C B$ and $\angle T A O=\angle O A C$. Thus $\angle M T B=\angle A T B=\angle A O B=\angle A C O+\angle O A C=\angle B^{\prime} B C+\angle T A O=\angle B^{\prime} B C+\angle T B O=$ $\angle T B M$ which immediately implies the given statement.
+
+4. Let $A$ be the number of 4 -tuples $(x, y, z, t)$ of positive integers smaller than $2006^{2006}$ such that
+
+$$
+x^{3}+y^{2}=z^{3}+t^{2}+1,
+$$
+
+and let $B$ the number of 4-tuples $(x, y, z, t)$ of positive integers smaller than $2006^{2006}$ such that
+
+$$
+x^{3}+y^{2}=z^{3}+t^{2} \text {. }
+$$
+
+Prove that $B>A$.
+
+Solution. For each natural number $k$, denote by $l_{k}$ the number of pairs $(x, y)$ such that $x^{3}+y^{2}=k$. Then $B=l_{2}^{2}+l_{3}^{2}+\cdots+l_{r}^{2}$ where $r$ is the maximal $k$ for which $l_{k} \neq 0$. Similarly, $A=l_{2} l_{3}+l_{3} l_{4}+\cdots+l_{r-1} l_{r}$. Now
+
+$$
+B=\frac{l_{2}^{2}}{2}+\frac{1}{2}\left(l_{2}^{2}+l_{3}^{2}\right)+\frac{1}{2}\left(l_{3}^{2}+l_{4}^{2}\right)+\cdots+\frac{1}{2}\left(l_{r-1}^{2}+l_{r}^{2}\right)+\frac{l_{r}^{2}}{2}>l_{2} l_{3}+l_{3} l_{4}+\cdots+l_{r-1} l_{r}
+$$
+
+We have used that $a^{2}+b^{2} \geq 2 a b$.
+
+5. Prove that the functional equations
+
+$$
+\begin{aligned}
+f(x+y) & =f(x)+f(y) \\
+\text { and } \quad f(x+y+x y) & =f(x)+f(y)+f(x y) \quad(x, y \in \mathbb{R})
+\end{aligned}
+$$
+
+are equivalent.
+
+Solution. Let us assume that $f(x+y)=f(x)+f(y)$ for all reals. In this case we trivially apply the equation to get $f(x+y+x y)=f(x+y)+f(x y)=f(x)+f(y)+f(x y)$. Hence the equivalence is proved in the first direction.
+
+Now let us assume that $f(x+y+x y)=f(x)+f(y)+f(x y)$ for all reals. Plugging in $x=y=0$ we get $f(0)=0$. Plugging in $y=-1$ we get $f(x)=-f(-x)$. Plugging in $y=1$ we get $f(2 x+1)=2 f(x)+f(1)$ and hence $f(2(u+v+u v)+1)=$ $2 f(u+v+u v)+f(1)=2 f(u v)+2 f(u)+2 f(v)+f(1)$ for all real $u$ and $v$. On the other hand, plugging in $x=u$ and $y=2 v+1$ we get $f(2(u+v+u v)+1)=f(u+(2 v+1)+u(2 v+1))=f(u)+2 f(v)+f(1)+f(2 u v+u)$. Hence it follows that $2 f(u v)+2 f(u)+2 f(v)+f(1)=f(u)+2 f(v)+f(1)+f(2 u v+u)$, i.e.,
+
+$$
+f(2 u v+u)=2 f(u v)+f(u) .
+$$
+
+Plugging in $v=-1 / 2$ we get $0=2 f(-u / 2)+f(u)=-2 f(u / 2)+f(u)$. Hence, $f(u)=2 f(u / 2)$ and consequently $f(2 x)=2 f(x)$ for all reals. Now (1) reduces to $f(2 u v+u)=f(2 u v)+f(u)$. Plugging in $u=y$ and $x=2 u v$, we obtain $f(x)+f(y)=f(x+y)$ for all nonzero reals $x$ and $y$. Since $f(0)=0$, it trivially holds that $f(x+y)=f(x)+f(y)$ when one of $x$ and $y$ is 0 .
+

BayArea/md/en-monthly/en-0607-mc1sol.md

@@ -0,0 +1,40 @@
+# Berkeley Math Circle <br> Monthly Contest 1 - Solutions 
+
+1. If $n$ is an integer, prove that the number
+
+$$
+1+n+n^{2}+n^{3}+n^{4}
+$$
+
+is not divisible by 4 .
+
+Solution. We will prove that $1+n+n^{2}+n^{3}+n^{4}$ is odd number and the required claim will follow immediately since no odd number is divisible by 4 . There are two possibilities for $n$ - it can be even or odd. In the case when $n$ is even, then $n+n^{2}+n^{3}+n^{4}$ is even as a sum of even numbers. Thus $1+n+n^{2}+n^{3}+n^{4}$ is odd number. On the other hand, if $n$ is odd then, all five numbers $1, n, n^{2}, n^{3}$, and $n^{4}$ are odd and hence their sum must be also odd.
+
+2. In how many different ways one can place 3 rooks on the cells of $6 \times 2006$ chessboard such that they don't attack each other?
+
+Solution. In order not to attack each other, the rooks no to rooks can share a row or a column. Since there are 6 columns in total, we have 20 possibilities for choosing 3 columns in which we will place our rooks. After choosing the columns, we will place our rooks one by one in these columns. The first rook can be placed in 2006 different ways. The second one, however can't be placed in the row that contains the first rook. Hence there are 2005 possibilities for the second rook. Similarly, the third rook can be placed in 2004 different ways. In total, there are $20 \cdot 2006 \cdot 2005 \cdot 2004$ ways to place 3 rooks in $6 \times 2006$ chessboard so that they don't attack each other.
+
+3. Given a square table $n \times n$, two players $A$ and $B$ are playing the following game: At the beginning all cells of the table are empty, and the players alternate playing with coins. Player $A$ has the first move, and in each of the moves a player will put a coin on some of the cells that doesn't contain a coin and is not adjacent to any of the cells that already contains a coin. The player who makes the last move is the winner. Which player has a winning strategy and what is the strategy?
+
+Remark. The cells are adjacent if they share an edge.
+
+Solution. If $n$ is even, $B$ has the winning strategy - he just needs to put the coin in the centrally symmetric cell to the cell where $A$ has put the coin in his prevoius move. Following this strategy, $B$ will always have a free cell to put his coin, and thus he can never loose. Since the game has to end, $A$ is the looser.
+
+If $n$ is odd, $A$ has a winning strategy: first he needs to put a coin in the central cell, and after that he puts the coin in the centrally symmetric cell to the cell where $B$ has put his coin.
+
+4. Given a circle $k$, let $A B$ be its diameter. An arbitrary line $l$ intersects the circle $k$ at the points $P$ and $Q$. If $A_{1}$ and $B_{1}$ are the feet of perpendiculars from $A$ and $B$ to $P Q$ prove that $A_{1} P=B_{1} Q$.
+
+Solution. Let $A_{2}$ and $B_{2}$ be the intersections of the lines $A A_{1}$ and $B B_{1}$ with the circle $k$. Furthermore, let $s$ be the line through the center of $k$ perpendicular to $l$. Then $A B_{2} B A_{2}$ is a rectangle (because $A A_{2} \| B B_{2}$ and $\angle A B_{2} B=90^{\circ}$ since $A B$ is a diameter) and $s$ is an axis of symmetry of that rectangle. $A_{1} B_{1} B_{2} A$ is also a rectangle and $s$ is also an axis of symmetry for that rectangle. Since $k$ is also symmetric with repsect to $s$, we conclude that $P$ and $Q$ are symmetric with respect to $s$. Thus $A_{1} Q=B_{1} P$.
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_00fc6b6d484a7faeafdeg-1.jpg?height=425&width=437&top_left_y=1495&top_left_x=1278)
+
+5. Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that $f(f(x)+y+1)=x+f(y)+1$ for every two integers $x$ and $y$.
+
+Solution. Substituting $x=0$ we get $f(f(0)+y+1)=f(y)+1$. Now we have $x+f(y)+1=f(f(x)+y+1)=$ $f[(f(x)+1)+y]=f[f(f(0)+x+1)+(y-1)+1]=f(y-1)+(f(0)+x+1)+1$ from where we conclude that
+
+$$
+f(y)=f(y-1)+f(0)+1 .
+$$
+
+Using the induction we obtain $f(n)=f(0)+n(f(0)+1)$ for all integers $n$. If we set $y=0$ in (1) we get $f(-1)=-1$ and substituting $y=-1$ in the original equation yields $f(f(x))=x$. If we apply $f$ to both sides of $f(n)=f(0)+n(f(0)+1)$ we get $n=f(0)+[f(0)+n(f(0)+1)](f(0)+1)$ which is equivalent to $n=f(0)(f(0)+2)+n(f(0)+1)^{2}$ and this can be satisfied for all integers $n$ if and only if $f(0)=0$ or $f(0)=-2$. In the first case we get $f(n)=n$ and in the second $f(n)=-n-2$. It is easy to verify that both functions satisfy the required conditions.
+

BayArea/md/en-monthly/en-0607-mc2sol.md

@@ -0,0 +1,53 @@
+# Berkeley Math Circle <br> Monthly Contest 2 - Solutions 
+
+1. Find all positive prime numbers $p$ and $q$ such that $p^{2}-q^{3}=1$.
+
+Remark. $p$ is prime if it has only two divisors: 1 and itself. The numbers $2,3,5,7,11,13$ are prime, but $1,4,6,8,9$ are not.
+
+Solution. If both of the numbers $p$ and $q$ are odd, then each of $p^{2}$ and $q^{3}$ has to be odd so their difference must be even. Hence, at least one of $p$ and $q$ has to be even. Since 2 is the only even prime number, and at the same time it is the smallest prime number, we must have $q=2$. Now we get $p=3$.
+
+2. A line $l$ and two points $A$ and $B$ are given in a plane in such a way that $A$ belongs to $l$ but $B$ doesn't. Construct the circle $k$ that passes through $B$ and touches $l$ at the point $A$.
+
+Solution. Let $m$ be the bisector of the segment $A B$. The center of the circle $k$ has to belong to $m$. Similarly, since the circle has to be tangent to $l$ at the point $A$ its center has to be located on the line $a$ perpendicular to $l$ that passes through $A$.
+
+The lines $m$ and $a$ are easy to construct and their intersection is the center $O$ of the circle. Now we have the center and the point $B$ hence the circle is determined.
+
+3. If the sum of digits in a decimal representation of a natural number $n$ is equal to 2006, prove that $n$ can't be a perfect square of an integer.
+
+Solution. The remainder of $n$ upon division by 3 is equal to the sum of its digits, i.e. 2006. Hence number $n$ has a remainder 2 upon division by 3 and no square can have that remainder.
+
+4. Let $\triangle A B C$ be a triangle such that $\angle A=90^{\circ}$. Determine whether it is possible to partition $\triangle A B C$ into 2006 smaller triangles in such a way that
+
+$1^{\circ}$ Each triangle in the partition is similar to $\triangle A B C$;
+
+$2^{\circ}$ No two triangles in the partition have the same area.
+
+Explain your answer!
+
+Solution. The required partition is always possible. We consider 2 cases:
+
+(a) The triangle is not isosceles - First we construct the perpendicular from the vertex of the right angle. The triangle is divided into two similar, but non-congruent triangle. Now we divide smaller triangle, and keep going until the total number of triangle becomes 2006 .
+
+(b) The triangle is isosceles - Repeat the procedure from the previous problem until we get 2001 triangles. Then we divide one of the smallest triangles into 6 smaller and noncongruent triangles as shown in the second picture.
+![](https://cdn.mathpix.com/cropped/2024_04_17_9df8cafc67695d71c626g-1.jpg?height=374&width=680&top_left_y=1355&top_left_x=1166)
+
+5. Let $S>0$. If $a, b, c, x, y, z$ are positive real numbers such that $a+x=b+y=c+z=S$, prove that
+
+$$
+a y+b z+c x<S^{2}
+$$
+
+Solution. Denote $T=S / 2$. One of the triples $(a, b, c)$ and $(x, y, z)$ has the property that at least two of its members are greater than or equal to $T$. Assume that $(a, b, c)$ is the one, and choose $\alpha=a-T, \beta=b-T$, and $\gamma=c-T$. We then have $x=T-\alpha$, $y=T-\beta$, and $z=T-\gamma$. Now the required inequality is equivalent to
+
+$$
+(T+\alpha)(T-\beta)+(T+\beta)(T-\gamma)+(T+\gamma)(T-\alpha)<4 T^{2}
+$$
+
+After simplifying we get that what we need to prove is
+
+$$
+-(\alpha \beta+\beta \gamma+\gamma \alpha)<T^{2}
+$$
+
+We also know that at most one of the numbers $\alpha, \beta, \gamma$ is negative. If all are positive, there is nothing to prove. Assume that $\gamma<0$. Now (1) can be rewritten as $-\alpha \beta-\gamma(\alpha+\beta)<T^{2}$. Since $-\gamma<T$ we have that $-\alpha \beta-\gamma(\alpha+\beta)<-\alpha \beta+T(\alpha+\beta)$ and the last term is less than $T$ since $(T-\alpha)(T-\beta)>0$.
+

BayArea/md/en-monthly/en-0607-mc3sol.md

@@ -0,0 +1,59 @@
+# Berkeley Math Circle Monthly Contest 3 - Solutions 
+
+1. Is the number $\left|2^{3000}-3^{2006}\right|$ bigger or smaller than $\frac{1}{2}$ ?
+
+Solution. Notice that $\left|2^{3000}-3^{2006}\right|$ is a non-negative integer, so it is either 0 or bigger than $1 / 2$. However, this number is not 0 since $2^{3000} \neq 3^{2006}$, hence $\left|2^{3000}-3^{2006}\right|>\frac{1}{2}$.
+
+2. If $a, b, c, d$ are positive real numbers such that $\frac{5 a+b}{5 c+d}=\frac{6 a+b}{6 c+d}$ and $\frac{7 a+b}{7 c+d}=9$, calculate $\frac{9 a+b}{9 c+d}$.
+
+Solution. Let $\frac{5 a+b}{5 c+d}=\frac{6 a+b}{6 c+d}=k$. Then $5 a+b=k(5 c+d)$ and $6 a+b=k(6 c+d)$. Subtracting these two equations gives $a=k c$. Now we can easily get that $b=k d$. From $\frac{7 a+b}{7 c+d}=\frac{7 k c+k d}{7 c+d}=k=9$ we get $\frac{9 a+b}{9 c+d}=9$ as well.
+
+3. Let $n>3$ be an integer which is not divisible by 3 . Two players $A$ and $B$ play the following game with $n \times n$ chocolate table. First, player $A$ has to chose and remove one piece of the chocolate, without breaking other pieces. After his move, player $B$ tries to partition the remaining chocolate into $3 \times 1$ (and $1 \times 3$ ) rectangles. If $B$ manages to do so, then he/she is the winner. Otherwise the winner is $A$. Determine which player has a winning strategy and describe the strategy.
+
+Solution. The player $A$ has a winning strategy. Imagine that the chocolate is painted in 3 colors, $R$ (ed), $G$ (reen), and $B$ (lue) as shown in the table
+
+$$
+\begin{array}{cccccccc}
+R & G & B & R & G & B & R & \cdots \\
+B & R & G & B & R & G & B & \cdots \\
+G & B & R & G & B & R & G & \cdots \\
+R & G & B & R & G & B & R & \cdots \\
+B & R & G & B & R & G & B & \cdots \\
+G & B & R & G & B & R & G & \cdots \\
+R & G & B & R & G & B & R & \cdots \\
+& \vdots & & & \vdots & & & \vdots
+\end{array}
+$$
+
+Notice that every $1 \times 3$ rectangle will contain exactly one red, one green, and one blue piece. Moreover the orignal table had equal number of green and blue pieces, and one red piece extra. If $A$ takes out any green or blue piece, player $B$ won't be able to fullfil the requirement, and $A$ is the winner.
+
+4. Given a triangle $A B C$, let $D$ be the point of the ray $B A$ such that $B D=B A+A C$. If $K$ and $M$ are points on the sides $B A$ and $B C$, respectively, such that the triangles $B D M$ and $B C K$ have the same areas, prove that $\angle B K M=\frac{1}{2} \angle B A C$.
+
+Solution. Since the area of the triangle $B D M$ is equal to $B D \cdot B M \cdot \sin \angle D B M$ and similar formula holds for the area of $\triangle B C K$ we immediately have that $B D \cdot B M=B C \cdot B K$. This implies that $\frac{B D}{B K}=\frac{B C}{B M}$ hence $K M \| C D$. Thus $\angle B K M=\angle B D C=\angle A D C$. The last angle is equal to $\frac{1}{2} \angle B A C$ since $\triangle D A C$ is isosceles.
+
+5. Determine the greatest real number $a$ such that the inequality
+
+$$
+x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2} \geq a\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{4}+x_{4} x_{5}\right)
+$$
+
+holds for every five real numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$.
+
+Solution. Note that
+
+$$
+x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}=\left(x_{1}^{2}+\frac{x_{2}^{2}}{3}\right)+\left(\frac{2 x_{2}^{2}}{3}+\frac{x_{3}^{2}}{2}\right)+\left(\frac{x_{3}^{2}}{2}+\frac{2 x_{4}^{2}}{3}\right)+\left(\frac{x_{4}^{2}}{3}+x_{5}^{2}\right) .
+$$
+
+Now applying the inequality $a^{2}+b^{2} \geq 2 a b$ we get
+
+$$
+x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2} \geq \frac{2}{\sqrt{3}}\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{4}+x_{4} x_{5}\right) \text {. }
+$$
+
+This proves that $a \geq \frac{2}{\sqrt{3}}$. In order to prove $a \leq \frac{2}{\sqrt{3}}$ it is enough to notice that for $\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)=(1, \sqrt{3}, 2, \sqrt{3}, 1)$ we have
+
+$$
+x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}=\frac{2}{\sqrt{3}}\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{4}+x_{4} x_{5}\right) \text {. }
+$$
+

BayArea/md/en-monthly/en-0607-mc4sol.md

@@ -0,0 +1,67 @@
+# Berkeley Math Circle Monthly Contest 4 - Solutions 
+
+1. Do there exist positive integers $x, y$, and $z$ such that $x^{2006}+y^{2006}=z^{2007}$ ? Explain your answer.
+
+Remark. If the answer is yes, you should give an example of such $x, y$, and $z$. If the answer is no, you should prove that no $x$, $y$, $z$ can satisfy the above equation.
+
+Solution. Yes, for example, $x=y=z=2$.
+
+2. Let $O$ be the intersection of the diagonals $A C$ and $B D$ of the convex quadrilateral $A B C D$. Let $S_{1}, S_{2}, S_{3}$, and $S_{4}$ denote the areas of the triangles $A B O, B C O, C D O$, and $D A O$.
+
+(a) Prove that $S_{1} \cdot S_{3}=S_{2} \cdot S_{4}$.
+
+(b) Does there exist a quadrilateral $A B C D$ such that $S_{1}, S_{2}, S_{3}$, and $S_{4}$ are consecutive positive integers in some order?
+
+## Solution.
+
+(a) Let $M$ and $N$ be feet of perpendiculars from $B$ and $D$ to $A C$. Then $S_{1}=A O \cdot B M / 2, S_{2}=C O \cdot B M / 2, S_{3}=$ $C O \cdot D N / 2$, and $S_{4}=A O \cdot D N / 2$. Now the desired statement follows immediately from the previous four relations.
+
+(b) We will prove that the answer to the question is no. Assume the opposite, i.e. that $n, n+1, n+2, n+3$ are the given areas, where $n$ is a positive integer. Since $S_{1} \cdot S_{3}=S_{2} \cdot S_{4}$ and $n<n+1<n+2<n+3$ we must have $n(n+3)=(n+1)(n+2)$ which is impossible.
+
+3. 2006 vertices of a regular 2007 -gon are red. The remaining vertex is green. Let $G$ be the total number of polygons whose one vertex is green and the others are red. Denote by $R$ the number of polygons whose all vertices are red. Which number is bigger, $R$ or $G$ ? Explain your answer.
+
+Solution. We will prove that $G \geq R$. For each polygon $\mathcal{P}$ with all red vertices we can correspond a polygon with one green vertex (namely we can add the green vertex to the set of vertices of $\mathcal{P}$ ). Thus $G \geq R$. However, $G>R$ since the triangles with one green vertex can't be corresponded to some polygon whose all vertices are red.
+
+4. A sequence of numbers $\left\{a_{n}\right\}$ is given by $a_{1}=1, a_{n+1}=2 a_{n}+\sqrt{3 a_{n}^{2}+1}$ for $n \geq 1$. Prove that each term of the sequence is an integer.
+
+Solution. From the definition of the sequence we see that $\left(a_{n+1}-2 a_{n}\right)^{2}=3 a_{n}^{2}+1$. After simplification we get
+
+$$
+a_{n+1}^{2}+a_{n}^{2}-4 a_{n} a_{n+1}=1
+$$
+
+Adding $3 a_{n+1}^{2}$ to both sides of the last equation gives us $\left(2 a_{n+1}-a_{n}\right)^{2}=3 a_{n+1}^{2}+1$ and after taking the square root of both sides we get
+
+$$
+\sqrt{3 a_{n+1}^{2}+1}=2 a_{n+1}-a_{n}
+$$
+
+On the other hand, from the definition of the sequence we see that
+
+$$
+\sqrt{3 a_{n+1}^{2}+1}=a_{n+2}-2 a_{n+1} \text {. }
+$$
+
+From (1) and (2) we conclude that $a_{n+2}=4 a_{n+1}-a_{n}$ wich together with $a_{1}=1, a_{2}=4$ implies that all terms are integers.
+
+5. A finite set of circles in the plane is called nice if it satisfies the following three conditions:
+
+(i) No two circles intersect in more than one point;
+
+(ii) For every point $A$ of the plane there are at most two circles passing through $A$;
+
+(iii) Each circle from the set is tangent to exactly 5 other circles form the set.
+
+Does there exist a nice set consisting of exactly
+
+(a) 2006 circles?
+
+(b) 2007 circles?
+
+## Solution.
+
+(a) The answer is yes. The following picture shows that there is a nice set consisting of exactly 12 circles. It is also possible to construct a nice set with 22 circle (the picture can be found below). Since $2006=158 \cdot 12+5 \cdot 22$ we can make a nice set of 2006 by making a union of 158 discjoint nice sets of 12 circles each, and 5 discjoint nice sets each of which contains 22 circles.
+![](https://cdn.mathpix.com/cropped/2024_04_17_4253385ade340a8acceag-2.jpg?height=1788&width=1312&top_left_y=341&top_left_x=446)
+
+(b) We will prove that nice set can't contain odd number of circles. Let $n$ be a total number of circles. We will count the number of pairs $(k, P)$ where $k$ is a circle in the set and $P$ a point at which $k$ touches another circle. For each circle $k$ there are exactly 5 such pairs, and hence the total number of pairs is $5 n$. For each point $P$ there are exactly 2 pairs corresponding to it. Hence the total number of pairs has to be even, but $5 n$ can't be even if $n$ is odd.
+

BayArea/md/en-monthly/en-0607-mc5sol.md

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+# Berkeley Math Circle <br> Monthly Contest 5 - Solutions 
+
+1. Do there exist 100 consecutive positive integers such that their sum is a prime number?
+
+Hint. What is the method for summing 100 consecutive numbers?
+
+Solution. No. Let $n, n+1, \ldots, n+99$ be any 100 consecutive positive integers. Then $n+(n+1)+(n+2)+\cdots+(n+99)=$ $100 n+(1+2+\cdots+99)$. However, $1+2+\cdots+99=(1+99)+(2+98)+(3+97)+\cdots+(49+51)+50=49 \cdot 100+50=$ $50(2 \cdot 49+1)=50 \cdot 99$. Thus $n+(n+1)+\cdots+(n+99)=100 n+50 \cdot 99=50(2 n+99)$ and this is not prime.
+
+2. Let $A B C D E$ be a convex pentagon. If $\alpha=\angle D A C, \beta=\angle E B D, \gamma=A C E, \delta=\angle B D A$, and $\epsilon=\angle B E C$, as shown in the picture, calculate the sum $\alpha+\beta+\gamma+\delta+\epsilon$.
+
+Solution. Let $M, N, P, Q, R$ denote the intersections of the lines $A C$ and $B E, A C$ and $B D, C E$ and $B D, D A$ and $C E, E B$ and $A E$, respectively. Since the sum of the internal angles of a triangle is $180^{\circ}$, from $\triangle A M R$ we get $\alpha=180^{\circ}-\angle A M R-\angle M R A$. We also know that $\angle A M R=$ $\angle B M N$ and we can denote these angles by $\angle M$. Analogously $\angle A R M=\angle E R Q=\angle R$. With similar equations for $\beta, \gamma, \delta$, and $\epsilon$ we get that $\alpha+\beta+\gamma+\delta+\epsilon=5 \cdot 180^{\circ}-$ $2(\angle M+\angle N+\angle P+\angle Q+\angle R)$. Since $\angle M, \angle N, \angle P, \angle Q$, and $\angle R$ are exterior angles of the pentagon $M N P Q R$ their sum has to be $360^{\circ}$ implying that $\alpha+\beta+\gamma+\delta+\epsilon=180^{\circ}$.
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_d90eaa74d4107324eb71g-1.jpg?height=477&width=479&top_left_y=641&top_left_x=1254)
+
+3. Bart has 17 and 19 dollar bills only.
+
+(a) Prove that these bills are fake.
+
+(b) Prove that there exists $m>0$ such that for each $n \geq m$ Bart can give to Lisa exactly $n$ dollars using his bills.
+
+Remark. Part (a) is worth 0 points but we would like to see your "proof".
+
+Solution. We can have $m=17 \cdot 19$. For each $n>17 \cdot 19$, we consider the set $S=\{n, n-17, n-2 \cdot 17, \ldots, n-18 \cdot 17\}$. If none of these numbers is divisible by 19, two of them will give the same residue upon division by 19. Indeed, there are 19 numbers and without 0 there are only 18 residues. Let $n-17 a$ and $n-17 b$ give the same residue mod 19. Assuming $a<b$ we get that $n-17 a-(n-17 b)=17(b-a)$ is divisible by 19 which is impossible.
+
+Hence one of the numbers from $S$ is divisible by 19 , say $n-k \cdot 17$. Then $n-k \cdot 17=l \cdot 19$ and $n=k \cdot 17+l \cdot 19$ which solves the problem.
+
+4. Does there exist a convex polygon that can be partitioned into non-convex quadrilaterals?
+
+Solution. The answer is no. Assume that, on the contrary it is possible to partition a polygon $P$ into non-convex quadrilaterals. Let $n$ be the number of quadrilaterals. Denote by $S$ the total sum of all internal angles of all the quadrilaterals. Since the sum of internal angles of each quadrilateral is $360^{\circ}$ we have $S=360^{\circ}$. However, each of the nonconvex angles has to be in the interior of $P$, hence the sum of angles around the vertex of that angle has to be $360^{\circ}$. This immediately gives $360^{\circ} n$ as the sum of angles around such vertices. Since those are not the only vertices (at least the vertices of $P$ will contribute to the sum $S$ ), we have that $S>360^{\circ}$ and this is a contradiction.
+
+5. The numbers from the table
+
+| $a_{11}$ | $a_{12}$ | $\cdots$ | $a_{1 n}$ |
+| :---: | :---: | :---: | :---: |
+| $a_{21}$ | $a_{22}$ | $\cdots$ | $a_{2 n}$ |
+| $\vdots$ | $\vdots$ |  | $\vdots$ |
+| $a_{n 1}$ | $a_{n 2}$ | $\cdots$ | $a_{n n}$ |
+
+satisfy the inequality
+
+$$
+\sum_{i=1}^{n}\left|a_{i 1} x_{1}+a_{i 2} x_{2}+\cdots+a_{i n} x_{n}\right| \leq M
+$$
+
+for every choice $x_{i}= \pm 1$. Prove that $\left|a_{11}\right|+\left|a_{22}\right|+\cdots+\left|a_{n n}\right| \leq M$.
+
+Solution. We will sum the given inequality over all possible choices for $\left(x_{1}, \ldots, x_{n}\right)$. Since the number of such choices is $2^{n}$ we obtain
+
+$$
+\sum_{i=1}^{n} \sum_{\left(x_{1}, \ldots, x_{n}\right)}\left|a_{i 1} x_{1}+a_{i 2} x_{2}+\cdots+a_{i n} x_{n}\right| \leq 2^{n} M
+$$
+
+Now we will consider the expressions $\sum_{\left(x_{1}, \ldots, x_{n}\right)}\left|a_{i 1} x_{1}+a_{i 2} x_{2}+\cdots+a_{i n} x_{n}\right|$. We will group summands in pairs - members of each pair will have all terms $x_{j}$ the same, except for the $i$-th. One member will have $x_{i}=1$ and the other $x_{i}=-1$. To each of those terms we will apply the inequality $\left|a_{i i}+B\right|+\left|a_{i i}-B\right| \geq\left|2 a_{i i}\right|$. More precisely,
+
+$$
+\begin{aligned}
+& \sum_{\left(x_{1}, \ldots, x_{n}\right)}\left|a_{i 1} x_{1}+a_{i 2} x_{2}+\cdots+a_{i n} x_{n}\right| \\
+= & \sum_{\left(x_{1}, \ldots, x_{i-1}, x_{i+1}, \ldots, x_{n}\right)}\left(\left|a_{i 1} x_{1}+a_{i 2} x_{2}+\cdots+a_{i} i+\cdots+a_{i n} x_{n}\right|+\left|a_{i 1} x_{1}+a_{i 2} x_{2}+\cdots-a_{i} i+\cdots+a_{i n} x_{n}\right|\right) \\
+\geq & \sum_{\left(x_{1}, \ldots, x_{i-1}, x_{i+1}, \ldots, x_{n}\right)}\left|2 a_{i i}\right|=2^{n}\left|a_{i i}\right| .
+\end{aligned}
+$$
+
+Now (1) implies that $2^{n} M \geq 2^{n}\left(\left|a_{11}\right|+\left|a_{22}\right|+\cdots+\left|a_{n n}\right|\right)$ which is equivalent to the inequality we have to prove.
+

BayArea/md/en-monthly/en-0607-mc6sol.md

@@ -0,0 +1,51 @@
+# Berkeley Math Circle <br> Monthly Contest 6 - Solutions 
+
+1. If $a$ and $b$ are two positive numbers not greater than 1 prove that
+
+$$
+\frac{a+b}{1+a b} \leq \frac{1}{1+a}+\frac{1}{1+b}
+$$
+
+When does the equality hold?
+
+Solution. If we multiply both sides of the inequality with $(1+a b)(1+a)(1+b)$ the required inequality becomes equivalent to
+
+$$
+\begin{gathered}
+(a+b)(1+a)(1+b) \leq(1+a)(1+a b)+(1+b)(1+a b) \Leftrightarrow \\
+(a+b)(1+a+b+a b) \leq 1+a b+a+a^{2} b+1+a b+b+a b^{2} \Leftrightarrow \\
+a+a^{2}+a b+a^{2} b+b+a b+b^{2}+a b^{2} \leq 1+a b+a+a^{2} b+1+a b+b+a b^{2} \Leftrightarrow \\
+a^{2}+b^{2} \leq 2 .
+\end{gathered}
+$$
+
+The last inequality is true and the equality holds if and only if $a=b=1$.
+
+2. A car is moving at a constant speed. Every 15 minutes it makes a $90^{\circ}$ turn to either left or right. If the car has started the trip at some point $A$, prove that it can return to $A$ only after an integer number of hours.
+
+Solution. It is important to notice that an hour consists of 60 minutes and $60 / 15=4$. That's why it is common to say that 15 minutes is a "quarter of an hour". We may assume that the car is traveling in a square grid where the size of each square is the distance that the car travels in 15 minutes. Assume further that we know what is "north", "east", "west", and "south". Let $n$, $e, w$, and $s$, denote respectively the number of north, east, west, and south moves. If the car has returned to $A$, then $n=s$ and $w=e$. However since the car at every vertex changes the direction from the set $\{$ north, south $\}$ to the set $\{$ east, west $\}$, or vice versa, we see that $n=w$. Thus $n=s=w=e$ and the total number of moves is $n+s+w+e=4 n$ which is divisible by 4 . Thus the total time of travel is $4 \cdot 15^{\prime} n=n$ hours.
+
+3. Let $M$ be an interior point of a parallelogram $A B C D$. Prove that $M A+M B+M C+M D$ is strictly less than the length of the perimeter of $A B C D$.
+
+Solution. Denote by $X$ and $Y$ the points of intersection of the segments $A B$ and $C D$ with the line through $M$ parallel to $B C$. Similarly, let $U$ and $V$ denote the points of intersection of the segments $A D$ and $B C$ with the line through $M$ parallel to $A B$. Then $M A<A U+U M=X M+U M, M B<M X+X B=M X+M V, M C<M V+V C=M V+M Y$, and $M D<M Y+Y D=M Y+M U$. Hence $M A+M B+M C+M D<2(X M+Y M)+2(U M+V M)=2 A D+2 A B=$ $A B+B C+C D+D A$.
+
+4. Prove that the product of 6 consecutive positive integer is never equal to $n^{5}$ for some positive integer $n$.
+
+Solution. It is easy to show that at least one of 6 consecutive numbers is relatively prime to each of the others ( 3 of these numbers are odd, exactly one of these odd numbers is divisible by 3 , and for each prime $p>3$ at most one is divisible by $p$ ). Assume that $x-2, x-1, x, x+1, x+2, x+3, x \geq 3$ are given numbers. The number relatively prime to the others has to be a perfect 5 th power. Let $F$ be the product of the other 5 numbers. We have:
+
+$P(x)=(x-2)(x-1) x(x+1)(x+2)=x^{5}-5 x^{3}+4 x \leq F \leq x^{5}+5 x^{4}+5 x^{3}-5 x^{2}-6 x=(x-1) x(x+1)(x+2)(x+3)=Q(x)$.
+
+Since for $x \geq 3$ we have $(x+1)^{5}>Q(x) \geq F \geq P(x)>(x-1)^{5}$ we conclude that $F=x^{5}$. However this is a contradiction since $x$ is relatively prime to $x-1$ and $x+1$ and at least one of these numbers is a factor of $F$.
+
+5. The point $K$ lies in the interior of the unit circle. Four lines are drawn through $K$ such that each two adjacent lines form an angle of $45^{\circ}$, as shown in the picture. In such a way the circle is divided into 8 regions. Four of these regions are colored such that no two colored regions share more than one point. Find the maximal and minimal possible value for the total area of colored regions.
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_2f0af4b96a9a083248abg-2.jpg?height=414&width=417&top_left_y=381&top_left_x=842)
+
+## Solution.
+
+We will show that the total area of colored regions is equal to $\frac{1}{2}$ the area of the circle no matter how the lines are drawn.
+
+First we will inscribe our circle in a square whose sides are parallel to 2 of these lines. If we color the regions of the square in the same way as the regions of the circle are colored we can easily prove that the total area of colored regions is equal to the area of non-colored regions for the square. Hence it remains to prove the equality of the areas of colored and uncolored regions that are outside the cirle. Now we will look at the pictures below. For each picture it is obvious that it has the same area of colored parts as the previous picture (as some colored parts are just moved from one location to the other).
+![](https://cdn.mathpix.com/cropped/2024_04_17_2f0af4b96a9a083248abg-2.jpg?height=342&width=756&top_left_y=1508&top_left_x=273)
+![](https://cdn.mathpix.com/cropped/2024_04_17_2f0af4b96a9a083248abg-2.jpg?height=978&width=730&top_left_y=866&top_left_x=1095)
+

BayArea/md/en-monthly/en-0607-mc7sol.md

@@ -0,0 +1,58 @@
+# Berkeley Math Circle Monthly Contest 7 - Solutions 
+
+1. Find all positive integers $n$ such that $n(n+1)$ is a perfect square.
+
+Solution. Since $n$ and $n+1$ are coprime numbers, if $n(n+1)$ is a perfect square, then each of $n$ and $n+1$ has to be a perfect square itself. However that is impossible since if $n=x^{2}$ and $n+1=y^{2}$ we would have $1=y^{2}-x^{2}=(y-x)(y+x)$ and 1 can't be expressed as a product of two different integers. Hence there are no such integers.
+
+2. Prove that
+
+$$
+A=\sqrt{4-2 \sqrt{3}}-\frac{\sqrt{3}+1}{\sqrt{3}-1}
+$$
+
+is an integer.
+
+Solution. $\sqrt{4-2 \sqrt{3}}=\sqrt{3^{2}-2 \sqrt{3}+\sqrt{1}^{2}}=\sqrt{(\sqrt{3}-1)^{2}}=|\sqrt{3}-1|=\sqrt{3}-1$. If multiply both numerator and the denominator of $\frac{\sqrt{3}+1}{\sqrt{3}-1}$ by $\sqrt{3}+1$ we get:
+
+$$
+\frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{(\sqrt{3}+1)^{2}}{\sqrt{3}^{2}-1^{2}}=\frac{4+2 \sqrt{3}}{2}=2+\sqrt{3}
+$$
+
+Thus $A=-1-2=-3$, which is an integer.
+
+3. An isolated island has the shape of a circle. Initially there are 9 flowers on the circumference of the island: 5 of the flowers are red and the other 4 are yellow. During the summer 9 new flowers grow on the circumference of the island according to the following rule: between 2 old flowers of the same color a new red flower will grow, between 2 old flowers of different colors, a new yellow flower will grow. During the winter, the old flowers die, and the new survive. The same phenomenon repeats every year.
+
+Is it possible (for some configuration of initial 9 flowers) to get all red flowers after finitely many years?
+
+Solution. The answer is "no". Assume that we got all red flowers in the year $n$ for the first time. Then in the year $n-1$ all the flowers were yellow. We will prove that this is impossible.
+
+Let's change the weird story into the one with the flowers labeled by 1 (instead of red) and 1 (instead of yellow). What really happens is that between two flowers $a$ and $b$, the new flower will grow and will be labeled by $a b$. Notice that the initial product of all numbers is 1 , and at the end of each winter the product of the numbers is 1 again, so it will never be equal to -1 hence it is impossible to get the configuration where all the flowers are yellow. This is a contradiction.
+
+4. Given a triangle $A B C$, a circle $k$ is tangent to the lines $A B$ and $A C$ at $B$ and $P$. Let $H$ be the foot of perpendicular from the center $O$ of $k$ to $B C$, and let $T$ be the intersection point of $O H$ and $B P$. Prove that $A T$ bisects the segment $B C$.
+
+Solution. Let $X$ be the intersection of $A T$ and $B C, S$ the intersection of $B C$ and $A O$, and $Y$ the intersection of $B P$ and $A O$. Since $B Y \perp S O$ and $O H \perp B S, T$ is the orthocenter of $\triangle B O S$ and $S T \perp B O$, hence $S T \| A B$. Thus
+
+$$
+S T: A B=T X: X A \text {. }
+$$
+
+However since $A B=A P$ and the triangles $S Y T$ and $A Y P$ are similar we get
+
+$$
+S T: A P=T Y: Y P .
+$$
+
+From (1) and (2) we now get $X Y \| A P$ and since $Y$ is the midpoint of $B P, X Y$ is the middle line of $\triangle C B P$ and $B X=X C$.
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_679f3d10fe27b2e4bdfbg-1.jpg?height=665&width=462&top_left_y=1712&top_left_x=1274)
+
+5. A society has 100 members and every two members are either friends or enemies. Prove that there are two persons from the society that have an even number of common enemies.
+
+Solution. Assume that for each two members the set of common enemies has an odd number of elements. Then the set of their common friends has also an odd number of elements. Fix a person $a$. Let $F=\left\{f_{1}, \ldots, f_{k}\right\}$ be the set of friends of the member $a$, and $E=\left\{e_{1}, \ldots, e_{99-k}\right\}$ the set of its enemies. First we will show that $k$ is even. Assume that $f_{1}$ has $m_{1}$ friends inside $F$, $f_{2}$ has $m_{2}$ friends inside $F$, etc. Then $m_{1}+m_{2}+\cdots+m_{k}$ is equal twice the number of all friendships inside $F$. However each of the numbers $m_{1}, \ldots, m_{k}$ must be odd because $m_{i}$ represents the number of common friends for $a$ and $f_{i}$. Since the sum of $k$ odd numbers ended up being even, we conclude that $k$ is even.
+
+Thus, every member has an even number of friends (the argument in the previous paragraph works for any member of the society, not just $a$ ). Now we will caclulate the parity of the number of friendships between members of $E$ and members of $F$. We will do that in two different ways and get the contradiction.
+
+For every $i, 1 \leq i \leq 99-k, e_{i}$ has to have an odd number of friends among $F$, because $e_{i}$ and $a$ have an odd number of common freinds. The number of elements of $E$ is odd hence the total number of freindships between $E$ and $F$ has to be odd.
+
+On the other hand, $b_{i}(1 \leq i \leq k)$ has an odd number of friends inside $F, a$ is also a friend of $b_{i}$, and the total number of friends of $b_{i}$ has to be even. Thus, $b_{i}$ has an even number of friends inside $E$. Hence the number of friendships between $F$ and $E$ is even. A contradiction!
+

BayArea/md/en-monthly/en-0607-mc8sol.md

@@ -0,0 +1,69 @@
+# Berkeley Math Circle Monthly Contest 8 - Solutions 
+
+1. Define
+
+$$
+A=1+\frac{1}{2+\frac{1}{3+\frac{1}{4+\ddots_{+} \frac{1}{2006+\frac{1}{2007}}}}} \text { and } B=1+\frac{1}{2+\frac{1}{3+\frac{1}{4+\ddots+\frac{1}{2005+\frac{1}{2006}}}}}
+$$
+
+Which of the two numbers is greater, $A$ or $B$ ? Explain your answer!
+
+Solution. We will determine the sign of $A-B$. If that number happens to be positive than $A>B$, otherwise $A<B$. For each $n$ such that $1 \leq n \leq 2005$, let us define
+
+$$
+A_{n}=n+\frac{1}{n+1+\frac{1}{n+2+\frac{1}{n+3+\ddots^{\prime+\frac{1}{2006+\frac{1}{2007}}}}}} \quad B=n+\frac{1}{n+1+\frac{1}{n+2+\frac{1}{n+3+} \cdot \frac{1}{2005+\frac{1}{2006}}}}
+$$
+
+Then $A_{1}=A, B_{1}=B$ and we always have that $A_{i}-B_{i}$ and $A_{i+1}-B_{i+1}$ have different signs. Indeed, since $A_{i}=i+\frac{1}{A_{i+1}}$ and $B_{i}=i+\frac{1}{B_{i+1}}$ we have $A_{i}-B_{i}=\frac{B_{i+1}-A_{i+1}}{A_{i+1} B_{i+1}}$. Therefore if $A_{i}-B_{i}>0$ then $B_{i+1}-A_{i+1}>0$ and vice versa. Thus $A_{1}-B_{1}, A_{3}-B_{3}, \ldots, A_{2005}-B_{2005}$ have the same signs. From
+
+$$
+A_{2005}-B_{2005}=\left(2005+\frac{1}{2006+\frac{1}{2007}}\right)-\left(2005+\frac{1}{2006}\right)<0
+$$
+
+we get $A<B$.
+
+2. A group of mathematicians is lost in a forest. The forest has a shape of an infinite strip that is 1 mile wide. Prove that they can choose a path that will guarantee them a way out and that is at most $2 \sqrt{2}$ miles long.
+
+Remark. The mathematicians have no device for orientation and no maps. All they know is that the forest is a region between two parallel lines 1 mile appart from each other. They can't see the end of the forest, unless they are at the edge. However, they can precisely follow any path they design, e.g. they can move along straight line, circle, etc.
+
+Solution. Starting from their initial point $A$, the mathematicians should first move $\sqrt{2}$ miles in any direction. If they didn't get to the exit, they have arrived at point $B$. Then they should turn by $90^{\circ}$ and walk for another $\sqrt{2}$ miles to the point $C$. We claim that they did reach the exit. If not, then the triangle $A B C$ is entirely in the forest. The triangle is rectangular, isosceles and its altitude is 1 mile. Thus the triangle can't be placed in the infinite strip of width 1 unless the vertices are on the edges. Contradiction.
+
+3. The numbers $1,8,4,0$ are the first four terms of the infinite sequence. Every subsequent term of the sequence is obtained as the last digit of the sum of previous four terms. Therefore the fifth term of the sequence is 3 , because $1+8+4+0=13$; the sixth term is 5 because $8+4+0+3=5$, and so on.
+
+(a) Will 2, 0, 0, 7 ever appear as a subsequence?
+
+(b) Will 1, 8, 4, 0 appear again as a subsequence?
+
+Explain your answer!
+
+## Solution.
+
+(a) Yes, very soon, in fact the next four terms (from 7 to 10th) are 2, 0, 0, 7 .
+
+(b) We will prove that 1,8,4,0 will be a subsequence again. Assume the contrary. Since there are only finitely many combinations of four digits (precisely $10^{4}$ ), and the sequence is infinite, some combination of four digits (say $(a, b, c, d)$ ) has to reappear. Assume that $\left(x_{n}, x_{n+1}, x_{n+2}, x_{n+3}\right)=(a, b, c, d)$ is the first occurence of $(a, b, c, d)$ and that $\left(x_{m}\right.$, $\left.x_{m+1}, x_{m+2}, x_{m+3}\right)=(a, b, c, d)$ is the second. Clearly $m>n$. However, $x_{n-1}$ and $x_{m-1}$ are uniquely determined and they have to be the same numbers. Thus $x_{n-1}=x_{m-1}, x_{n-2}=x_{m-2}$, and so on. This means that $x_{1}=x_{m-n+1}$, $x_{2}=x_{m-n+2}, x_{3}=n_{m-n+3}$, and $x_{4}=x_{m-n+4}$, which is a contradiction.
+
+4. Let $X, Y$, and $Z$ be the points on the sides $B C, C A$, and $A B$ of the triangle $A B C$, such that $\triangle X Y Z \sim \triangle A B C(\varangle X=\varangle A$, $\varangle Y=\varangle B)$. Prove that the orthocenter of $\triangle X Y Z$ coincides with the circumcenter of $\triangle A B C$.
+
+Solution. Let $x, y$, and $z$ be the points passing through $X, Y$, and $Z$, parallel to $Y Z, Z X$, and $X Y$. Let $P, Q$, and $R$ be the intersection point of the lines $y$ and $z ; z$ and $x ; x$ and $y$, respectively. Then we have $\triangle P Q R \sim \triangle X Y Z$. The points $X, Y$, and $Z$ are respectively the midpoints of $R Q, Q P$, and $P R$. Let $M$ be the orthocenter of $\triangle X Y Z$. Obviously, $M$ is the circumcenter of $\triangle P Q R$ and $\varangle Z M Y=180^{\circ}-\varangle X=180^{\circ}-\varangle P=180^{\circ}-\varangle A$. Hence the points $P, A, Z, M$, and $Y$ belong to a circle. In a similar way we prove that the points $Z, B, R, X$ and $M$ belong to a circle. Then $\varangle P M A=\varangle P Z A=\varangle B Z R=\varangle B M R$. Since $M R=M P$ and $\varangle P A M=\varangle B R M=90^{\circ}$, we conclude that $M A=M B$. Analogously we conclude that $M A=M C$ implying that $M$ is the circumcenter of $\triangle A B C$.
+
+5. Let $n>1$ be an odd integer. Prove that every integer $l$ satisfying $1 \leq l \leq n$ can be represented as a sum or difference of two integers each of which is less than $n$ and relatively prime to $n$.
+
+Solution. We will use the following lemma (it is known as The Chinese Remainder Theorem).
+
+Lemma. Let $m_{1}, m_{2}, \ldots, m_{k}$ be different relatively prime numbers. If $q_{1}, q_{2}, \ldots, q_{k}$ are arbitrary non-negative integers then there exists a natural number $x$ less than $m_{1} m_{2} \ldots m_{k}$ such that
+
+$$
+\begin{aligned}
+x & \equiv q_{1}\left(\bmod m_{1}\right) \\
+x & \equiv q_{2}\left(\bmod m_{2}\right) \\
+& \vdots \\
+x & \equiv q_{k}\left(\bmod m_{k}\right) .
+\end{aligned}
+$$
+
+Proof of the Lemma. We can assume that $q_{i}<m_{i}$ for $1 \leq i \leq k$. We will prove this by induction. For $k=2$ we consider the numbers $x_{1}=q_{1}, x_{2}=m_{1}+q_{1}, x_{3}=2 m_{1}+q_{1}, \ldots, x_{m_{2}}=\left(m_{2}-1\right) m_{1}+q_{1}$. If two of the number $x_{i}$ and $x_{j}$ give the same remainder upon division by $m_{2}$ we have that $x_{j}-x_{i}$ is divisible by $m_{2}$ which is impossible since that difference is equal to $(j-i) m_{1}$. Thus all $x_{i}$ s give different remainders modulo $m_{2}$ and one of them has to give a remainder $q_{2}$. This finishes the proof for the case $k=2$.
+
+Assume now that the statement holds for $k$ and we want to prove it for $k+1$. Applying the inductional hypothesis $q_{1}, \ldots, q_{k}$ we find a number $x^{\prime}$ such that $x^{\prime} \equiv q_{i}\left(\bmod m_{i}\right)$ for $1 \leq i \leq m_{i}$. Now by the case $k=2$ we get that there is an $x$ such that $x \equiv x^{\prime}\left(\bmod m_{1} \cdots m_{k}\right)$ and $x \equiv q_{k+1}\left(\bmod m_{k+1}\right)$. Such $x$ satisfies the required condition.
+
+Let $n=p_{1}^{\alpha_{1}} \ldots p_{k}^{\alpha_{k}}$, where $p_{1}, \ldots, p_{k}$ are prime numbers and $\alpha_{1}, \ldots \alpha_{k}$ positive integers. Assume that $l$ is given and that $l \equiv t_{i}\left(\bmod p_{i}\right)$. Since $p_{i}>2$ we can choose $s_{i}$ such that $s_{i} \not \equiv 0\left(\bmod p_{i}\right)$ and $s_{i} \not \equiv t_{i}\left(\bmod p_{i}\right)$. By the Chinese Remainder Theorem there exists an integer $s<p_{1} \ldots p_{k}$ such that $s \equiv s_{i}\left(\bmod p_{i}\right)$ for every $i=1, \ldots, k$. If $s<l$ then choose $a=s$ and $b=l-s$. If $s>l$ then we can choose $a=s$ and $b=s-s$. It is easy to verify that such $a$ and $b$ satisfy the conditions of the problem.
+

BayArea/md/en-monthly/en-0708-comp1s.md

@@ -0,0 +1,44 @@
+# Berkeley Math Circle <br> Monthly Contest 1 - Solutions 
+
+1. If $a$ and $b$ are positive integers prove that
+
+$$
+a+b \leq 1+a b \text {. }
+$$
+
+Solution. Since $a$ and $b$ are positive integers we know that $a \geq 1$ and $b \geq 1$. This implies that $(a-1)(b-1) \geq 0$ hence $a b-a-b+1 \geq 0$ which is equivalent to the given inequality.
+
+2. A man has three pets: a mouse, a cat, and a dog. If the man leaves the cat and the dog alone, then the dog would kill the cat. If the man leaves the cat and the mouse alone, the cat would eat the mouse. One day the man decided to take his animals to the other side of the river. However, he has a small boat in which he can fit only one of the animals at a time. Show that it is possible for the man to take all of his animals to the other side of the river safely.
+
+Solution. Let $A$ be the side of the river where the trip starts, and $B$ the other side. In the first trip the man should take the cat to the side $B$. He would leave the mouse and the dog together, but that is not dangerous. Then he will return back for a dog and take it to the side $B$. However when he brings the dog, then he returns the cat from $B$ to $A$. Then he takes the mouse from $A$ to $B$ and comes back to the side $A$ to finally take the cat.
+
+3. On a small piece of paper two line segments $s_{1}$ and $s_{2}$ are drawn as shown on the picture. The extensions of $s_{1}$ and $s_{2}$ eventually intersect at a point $P$ that doesn't belong to the piece of paper. If $D$ is an arbitrary point marked on the paper, show how to construct a segment of the line connecting $D$ and $P$ using just a straight edge and a compass and performing all constructions on the given piece of paper.
+
+Solution. Let $m$ be an arbitrary line through $D$ that intersects $s_{1}$ and $s_{2}$ in $M_{1}$ and $M_{2}$ and let $n$ be an arbitrary line parallel to $m$ that intersects $s_{1}$ and $s_{2}$ at $N_{1}$ and $N_{2}$. If we find a point $E$ on $n$ such that $M_{1} D: D M_{2}=N_{1} E: E N_{2}$ then we will be sure that $s_{1}, s_{2}$, and $E D$ pass through the same point (because of the Thales' theorem). In order to construct $E$ we first construct the point $D^{\prime}$ on $M_{1} N_{2}$ such that $D D^{\prime} \| s_{2}$. Then $M_{1} D: D M_{2}=M_{1} D^{\prime}: D^{\prime} N_{2}$ because of the similarity of the triangles $M_{1} D D^{\prime}$ and $M_{1} M_{2} N_{2}$. Similarly, $E$ is the point of $n$ such that $E D^{\prime} \| s_{1}$ (now the similar triangles are $N_{2} E D^{\prime}$ and $N_{2} N_{1} M_{1}$ ).
+
+4. Denote by $f(n)$ the integer obtained by reversing the digits of a positive integer $n$. Find the greatest integer that is certain to divide $n^{4}-f(n)^{4}$ regardless of the choice of $n$.
+
+Solution. The answer is 99 . Let $x=\overline{d_{n-1} \ldots d_{2} d_{1} d_{0}}$, i.e. $x=10^{n-1} d_{n-1}+\cdots+10^{2} d_{2}+10 d_{1}+d_{0}$. Then $y=f(x)=10^{n-1} d_{0}+10^{n-2} d_{1}+\cdots+10 d_{n-2}+d_{n-1}$.
+
+Let us show separately that $9 \mid x^{4}-y^{4}$ and that $11 \mid x^{4}-y^{4}$. Since $10 \equiv 1(\bmod 9), x \equiv y \equiv d_{n-1}+d_{n-2}+\cdots+d_{1}+d_{0}(\bmod 9)$, so $9 \mid x-y$ and therefore also $9 \mid(x-y)\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)=x^{4}-y^{4}$. Since $10 \equiv-1(\bmod 11), x \equiv d_{n-1}-d_{n-2}+$ $\cdots \mp d_{1} \pm d_{0}(\bmod 11)$ and $y \equiv d_{0}+d_{1}+\cdots \mp d_{n-2} \pm d_{n-1}(\bmod 11)$, i.e. $x \equiv \pm y$. In either case $x^{4} \equiv y^{4} \bmod 11$, so $11 \mid x^{4}-y^{4}$. Since 9 and 11 are relatively prime, their product 99 must divide $x^{4}-y^{4}$.
+
+To see that there is no larger integer that always divides $n^{4}-f(n)^{4}$, let $n=10$, so that $f(n)=01=1, n^{4}-f(n)^{4}=9999=$ $3^{2} \cdot 11 \cdot 101$, and let $n=21$, so that $f(n)=12$ and $n^{4}-f(n)^{4}=173745=3^{5} \cdot 5 \cdot 11 \cdot 13$. The greatest common divisor of these two values of $n^{4}-f(n)^{4}$ is $3^{2} \cdot 11=99$, so there is no larger integer certain to divide $n^{4}-f(n)^{4}$.
+
+Remark. From the solution it is easy to see that 99 is even the common divisor of the numbers of the form $n^{2}-f(n)^{2}$.
+
+5. Given a polynomial $P(x)$ with integer coefficients, assume that for every positive integer $n$ we have $P(n)>n$. Consider the sequence
+
+$$
+x_{1}=1, x_{2}=P\left(x_{1}\right), \ldots, x_{n}=P\left(x_{n-1}\right), \ldots
+$$
+
+If for every positive integer $N$ there exists a member of the sequence divisible by $N$, prove that $P(x)=x+1$.
+
+Solution. Assume the contrary. The polynomial $Q(x)=P(x)-x$ is non-decreasing for all $x$ greater than some $M$, as otherwise it wouldn't be positive, and it has to be bigger than 1 for each $n$.
+
+If there are infinitely many integers $n$ such that $P(n)=n+1$ then we would have $P(x)=x+1$ (nonzero polynomial $P(x)-x-1$ can have only finitely many zeroes). Therefore there exists an index $k \in \mathbb{N}$ such that $x_{k}>M$ and $P\left(x_{l}\right)-x_{l} \geq 2$ for $l \geq k$.
+
+Assume that $d=x_{l+1}-x_{l} \geq 2$. Then $d$ is a divisor of $P\left(x_{l+1}\right)-P\left(x_{l}\right)$ i.e. of $x_{l+2}-x_{l+1}$. Since, if $a \equiv b(\bmod d)$, then $P(a) \equiv P(b)$, by induction all the numbers $x_{l}, x_{l+1}, \ldots$ have the same remainder modulo $d$. If that remainder is not zero than no term of the sequence could be divisible by $d^{m}$ for sufficiently large $m$ which would contradict the assumptions.
+
+Hence $x_{l}$ is divisible by $x_{l+1}-x_{l}$ for all $l \geq k$. Let $x_{l}=c_{l}\left(P\left(x_{l}\right)-x_{l}\right)$. Then $P\left(x_{l}\right)=\frac{c_{l}+1}{c_{l}} x_{l}$ holds for infinitely many pairs of integers $\left(x_{l}, c_{l}\right)$ such that $x_{l} \rightarrow \infty$. Because $\frac{c_{l}+1}{c_{l}} x_{l}=\left(1+\frac{1}{c_{l}}\right) x_{l} \leq 2 x_{l}$, the degree of $P$ cannot exceed 1 so that either $P(x)=2 x$ or $P(x)=x$. Neither of these polynomials satisfy the given condition, and this is a contradiction.
+

BayArea/md/en-monthly/en-0708-comp2s.md

@@ -0,0 +1,93 @@
+# Berkeley Math Circle Monthly Contest 2 - Solutions 
+
+1. Four friends, One, Two, Five, and Ten are located on one side of the dark tunnel, and have only one flashlight. It takes one minute for person One to walk through the tunnel, two minutes for Two, five for Five, and ten for Ten. The tunnel is narrow and at most two people can walk at the same time with the flashlight. Whenever two people walk together they walk at the speed of the slower one. Show that all four friends can go from one side of the tunnel to the other one in 17 minutes.
+
+Remark. Your explanation should be something like this: The friends $X$ and $Y$ first go through the tunnel using the flashlight, then $X$ returns with the flashlight to the other side,...
+
+Solution. Friends One and Two should walk to the other side. It will take them 2 minutes. Then, One returns - this will last additional 1 minute. Then, Ten and Five go to the other side - this will take 10 minutes, but Two should return the flashlight (2 additional minutes), and together with One, go through the tunnel for the last time - additional 2 minutes. It remains to notice that the time spent is exactly 17 minutes.
+
+2. The integers from 1 to 16 are arranged in a $4 \times 4$ array so that each row, column and diagonal adds up to the same number.
+
+(a) Prove that this number is 34 .
+
+(b) Prove that the four corners also add up to 34.
+
+## Solution.
+
+(a) Add up all the numbers in the square in two ways. On the one hand, it consists of four rows, each adding to the common sum $S$, so the entire square adds to $4 S$. But the numbers in the square are also the integers from 1 to 16 , whose sum is
+
+$$
+\frac{16 \cdot 17}{2}=152
+$$
+
+Hence
+
+$$
+\text { Sum of all numbers in square }=4 S=152
+$$
+
+whence $S=34$.
+
+(b) Denote the cells of the table as shown below.
+
+| $a$ | $b$ | $c$ | $d$ |
+| :---: | :---: | :---: | :---: |
+| $e$ | $f$ | $g$ | $h$ |
+| $i$ | $j$ | $k$ | $m$ |
+| $n$ | $p$ | $q$ | $r$ |
+
+Adding the equations
+
+$$
+\begin{aligned}
+a+b+c+d & =34 \\
+n+p+q+r & =34 \\
+a+f+k+r & =34 \\
+n+j+g+d & =34 \\
+-b-f-j-p & =-34 \\
+-c-g-k-q & =-34
+\end{aligned}
+$$
+
+yields $2 a+2 d+2 n+2 r=68$, or $a+d+n+r=34$.
+
+3. Let $A_{1}, B_{1}, C_{1}$ be the points on the sides $B C, C A, A B$ (respectively) of the triangle $A B C$. Prove that the three circles circumscribed about the triangles $\triangle A B_{1} C_{1}, \triangle B C_{1} A_{1}$, and $\triangle C A_{1} B_{1}$ intersect at one point.
+
+Solution. Denote by $\alpha$, $\beta$, and $\gamma$ the angles of $\triangle A B C$. Assume that the circumcircles of $\triangle A B_{1} C_{1}$ and $B C_{1} A_{1}$ intersect at the point $M$. Assume that $M$ is in the interior of $\triangle A B C$ (the other cases are similar). By the properties of the inscribed quadrilaterals $A B_{1} M C_{1}$ and $B A_{1} M C_{1}$ we get: $\angle C_{1} M B_{1}=180^{\circ}-\angle C_{1} A B_{1}=180^{\circ}-\alpha$ and $\angle C_{1} M A_{1}=180^{\circ}-$ $\angle C_{1} B A_{1}=180^{\circ}-\beta$ hence $\angle B_{1} M A_{1}=360^{\circ}-\left(\angle C_{1} M B_{1}+\angle C_{1} M A_{1}\right)=360^{\circ}-\left[360^{\circ}-(\alpha+\beta)\right]=\alpha+\beta$. Therefore $\angle B_{1} M A_{1}+\angle B_{1} C A_{1}=\angle B_{1} M A_{1}+\gamma=\alpha+\beta+\gamma=180^{\circ}$. Thus the points $B_{1}, M, A_{1}$, and $C$ belong to a circle.
+
+4. A Mystic Four Calculator has a four-digit display and four buttons. The calculator works as follows: Pressing button 1 replaces the number in the display with 1; Pressing button 2 divides the number in the display by 2; Pressing button 3 subtracts 3 from the number in the display; Pressing button 4 multiplies the number in the display by 4 .
+
+Initially the display shows 0 . Any operation yielding a negative, fractional, or five-digit answer is ignored.
+
+(a) Can 2007 appear in the display?
+
+(b) Can 2008 appear in the display?
+
+## Solution.
+
+(a) No. Notice that if the number on the display is not divisible by 3 , then none of the operation can have as a result a number divisible by 3 . At the start, 1 is the only button producing a result, so we are required to press it at some point. After that the number will never be divisible by 3 again, and 2007 is divisible by 3.
+
+(b) Yes. For instance, press 1 so that the display shows 1, press 4 six times so that the display shows 4096, and press 3696 times so that the display shows 2008.
+
+5. A number bracelet in base $m$ is made by choosing two nonnegative integers less than $m$ (not both 0 ) and continuing in a clockwise loop, each succeeding number being the mod $m$ sum of its two predecessors. The figure is closed up as soon as it starts to repeat. The figure to the right shows two number bracelets in base 10 , starting with the pairs $(1,3)$, and $(2,2)$, respectively. Prove that the lengths of all number bracelets in a given base are divisors of the length of the number bracelet beginning with $(0,1)$.
+
+Solution. The elements of a number bracelet $N$ will be denoted $N_{0}, N_{1}, N_{2}, \ldots ; N_{0}$ and $N_{1}$ being the starting numbers. Let $F$ be the bracelet with starting numbers $F_{0}=0, F_{1}=1$. All congruences are modulo $m$ unless otherwise noted.
+
+Since there are only $m^{2}$ possible pairs $\left(N_{k}, N_{k+1}\right.$ ), there must be repetition, so that
+
+$$
+\left(N_{k}, N_{k+1}\right)=\left(N_{p+k}, N_{p+k+1}\right)
+$$
+
+for some $p \geq 1$ and $k$. If this is true for one $k$, it must be true for the next $k$, since $N_{k+1}=N_{p+k+1}$ by hypothesis and $N_{k+2} \equiv N_{k}+N_{k+1}=N_{p+k}+N_{p+k+1} \equiv N_{p+k+2}$, and so it is true for all $k$ bigger than that $k$. It must also be true for the previous $k$, since $N_{k}=N_{p+k}$ by hypothesis and $N_{k-1} \equiv N_{k+1}-N_{k}=N_{p+k+1}-N p+k \equiv N_{p+k-1}$, and so it is true for all $k$ smaller than that $k$. Therefore ili is true for all $k$. Let $P_{N}$, the period of $N$, be the smallest $p$ such that is true for some $k$ (and therefore for all $k$ ).
+
+It is clear that if $0 \leq i, j<P(N), i \neq j$, then $\left(N_{i}, N_{i+1}\right) \neq\left(N_{j}, N_{j}+1\right)$. Therefore, among the terms $N_{0}$ through $N_{P_{N}}$, there is no repetition even of pairs of adjacent terms. But because of property il of $P_{N},\left(N_{0}, N_{1}\right)=\left(N_{P_{N}}\right),\left(N_{P_{N}+1}\right)$ and the repetition will continue. Thus the number bracelet $N$ consists of a single loop of $P_{N}$ elements.
+
+We will prove by induction that for all $n \geq 1$,
+
+$$
+N_{n} \equiv N_{0} F_{n-1}+N_{1} F_{n} .
+$$
+
+The case $n=1$ is trivial, and $n=2$ follows directly from the definition of $N$. If $\boldsymbol{v}$ is true for $n=k$ and $n=k-1$, $N_{k+1} \equiv N_{k}+N_{k-1} \equiv N_{0} F_{k-1}+N_{1} F_{k}+N_{0} F_{k-2}+N_{1} F_{k-1}=N_{0} F_{k}+N_{1} F_{k+1}$ so it is also true for $n=k+1$. It is clear from the circularity of the number bracelet that $N_{k+R}=N_{k}$ for all $k>1$ if and only if $P_{N} \mid R$. Since $N_{k+P_{F}}=$ $N_{0} F_{P_{F}+k-1}+N_{1} F_{P_{F}+k}=N_{0} F_{k-1}+N_{1} F_{k}=N_{k}$ for all $k>1, P_{N} \mid P_{F}$. This completes the proof.
+

BayArea/md/en-monthly/en-0708-comp3s.md

@@ -0,0 +1,75 @@
+# Berkeley Math Circle Monthly Contest 3 - Solutions 
+
+1. Given 8 oranges on the table, 7 of them have exactly the same weight and the 8th is a little bit lighter. You are given a balance that can measure oranges against each other and you are allowed to use the balance at most twice! How can you determine which one of the oranges is lighter then the others? Explain your answer!
+
+Remark. All oranges look the same and the difference in the weight of the lighter orange is not big enough for you to distinguish it without using the balance. The balance doesn't have any weights or numbers. If you put some oranges on each side of the balance, you can only tell which side (if any) is heavier.
+
+Solution. First we put 3 oranges on the left and 3 on the right-hand side of the balance. In the case that balance shows equal weights, one of the remaining two oranges is lighter and in the second measurment we can easily see which one. However if one side (say left) is lighter, then we know that the orange we are looking for is one of the three that appear on the left side. Let us denote these oranges by $A, B, C$. In the second measurment we can measure $A$ against $B$. If one of them is lighter than the other, we are done (as the lighter orange is the one we are looking for). If the weights of $A$ and $B$ are the same, then $C$ is the lighter orange.
+
+2. Find all prime numbers $p$ such that $p^{2}+8$ is prime number, as well.
+
+Remark. A number $p$ is prime if it has exactly 2 divisors: 1 and $p$. Numbers $2,3,5,7,11,13, \ldots$ are prime, while 4 and 2007 are not.
+
+Hint. Write down first several prime numbers (hint - you can copy them from the paragraph above), calculate $p^{2}+8$ for them, and look at those that happen to be composite. Notice further that they all have a common divisor.
+
+Solution. For $p=3$ we have $p^{2}+8=17$, which is prime. If $p \neq 3$ then $p$ is not divisible by 3 . The remainder of $p$ when divided by 3 is either 1 or 2 . This means that $p=3 k+1$ for some integer $k$, or $p=3 l+1$ for some integer $l$. In the first case we get $p^{2}=9 k^{2}+6 k+1$ and in the second, $p^{2}=9 l^{2}-6 l+1$. In both cases $p^{2}$ gives a remainder 1 upon division by 3 . Hence $p^{2}+8$ is divisible by 3 for all prime numbers different than 3 .
+
+3. $p$ is a prime number such that the period of its decimal reciprocal is 200 . That is,
+
+$$
+\frac{1}{p}=0 . X X X X \ldots
+$$
+
+for some block of 200 digits $X$, but
+
+$$
+\frac{1}{p} \neq 0 . Y Y Y Y \ldots
+$$
+
+for all blocks $Y$ with less than 200 digits. Find the 101st digit, counting from the left, of $X$.
+
+Solution. Let $X$ be a block of $n$ digits and let $a=0 . X \ldots$ Then $10^{n} a=X . X \ldots$. Subtracting the previous two equalities gives us $\left(10^{n}-1\right) a=X$, i.e. $a=\frac{X}{10^{n}-1}$.
+
+Then the condition that $a=\frac{1}{p}$ reduces to $\frac{1}{p}=\frac{X}{10^{n}-1}$ or $p X=10^{n}-1$. For a given $p$ and $n$, such an $X$ exists if and only if $p$ divides $10^{n}-1$. Thus $p$ divides $10^{200}-1$ but not $10^{n}-1,1 \leq n \leq 199$. Note that $10^{200}-1$ can be factored in this way:
+
+$$
+\begin{aligned}
+10^{200}-1 & =\left(10^{100}\right)^{2}-1 \\
+& =\left(10^{100}-1\right)\left(10^{100}+1\right) .
+\end{aligned}
+$$
+
+Since $p$ is prime and does not divide $10^{100}-1$, it must divide $10^{100}+1$, so that $10^{100}+1=k p$ for an integer $k$ and $X=\frac{\left(10^{100}-1\right)\left(10^{100}+1\right)}{p}=\left(10^{100}-1\right) k$.
+
+If $p=2,3,5$, or 7 , the fraction $\frac{1}{p}$ either terminates or repeats less than 200 digits. Therefore $p>10$ and $k<\frac{10^{100}}{p}<10^{99}$. Now let us calculate the 101st digit of $X=10^{100} k-k$, i.e. the digit representing multiples of $10^{99}$. Since $10^{100} k$ is divisible by $10^{100}$, its $10^{99} \mathrm{~s}$ digit and all later digits are 0 . Since $k<10^{99}$, $k$ does not contribute a digit to the $10^{99} \mathrm{~s}$ place, but it generates a borrow to this place, changing it into a 9. Thus the 101st digit of $X$ is a 9 .
+
+4. Let $A B C D$ be a trapezoid such that $A B \| C D$ and let $P$ be the point on the extension of the diagonal $A C$ such that $C$ is between $A$ and $P$. If $X$ and $Y$ are midpoints of the segments $A B$ and $C D$, and $M, N$ intersection points of the lines $P X, P Y$ with $B C, D A$ (respectively) prove that $M N$ is parallel to $A B$.
+
+Solution. Notice that $\frac{B M}{M C}=\frac{S_{\triangle P M B}}{S_{\triangle P M C}}$. Since $S_{\triangle P M B}+S_{\triangle X M B}=S_{\triangle P X B}=S_{\triangle P X A}=S_{\triangle P C M}+S_{\triangle A C M}+S_{\triangle A X M}$, we get that $S_{\triangle P M B}=S_{\triangle P M C}+S_{\triangle A C M}$. Hence,
+
+$$
+\frac{B M}{M C}=\frac{S_{\triangle P M C}+S_{\triangle A C M}}{S_{\triangle P M C}}=1+\frac{S_{\triangle A C M}}{S_{\triangle P M C}}=1+\frac{A C}{C P}
+$$
+
+Similarly as above we will use that $\frac{A N}{N D}=\frac{S_{\triangle P N A}}{S_{\triangle P D N}}$. Since $Y$ is the midpoint of $C D$, we have: $S_{\triangle P N C}=S_{\triangle P Y C}+S_{\triangle Y C N}=$ $S_{\triangle P D Y}+S_{\triangle D Y N}=S_{\triangle P D N}$. From $S_{\triangle P N A}=S_{\triangle P N C}+S_{\triangle C N A}$ we get
+
+$$
+\frac{A N}{N D}=\frac{S_{\triangle P N A}}{S_{\triangle P D N}}=\frac{S_{\Delta P N C}+S_{\triangle C N A}}{S_{\triangle P N C}}=1+\frac{S_{\Delta C N A}}{S_{\triangle P N C}}=1+\frac{A C}{C P} .
+$$
+
+From (1) and (2) it follows that $\frac{A N}{N D}=\frac{M B}{M C}$, which implies $N M\|A B\| C D$.
+
+5. Let $0<a_{0} \leq a_{1} \leq \cdots \leq a_{n}$. If $z$ is a complex number such that $a_{0} z^{n}+a_{1} z^{n-1}+\cdots+a_{n}=0$ prove that $|z| \geq 1$.
+
+Solution. Assume that $|z|<1$. If $a_{0} z^{n}+a_{1} z^{n-1}+\cdots+a_{n}=0$ then $a_{0} z^{n+1}+a_{1} z^{n}+\cdots+a_{n} z=0$ and subtracting these two equations leads to $a_{0} z^{n+1}+\left(a_{1}-a_{0}\right) z^{n}+\cdots+\left(a_{n}-a_{n-1}\right) z-a_{n}=0$, or equivalently $a_{n}=a_{0} z^{n+1}+\left(a_{1}-a_{0}\right) z^{n}+$ $\cdots+\left(a_{n}-a_{n-1}\right) z$ hence
+
+$$
+\begin{aligned}
+\left|a_{n}\right| & =\left|a_{0} z^{n+1}+\left(a_{1}-a_{0}\right) z^{n}+\cdots+\left(a_{n}-a_{n-1}\right) z\right| \\
+& \leq a_{0}|z|^{n+1}+\left(a_{1}-a_{0}\right)|z|^{n}+\cdots+\left(a_{n}-a_{n-1}\right)|z| \\
+& <a_{0}+\left(a_{1}-a_{0}\right)+\cdots+\left(a_{n}-a_{n-1}\right)=a_{n},
+\end{aligned}
+$$
+
+which is impossible. Thus $|z| \geq 1$.
+

BayArea/md/en-monthly/en-0708-comp4s.md

@@ -0,0 +1,36 @@
+# Berkeley Math Circle Monthly Contest 4 - Solutions 
+
+1. There are 10 bags full of coins. All coins look the same and all wight 10 grams, except the coins from one bag that are fake and all weight 9 grams. Given a scale, how could you tell which bag has the wrong coins in just one measurement? Explain your answer!
+
+Solution. Take one coin from the first bag, 2 from the second, 3 from the third, . . , 10 from the 10th and place them on the scale. The total mass shown by the scale should be $(1+2+\cdots+10) \cdot 10-x$ grams, where $x$ is the number of fake coins (each coin weighted 10 grams, except for the fake ones that are one gram lighter). Since the scale will show $550-x$ grams we can see how many coins are fake and that is enough to deduce from which bag those coines were taken.
+
+2. Find all prime numbers $p$ such that $p^{2}+2007 p-1$ is prime as well.
+
+Hint. Except for 3, prime numbers are not divisible by 3 . Hence if $p$ is not equal to 3 then either $p=3 k+1$ or $p=3 k-1$ for some integer $k$. If you wish you may use lists of prime numbers from the internet (e.g. www.imomath.com/primes
+
+Solution. If $p=3$, then $p^{2}+2007 p-1=6029$ which is a prime. For $p \neq 3$, we know that $p=3 k \pm 1$ hence $p^{2}+2007 p-1=$ $9 k^{2} \pm 6 k+1+2007 p-1=9 k^{2} \pm 6 k+2007 p$ which is divisible by 3 and can't be prime. Thus the only such prime number is $p=3$.
+
+3. The sequence of numbers $1,2,3, \ldots, 100$ is written on the blackboard. Between each two consecutive numbers a square box is drawn. Player $A$ starts the game and the players $A$ and $B$ alternate the moves. In each turn a player choses an empty box and places "+" or "." sign in it. After all the boxes are filled the expression on the blackboard is evaluated and if the result is an odd number the winner is $A$. Otherwise the winner is $B$. Determine which of the players has a winning strategy and what the strategy is.
+
+Solution. $A$ has the winning strategy. She should first place the sign "+" between the numbers 1 and 2 . After that she should group the square boxes into pairs: each pair consisting of two boxes adjacent to the same odd number. Then the player $A$ should make sure that there is at least one sign "." in each pair of boxes. Since $\cdot$ has the priority over + , after all products of numbers are calculated, 1 will be the only odd summand in the whole expression, hence the sum is odd. $A$ can achieve this goal by placing $\cdot$ sign in the box of the pair where $B$ has previously put his sign.
+
+4. The sum of the squares of five real numbers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ equals 1 . Prove that the least of the numbers $\left(a_{i}-a_{j}\right)^{2}$, where $i, j=1,2,3,4,5$ and $i \neq j$, does not exceed $1 / 10$.
+
+Solution. Assume w.l.o.g. that $a_{1} \leq a_{2} \leq a_{3} \leq a_{4} \leq a_{5}$. If $m$ is the least value of $\left|a_{i}-a_{j}\right|, i \neq j$, then $a_{i+1}-a_{i} \geq m$ for $i=1,2, \ldots, 5$, and consequently $a_{i}-a_{j} \geq(i-j) m$ for any $i, j \in\{1, \ldots, 5\}, i>j$. Then it follows that
+
+$$
+\sum_{i>j}\left(a_{i}-a_{j}\right)^{2} \geq m^{2} \sum_{i>j}(i-j)^{2}=50 m^{2}
+$$
+
+On the other hand, by the condition of the problem,
+
+$$
+\sum_{i>j}\left(a_{i}-a_{j}\right)^{2}=5 \sum_{i=1}^{5} a_{i}^{2}-\left(a_{1}+\cdots+a_{5}\right)^{2} \leq 5
+$$
+
+Therefore $50 m^{2} \leq 5$; i.e., $m^{2} \leq \frac{1}{10}$.
+
+5. Let $A B C D$ be a parallelogram. A variable line $l$ passing through the point $A$ intersects the rays $B C$ and $D C$ at points $X$ and $Y$, respectively. Let $K$ and $L$ be the centers of the excircles of triangles $A B X$ and $A D Y$, touching the sides $B X$ and $D Y$, respectively. Prove that the size of angle $K C L$ does not depend on the choice of the line $l$.
+
+Solution. Since $\angle A D L=\angle K B A=180^{\circ}-\frac{1}{2} \angle B C D$ and $\angle A L D=\frac{1}{2} \angle A Y D=\angle K A B$, triangles $A B K$ and $L D A$ are similar. Thus $\frac{B K}{B C}=\frac{B K}{A D}=\frac{A B}{D L}=\frac{D C}{D L}$, which together with $\angle L D C=\angle C B K$ gives us $\triangle L D C \sim \triangle C B K$. Therefore $\angle K C L=360^{\circ}-\angle B C D-(\angle L C D+\angle K C B)=360^{\circ}-\angle B C D-(\angle C K B+\angle K C B)=180^{\circ}-\angle C B K$, which is constant.
+

BayArea/md/en-monthly/en-0708-comp5s.md

@@ -0,0 +1,45 @@
+# Berkeley Math Circle <br> Monthly Contest 5 - Solutions 
+
+1. The rectangle $M N P Q$ is inside the rectangle $A B C D$. The portion of the rectangle $A B C D$ outside of $M N P Q$ is colored in green. Using just a straightedge construct a line that divides the green figure in two parts of equal areas.
+
+Solution. The line passing through the centers of $A B C D$ and $M N P Q$ divides each of the rectangles in two pieces of equal areas. Hence that line will divide the grean part into equal areas. The line can be easily constructed using only a straightedge because we can construct the centers of the given rectangles as intersections of their diagonals.
+
+2. Determine the positive real numbers $a$ and $b$ satisfying $9 a^{2}+16 b^{2}=25$ such that $a \cdot b$ is maximal. What is the maximum of $a \cdot b$ ? Explain your answer!
+
+Hint. If $x$ and $y$ are any two real numbers then $x^{2}+y^{2} \geq 2 x y$.
+
+Solution. Applying the inequality $x^{2}+y^{2} \geq 2 x y$ on $x=3 a$ and $y=4 b$ we get $25=(3 a)^{2}+(4 b)^{2} \geq 2 \cdot 3 a \cdot 4 b=24 a b$. Hence $a b \leq \frac{25}{24}$. The equality is attained for $x=y$ or equivalently for $3 a=4 b$. In that case $25=(3 a)^{2}+(4 b)^{2}=2 \cdot 9 a^{2}$ hence $a=\frac{5^{24}}{3 \sqrt{2}}$. Now we have $b=\frac{3}{4} a=\frac{5}{4 \sqrt{2}}$.
+
+3. Find all pairs of integers $(x, y)$ for which $x^{2}+x y=y^{2}$.
+
+Solution. The only such pair is $(0,0)$. If $x=0$, we easily get $y=0$ which satisfies the equation. Otherwise dividing through by $x^{2}$ we get
+
+$$
+1+\frac{y}{x}=\left(\frac{y}{x}\right)^{2} .
+$$
+
+This implies $\left(\frac{y}{x}\right)^{2}-\frac{y}{x}-1=0$ and the quadratic formula gives us $\frac{y}{x}=\frac{1 \pm \sqrt{5}}{2}$. Hence $\frac{2 y}{x}=1 \pm \sqrt{5}$ or equivalently $\frac{2 y-x}{x}= \pm \sqrt{5}$. Since $\sqrt{5}$ is irrational, this is a contradiction.
+
+4. Let $n$ be a positive integer. Prove that there exist distinct positive integers $x, y, z$ such that
+
+$$
+x^{n-1}+y^{n}=z^{n+1} \text {. }
+$$
+
+Solution. One solution is
+
+$$
+x=2^{n^{2}} 3^{n+1}, \quad y=2^{n^{2}-n} 3^{n}, \quad z=2^{n^{2}-2 n+2} 3^{n-1} .
+$$
+
+5. Let $A B C$ be a triangle such that $\angle A=90^{\circ}$ and $\angle B<\angle C$. The tangent at $A$ to its circumcircle $\omega$ meets the line $B C$ at $D$. Let $E$ be the reflection of $A$ across $B C, X$ the foot of the perpendicular from $A$ to $B E$, and $Y$ the midpoint of $A X$. Let the line $B Y$ meet $\omega$ again at $Z$. Prove that the line $B D$ is tangent to the circumcircle of triangle $A D Z$.
+
+Solution.Let $M$ be the point of intersection of $A E$ and $B C$, and let $N$ be the point on $\omega$ diametrically opposite $A$. Since $\angle B<\angle C$, points $N$ and $B$ are on the same side of $A E$.
+
+Furthermore, $\angle N A E=\angle B A X=90^{\circ}-\angle A B E$; hence the triangles $N A E$ and $B A X$ are similar. Consequently, $\triangle B A Y$ and $\triangle N A M$ are also similar, since $M$ is the midpoint of $A E$. Thus $\angle A N Z=\angle A B Z=\angle A B Y=\angle A N M$, implying that $N, M, Z$ are collinear. Now we have $\angle Z M D=$ $90^{\circ}-\angle Z M A=\angle E A Z=\angle Z E D$ (the last equality because $E D$ is tangent to $\omega$ ); hence $Z M E D$ is a cyclic quadri-
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_b17da2034ebacbc8658cg-1.jpg?height=280&width=552&top_left_y=1670&top_left_x=1226)
+lateral. It follows that $\angle Z D M=\angle Z E A=\angle Z A D$, which is enough to conclude that $M D$ is tangent to the circumcircle of $A Z D$.
+
+Remark. The statement remains valid if $\angle B \geq \angle C$.
+

BayArea/md/en-monthly/en-0708-comp6s.md

@@ -0,0 +1,85 @@
+# Berkeley Math Circle <br> Monthly Contest 6 - Solutions 
+
+1. Ten different points are marked on a circle. Two players $A$ and $B$ play the following game. $A$ moves first and the players alternate their moves. In each of the moves a player connects two of the points with a straight line segment. A player whose segment crosses a segment previously drawn will lose the game. Which player has a winning strategy and what is the strategy.
+
+Solution. Notice that a player can draw a line segment if and only if the 10-gon is not partitioned into triangles. Since there is a total of 17 segments for any partition of 10-gon into triangles, the first player will win the game no matter how he plays.
+
+2. Prove that no integer greater than 2008 can be equal to the sum of squares of its digits.
+
+Solution. Let $n=\overline{a_{k} a_{k-1} \ldots a_{0}}$ be an integer equal to the sum of squares of its own digits. Then $n=10^{k} a_{k}+10^{k-1} a_{k-1}+$ $\cdots+a_{0}$. On the other hand $a_{0}^{2}+a_{1}^{2}+\cdots+a_{k}^{2} \leq 9^{2} \cdot(k+1)=81 \cdots(k+1)<10^{k}$ for $k \geq 4$ (the last inequality is easy to prove by induction because it holds for $k=4$ and if it holds for some $k$ then $81 \cdot(k+2)=81 \cdot(k+1)+81 \leq 10^{k}+81<$ $10^{k}+10^{k}<10^{k+1}$ ).
+
+3. If $x \geq 4$ is a real number prove that
+
+$$
+\sqrt{x}-\sqrt{x-1} \geq \frac{1}{x}
+$$
+
+Solution. Notice that
+
+$$
+\sqrt{x}-\sqrt{x-1}=\frac{(\sqrt{x}-\sqrt{x-1}) \cdot(\sqrt{x}+\sqrt{x-1})}{\sqrt{x}+\sqrt{x-1}}=\frac{1}{\sqrt{x}+\sqrt{x-1}}
+$$
+
+Now the required inequality is equivalent to $x \geq \sqrt{x}+\sqrt{x-1}$ or after dividing both sides by $\sqrt{x}$ :
+
+$$
+\sqrt{x} \geq 1+\sqrt{1-\frac{1}{x}}
+$$
+
+The last inequality holds for $x \geq 4$ because $1+\sqrt{1-\frac{1}{x}} \leq 2 \leq \sqrt{x}$.
+
+4. Wally has a very unusual combination lock number. It has five digits, all different, and is divisible by 111 . If he removes the middle digit and replaces it at the end, the result is a larger number that is still divisible by 111. If he removes the digit that is now in the middle and replaces it at the end, the result is a still larger number that is still divisible by 111. What is Wally's combination lock number? Explain your answer!
+
+Solution. The solution is 74259 . The numbers 74259, 74592, and 74925 are all divisible by 111 . Denote the original number by $\overline{a b c d e}$ (the line prevents confusion with $a \cdot b \cdot c \cdot d \cdot e$ ). Then we have
+
+$$
+\begin{aligned}
+& 111 \mid \overline{a b c d e} \\
+& 111 \mid \overline{a b d e c}
+\end{aligned}
+$$
+
+Subtracting,
+
+$$
+\begin{gathered}
+111 \mid \overline{a b d e c}-\overline{a b c d e} \\
+111 \mid \overline{d e c}-\overline{c d e}
+\end{gathered}
+$$
+
+Since the number $\overline{d e c}-\overline{c d e}=90 d+9 e-99 c$ is divisible by 9 we get $33|\overline{d e c}-\overline{c d e} .333| \overline{d e c}-\overline{c d e}$. Since it is given that $\overline{a b d e c}>\overline{a b c d e}, \overline{c d e} \leq \overline{d e c}-333$.
+
+However, we could have done the above logic with $d, e, c$ instead of $c, d, e$ and gotten $\overline{d e c} \leq \overline{e c d}-333$. Consequently $\overline{c d e} \leq \overline{e c d}-666$. Since obviously $\overline{e c d} \leq 999, \overline{c d e}$ is one of the multiples of 37 up to 333:
+
+$$
+000037074111148185222259296333
+$$
+
+We can immediately eliminate $000,111,222$, and 333 , since we know that the digits are all different. We can also eliminate 074, 185, and 296, since $\overline{e c d}>\overline{d e c}$. The three remaining choices are all of the form $111 \cdot k+37$. So
+
+$$
+\begin{aligned}
+11 \mid \overline{a b c d e} & \\
+& =\overline{a b 000}+\overline{c d e} \\
+& =999 \cdot \overline{a b}+\overline{a b}+111 \cdot k+37 \\
+11 \mid \overline{a b}+37 &
+\end{aligned}
+$$
+
+whence $\overline{a b}=74$. This leaves only three possibilities for Wally's combination lock number: 74037, 74148, and 74259, of which only the last has all unlike digits.
+
+5. Let $A_{0}, A_{1}, \ldots, A_{n}$ be points in a plane such that
+
+(i) $A_{0} A_{1} \leq \frac{1}{2} A_{1} A_{2} \leq \cdots \leq \frac{1}{2^{n-1}} A_{n-1} A_{n}$ and
+
+(ii) $0<\measuredangle A_{0} A_{1} A_{2}<\measuredangle A_{1} A_{2} A_{3}<\cdots<\measuredangle A_{n-2} A_{n-1} A_{n}<180^{\circ}$,
+
+where all these angles have the same orientation. Prove that the segments $A_{k} A_{k+1}, A_{m} A_{m+1}$ do not intersect for each $k$ and $n$ such that $0 \leq k \leq m-2<n-2$.
+
+Solution. Suppose that $A_{k} A_{k+1} \cap A_{m} A_{m+1} \neq \emptyset$ for some $k, m>k+1$. Without loss of generality we may suppose that $k=0, m=n-1$ and that no two segments $A_{k} A_{k+1}$ and $A_{m} A_{m+1}$ intersect for $0 \leq k<m-1<n-1$ except for $k=0$, $m=n-1$. Also, shortening $A_{0} A_{1}$, we may suppose that $A_{0} \in A_{n-1} A_{n}$. Finally, we may reduce the problem to the case that $A_{0} \ldots A_{n-1}$ is convex: Otherwise, the segment $A_{n-1} A_{n}$ can be prolonged so that it intersects some $A_{k} A_{k+1}, 0<k<n-2$. If $n=3$, then $A_{1} A_{2} \geq 2 A_{0} A_{1}$ implies $A_{0} A_{2}>A_{0} A_{1}$, hence $\angle A_{0} A_{1} A_{2}>\angle A_{1} A_{2} A_{3}$, a contradiction.
+
+Let $n=4$. From $A_{3} A_{2}>A_{1} A_{2}$ we conclude that $\angle A_{3} A_{1} A_{2}>\angle A_{1} A_{3} A_{2}$. Using the inequality $\angle A_{0} A_{3} A_{2}>\angle A_{0} A_{1} A_{2}$ we obtain that $\angle A_{0} A_{3} A_{1}>\angle A_{0} A_{1} A_{3}$ implying $A_{0} A_{1}>A_{0} A_{3}$. Now we have $A_{2} A_{3}<A_{3} A_{0}+A_{0} A_{1}+A_{1} A_{2}<$ $2 A_{0} A_{1}+A_{1} A_{2} \leq 2 A_{1} A_{2} \leq A_{2} A_{3}$, which is not possible.
+
+Now suppose $n \geq 5$. If $\alpha_{i}$ is the exterior angle at $A_{i}$, then $\alpha_{1}>\cdots>\alpha_{n-1}$; hence $\alpha_{n-1}<\frac{360^{\circ}}{n-1} \leq 90^{\circ}$. Consequently $\angle A_{n-2} A_{n-1} A_{0} \geq 90^{\circ}$ and $A_{0} A_{n-2}>A_{n-1} A_{n-2}$. On the other hand, $A_{0} A_{n-2}<A_{0} A_{1}+A_{1} A_{2}+\cdots+A_{n-3} A_{n-2}<$ $\left(\frac{1}{2^{n-2}}+\frac{1}{2^{n-3}}+\cdots+\frac{1}{2}\right) A_{n-1} A_{n-2}<A_{n-1} A_{n-2}$, which contradicts the previous relation.
+

BayArea/md/en-monthly/en-0708-comp7s.md

@@ -0,0 +1,51 @@
+# Berkeley Math Circle Monthly Contest 7 - Solutions 
+
+1. Find all positive prime numbers $p$ such that $p+2$ and $p+4$ are prime as well.
+
+Hint. Show that for most prime numbers $p$, either $p+2$ or $p+4$ is divisible by 3 .
+
+Solution. For $p=3, p+2=5, p+4=7$ and these are obviously prime. For $p>3$, we know that $p$ is not divisible by 3 . The remainder of $p$ when divided by 3 can be either 1 or 2 . If it is one, then $p+2$ is divisible by 3 , if it is 2 , then $p+4$ is divisible by 3 . Hence $p=3$ is the only solution.
+
+2. Let $P$ be the point inside the square $A B C D$ such that $\triangle A B P$ is equilateral. Calculate the angle $\angle C P D$. Explain your answer!
+
+Solution. The triangle $D A P$ is isosceles because $A D=A P$ hence $\angle A D P=\angle A P D=\frac{180^{\circ}-\angle D A P}{2}=75^{\circ}$. Hence $\angle P D C=15^{\circ}$. Similarly $\angle D C P=15^{\circ}$ and hence $\angle C P D=180^{\circ}-2 \cdot 15^{\circ}=150^{\circ}$.
+
+3. Find at least one non-zero polynomial $P(x, y, z)$ such that $P(a, b, c)=0$ for every three real numbers that satisfy $\sqrt[3]{a}+\sqrt[3]{b}=$ $\sqrt[3]{c}$
+
+Remark. Polynomial in three variables refers to any expression built from $x, y, z$ and numerlas using only addition, subtraction, and multiplication. Parentheses or positive integer exponents, as in $x(y+z)^{2}$ are allowed since this can be expanded to $x y y+2 x y z+x z z$.
+
+Solution. Cube both sides of the condition $\sqrt[3]{x}+\sqrt[3]{y}=\sqrt[3]{z}$ :
+
+$$
+\begin{gathered}
+x+3 \sqrt[3]{x} \sqrt[3]{x} \sqrt[3]{y}+3 \sqrt[3]{x} \sqrt[3]{y} \sqrt[3]{y}+y=z \\
+3 \sqrt[3]{x} \sqrt[3]{x} \sqrt[3]{y}+\sqrt[3]{x} \sqrt[3]{y} \sqrt[3]{y}=z-x-y \\
+3 \sqrt[3]{x} \sqrt[3]{y}(\sqrt[3]{x}+\sqrt[3]{y})=z-x-y
+\end{gathered}
+$$
+
+But since $\sqrt[3]{x}+\sqrt[3]{y}=\sqrt[3]{z}$
+
+$$
+\begin{gathered}
+3 \sqrt[3]{x} \sqrt[3]{y} \sqrt[3]{z}=z-x-y \\
+27 x y z=(z-x-y)^{3}
+\end{gathered}
+$$
+
+Hence $P(x, y, z)=27 x y z-(z-x-y)^{3}$ is one such polynomial.
+
+4. If $f(1)=1$ and $f(1)+f(2)+\cdots+f(n)=n^{2} f(n)$ for every integer $n \geq 2$, evaluate $f(2008)$.
+
+Solution. $n^{2} f(n)-f(n)=f(1)+f(2)+\cdots+f(n-1)=(n-1)^{2} f(n-1)$ hence $f(n)=\frac{(n-1)^{2}}{n^{2}-1} f(n-1)=\frac{n-1}{n+1} f(n-1)$. Thus $f(2008)=\frac{2007}{2009} f(2007)=\frac{2007}{2009} \cdot \frac{2006}{2008} f(2006)=\frac{2007}{2009} \cdot \frac{2006}{2008} \cdot \frac{2005}{2007} f(2005)=\cdots=\frac{2007 !}{2009 \cdot 2008 \cdots 4 \cdot 3} f(1)=\frac{2}{2009 \cdot 2008}$.
+
+5. Given five vertices of a regular heptagon, construct the two remaining vertices using straightedge alone.
+
+Solution. Let $A, B, C, D$, and $E$ be the known vertices and $F$ and $G$ the unknown vertices. The arrangement of $A, B, C, D$, and $E$ depends on the relative positions of $F$ and $G$ as shown in the diagram. The following construction applies to all three cases.
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_a0d8313440f2d86f15f5g-1.jpg?height=171&width=625&top_left_y=1819&top_left_x=744)
+
+By connecting the intersections of $B C$ with $D E$ and $B D$ with $C E$ one finds $l$, the line of symmetry through $G$. Let $A B$ and $A C$ meet $l$ in $H$ and $I$ respectively. By symmetry about $l, H E$ and $I D$ both pass through $F$. Now perform the same construction on $B, C, D, E, F$ to get $G$.
+
+Remark. In general, given five consecutive vertices $A, B, C, D, E$ of a regular polygon or regular star polygon with at least seven sides, the next $F$ can be found by this construction, except that $l$ will not necessarily pass through $G$, although it is always the perpendicular bisector of $C D, B E$ and $A F$ and is a line of symmetry of the polygon.
+

BayArea/md/en-monthly/en-0809-comp1s.md

@@ -0,0 +1,83 @@
+# Berkeley Math Circle <br> Monthly Contest 1 - Solutions 
+
+1. If $x$ and $y$ are integers such that
+
+$$
+x^{2} y^{2}=x^{2}+y^{2} \text {, }
+$$
+
+prove that $x=y=0$.
+
+Solution. We move all the terms to the left side of the equation and add 1.
+
+$$
+\begin{array}{r}
+x^{2} y^{2}-x^{2}-y^{2}+1=1 \\
+\left(x^{2}-1\right)\left(y^{2}-1\right)=1
+\end{array}
+$$
+
+Since the factors are integers, they must be both 1 or both -1 . If both are 1 , we get $x^{2}=2$ which is impossible for integer $x$. If both are -1 , we get $x^{2}=0$ and $y^{2}=0$, so both $x$ and $y$ are zero.
+
+2. The country of Squareland is shaped like a square and is divided into 64 congruent square cities. We want to divide Squareland into states and assign to each state a capital city so that the following rules are satisfied:
+
+(a) Every city lies entirely within one state.
+
+(b) Given any two states, the numbers of cities in them differ by at most 1 .
+
+(c) Any city in a state shares at least one corner with the state's capital.
+
+What is the smallest possible number of states?
+
+Solution. In the diagram below, no city shares a corner with any two of the cities marked X. Therefore the nine X's are
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_921110140cad74839444g-1.jpg?height=615&width=618&top_left_y=847&top_left_x=1187)
+in nine different states. The diagram at right shows that nine states are also sufficient ( $*$ denotes capital).
+![](https://cdn.mathpix.com/cropped/2024_04_17_921110140cad74839444g-1.jpg?height=614&width=1398&top_left_y=1576&top_left_x=300)
+
+3. Eight unit squares are glued together to make two groups of four, and the groups are pressed together so as to meet in three points $A, B, C$ as shown in the diagram. Find the distance $A B$.
+
+Solution. Since $A D=1=B E, \angle A D B=90=\angle B E C$, and $\angle A B D=90-\angle E B C=\angle B C E$, triangles $A D B$ and $B E C$ are congruent. Let $A B=B C=x$. Notice that $D B=2-B C=2-x$. We apply the Pythagorean theorem to triangle $A B D$ :
+
+$$
+\begin{aligned}
+1+(2-x)^{2} & =x^{2} \\
+1+4-4 x+x^{2} & =x^{2} \\
+5 & =4 x \\
+\frac{5}{4} & =x
+\end{aligned}
+$$
+
+Thus $A B=\frac{5}{4}=1 \frac{1}{4}=1.25$ (all three answers are correct).
+
+4. Let $a, b, c$ be positive real numbers satisfying $a b c=1$. Prove that
+
+$$
+a(a-1)+b(b-1)+c(c-1) \geq 0 .
+$$
+
+Solution. At least two of $a, b, c$ are either not less than 1 or not greater than 1. Assume that $a$ and $b$ are on the same side of 1 . Next, transform the inequality as follows:
+
+$$
+\begin{aligned}
+a(a-1)+b(b-1)+c(c-1) & \stackrel{?}{\geq} 0 \\
+a(a-1)+b(b-1)+c^{2}\left(1-\frac{1}{c}\right) & \stackrel{?}{\geq} 0 \\
+a(a-1)+b(b-1)+c^{2}(1-a b) & \stackrel{?}{\geq} 0 \\
+a(a-1)+b(b-1)-c^{2}(a-1)-c^{2}(a b-a) & \stackrel{?}{\geq} 0 \\
+\left(a-c^{2}\right)(a-1)+\left(b-c^{2} a\right)(b-1) & \stackrel{?}{\geq} 0 \\
+\left(a-\frac{1}{a^{2} b^{2}}\right)(a-1)+\left(b-\frac{1}{a^{2} b}\right)(b-1) & \stackrel{?}{\geq} 0
+\end{aligned}
+$$
+
+Using the hypotheses concerning $a$ and $b$, it is not hard to see that the four factors in parentheses are all nonnegative or all nonpositive, and therefore the left side is nonnegative.
+
+5. The positive integers from 1 to 100 are written, in some order, in a $10 \times 10$ square. In each row, the five smallest numbers are crossed out. In each column, the five largest numbers (including those that have already been crossed out) are circled. Prove that at least 25 numbers will be circled but not crossed out.
+
+Solution. We generalize to the following:
+
+Let $m \geq p, n \geq q$ be integers. Each square of an $m$-row, $n$-column grid is filled with a different positive integer.
+
+Then at least $p q$ numbers are among the $p$ largest numbers in their columns and the $q$ largest numbers in their rows.
+
+We prove this latter statement by induction on $m+n$. The base case, when $m=1$ or $n=1$, is trivial. Given an $m \times n$ grid with $m \geq 2$ and $n \geq 2$, let $L$ be the largest number that is one of the largest in its row or its column, but not both. Assume without loss of generality that $L$ is among the largest in its column and not among the largest in its row, so the $q$ largest numbers in $L$ 's row are all larger than $L$ and all among the largest in their columns. Temporarily remove this row from the matrix. By the induction hypothesis, at least $(p-1) q$ numbers are now among the $q$ largest numbers in their rows and the $p-1$ largest numbers in their columns. These are also among the $p$ largest numbers in their columns in the original matrix. Adding the $q$ largest numbers in the deleted row, we find $(p-1) q+q=p q$ numbers satisfying the original condition. The problem follows by taking $m=n=10$ and $p=q=5$.
+

BayArea/md/en-monthly/en-0809-comp2s.md

@@ -0,0 +1,85 @@
+# Berkeley Math Circle Monthly Contest 2 - Solutions 
+
+1. A building has seven rooms numbered 1 through 7, all on one floor, and any number of doors connecting these rooms. These doors may be one-way, admitting motion in only one of the two directions, or two-way. In addition, there is a two-way door between room 1 and the outside, and a treasure in room 7. Your object is to choose the arrangement of the rooms and the locations of the doors in such a way that
+
+(a) it is possible to enter room 1 , reach the treasure, and make it back outside,
+
+(b) the minimum number of steps required to to this (each step consisting of walking through a door) is as large as possible.
+
+Solution. On the way to the treasure, no room need be entered twice; otherwise the path could be shortened by skipping the loop. Thus, the minimal path to the treasure, if it exists, is at most 7 steps long. Similarly, the minimal path from the treasure to the outside is at most 7 steps long, so the total number of steps cannot exceed 14. The arrangement of the rooms in a line, with 1 and 7 at opposite ends and two-way doors between all adjacent rooms, shows that 14 is attainable.
+
+2. Prove that there is exactly one way to place circles in four of the blank squares of the cross-equation puzzle at right such that, no matter what natural numbers are placed in the circled squares, the five uncircled squares can be filled with natural numbers that make the three horizontal and three vertical equations true.
+
+| $a$ | + | $b$ | $=$ | $c$ |
+| :---: | :---: | :---: | :---: | :---: |
+| + |  | $=$ |  | + |
+| $d$ | $=$ | $e$ | + | $f$ |
+| $=$ |  | + |  | $=$ |
+| $g$ | + | $h$ | $=$ | $k$ |
+
+Solution. If $a, e, f$, and $h$ are circled, the equations
+
+$$
+b=e+h, d=e+f, g=d+a, c=b+a
+$$
+
+define $b, d, g$, and $c$ in such a way that the first two horizontal and first two vertical equations are satisfied. Then since
+
+$$
+g+h=d+a+h=e+f+a+h=b+f+a=c+f
+$$
+
+a value of $k$ can be found which completes the puzzle. Note that $b, c, d, g$, and $k$ are the sum of two other numbers in the puzzle. If any of them is circled and then filled with a 1 , the puzzle will be unsolvable since no two natural numbers add to 1 . Thus $a, e, f$, and $h$ is the only permissible circled quadruplet.
+
+3. A number is called a $j$-half if it leaves a remainder of $j$ when divided by $2 j+1$.
+
+(a) Prove that for any $k$, there is a number which is simultaneously a $j$-half for $j=1,2, \ldots, k$.
+
+(b) Prove that there is no number which is a $j$-half for all positive integers $j$.
+
+## Solution.
+
+(a) Note that a number $n$ is a $j$-half if and only if $2 n$ has a remainder of $2 j$ when divided by $2 j+1$, which happens exactly when $2 n+1$ is divisible by $2 j+1$. Thus
+
+$$
+n=\frac{3 \cdot 5 \cdot 7 \cdots(2 k+1)-1}{2}
+$$
+
+which is clearly an integer, is a $j$-half for $j=1,2,2, \ldots, k$.
+
+(b) If $n$ is any positive integer, take $j>n$. Then $n$ obviously leaves a remainder of $n$ when divided by $2 n+1$, not $j$ as was desired.
+
+4. Let $A O B$ be a 60 -degree angle. For any point $P$ in the interior of $\angle A O B$, let $A^{\prime}$ and $B^{\prime}$ be the feet of the perpendiculars from $P$ to $A O$ and $B O$ respectively. Denote by $r$ and $s$ the distances $O P$ and $A^{\prime} B^{\prime}$. Find all possible pairs of real numbers $(r, s)$.
+
+Solution. Extend $A^{\prime} P$ to meet $O B$ at $Z$. Notice that, because $\angle O A^{\prime} P$ and $\angle O B^{\prime} P$ are both right, the circle with diameter $O P$ passes through $O, P, A^{\prime}$, and $B^{\prime}$. Thus $\angle B^{\prime} O P=$ $\angle B^{\prime} A^{\prime} P$ since both intercept the same arc on this circle, and $\triangle Z O P \sim \triangle Z A^{\prime} B^{\prime}$ by AA. We get
+
+$$
+\frac{s}{r}=\frac{B^{\prime} A^{\prime}}{O P}=\frac{Z A^{\prime}}{Z O}=\frac{\sqrt{3}}{2}
+$$
+
+because $Z O A^{\prime}$ is a $30-60-90$ triangle. Since $r$ can obviously take on any value, the possibilities for $(r, s)$ are $\left(r, \frac{r \sqrt{3}}{2}\right)$ for
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_2fae5e31928f586282e5g-2.jpg?height=423&width=287&top_left_y=274&top_left_x=1060)
+every positive real $r$.
+
+5. Prove that for every positive integer $n$, there is an integer $x$ such that $x^{2}-17$ is divisible by $2^{n}$.
+
+Solution. We prove this by induction on $n$. If $n=1,2$, or 3 , then $x=1$ works. Suppose that $x^{2}-17$ is divisible by $2^{n}$ and $n \geq 3$. We seek to find $y$ such that $y^{2}-17$ is divisible by $2^{n+1}$. Let
+
+$$
+x^{2}-17=k \cdot 2^{n} \text {. }
+$$
+
+If $k$ is even, we are done since $x^{2}-17$ is divisible by $2^{n+1}$. If $k$ is odd, $k=2 m+1$, we have
+
+$$
+\begin{aligned}
+\left(x+2^{n-1}\right)^{2}-17 & =x^{2}+2 \cdot x \cdot 2^{n-1}+2^{2(n-1)}-17 \\
+& =x^{2}-17+x \cdot 2^{n}+2^{2 n-2} \\
+& =(2 m+1) \cdot 2^{n}+x \cdot 2^{n}+2^{2 n-2} \\
+& =2^{n}(x+1)+2^{n+1}\left(m+2^{n-3}\right)
+\end{aligned}
+$$
+
+Since $x$ is obviously odd and $n \geq 3$, this is a multiple of $2^{n+1}$, and $y=x+2^{n-1}$ works.
+

BayArea/md/en-monthly/en-0809-comp3s.md

@@ -0,0 +1,66 @@
+# Berkeley Math Circle <br> Monthly Contest 2 - Solutions <br> Due November 4, 2008 
+
+1. Find all positive integers $p$ such that $p, p+4$, and $p+8$ are all prime.
+
+Solution. If $p=3$, then $p+4=7$ and $p+8=11$, both prime. If $p \neq 3$, then $p$ is not a multiple of 3 and is therefore of one of the forms $3 k+1,3 k+2(k \geq 0)$ If $p=3 k+1$, then $p+8=3 k+9=3(k+3)$, which is not prime since $k+3>1$. If $p=3 k+2$, then $p+4=3 k+6=3(k+2)$, which is not prime since $k+2>1$. Thus $p=3$ is the only solution with all three numbers prime.
+
+2. Each vertex of a regular heptagon is colored either red or blue. Prove that there is an isosceles triangle with all its vertices the same color.
+
+Solution. Denote the vertices of the heptagon by $A B C D E F G$. Since an alternating arrangement cannot be continued all the way around the heptagon, two adjacent vertices must be the same color, say $A$ and $B$. If any of $C, E, G$ shares this color, we are done since triangles $A B C, A B E$, and $A B G$ are all isosceles. On the other hand, if $C, E$, and $G$ are all of the opposite color, we are also done because triangle $C E G$ is isosceles. Thus in all cases we can find an isosceles triangle.
+
+3. Let $a, b$, and $c$ be positive real numbers satisfying $a^{b}>b^{a}$ and $b^{c}>c^{b}$. Does it follow that $a^{c}>c^{a}$ ?
+
+Solution. Yes. We have
+
+$$
+\left(a^{c}\right)^{b}=\left(a^{b}\right)^{c}>\left(b^{a}\right)^{c}=\left(b^{c}\right)^{a}>\left(c^{b}\right)^{a}=\left(c^{a}\right)^{b} ;
+$$
+
+the desired inequality follows by taking the $b$ th root.
+
+4. Let $n$ be a positive integer and let $S$ be the set $\{1,2, \ldots, n\}$. Define a function $f: S \rightarrow S$ by
+
+$$
+f(x)= \begin{cases}2 x & \text { if } 2 x \leq n, \\ 2 n-2 x+1 & \text { otherwise. }\end{cases}
+$$
+
+Define $f^{2}(x)=f(f(x)), f^{3}(x)=f(f(f(x)))$, and so on. If $m$ is a positive integer satisfying $f^{m}(1)=1$, prove that $f^{m}(k)=k$ for all $k \in S$.
+
+Solution. First note that
+
+$$
+f(x) \equiv \pm 2 x \quad \bmod 2 n+1
+$$
+
+It follows that
+
+$$
+f^{p}(x) \equiv \pm 2^{p} x \quad \bmod 2 n+1
+$$
+
+Thus if $f^{m}(1)=1,2^{m} \equiv \pm 1$ and so, for any $k \in S$,
+
+$$
+f^{m}(k) \equiv \pm 2^{m} k \equiv \pm k \quad \bmod 2 n+1
+$$
+
+that is, $f^{m}(k) \pm k=j(2 n+1)$ for some integer $j$ and some choice of the sign. Since
+
+$$
+0<1+1 \leq f^{m}(k)+k \leq n+n<2 n+1,
+$$
+
+the plus sign is invalid. Thus the minus sign holds, and since
+
+$$
+-(2 n+1)<1-n \leq f^{m}(k)-k \leq n-1<2 n+1,
+$$
+
+we get $j=0$, i.e. $f^{m}(k)=k$.
+
+5. This problem was invalid on the contest. Correct formulation as of December 9. Let $\omega_{1}, \omega_{2}$, and $\omega_{3}$ be three circles passing through the origin $O$ of the coordinate plane but not tangent to each other or to either axis. Denote by $\left(x_{i}, 0\right)$ and $\left(0, y_{i}\right)$, $1 \leq i \leq 3$, the respective intersections (besides $O$ ) of circle $\omega_{i}$ with the $x$ and $y$ axes. Prove that $\omega_{1}, \omega_{2}$, and $\omega_{3}$ have a common point $P \neq O$ if and only if the points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$, and $\left(x_{3}, y_{3}\right)$ are collinear.
+
+Solution. First of all, note that if a circle passes through $(0,0),\left(x_{i}, 0\right)$, and $\left(0, y_{i}\right)$, its center must be $\left(\frac{x_{i}}{2}, \frac{y_{i}}{2}\right)$, the midpoint of the side opposite the right angle at $O$. Also note that the three points $\left(\frac{x_{i}}{2}, \frac{y_{i}}{2}\right)$ are related to $\left(x_{i}, y_{i}\right)$ by a dilation about $O$; thus the former three will be collinear if and only if the latter three are. It suffices to prove that the circles have a second common point if and only if their centers are collinear.
+
+If the centers are collinear, all three circles are symmetric about the line of centers. Thus the reflection of $O$ about this line is a second common point of the three circles. Conversely, assume that the circles have two common points, $O$ and $P$. Then all three centers lie on the perpendicular bisector of $O P$.
+

BayArea/md/en-monthly/en-0809-comp4s.md

@@ -0,0 +1,173 @@
+# Berkeley Math Circle <br> Monthly Contest 2 - Solutions <br> Due November 4, 2008 
+
+1. Each square of a $100 \times 100$ grid is colored black or white so that there is at least one square of each color. Prove that there is a point with is a vertex of exactly one black square.
+
+Solution. Locate the uppermost row that has at least one black square. Then, within that row, find the leftmost black square. By construction, all squares above and/or to the left of that square are white. Therefore, the upper left corner of that square will solve the problem.
+
+2. Let $a$ and $b$ be nonzero real numbers. Prove that at least one of the following inequalities is true:
+
+$$
+\begin{aligned}
+& \left|\frac{a+\sqrt{a^{2}+2 b^{2}}}{2 b}\right|<1 \\
+& \left|\frac{a-\sqrt{a^{2}+2 b^{2}}}{2 b}\right|<1
+\end{aligned}
+$$
+
+Solution. Assume for the sake of contradiction that both inequalities are false, that is,
+
+$$
+1 \leq\left|\frac{a+\sqrt{a^{2}+2 b^{2}}}{2 b}\right| \text { and } 1 \leq\left|\frac{a-\sqrt{a^{2}+2 b^{2}}}{2 b}\right| .
+$$
+
+Multiplying these inequalities together yields
+
+$$
+1 \leq\left|\frac{a+\sqrt{a^{2}+2 b^{2}}}{2 b} \cdot \frac{a-\sqrt{a^{2}+2 b^{2}}}{2 b}\right|=\left|\frac{a^{2}-\left(a^{2}+2 b^{2}\right)}{4 b^{2}}\right|=\left|\frac{-2 b^{2}}{4 b^{2}}\right|=\frac{1}{2},
+$$
+
+a contradiction.
+
+3. In acute triangle $A B C$, the three altitudes meet at $H$. Given that $A H=B C$, calculate at least one of the angles of $\triangle A B C$.
+
+Answer: $\angle A=45^{\circ}$ (angles $B$ and $C$ cannot be determined).
+
+Solution: In the diagram, $\angle A F H=\angle C F B$ (both are right angles) and $\angle F A H=\angle F C B$ (both are complementary to $\angle A B C$ ). We get $\triangle A F H \cong \triangle C F B$ by AAS. It follows that $A F=F C$, so $\triangle A F C$ is right isosceles. We find that $\angle B A C=45^{\circ}$.
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_1c1fa8f951f9f7756383g-1.jpg?height=455&width=377&top_left_y=1667&top_left_x=1037)
+
+4. Let $x$ be an integer greater than 2 . Prove that the binary representation of $x^{2}-1$ has at least three consecutive identical digits (000 or 111).
+
+Solution. We consider three cases:
+
+- $x$ is odd. Then $x+1$ and $x-1$ are consecutive even integers, and thus one of them must be divisible by 4 , the other only by 2 . So $x^{2}-1=(x+1)(x-1)$ is divisible by 8 ; since $x \geq 3, x^{2}-1$ has four or more digits, the last three of which are 0 's.
+- $x$ is even but $\frac{x}{2}$ is odd. By the same reasoning, $\left(\frac{x}{2}\right)^{2}-1=8 k$ for some integer $k$, so
+
+$$
+x^{2}-1=4\left(\frac{x}{2}\right)^{2}-1=4(8 k+1)-1=32 k+3
+$$
+
+Since $x \geq 6$, this number has six or more digits, the last five of which are 00011 . Thus we also get three consecutive zeros in this case.
+
+- $x$ and $\frac{x}{2}$ are both even, that is, $x$ is divisible by 4 . Then $x^{2}$ is divisible by 16 . Since $x \geq 4, x^{2}-1$ has at least four digits, the last four of which are 1's.
+
+5. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying
+
+$$
+f(x(1+y))=f(x)(1+f(y))
+$$
+
+for all $x, y \in \mathbb{R}$.
+
+Solution. The answers are $f(x)=0$ and $f(x)=x$. It is easily checked that both of these satisfy the equation.
+
+If $f(x)=c$, a constant, for all $x$, we must have
+
+$$
+\begin{aligned}
+& c=c(1+c) \\
+& c=c+c^{2} \\
+& 0=c^{2} \\
+& 0=c,
+\end{aligned}
+$$
+
+and the solution $f(x)=0$ has already been noted. Thus we may assume that $f$ is not a constant.
+
+Plugging $y=-1$ into (3) gives
+
+$$
+\begin{aligned}
+f(x(1-1)) & =f(x)(1+f(-1)) \\
+f(0) & =f(x)(1+f(-1))
+\end{aligned}
+$$
+
+If $f(-1) \neq-1$, we have $f(x)=\frac{f(0)}{1+f(-1)}$, a constant. Thus we may assume that $f(-1)=-1$. It follows that $f(0)=0$. Next we try to find $u:=f(1)$. Plugging $x=1$ and $y=-2$ into (3) gives
+
+$$
+\begin{aligned}
+f(1(1-2)) & =f(1)(1+f(-2)) \\
+f(-1) & =f(1)(1+f(-2)) \\
+-1 & =u(1+f(-2)) .
+\end{aligned}
+$$
+
+On the other hand, with $x=-1$ and $y=1$ in (3), we get
+
+$$
+\begin{aligned}
+f(-(1+1)) & =f(-1)(1+f(1)) \\
+f(-2) & =-(1+u)
+\end{aligned}
+$$
+
+and plugging this into (4),
+
+$$
+\begin{aligned}
+-1 & =u(1-(1-u)) \\
+& =u(-u) \\
+1 & =u^{2}
+\end{aligned}
+$$
+
+so $u=1$ or $u=-1$. If $u=-1$, using $y=1$ in (3) yields
+
+$$
+\begin{aligned}
+f(x(1+1)) & =f(x)(1+f(1)) \\
+f(2 x) & =f(x)(1-1)=0
+\end{aligned}
+$$
+
+and $f$ becomes a constant. Thus we must have $f(1)=1$. Using $x=1$ in (3),
+
+$$
+f(1+y)=f(1)(1+f(y))=1+f(y) \text {. }
+$$
+
+We can use this identity to transform (3):
+
+$$
+\begin{aligned}
+f(x(1+y)) & =f(x)(1+f(y)) \\
+& =f(x) f(1+y)
+\end{aligned}
+$$
+
+and letting $z=1+y$, we get
+
+$$
+f(x z)=f(x) f(z)
+$$
+
+for all $x$ and $z$. Now we transform (3) in yet another way:
+
+$$
+\begin{aligned}
+f(x(1+y)) & =f(x)(1+f(y)) \\
+f(x+x y) & =f(x)+f(x) f(y) \\
+& =f(x)+f(x y)
+\end{aligned}
+$$
+
+letting $z=x y$,
+
+$$
+f(x+z)=f(x)+f(z)
+$$
+
+whenever $x \neq 0$. Since this equation is trivially true when $x=0$, we can use it for all $x$ and $z$. Repeatedly applying it to the known values $f(1)=1, f(-1)=-1$, we find $f(x)=x$ whenever $x$ is an integer. Then, if $\frac{m}{n}$ is any rational number, plugging $x=\frac{m}{n}, z=n$ into (5) shows that $f(x)=x$ for rational $x$ as well. Setting $x=y$ in (5) gives
+
+$$
+f\left(x^{2}\right)=(f(x))^{2},
+$$
+
+so if $x$ is nonnegative, so is $f(x)$. By (6), if $x \geq y$, then $f(x) \geq f(y)$. Suppose that $f(a)>a$ for some irrational $a$. Then there is a rational $p$ with $f(a)>p>a$, and
+
+$$
+f(a) \leq f(p)=p<f(a),
+$$
+
+a contradiction. Similarly if $f(a)<a$, we get a contradiction. Thus $f(x)=x$ for all $x$, the second of the two solutions to the functional equation.
+

BayArea/md/en-monthly/en-0809-comp5s.md

@@ -0,0 +1,99 @@
+# Berkeley Math Circle <br> Monthly Contest 5 - Solutions 
+
+1. In how many ways can each square of a $2 \times 9$ board be colored red, blue, or green so that no two squares that share an edge are the same color? a $100 \times 100$ grid is colored black or white so that there is at least one square of each color. Prove that there is a point with is a vertex of exactly one black square.
+
+Solution. Orient the board so that it consists of 9 rows of 2 squares each. The upper left square can be colored in 3 ways; the square to its right, in 2 ways. We continue from top to bottom, coloring one row at a time. If $\mathrm{X}$ and $\mathrm{Y}$ denote the (obviously different) colors of one row and $\mathrm{Z}$ the color not appearing in that row, the possibilities of the row below are limited to YX, YZ, and ZX. Thus there are 3 ways to color each of the eight rows below the first. The total number of colorings of the board is
+
+$$
+(3 \cdot 2) \cdot 3^{8}=39366
+$$
+
+2. Prove that the sum of the 2009th powers of the first 2009 positive integers is divisible by 2009 .
+
+Solution. Using the factoring formula
+
+$$
+x^{2009}+y^{2009}=(x+y)\left(x^{2008}-x^{2007} y+x^{2006} y^{2}-\cdots-x y^{2007}+y^{2008}\right. \text {, }
+$$
+
+we find that each of the numbers
+
+$$
+0^{2009}+2009^{2009}, 1^{2009}+2008^{2009}, 2^{2009}+2007^{2009}, \ldots, 1004^{2009}+1005^{2009}
+$$
+
+is divisible by 2009. Consequently so is their sum.
+
+3. If real numbers $a, b, c, d$ satisfy
+
+$$
+\frac{a+b}{c+d}=\frac{b+c}{a+d} \neq-1,
+$$
+
+prove that $a=c$.
+
+Solution. Assume for the sake of contradiction that $a \neq c$. Cross-multiplying,
+
+$$
+\begin{aligned}
+(a+b)(a+d) & =(c+b)(c+d) \\
+a^{2}+a b+a d+b d & =c^{2}+b c+c d+b d \\
+a^{2}-c^{2}+a b-b c+a d-c d & =0 \\
+(a-c)(a+c)+(a-c) b+(a-c) d & =0 \\
+a+c+b+d & =0 \\
+b+c & =-a-d .
+\end{aligned}
+$$
+
+Because we are given that $a+d \neq 0$, we obtain
+
+$$
+\frac{b+c}{a+d}=-1
+$$
+
+a contradiction.
+
+4. In triangle $A B C$, the bisector of $\angle B$ meets the circumcircle of $\triangle A B C$ at $D$. Prove that
+
+$$
+B D^{2}>B A \cdot B C
+$$
+
+Solution: In the diagram, $\angle A B D=\angle D B C$ (angle bisector) and $\angle B A C=\angle B D C$ (both intercept arc $B C$ ), so $\triangle B A E \sim \triangle B D C$. We get
+
+$$
+\begin{aligned}
+\frac{B A}{B E} & =\frac{B D}{B C} \\
+B D \cdot B E & =B A \cdot B C,
+\end{aligned}
+$$
+
+from which the desired inequality follows since $B E<B D$.
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_63121b3605f2196726f5g-2.jpg?height=556&width=558&top_left_y=112&top_left_x=1030)
+
+5. A calculator has a switch and four buttons with the following functions:
+
+- Flipping the switch from the down to the up position adds 1 to the number in the display.
+- Flipping the switch from the up to the down position subtracts 1 from the number in the display.
+- Pressing the red button multiplies the number in the display by 3 .
+- If the number in the display is divisible by 3 , pressing the yellow button divides it by 3 ; otherwise the yellow button is nonfunctional.
+- Pressing the green button multiplies the number in the display by 5 .
+- If the number in the display is divisible by 5 , pressing the blue button divides it by 5 ; otherwise the blue button is nonfunctional.
+
+Solution. Assume that not every positive integer can appear in the display and let $N$ be the smallest number that cannot. We first note that since the buttons do not change the parity of the display, it must be even when the switch is down and odd when the switch is up. Consequently $N$ is even, for if it were odd we could obtain $N$ from the displayable number $N-1$ by flipping the switch. $N$ cannot be divisible by 3 , for otherwise we could get it from the displayable number $\frac{N}{3}$. If $N$ were $2 \bmod 3$, then multiplying by 3 and flipping the switch would suffice to create $N$ from the smaller integer $\frac{N+1}{3}$. We conclude that $N \equiv 1$ $\bmod 3$, so $N=6 n+4$ for some integer $n$. Because
+
+$$
+6 n+4=\frac{3(10 n+7)-1}{5},
+$$
+
+neither $10 n+7$ nor $10 n+6$, which differs from it only in the position of the switch, is obtainable. Similarly, since
+
+$$
+6 n+4=\frac{3(10 n+8)+1}{5}-1,
+$$
+
+$10 n+8$ is not obtainable. Of the three consecutive integers $10 n+6,10 n+7$, and $10 n+8$, at least one is a multiple of 3 . Since one-third of that number is necessarily less than $6 n+4$, it can be displayed, and we have a contradiction.
+
+Remark. In this solution it is noticeable that the green button is never used.
+

BayArea/md/en-monthly/en-0809-comp6s.md

@@ -0,0 +1,110 @@
+# Berkeley Math Circle <br> Monthly Contest 6 - Solutions 
+
+1. Let $p, q$, and $r$ be distinct primes. Prove that $p+q+r+p q r$ is composite.
+
+Solution. Note that at most one of $p, q$, and $r$ can equal 2. If $p=2$, then $q$ and $r$ are both odd, but $p q r$ is even. Therefore their sum is even. Similarly, if $q$ or $r$ is 2, then $p+q+r+p q r$ is even. Finally, if $p, q$, and $r$ are all odd, then so is $p q r$, and $p+q+r+p q r$ is even. Thus $p+q+r+p q r$ is always divisible by 2 , and since it is obviously greater than 2 , it must be composite.
+
+2. The sequence
+
+$$
+5,9,49,2209, \ldots
+$$
+
+is defined by $a_{1}=5$ and $a_{n}=a_{1} a_{2} \cdots a_{n-1}+4$ for $n>1$. Prove that $a_{n}$ is a perfect square for $n \geq 2$.
+
+Solution. This is clear for $n=2$. We use the relation
+
+$$
+\begin{gathered}
+a_{n-1}=a_{1} a_{2} \cdots a_{n-2}+4 \\
+a_{1} a_{2} \cdots a_{n-2}=a_{n-1}-4
+\end{gathered}
+$$
+
+for $n \geq 3$ to transform $a_{n}$ :
+
+$$
+\begin{aligned}
+a_{n} & =a_{1} a_{2} \cdots a_{n-2} a_{n-1}+4 \\
+& =\left(a_{n-1}-4\right) a_{n-1}+4 \\
+& =a_{n-1}^{2}-4 a_{n-1}+4 \\
+& =\left(a_{n-1}-2\right)^{2} .
+\end{aligned}
+$$
+
+This is clearly the square of an integer.
+
+3. The integers from 1 to 13 are arranged around several rings such that every number appears once and every ring contains at least one two-digit number. Prove that there exist three one-digit numbers adjacent to one another on one ring.
+
+Solution. For any $n \in S=1,2, \ldots, 13$, define $f(n)$ to be the number immediately clockwise of $n$ on the same ring, where $f(n)=n$ if $n$ lies on a one-element ring. Notice that $f$ is a bijective function, since every number is immediately clockwise of exactly one number. Notice that there are four numbers $n \in S$ such that $n$ has two digits, four numbers $n$ such that $f(n)$ has two digits, and four numbers $n$ such that $f(f(n))$ has two digits. This leaves at least $13-4-4-4=1$ number $n$ such that $n, f(n)$, and $f(f(n))$ are all one-digit numbers. No two of them can be equal, or else the ring containing them would have no two-digit numbers, so $n, f(n)$, and $f(f(n))$ are the desired adjacent one-digit numbers.
+
+4. Let $A B C$ be a triangle with $\angle A B C=90^{\circ}$. Points $D$ and $E$ on $A C$ and $B C$ respectively satisfy $B D \perp A C$ and $D E \perp B C$. The circumcircle of $\triangle C D E$ intersects $A E$ at two points, $E$ and $F$. Prove that $B F \perp A E$.
+
+Solution: By Power of a Point,
+
+$$
+A F \cdot A E=A D \cdot A C
+$$
+
+because $\triangle A B C$ is right,
+
+$$
+A D \cdot A C=A B^{2}
+$$
+
+Combining,
+
+$$
+A F \cdot A E=A B^{2} .
+$$
+
+On the other hand, if $F^{\prime}$ is the foot of the altitude from $B$ to $A E$, then
+
+$$
+A F^{\prime} \cdot A E=A B^{2}
+$$
+
+Consequently $A F^{\prime}=A F$ and $F^{\prime}=F$.
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_9791c04439a140e31becg-1.jpg?height=580&width=331&top_left_y=1914&top_left_x=1081)
+
+Remark. The right angle at $D$ is a red herring. The proof is valid for any segment $D E$ parallel to leg $A B$.
+
+5. Let $a_{1}, a_{2}, \ldots, a_{n}$ be distinct integers. Prove that there do not exist two nonconstant integer-coefficient polynomials $p$ and $q$ such that
+
+$$
+\left(x-a_{1}\right)\left(x-a_{2}\right) \cdots\left(x-a_{n}\right)-1=p(x) q(x)
+$$
+
+for all $x$.
+
+Solution. Assume for the sake of contradiction that $p$ and $q$ exist. If we substitute $x=a_{i}$ for $i=1, \ldots, n$, the left side of (1) becomes -1 . Since $p\left(a_{i}\right)$ and $q\left(a_{i}\right)$ are both integers, we either have
+
+$$
+p\left(a_{i}\right)=1, \quad q\left(a_{i}\right)=-1
+$$
+
+or
+
+$$
+p\left(a_{i}\right)=-1, \quad q\left(a_{i}\right)=1
+$$
+
+In either case,
+
+$$
+(p+q)\left(a_{i}\right)=0
+$$
+
+Thus $p+q$ is a polynomial with $n$ distinct roots $a_{1}, a_{2}, \ldots, a_{n}$. Such a polynomial must have degree at least $n$-unless it is the zero polynomial.
+
+Case 1. $p+q$ has degree $\geq n$. Then one of $p$ and $q$ has degree $\geq n$, and the other, of course, has degree $\geq 1$. It follows that $p q$ has degree $\geq n+1$, a contradiction since the left side of (1) has degree $n$.
+
+Case 2. $p(x)+q(x)=0$ for all $x$. Substituting $q=-p$ into (1),
+
+$$
+\left(x-a_{1}\right)\left(x-a_{2}\right) \cdots\left(x-a_{n}\right)-1=-p(x)^{2} \text {. }
+$$
+
+Now substitute an integer for $x$ that is so huge that each of the factors $\left(x-a_{i}\right)$ is greater than 1 . This makes the left side strictly positive. Since the right side is negative or zero, we have a contradiction.
+

BayArea/md/en-monthly/en-0809-comp7s.md

@@ -0,0 +1,174 @@
+# Berkeley Math Circle <br> Monthly Contest 7 - Solutions 
+
+1. In the sequence
+
+$$
+77492836181624186886128 \ldots,
+$$
+
+all of the digits except the first two are obtained by writing down the products of pairs of consecutive digits. Prove that infinitely many 6 s appear in the sequence.
+
+Solution. Since 868 appears in the sequence and $8 \cdot 6=6 \cdot 8=48,4848$ appears later in the sequence. Then $4 \cdot 8=32$, so 3232 appears somewhere later. And so on:
+
+$$
+868 \rightarrow 4848 \rightarrow 3232 \rightarrow 666 \rightarrow 3636 \rightarrow 1818 \rightarrow 888 \rightarrow 6464 \rightarrow 2424 \rightarrow 888 \ldots
+$$
+
+It is clear that the three last sequences repeat indefinitely. Therefore 6464 appears infinitely many times, and in particular 6 appears infinitely many times.
+
+2. Let $k$ be a rational number greater than 1 (correction by Fengning Ding). Prove that there exist positive integers $a, b, c$ satisfying the equations
+
+$$
+\begin{aligned}
+a^{2}+b^{2} & =c^{2} \\
+\frac{a+c}{b} & =k .
+\end{aligned}
+$$
+
+Solution. Let $k=x / y$, where $x$ and $y$ are positive integers. We find that $x>y$. It suffices to note that
+
+$$
+a=x^{2}-y^{2}, \quad b=2 x y, \quad c=x^{2}+y^{2}
+$$
+
+are positive integers satisfying both of the given equations.
+
+3. Four congruent circles are tangent to each other and to the sides of a triangle as shown.
+
+(a) Prove that $\angle A B C=90^{\circ}$.
+
+(b) If $A B=3$ and $B C=4$, find the radius of the circles.
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_4b41c25c94bf9322a4e3g-1.jpg?height=415&width=528&top_left_y=1191&top_left_x=1037)
+
+Solution. Denote the centers of the four circles by $D, E, F, G$ and their tangency points with the sides of $\triangle A B C$ by $T, U, V, W, X, Y, Z$ as shown in the diagram. Segments $T D$ and $U E$ are congruent (because they are radii of congruent circles) and parallel (because they are perpendicular to $A B$ ). Therefore $D T U E$ is a parallelogram, in fact a rectangle since $\angle D T U=90^{\circ}$. Similarly, quadrilaterals $E V W F, F X Y G$, and $G Y Z D$ are rectangles. We find that $D, G$, and $F$ are collinear; we also have $\triangle D E F \sim \triangle A B C$, since corresponding sides are parallel.
+
+For part (a), we note that $D G=E G=F G$, so $\angle D E F$ is inscribed in a semicircle centered at $G$. If $\angle D E F$ is right, so is $\angle A B C$.
+
+For part (b), let $r$ be the desired radius. Noting $A C=5$ by Pythagoras, we have
+
+$$
+D F=4 r, \quad D E=\frac{3}{5} \cdot 4 r=\frac{12 r}{5}, \quad E F=\frac{4}{5} \cdot 4 r=\frac{16 r}{5}
+$$
+
+We now compute $A B+B C-A C=3+4-5=2$ in another way:
+
+$$
+\begin{aligned}
+A B+B C-A C & =(A T+D E+U B)+(B V+E F+W C)-(C X+F D+Z A) \\
+& =(A T-A Z)+(W C-X C)+D E+U B+B V+E F-F D \\
+& =0+0+\frac{12 r}{5}+r+r+\frac{16 r}{5}-4 r \\
+& =\frac{(12+5+5+16-20) r}{5}=\frac{18 r}{5} .
+\end{aligned}
+$$
+
+Thus $r=2 \cdot 5 / 18=5 / 9$.
+
+4. Find all pairs $(a, b)$ of positive integers such that
+
+$$
+1+5^{a}=6^{b} .
+$$
+
+Solution (Based on work by Fengning Ding) The only solution is $(1,1)$.
+
+It is clear that if $b=1$ then $a=1$, and that $(1,1)$ is a solution. Consequently assume $b>1$. Then $6^{b}$ is divisible by 4 . On the other hand, since $5^{a} \equiv 1^{a}=1 \bmod 4$ for all $a$, the left side is $2 \bmod 4$. Thus there are no solutions for $b>1$.
+
+5. The tower function of twos, $T(n)$, is defined by $T(1)=2$ and $T(n+1)=2^{T(n)}$ for $n \geq 1$. Prove that $T(n)-T(n-1)$ is divisible by $n$ ! for $n \geq 2$.
+
+Solution. Let $U(n)=T(n)-T(n-1)$. Note that
+
+$$
+U(n)=T(n)-T(n-1)=2^{T(n-1)}-2^{T(n-2)}=2^{T(n-2)}\left(2^{T(n-1)-T(n-2)}-1\right)=T(n-1)\left(2^{U(n-1)}-1\right)
+$$
+
+We first prove two lemmas, the second a refinement of the first.
+
+Lemma 1. $U(n-1) \mid U(n)$.
+
+Proof. By induction on $n$. It is clear when $n=2$ and $n=3$. Suppose that $U(n-2) \mid U(n-1), n \geq 3$. Then
+
+$$
+\frac{U(n)}{U(n-1)}=\frac{T(n-1)}{T(n-2)} \cdot \frac{2^{U(n-1)}-1}{2^{U(n-2)}-1} .
+$$
+
+The first fraction is clearly an integer. Since $a \mid b$ implies $2^{a}-1 \mid 2^{b}-1$, the second fraction is an integer as well.
+
+Lemma 2. If $p$ is a prime and $U(k)$ is divisible by $p^{r}, r \geq 1$, then $U(k+1)$ is divisible by $p^{r+1}$.
+
+Proof. If $p=2$, the $r$ factors of 2 in $U(k)$ must be contained in $T(k-1)$. Since $T(k)>T(k-1), T(k)$ has at least $r+1$ factors of 2 , and so does $U(k+1)$.
+
+If $p>2$, we use induction on $n$. Let $n_{0}$ be the smallest value such that $p \mid U\left(n_{0}\right)$. Let $r_{0}$ be the number of factors of $p$ in $U\left(n_{0}\right)$. Since $T\left(n_{0}-1\right)$ is a power of 2 , the $r_{0}$ factors are all contained in $2^{U\left(n_{0}-1\right)}-1$, i.e.
+
+$$
+2^{U\left(n_{0}-1\right)} \equiv 1 \bmod p^{r_{0}} .
+$$
+
+By the preceding lemma, $U\left(n_{0}\right)$ is divisible by $U\left(n_{0}-1\right)$; however, by definition of $n_{0}, U\left(n_{0}\right)$ is divisible by $p$ while $U\left(n_{0}-1\right)$ is not. Consequently
+
+$$
+U\left(n_{0}\right)=U\left(n_{0}-1\right) \cdot p \cdot J
+$$
+
+where $J$ is an integer. We find that
+
+$$
+\begin{aligned}
+2^{U\left(n_{0}-1\right)} & \equiv 1 \bmod p^{r_{0}} \\
+2^{U\left(n_{0}-1\right) \cdot J} & \equiv 1 \bmod p^{r_{0}} \\
+2^{U\left(n_{0}\right)}=2^{U\left(n_{0}-1\right) \cdot J \cdot p} & \equiv 1 \bmod p^{r_{0}+1}
+\end{aligned}
+$$
+
+where we have used the fact that $a \equiv b \bmod p^{r}$ implies $a^{p} \equiv b^{p} \bmod p^{r+1}$. We conclude that $p^{r_{0}+1}\left|2^{U\left(n_{0}\right)}-1\right| U\left(n_{0}+1\right)$. This completes the base case. The induction step is similar, except that $n_{0}$ and $r_{0}$ are replaced by $n$ and $r$, and equation (3) follows from the induction hypothesis rather than the definition of $n_{0}$.
+
+We prove the following statement by strong induction for $n \geq 2$ :
+
+Lemma 3. $U(n)$ is divisible by all prime powers less than $3 \cdot 2^{n-2}-1$.
+
+Proof. We can verify the statement by hand for $n=2$ and $n=3$. Thus we suppose that $U(k)=T(k)-T(k-1)$ is divisible by $k$ ! for $k<n$, where $n \geq 4$. We prove the second assertion first. Let $p^{r}$ be a prime power less than $3 \cdot 2^{n-2}-1$. If $r>1$, we note that
+
+$$
+p^{r-1} \leq \frac{p^{r}}{2} \leq \frac{3 \cdot 2^{n-2}-2}{2}=3 \cdot 2^{n-3}-1
+$$
+
+furthermore, at least one of these inequalities is strict, since the first becomes equality only if $p=2$, and $2^{n}-1<3 \cdot 2^{n-2}<2^{n}$ when $n \geq 4$. Thus by the induction hypothesis, $p^{r-1}$ divides $U_{n-1}$, and we are done by Lemma 2 . Now we assume that $r=1$. We may assume that $p \geq 5$ since $U_{3}$ is already divisible by 2 and 3 . Then $p-1$ is composite, and we can resolve it into prime powers
+
+$$
+p-1=q_{1}^{a_{1}} q_{2}^{a_{2}} \cdots q_{s}^{a_{s}}
+$$
+
+where each factor $q_{i}^{a_{i}}$ is at most
+
+$$
+\frac{p-1}{2}<\frac{3 \cdot 2^{n-2}-2}{2}=3 \cdot 2^{n-3}-1 .
+$$
+
+We infer that each $q_{i}^{a_{i}}$ divides $U_{n-1}$, and therefore $p-1$ divides $U_{n-1}$. We now have
+
+$$
+p\left|2^{p-1}-1\right| 2^{U_{n-1}}-1 \mid U_{n}
+$$
+
+as desired.
+
+We are now ready to solve the problem. Assume that $n \geq 4$; the cases $n=2$ and $n=3$ are easily checked by hand. Let $p$ be a prime dividing $n !$, that is, $p \leq n$. The number of factors of $p$ in $n$ ! is
+
+$$
+R=\sum_{i=1}^{\infty}\left\lfloor\frac{n}{p^{i}}\right\rfloor<\sum_{i=1}^{\infty} \frac{n}{p^{i}}=n \sum_{i=1}^{\infty} \frac{1}{p^{i}}=n \cdot \frac{1 / p}{1-1 / p}=\frac{n}{p-1}
+$$
+
+SO
+
+$$
+R \leq \frac{n-1}{p-1}
+$$
+
+We observe that $p \mid U(p)$ : by hand calculation if $p=2$, and because $p<3 \cdot 2^{p-2}-1$ if $p>3$. By Lemma 2, applied $n-p$ times,
+
+$$
+p^{n-p+1} \mid U(n) .
+$$
+
+Since $n-p+1 \geq \frac{n-1}{p-1}$ if $n \geq p$, we have $p^{R} \mid U(n)$, as desired.
+

BayArea/md/en-monthly/en-0809-comp8s.md

@@ -0,0 +1,168 @@
+# Berkeley Math Circle <br> Monthly Contest 8 - Solutions 
+
+1. Determine whether there exists a natural number having exactly 10 divisors (including itself and 1 ), each ending in a different digit.
+
+Solution. The answer is no.
+
+Suppose that such a number $n$ exists. Since $n$ has a divisor ending in $5, n$ is divisible by 5 . Then for each divisor $d$ of $n$ not divisible by $5,5 d$ also divides $n$. This contradicts the assumption that $n$ has eight divisors ending in $1,2,3,4,6,7,8$, or 9 and only two ending in 5 or 0 .
+
+2. Two bikers, Bill and Sal, simultaneously set off from one end of a straight road. Neither biker moves at a constant rate, but each continues biking until he reaches one end of the road, at which he instantaneously turns around. When they meet at the opposite end from where they started, Bill has traveled the length of the road eleven times and Sal seven times. Find the number of times the bikers passed each other moving in opposite directions.
+
+Solution. Define a pass $(\mathrm{P})$ to be an time when Bill and Sal pass one another moving in opposite directions and a turn (T) to be a time when one of the bikers turns around. If both bikers turn around simultaneously, we may alter their speeds slightly, causing one turn to happen before the other, without affecting the number of passes. We distinguish three states:
+
+A. The bikers are moving in the same direction.
+
+B. The bikers are moving toward each other.
+
+C. The bikers are moving away from one another.
+
+We notice that the bikers can only change state by turning or passing. Moreover, there are only three possible state changes:
+
+- From state A, one biker reaches the end and turns around, creating state B.
+- From state B, the bikers pass one another, creating state C.
+- From state $\mathrm{C}$, one biker reaches the end and turns around, creating state A.
+
+Since there are no other possibilities and we begin at state A, the states must follow the sequence ABCABC $\cdots \mathrm{ABCA}$ and the turns and passes TPTTPT ... TPT. We are given that Bill makes ten turns and Sal six, or 16 in all. Consequently there are eight passes.
+
+3. A position of the hands of a (12-hour, analog) clock is called valid if it occurs in the course of a day. For example, the position with both hands on the 12 is valid; the position with both hands on the 6 is not. A position of the hands is called bivalid if it is valid and, in addition, the position formed by interchanging the hour and minute hands is valid. Find the number of bivalid positions.
+
+Solution. Let $h$ and $m$ denote the respective angles of the hour and minute hands, measured clockwise in degrees from 12 o'clock $(0 \leq h, m<360)$. Since the minute hand moves twelve times as fast as the hour hand, we have
+
+$$
+m=12 h-360 a
+$$
+
+with $a$ an integer, for any valid time. Conversely, it is clear that any ordered pair $(h, m)$ satisfying (1) represents the valid time of $h / 30$ hours after 12 o'clock.
+
+The condition that a time be valid after switching the hour and minute hands is, of course,
+
+$$
+h=12 m-360 b \text {, }
+$$
+
+with $b$ an integer. Thus the problem of finding bivalid times is reduced to finding pairs $(h, m)$ satisfying (1) and (2). Note that if $h$ is known, $m$ is uniquely determined from (1) and the condition $0 \leq m<360$. Note also that the truth of (2) is not
+affected by increasing or decreasing $m$ by 360 , as this simply increases or decreases $b$ by 12 , giving a new integer value $b^{\prime}$. Consequently we substitute $12 h$ for $m$ in (2), getting
+
+$$
+\begin{gathered}
+h=144 h-360 b^{\prime} \\
+143 h=360 b^{\prime} \\
+h=\frac{360 b^{\prime}}{143} .
+\end{gathered}
+$$
+
+This equation has 143 solutions in the range $0 \leq h<360$, each of which gives rise to one solution of (1) and (2). We conclude that there are 143 bivalid positions.
+
+4. Let $A B C$ be a triangle, and let $M$ and $N$ be the respective midpoints of $A B$ and $A C$. Suppose that
+
+$$
+\frac{C M}{A C}=\frac{\sqrt{3}}{2}
+$$
+
+Prove that
+
+$$
+\frac{B N}{A B}=\frac{\sqrt{3}}{2}
+$$
+
+Solution. Let $L$ be the midpoint of $B C$. Let $B C=2 a, A C=2 b$, and $A B=2 c$. Applying the parallelogram law to $C L M N$ gives
+
+$$
+C M^{2}+c^{2}=a^{2}+b^{2}+a^{2}+b^{2}
+$$
+
+or
+
+$$
+C M^{2}=2 a^{2}+2 b^{2}-c^{2} .
+$$
+
+Squaring both sides of the given equation and substituting yields
+
+$$
+\frac{2 a^{2}+2 b^{2}-c^{2}}{4 b^{2}}=\frac{3}{4}
+$$
+
+which simplifies to
+
+$$
+b^{2}+c^{2}=2 a^{2}
+$$
+
+This equation is equivalent to $C M / A C=\sqrt{3} / 2$. Because it is symmetric in $b$ and $c$, we conclude that it is also equivalent to $B N / A B=\sqrt{3} / 2$.
+
+5. Let $M_{n}$ be the number of integers $N$ such that
+
+(a) $0 \leq N<10^{n}$;
+
+(b) $N$ is divisible by 4 ;
+
+(c) The sum of the digits of $N$ is also divisible by 4 .
+
+Prove that $M_{n} \neq 10^{n} / 16$ for all positive integers $n$.
+
+Solution: $\quad$ Since $10^{n} / 16$ is not an integer for $n=1,2,3$, we may assume that $n \geq 4$. Let
+
+$$
+\tilde{M}_{n}=M_{n}-\frac{10^{n}}{16}
+$$
+
+We note that numbers whose hundreds digit is between 2 and 9 inclusive make no total contribution to $\tilde{M}_{n}$ for $n \geq 3$. This is because, given the remaining digits, divisibility by 4 (which depends only on the tens and units digits) occurs one-fourth of the time, and exactly two of the eight choices for the hundreds digit brings the total of the digits to a multiple of 4. Similarly, numbers with hundreds digit 0 or 1 but thousands digit at least 2 make no contribution to $\tilde{M}_{n}$. (We require the hundreds digit to be less than 2 to avoid double counting.) Continuing in this way, we find that only the $25 \cdot 2^{n}$ numbers with all but the last two digits equal to 0 or 1 can make $\tilde{M}_{n}$ nonzero. Let $L_{n}$ be the number of such numbers satisfying conditions (a), (b), and (c). We thus have
+
+$$
+\tilde{M}_{n}=L_{n}-\frac{25 \cdot 2^{n}}{16}
+$$
+
+Now let $f(i, k)$ be the number of binary strings of length $k$ whose sum is congruent to $i \bmod k$. Since there are respectively $9,4,6,6$ multiples of 4 less than 100 with digit sums $\equiv 0,1,2,3 \bmod 4$, we have
+
+$$
+L_{n}=9 f(0, n-2)+4 f(3, n-2)+6 f(2, n-2)+6 f(1, n-2) \text {. }
+$$
+
+If we let $\tilde{f}(i, k)=f(i, k)-2^{n} / 4$, we can subtract $25 \cdot 2^{n} / 16$ from both sides of the above equation to get
+
+$$
+\tilde{M}_{n}=9 \tilde{f}(0, n-2)+4 \tilde{f}(3, n-2)+6 \tilde{f}(2, n-2)+6 \tilde{f}(1, n-2)
+$$
+
+On the other hand, dividing a binary $k$-string with sum $i$ into a $k-1$-string and a single bit and noting that the sum of the $k-1$-string is either $i$ or $i-1$, we get the recurrence
+
+$$
+f(i, k)=f(i, k-1)+f(i-1, k-1)
+$$
+
+which is equivalent to
+
+$$
+\tilde{f}(i, k)=\tilde{f}(i, k-1)+\tilde{f}(i-1, k-1) .
+$$
+
+Together with the boundary conditions
+
+$$
+\begin{array}{ll}
+f(0,0)=1, & f(1,0)=f(2,0)=f(3,0)=0, \\
+\tilde{f}(0,0)=\frac{3}{4}, & \tilde{f}(1,0)=\tilde{f}(2,0)=\tilde{f}(3,0)=-\frac{1}{4},
+\end{array}
+$$
+
+this enables us to calculate values of $\tilde{f}(i, k)$ and therefore of $\tilde{M}_{n}$. It is trivial to prove by induction that for $k \geq 1$,
+
+$$
+\tilde{f}(i, k)=-\tilde{f}(i+2, k),
+$$
+
+from which it follows after a short computation using (4) that
+
+$$
+\tilde{f}(i, k+4)=-4 \tilde{f}(i, k), \quad k \geq 1
+$$
+
+Now we substitute this equation into (3), yielding
+
+$$
+\tilde{M}_{n+4}=-4 \tilde{M}_{n}, \quad n \geq 3
+$$
+
+This implies that $\tilde{M}_{n+4}$ is nonzero if $\tilde{M}_{n}$ is. We have already noted that $\tilde{M}_{3}$ is nonzero because it is not an integer. Computing $\tilde{M}_{4}=2, \tilde{M}_{5}=-1$, and $\tilde{M}_{6}=-6$ by hand using (5), (4), and (3) completes the proof that $\tilde{M}_{n}$ is nonzero for all $n \geq 1$.
+

BayArea/md/en-monthly/en-1011-comp1s.md

@@ -0,0 +1,125 @@
+# Berkeley Math Circle <br> Monthly Contest 1 - Solutions 
+
+1. Are there natural numbers $a$ and $b$ that make the equation
+
+$$
+2 a^{2}+1=4 b^{2}
+$$
+
+true? (The natural numbers are the counting numbers, $1,2,3, \ldots$ )
+
+Solution. The answer is no. Suppose that $a$ and $b$ are any natural numbers. Then $2 a^{2}$ is even, and so is $4 b^{2}$. But an even number plus 1 will be odd, not even, so the equation can never be made to work, i.e. there are no natural numbers that satisfy the equation.
+
+2. Given three points $A, B, C$ known to lie on a circle, prove that one can reconstruct the original circle with a straightedge and compass.
+
+Solution. Here is a suitable procedure: First draw circles centered at $A$ and $B$ with radius $A B$ and join their two points of intersection to create the perpendicular bisector of $A B$. This line contains all points that have the same distance from $A$ and $B$, so the center of a circle through $A$ and $B$ lies on it. Then, similarly construct the perpendicular bisector of $A C$. Because $A, B$, and $C$ are known to lie on a circle, the two perpendicular bisectors are not parallel (or else no point could have the same distance from $A, B$, and $C$ ), so they must meet at a point $O$. Draw a circle centered at $O$ with radius $O A$.
+
+Since $O$ lies on both perpendicular bisectors, this circle passes through $B$ and $C$. Finally, any other circle through the same three points would have to have its center on both perpendicular bisectors, and hence at $O$. Its radius must equal $O A$, implying that it coincides with the constructed circle.
+
+3. (a) You are given the expression
+
+$$
+1 \diamond 2 \diamond 3 \diamond 4 \diamond 5 \diamond 6 \diamond 7 \text {. }
+$$
+
+Determine whether it is possible to replace one of the symbols $\diamond$ with $=$ and the other symbols $\diamond$ with + or - so as to end up with a correct equation.
+
+Solution. It is possible; one of many solutions is
+
+$$
+1=2+3+4+5-6-7
+$$
+
+(b) The same question for the expression
+
+$$
+1 \diamond 2 \diamond 3 \diamond 4 \diamond 5 \diamond 6
+$$
+
+Solution. It is impossible. Suppose for the sake of contradiction that we have a solution. Change the $=$ sign to $\mathrm{a}+$. We are now adding together two equal integers, so the sum must be even. Now change each - sign, one by one, to $a+$. At each such step, we increase the value of the expression by twice the value of the number after the changed sign. This keeps the total even. But at the end of all these operations, we have changed all the signs to + , so the result is
+
+$$
+1+2+3+4+5+6
+$$
+
+which is 21 , an odd number. Therefore, our assumption is false; i.e. there is no way to place the signs as required.
+
+4. Two villages, $A$ and $B$, lie on opposite sides of a straight river in the positions shown. There will be a market in village $B$, and residents of village $A$ wish to attend. The people of village $A$ would like to build a bridge across and perpendicular to the river so that the total route walked by residents of $A$ to the bridge, across the bridge, and onward to $B$ is as short as possible. How can they find the exact location of the bridge, and how long will the route be?
+
+Solution. For convenience, we label the point $X$, the foot of the perpendicular from $A$ to the river. We also label $Y$, the point where the perpendicular $B Y$ to the river through $B$ meets the parallel $A Y$ to the river through $A$. The key construction is to translate $B$ southward $1 \mathrm{~km}$ to point $E$. Let $A C D B$ be any route. Then segments $D C$ and $B E$ are congruent and parallel,
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_54f31c14ab607b7a972cg-1.jpg?height=404&width=396&top_left_y=1928&top_left_x=1553)
+making a parallelogram $D C E B$. The total route is $A C+C D+D B=A C+C E+E B$. But by the triangle inequality, $A C+C E \geq A E$ so the total route cannot be shorter than $A E+E B=5+1=6 \mathrm{~km}$. (continued)
+
+The residents of $A$ can ensure a 6 -kilometer route by putting the start $C$ of the bridge at the point where $A E$ crosses the lower bank of the river. In this situation, since $A, C$, and $E$ lie on a line, the triangle inequality is an equality, so the total route is $A C+C D+D B=A C+C E+E B=A E+E B=5+1=6$.
+
+Remark. Using the fact that triangles $A C X$ and $E A Y$ are similar if $A, C$, and $E$ are collinear, it is possible to calculate that the optimal location of point $C$ is $0.75 \mathrm{~km}$ east of $X$.
+
+5. We have four bowls labeled $A, B, C, D$ in a row, and we have four indistinguishable marbles which we can initially distribute among the bowls any way we like. A move consists of transferring one marble from a bowl to one of the adjacent bowls in the row. Is it possible to perform a succession of moves in which every distribution of the marbles appears exactly once?
+
+Solution. The answer is no. Call a position of the marbles "even" if bowls $A$ and $C$ have an even total number of marbles and "odd" otherwise. It is easy to see that any move changes an even position to an odd position and vice versa. However, note that, for any nonnegative integer $k$, the number of ways to distribute $k$ marbles among two bowls is $k+1$ (as the first bowl can hold any number of marbles from 0 to $k$ ). Therefore it is easy to count the number of even and odd positions:
+
+| Marbles in $A$ and $C$ | Marbles in $B$ and $D$ | Results |
+| :--- | :--- | :--- |
+| 4, in 5 ways | 0, in 1 way | $5 \cdot 1=5$ even positions |
+| 3, in 4 ways | 1, in 2 ways | $4 \cdot 2=8$ odd positions |
+| 2, in 3 ways | 2, in 3 ways | $3 \cdot 3=9$ even positions |
+| 1, in 2 ways | 3, in 4 ways | $2 \cdot 4=8$ odd positions |
+| 0, in 1 way | 4, in 5 ways | $1 \cdot 5=5$ even positions |
+
+The totals are 19 even positions and 16 odd. It is not possible to visit all 19 even positions without the 18 odd positions separating them containing some duplicates.
+
+6. In the interior of a triangle $A B C$ with area 1 , points $D, E$, and $F$ are chosen such that $D$ is the midpoint of $A E, E$ is the midpoint of $B F$, and $F$ is the midpoint of $C D$. Find the area of triangle $D E F$.
+
+Solution. Let $x$ be the area of $\triangle D E F$. Comparing triangles $A B E$ and $D E F$, we find that base $A E$ is twice base $D E$ but, since $E$ bisects $B F$, the heights to these bases are equal. Thus $\triangle A B E$ has area $2 x$. Symmetrically, triangles $B C F$ and $C A D$ have area $2 x$. Since these four triangles fill up $\triangle A B C$, we have $1=x+2 x+2 x+2 x=7 x$, so $x=1 / 7$.
+
+7. Let $a, b$, and $c$ be positive real numbers such that
+
+$$
+a b c=1
+$$
+
+and
+
+Prove that for every positive integer $n$,
+
+$$
+a+b+c>\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \text {. }
+$$
+
+Using $a b c=1$, we transform the given condition to
+
+$$
+a^{n}+b^{n}+c^{n}>\frac{1}{a^{n}}+\frac{1}{b^{n}}+\frac{1}{c^{n}} .
+$$
+
+$$
+a+b+c>b c+c a+a b
+$$
+
+or
+
+$$
+-b c-c a-a b+a+b+c>0 .
+$$
+
+We then add $a b c-1(=0)$ to the left side, getting
+
+$$
+a b c-b c-c a-a b+a+b+c-1>0
+$$
+
+which factors as
+
+$$
+(a-1)(b-1)(c-1)>0 .
+$$
+
+In exactly the same way we transform the condition to be proved to
+
+$$
+\left(a^{n}-1\right)\left(b^{n}-1\right)\left(c^{n}-1\right)>0 .
+$$
+
+However, for any positive real $x$, the numbers $x-1$ and $x^{n}-1$ are both positive, both negative, or both zero. Consequently (1) and (2) are equivalent.
+

BayArea/md/en-monthly/en-1011-comp2s.md

@@ -0,0 +1,149 @@
+# Berkeley Math Circle Monthly Contest 2 - Solutions 
+
+1. Arrange the following number from smallest to largest: $2^{1000}, 3^{750}, 5^{500}$.
+
+Solution. Since all the numbers are positive, taking the 250th root of each number will not change their ordering. The resulting numbers are $2^{4}=16,3^{3}=27$, and $5^{2}=25$. These have the ordering $2^{4}<5^{2}<3^{3}$, so the ordering of the original numbers is $2^{1000}<5^{500}<3^{750}$.
+
+2. In the diagram at right, $A B C D$ is a parallelogram. Given that the distances from $A, B$, and $C$ to line $\ell$ are, respectively, 3,1 , and 5 , find the distance from $D$ to $\ell$ and prove that your answer is correct.
+
+Remark. In the contest, the labels on the points $A$ and $C$ were switched. This makes the figure grossly out of scale but does not change the solution.
+
+Solution. As shown, label the feet of the perpendiculars from $A, B, C, D$ to $\ell$ by $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$. Also, drop perpendiculars $B X$ and $C Y$ to $A A^{\prime}$ and $D D^{\prime}$ respectively. We find that triangles $A B X$ and $D C Y$ are similar since their corresponding sides are parallel, and in fact congruent since $A B$ and $C D$ are opposite sides of the parallelogram. Thus $A X=D Y$. But $A X=A A^{\prime}-$ $X A^{\prime}=A A^{\prime}-B B^{\prime}=3-1=2$, and hence $D D^{\prime}=D Y+Y D^{\prime}=A B+C C^{\prime}=$
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_89d84300f5630edac230g-1.jpg?height=472&width=572&top_left_y=534&top_left_x=1335)
+$2+5=7$.
+
+3. Ten cups lie upside down in a line. It is known that pennies lie under two of the cups which are consecutive in the line. Choosing several of the cups, you may ask for the total number of coins under them. Is it possible to determine the positions of the pennies by asking two such questions, without knowing the answer to the first question before making the second?
+
+Solution. The answer is yes; here is one strategy that works. On the first turn choose the cups 1, 4, 5, 6, 7, and on the second turn choose $1,6,7,8,9$. Depending on the locations of the coins, the answers will be:
+
+| Coins | $1,4,5,6,7$ | $1,6,7,8,9$ |
+| :---: | :---: | :---: |
+| 1,2 | 1 | 1 |
+| 2,3 | 0 | 0 |
+| 3,4 | 1 | 0 |
+| 4,5 | 2 | 0 |
+| 5,6 | 2 | 1 |
+| 6,7 | 2 | 2 |
+| 7,8 | 1 | 2 |
+| 8,9 | 0 | 2 |
+| 9,10 | 0 | 1 |
+
+Since each possible location for the coins gives a different pair of answers, the answers are enough to reconstruct the location of the coins.
+
+4. The sequences $\left\{x_{n}\right\}$ and $\left\{y_{n}\right\}$ are defined by $x_{0}=2, y_{0}=1$ and, for $n \geq 0$,
+
+$$
+x_{n+1}=x_{n}^{2}+y_{n}^{2} \quad \text { and } \quad y_{n+1}=2 x_{n} y_{n}
+$$
+
+Find and prove an explicit formula for $x_{n}$ in terms of $n$.
+
+Solution. If we add the formulas for $x_{n+1}$ and $y_{n+1}$ together, we get
+
+$$
+x_{n+1}+y_{n+1}=x_{n}^{2}+y_{n}^{2}+2 x_{n} y_{n}=\left(x_{n}+y_{n}\right)^{2} \text {. }
+$$
+
+Thus, increasing $n$ by 1 raises $x_{n}+y_{n}$ to the power 2 . Since $x_{0}+y_{0}=3$, we get
+
+$$
+x_{n}+y_{n}=\left(\cdots\left(3^{2}\right)^{2} \cdots\right)^{2}[n \text { squarings }]=3^{2^{n}} .
+$$
+
+Similarly, we can subtract the formulas for $x_{n+1}$ and $y_{n+1}$ to get
+
+$$
+x_{n+1}-y_{n+1}=x_{n}^{2}+y_{n}^{2}-2 x_{n} y_{n}=\left(x_{n}-y_{n}\right)^{2} .
+$$
+
+However, $x_{0}-y_{0}=1$. Since $1^{2}=1$, we get
+
+$$
+x_{n}-y_{n}=1
+$$
+
+for all $n$.
+
+Finally, we add together equations (1) and (2), getting
+
+$$
+2 x_{n}=3^{2^{n}}+1
+$$
+
+which yields an explicit formula
+
+$$
+x_{n}=\frac{3^{2^{n}}+1}{2} \text {. }
+$$
+
+5. Let $A B C D$ be a square. Consider four circles $k_{1}, k_{2}, k_{3}, k_{4}$ which pass respectively through $A$ and $B, B$ and $C, C$ and $D, D$ and $A$, and whose centers are outside the square. Circles $k_{4}$ and $k_{1}, k_{1}$ and $k_{2}, k_{2}$ and $k_{3}, k_{3}$ and $k_{4}$ intersect respectively at $L$, $M, N, P$ inside the square. Prove that quadrilateral $L M N P$ can be inscribed in a circle.
+
+Solution. We use a standard theorem that states that a convex quadrilateral $W X Y Z$ can be inscribed in a circle if and only if $\angle W+\angle Y=180$. We have
+
+$$
+\begin{aligned}
+\angle P L M+\angle M N P & =(360-\angle M L A-\angle A L P)+(360-\angle P N C-\angle C N M) \\
+& =(180-\angle M L A)+(180-\angle A L P)+(180-\angle P N C)+(180-\angle C N M) \\
+& =\angle A B M+\angle P D A+\angle C D P+\angle M B C \\
+& =(\angle A B M+\angle M B C)+(\angle P D A+\angle C D P) \\
+& =90+90=180,
+\end{aligned}
+$$
+
+as desired.
+
+Remark. This problem works for any inscribed quadrilateral $A B C D$. It was restricted to squares to avoid the annoying issues of multiple diagram configurations and nonconvex quadrilaterals.
+
+6. Seven people are sitting around a circular table, not necessarily equally spaced from each other. Several vases are standing on the table. We say that two people can see each other if there are no vases on the line segment connecting them. (Treat both people and vases as points.)
+
+(a) If there are 12 vases, prove that there are two people who can see each other.
+
+(b) If there are 13 vases, prove that one can rearrange the people and the vases in such a way that no two people can see each other.
+
+Solution. (a) Assign each of the 21 line segments connecting two people a "weight" as follows: each side of the heptagon $(A B, B C, C D$, etc.) gets a weight of 1 ; each "skip-one" diagonal $(A C, B D$, etc.) gets a weight of $1 / 2$; each "skip-two" diagonal ( $A D, B E$, etc.) gets a weight of $1 / 3$.
+
+Claim 1. The total weight of all the lines of sight (segments connecting two people) blocked off by any vase is at most 1.
+
+Proof. If the vase does not lie on any line of sight, the total weight is clearly 0 . If the vase lies on a side, the total weight is 1 because the sides do not intersect any other sides or diagonals. If the vase lies on a skip-one diagonal, say $A C$, any other diagonal through the vase must have $B$ as an endpoint, so there is at most one such diagonal and the total weight is at most $1 / 2+1 / 2=1$. Finally, if the vase lies on a skip-two diagonal, say $A D$, but not on any skip-one diagonal, any other diagonal passing through the vase is also a skip-two diagonal. There can be at most two such diagonals (since they must have an endpoint at $B$ or $C$ ), so the total weight is at most $1 / 3+1 / 3+1 / 3=1$.
+
+We now see that the total weight of all the lines of sight is $7+7 / 2+7 / 3=12 \frac{5}{6}$, so it is impossible to block them all off with 12 vases.
+
+(b) Let six people sit at the vertices of a regular hexagon $A B C D E F$, and let the seventh person $G$ sit on the arc $A F$ such that $C G$ passes through the intersection of $A E$ and $B F$. Place one vase at the center of the table (hence on $A D, B E$, and $C F$ ) and one at the intersection of $A E, B F$, and $C G$. Then place vases on the intersections of: $A C$ and $B D ; C E$ and $D G ; D F$ and $E G ; A F$ and $B G$. Finally, put one vase at the midpoint of each side of heptagon $A B C D E F G$. It is easy to see that no two people can see each other.
+
+7. Let $x, y$, and $z$ be positive integers satisfying $x y=z^{2}+1$. Prove that there are integers $a, b, c$, and $d$ such that $x=a^{2}+b^{2}$, $y=c^{2}+d^{2}$, and $z=a c+b d$.
+
+Solution. We will use strong induction on $z$. Assume first that $z=1$. Then $x y=1^{2}+1=2$, so $x$ and $y$ are 1 and 2 in some order. If $x=2$ and $y=1$, then there is a solution $a=b=c=1, d=0$; if $x=1$ and $y=2$, we may take $a=c=d=1$, $b=0$.
+
+Now assume that $z>1$ and that, whenever $Z<z$, the equation $X Y=Z^{2}+1$ implies that there are integers $A, B, C$, and $D$ such that $X=A^{2}+B^{2}, Y=C^{2}+D^{2}$, and $Z=A C+B D$. We note that neither $x$ nor $y$ can equal $z$, for then $x y$ and $z^{2}$ would be two multiples of $z$ although their difference is 1 . We note that $x$ and $y$ cannot be both greater or both less than $z$ since
+
+$$
+(z-1)^{2}<z^{2}+1<(z+1)^{2} .
+$$
+
+Thus, because the symmetry of the problem allows us to switch $x$ and $y$, we can assume that $x>z>y$.
+
+Let $X=x+y-2 z, Y=y$, and $Z=z-y$. Multiplying out shows that the equation $X Y=Z^{2}+1$ is satisfied. Also, $Z<z$, so we have integers $A, B, C$, and $D$ such that the following equations hold:
+
+$$
+\begin{aligned}
+& X=A^{2}+B^{2} \\
+& Y=C^{2}+D^{2} \\
+& Z=A C+B D
+\end{aligned}
+$$
+
+Let $a=A+C, b=B+D, c=C$, and $d=D$. Equation (4) already tells us that $y=c^{2}+d^{2}$. Since $Y+Z=y+z-y=z$, we can add equation (4) and (5) together to get
+
+$$
+z=Y+Z=A C+C^{2}+B D+D^{2}=(A+C) C+(B+D) D=a c+b d
+$$
+
+Finally, since $X+Y+2 Z=x+y-2 z+y+2 z-2 y=x$, we add (3), (4), and twice (5) to get
+
+$$
+x=X+2 Z+Y=A^{2}+2 A C+C^{2}+B^{2}+2 B D+D^{2}=(A+C)^{2}+(B+D)^{2}=a^{2}+b^{2} .
+$$
+
+This completes the induction.
+

BayArea/md/en-monthly/en-1011-comp3s.md

@@ -0,0 +1,157 @@
+# Berkeley Math Circle Monthly Contest 3 - Solutions 
+
+1. You are given an $m \times n$ chocolate bar divided into $1 \times 1$ squares. You can break a piece of chocolate by splitting it into two pieces along a straight line that does not cut through any of the $1 \times 1$ squares. What is the minimum number of times you have to break the bar in order to separate all the $1 \times 1$ squares?
+
+Solution. We note that the number of separate pieces of chocolate increases by 1 at each cut. We begin with 1 piece and end with $m n$ pieces, so we must make $m n-1$ cuts. Thus $m n-1$ is the minimum (and also the maximum) number of cuts necessary to separate all the $1 \times 1$ squares.
+
+2. Let $n$ be a positive integer. Prove that the $n$th prime number is greater than or equal to $2 n-1$.
+
+Solution. We can verify that the first prime, 2 , is greater than $2 \cdot 1-1=1$ and the second prime, 3 , is equal to $2 \cdot 2-1$. From then, on, since all primes except 2 are odd, the difference between consecutive primes is at least 2 . Therefore, for $n \geq 3$,
+
+$$
+\begin{aligned}
+n \text {th prime } & \geq 3+\overbrace{2+2+\cdots+2}^{n-2 \text { differences }} \\
+& =3+2(n-2) \\
+& =3+2 n-4 \\
+& =2 n-1 .
+\end{aligned}
+$$
+
+3. Given the hypotenuse and the difference of the two legs of a right triangle, show how to reconstruct the triangle with ruler and compass.
+
+Solution. Here is one construction. Let $A B=d$ be the segment representing the difference of the two legs. Extend $A B$ to $C$ and raise a perpendicular $B D$ to $A C$ at $B$. Bisect angle $D B C$ to make ray $B E$ with $\angle E B C=45^{\circ}$. Now set the compass to the length $c$ of the hypotenuse and draw a circle $k$ of radius $c$ centered at $A$. Because the hypotenuse of a right triangle is longer than the difference of the legs, $c>A B$, and thus $k$ will intersect ray $B E$ at a point $X$. Drop a perpendicular $X Y$ from $X$ to $A C$. Then $\triangle A X Y$ is a right triangle with hypotenuse $A X$, and the difference $A Y-X Y$ of the legs equals $A Y-B Y=A B$ since $B X Y$ is an isosceles right triangle.
+
+If there were any other right triangle satisfying the same specifications, we could put it in the position $A X^{\prime} Y^{\prime}$ with the smallest angle at $A$, the longer leg $A Y^{\prime}$ on ray $A C$, and $X^{\prime}$ above $Y^{\prime}$. Then $X^{\prime}$ would lie on $k$ because $A X^{\prime}=c$; also, since $A X^{\prime}-X^{\prime} Y^{\prime}=A B=$
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_bf95a5ca2d32fe9fcb20g-1.jpg?height=385&width=448&top_left_y=1165&top_left_x=1462)
+$A Y^{\prime}-B Y^{\prime}, B X^{\prime} Y^{\prime}$ is an isosceles right triangle, and so $X^{\prime}$ lies on $\overrightarrow{B E}$. However, since $B$ is inside $k, k$ and $\overrightarrow{B E}$ can only intersect once. Thus $X^{\prime}=X$ and $Y^{\prime}=Y$.
+
+4. Show that each number in the sequence
+
+$$
+49, \quad 4489, \quad 444889, \quad 44448889, \quad \ldots
+$$
+
+is a perfect square.
+
+Solution. Let $x_{n}$ denote the $n$th number in the sequence. If we multiply $x_{n}$ by 9 by the standard method, we see that the calculations $9 \cdot 8+8=80$ and $9 \cdot 4+4=40$ occur repeatedly:
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_bf95a5ca2d32fe9fcb20g-1.jpg?height=182&width=425&top_left_y=1947&top_left_x=839)
+
+The result is $9 x_{n}=4 \cdot 10^{2 n}+4 \cdot 10^{n}+1=\left(2 \cdot 10^{n}+1\right)^{2}$, so
+
+$$
+x_{n}=\left(\frac{2 \cdot 10^{n}+1}{3}\right)^{2} \text {. }
+$$
+
+To see that the expression in parentheses is an integer, we may note that the sum of the digits of the numerator $200 \cdots 001$ is 3 , a multiple of 3 .
+
+5. Let $\left\{a_{1}, a_{2}, a_{3}, \ldots\right\}$ be a sequence of real numbers such that for each $n \geq 1$,
+
+$$
+a_{n+2}=a_{n+1}+a_{n}
+$$
+
+Prove that for all $n \geq 2$, the quantity
+
+$$
+\left|a_{n}^{2}-a_{n-1} a_{n+1}\right|
+$$
+
+does not depend on $n$.
+
+Solution. It suffices to prove that increasing $n$ to $n+1$ does not change the value, i.e. that for $n \geq 2$,
+
+$$
+\left|a_{n}^{2}-a_{n-1} a_{n+1}\right|=\left|a_{n+1}^{2}-a_{n} a_{n+2}\right| .
+$$
+
+We will prove more specifically that
+
+$$
+a_{n}^{2}-a_{n-1} a_{n+1}=-\left(a_{n+1}^{2}-a_{n} a_{n+2}\right)
+$$
+
+by transforming:
+
+$$
+\begin{aligned}
+a_{n}^{2}-a_{n-1} a_{n+1} & \stackrel{?}{=}-\left(a_{n+1}^{2}-a_{n} a_{n+2}\right) \\
+a_{n}^{2}-a_{n-1} a_{n+1} & \stackrel{?}{=}-a_{n+1}^{2}+a_{n} a_{n+2} \\
+a_{n+1}^{2}-a_{n-1} a_{n+1} & \stackrel{?}{=} a_{n} a_{n+2}-a_{n}^{2} \\
+a_{n+1}\left(a_{n+1}-a_{n-1}\right) & \stackrel{?}{=} a_{n}\left(a_{n+2}-a_{n}\right) .
+\end{aligned}
+$$
+
+Using the given relation $a_{n+2}=a_{n+1}+a_{n}$, we see that the right side equals $a_{n} \cdot a_{n+1}$. Replacing $n$ by $n-1$ in the given relation gives $a_{n+1}=a_{n}+a_{n-1}$, so the left side equals $a_{n+1} \cdot a_{n}$ and thus the equality is true.
+
+6. The inscribed circle of a triangle $A B C$ touches the sides $B C, C A, A B$ at $D, E$, and $F$ respectively. Let $X, Y$, and $Z$ be the incenters of triangles $A E F, B F D$, and $C D E$, respectively. Prove that $D X, E Y$, and $C Z$ meet at one point.
+
+Solution. Consider the midpoint $M$ of arc $E F$ on the incircle of $\triangle A B C$. Angles $A F M$ and $M F E$ are equal since they intercept equal arcs $F M$ and $M E$, and so $M$ is on the bisector of $\angle A F E$. Similarly, $M$ is on the bisector of $\angle F E A$, and therefore $M$ coincides with $X$. Moreover, angles $F D X$ and $X D E$ are equal since they intercept equal arcs $F X$ and $X E$, and so $D X$ is the angle bisector of $\angle D$ in $\triangle D E F$. Similarly, $E Y$ and $F Z$ are the other two angle bisectors in $\triangle D E F$. But the three angle bisectors in a triangle always meet!
+
+7. Define a sequence $a_{0}, a_{1}, a_{2}, \ldots$ in the following way: $a_{0}=0$, and for $n \geq 0$,
+
+$$
+a_{n+1}=a_{n}+5^{a_{n}} .
+$$
+
+Let $k$ be any positive integer. Prove that the remainders when $a_{0}, a_{1}, \ldots, a_{2^{k-1}}$ are divided by $2^{k}$ are all different.
+
+Remark. It was intended to prove that $a_{0}, a_{1}, \ldots, a_{2^{k}-1}$ (not just up to $a_{2^{k-1}}$ ) have different remainders. This typo does not affect the truth of the problem, and we will prove the strengthened statement.
+
+Solution. We begin with a simple numerical lemma.
+
+Lemma 1. For all $k \geq 0,5^{2^{k}}-1$ is divisible by $2^{k+2}$.
+
+Proof. By induction. For $k=0$, the statement may be checked directly. To step from $k$ to $k+1$, we write
+
+$$
+5^{2^{k+1}}-1=\left(5^{2^{k}}\right)^{2}-1=\left(5^{2^{k}}-1\right)\left(5^{2^{k}}+1\right)
+$$
+
+and note that the first factor is divisible by $2^{k+2}$ by the induction hypothesis and the second factor is clearly divisible by 2 , so the product is divisible by $2^{k+3}$.
+
+Now we turn to the main lemma of the proof.
+
+Lemma 2. Fix $k \geq 0$. For all $r \geq 0$, the difference
+
+$$
+a_{r+2^{k}}-a_{r}
+$$
+
+is divisible by $2^{k}$, not divisible by $2^{k+1}$, and independent of $r$ mod $2^{k+2}$.
+
+Proof. By induction. For $k=0$,
+
+$$
+a_{r+1}-a_{r}=5^{a_{r}}
+$$
+
+is clearly divisible by 1 , not divisible by 2 , and congruent to the constant value $1^{a_{r}}=1 \bmod 4$. Now assume that the statement is true for $k$; we will prove it for $k+1$. We know that $a_{r+2^{k}}-a_{r}$ has the form $2^{k} \cdot m$ for $m$ odd. The difference $a_{r+2^{k+1}}-a_{r+2^{k}}$ has the same value, $2^{k} \cdot m$, to the modulus $2^{k+2}$. Therefore, $\bmod 2^{k+2}$,
+
+$$
+a_{r+2^{k+1}}-a_{r}=\left(a_{r+2^{k+1}}-a_{r+2^{k}}\right)+\left(a_{r+2^{k}}-a_{r}\right) \equiv 2^{k} \cdot m+2^{k} \cdot m=2^{k+1} m
+$$
+
+so this difference is divisible by $2^{k+1}$ but not $2^{k+2}$.
+
+It remains to prove that $a_{r+2^{k+1}}-a_{r}$ is constant $\bmod 2^{k+3}$, i.e. that $2^{k+3}$ divides the difference between two values for consecutive values of $r$ :
+
+$$
+\begin{aligned}
+\left(a_{r+2^{k+1}+1}-a_{r+1}\right)-\left(a_{r+2^{k+1}}-a_{r}\right) & =\left(a_{r+2^{k+1}+1}-a_{r+2^{k+1}}\right)-\left(a_{r+1}-a_{r}\right) \\
+& =5^{a_{r+2^{k+1}}}-5^{a_{r}} \\
+& =5^{a_{r}}\left(5^{a_{r+2^{k+1}}-a_{r}}-1\right) .
+\end{aligned}
+$$
+
+We have already proved that $a_{r+2^{k+1}}-a_{r}$ is divisible by $2^{k+1}$. By Lemma $1,5^{2^{k+1}} \equiv 1 \bmod 2^{k+3}$, so $5^{a_{r+2^{k+1}}-a_{r}}$ is also 1 $\bmod 2^{k+3}$, completing the proof.
+
+To solve the problem, suppose that $r$ and $s$ are two nonnegative integers such that $0 \leq r<s \leq 2^{k}-1$ and $a_{r} \equiv a_{s} \bmod 2^{k}$. Let $s-r=2^{\ell} \cdot m$, where $m$ is odd. Mod $2^{\ell+1}, a_{t+2^{\ell}}-a_{t} \equiv 2^{\ell}$ for all $t \geq 0$ and therefore
+
+$$
+a_{s}-a_{r} \equiv\left(a_{r+2^{\ell}}-a_{r}\right)+\left(a_{r+2 \cdot 2^{\ell}}-a_{r+2^{\ell}}\right)+\cdots+\left(a_{r+m \cdot 2^{\ell}}-a_{r+(m-1) 2^{\ell}}\right) \equiv m \cdot 2^{\ell} \not \equiv 0,
+$$
+
+implying that $\ell+1 \geq k+1$, i.e. $\ell \geq k$. Thus $s-r \geq 2^{k}$, which is a contradiction.
+

BayArea/md/en-monthly/en-1011-comp4s.md

@@ -0,0 +1,120 @@
+# Berkeley Math Circle <br> Monthly Contest 4 - Solutions 
+
+1. For an arrangement of the digits 0 through 9 around a circle, a number is called a neighbor sum if it is the sum of some two adjacent digits in the arrangement. For example, the arrangement
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_b1ffea3d96ee79795ed2g-1.jpg?height=179&width=279&top_left_y=363&top_left_x=912)
+
+has five neighbor sums: $4,7,8,11$, and 14 . What is the minimal possible number of neighbor sums, given that each digit must be used just once?
+
+Solution. The answer is 3 . It can be achieved using the arrangement
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_b1ffea3d96ee79795ed2g-1.jpg?height=176&width=280&top_left_y=733&top_left_x=909)
+
+in which the only neighbor sums are 5,9 , and 10 . To prove that no smaller number is possible, first consider the digit 0 . It is adjacent to two different numbers between 1 and 9 inclusive, yielding two neighbor sums in the range $1, \ldots, 9$. Now consider the digit 9 . It is adjacent to two numbers and at least one of them is not 0 . Thus 9 has a neighbor between 1 and 8 inclusive, and thus there is a neighbor sum in the range $10, \ldots, 17$. Since these ranges do not overlap, there are altogether at least three different neighbor sums in the range $1, \ldots, 17$.
+
+2. A gadget has four dials in a row, each of which can be turned to point to one of three numbers: 0 (left), 1 (up) or 2 (right). Initially the dials are in the respective positions $2,0,1,0$, so that the gadget reads "2010." You may perform the following operation: choose two adjacent dials pointing at different numbers, and turn them to point to the third number. For example, taking the first two dials, you could change "2010" to "1110." Is it possible to perform a sequence of such operations so that the gadget reads "2011"?
+
+Solution. The answer is no. We notice that initially the sum of the numbers on the dials is 3 . We claim that after each operation, the sum of the numbers on the dials remains a multiple of 3 . To see this, consider the three possible types of moves:
+
+(a) Changing a 0 and a 2 to two 1 's does not change the digit sum.
+
+(b) Changing a 0 and a 1 to two 2's increases the sum by 3 .
+
+(c) Changing a 1 and a 2 to two 0 's decreases the sum by 3 .
+
+Thus, the sum always goes up or down by multiples of 3 , and thus we cannot reach the position 2011 in which the sum is 4 .
+
+3. Suppose that $A B C$ and $A^{\prime} B^{\prime} C^{\prime}$ are two triangles such that $\angle A=\angle A^{\prime}, A B=A^{\prime} B^{\prime}$, and $B C=B^{\prime} C^{\prime}$. Suppose also that $\angle C=90^{\circ}$. Prove that triangles $A B C$ and $A^{\prime} B^{\prime} C^{\prime}$ are congruent.
+
+Solution. Construct a point $D$ on ray $A C$ such that $A D=A^{\prime} C^{\prime}$. Then $\triangle A B D \cong \triangle A^{\prime} B^{\prime} C^{\prime}$ by SAS. If $D=C$, then we are done, so assume that $D \neq C$. We have $B D=B^{\prime} C^{\prime}$ and also $B C=B^{\prime} C^{\prime}$, so right triangle $B C D$ has hypotenuse $B D$ equal to leg $B C$, which is a contradiction.
+
+4. Evaluate the sum
+
+$$
+\frac{1}{1+\tan 1^{\circ}}+\frac{1}{1+\tan 2^{\circ}}+\frac{1}{1+\tan 3^{\circ}}+\cdots+\frac{1}{1+\tan 89^{\circ}}
+$$
+
+(The tangent $(\tan )$ of an angle $\alpha$ is the ratio $B C / A C$ in a right triangle $A B C$ with $\angle C=90^{\circ}$ and $\angle A=\alpha$, and its value does not depend on the triangle used.)
+
+Solution. By examining a triangle with angles $x, 90-x$, and 90 , it is not hard to see the trigonometric identity
+
+$$
+\tan \left(90^{\circ}-x\right)=\frac{1}{\tan x} .
+$$
+
+Let us pair up the terms
+
+$$
+\frac{1}{1+\tan x} \text { and } \frac{1}{1+\tan \left(90^{\circ}-x\right)}
+$$
+
+for $x=1,2,3, \ldots, 44$. The sum of such a pair of terms is
+
+$$
+\begin{aligned}
+\frac{1}{1+\tan x}+\frac{1}{1+\tan \left(90^{\circ}-x\right)} & =\frac{1}{1+\tan x}+\frac{1}{1+\frac{1}{\tan x}} \\
+& =\frac{1}{1+\tan x}+\frac{\tan x}{\tan x+1} \\
+& =\frac{1+\tan x}{1+\tan x}=1
+\end{aligned}
+$$
+
+There is one more term in the sum, the middle term:
+
+$$
+\frac{1}{1+\tan 45^{\circ}}=\frac{1}{1+1}=\frac{1}{2} .
+$$
+
+Therefore the sum is $44(1)+1 / 2=89 / 2$.
+
+5. Several positive integers are written on the blackboard. You can erase any two numbers and write their greatest common divisor (GCD) and least common multiple (LCM) instead. Prove that eventually the numbers will stop changing.
+
+Solution. First of all, since
+
+$$
+a \cdot b=\operatorname{gcd}(a, b) \cdot \operatorname{lcm}(a, b)
+$$
+
+for all positive integers $a$ and $b$, the product of the numbers on the board does not change.
+
+Second, we claim that at each step in which the numbers change, their sum increases. Suppose we pick two numbers $a$ and $b$ $(a \leq b)$ and replace them with $g=\operatorname{gcd}(a, b)$ and $\ell=\operatorname{lcm}(a, b)$. If $g=a$, then $\ell=b$ and the numbers do not change, so $g<a$ and therefore $g<b$. The change in the sum of the numbers is
+
+$$
+(g+\ell)-(a+b)=\frac{g^{2}+g \ell-g a-g b}{g}=\frac{g^{2}+a b-g a-g b}{g}=\frac{(a-g)(b-g)}{g}>0 .
+$$
+
+Since each number on the blackboard cannot exceed the product of the numbers that were originally there, the sum cannot increase unboundedly, and thus the numbers must stop changing.
+
+6. For all positive integers $n$, prove that
+
+$$
+\sum_{k=1}^{n} \phi(k)\left\lfloor\frac{n}{k}\right\rfloor=\frac{n(n+1)}{2}
+$$
+
+(For a positive integer $n, \phi(n)$ denotes the number of positive integers less than or equal to $n$ and relatively prime to $n$. For a real number $x,\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$.)
+
+Solution. Consider the fractions $a / b$, where $a$ and $b$ range over integers such that $1 \leq a \leq b \leq n$. We will count these fractions in two ways:
+
+(a) By unreduced form. For each denominator $b$, there are $b$ possible numerators $a=1,2, \ldots, b$, so the total number of fractions is
+
+$$
+1+2+\cdots+n=\frac{n(n+1)}{2}
+$$
+
+(b) By reduced form. Suppose a fraction $a / b$ has been reduced to $c / d$. Given the denominator $d$, there are $\phi(d)$ possible numerators $c$ such that $c \leq d$ and $c / d$ is in lowest terms. To get from $c / d$ back to $a / b$, we must multiply numerator and denominator by a positive integer $k$ such that $k c \leq k d \leq n$. Since $k c \leq k d$ always holds if $c \leq d$, the choice of $k$ is limited only by the inequality $k d \leq n$, which has $\lfloor n / d\rfloor$ solutions. Thus there are $\phi(d)\lfloor n / d\rfloor$ fractions $a / b$ for a given choice of $d$, so the total number of fractions is
+
+$$
+\sum_{d=1}^{n} \phi(d)\left\lfloor\frac{n}{d}\right\rfloor
+$$
+
+Since both methods of counting must yield the same answer, the identity follows.
+
+7. We are given a right isosceles triangle with legs of length 1 inside which every point (including vertices, points on sides, and all points in the interior) is colored red, yellow, green, or blue. Prove that there are two points of the same color such that the distance between them is at least $2-\sqrt{2}$.
+
+Solution.
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_b1ffea3d96ee79795ed2g-3.jpg?height=330&width=328&top_left_y=1212&top_left_x=888)
+
+Define points $D, E, F, G$ on the sides of the triangle as shown such that $A D=A F=G B=B E=2-\sqrt{2}$. Then $C D=C E=\sqrt{2}-1$, and since $\triangle C D E$ is isosceles and right, $D E=\sqrt{2}(\sqrt{2}-1)=2-\sqrt{2}$. Now segments $D E$ and $A F$ are equal and parallel, so $A D E F$ is a parallelogram and $E F=A D=2-\sqrt{2}$. Similarly, $D G=2-\sqrt{2}$.
+
+Assume that the conclusion is false, i.e. any two points at a distance of at least $2-\sqrt{2}$ are different colors. Since $\triangle A B C$ has all side lengths greater than $2-\sqrt{2}$, all of its vertices are different colors. Without loss of generality, assume that $A$ is red, $B$ is yellow, and $C$ is green. Points $F$ and $G$, which are at distances of at least $2-\sqrt{2}$ from all three vertices, must both be blue. Point $E$, which is at a distance of at least $2-\sqrt{2}$ from $A, F$, and $B$, must be green. Similarly, $D$ is green. But $D$ and $E$ are a distance of $2-\sqrt{2}$ from each other, so we have a contradiction.
+

BayArea/md/en-monthly/en-1011-comp5s.md

@@ -0,0 +1,126 @@
+# Berkeley Math Circle <br> Monthly Contest 5 - Solutions 
+
+1. A house has several rooms. There are also several doors, each of which connects either one room to another or a room to the outside. Suppose that every room has an even number of doors leaving it. Prove that the number of outside entrance doors is even as well.
+
+Solution. Every door has two "sides," one toward one room and one toward either another room or the outside. Clearly the total number of sides, being twice the number of doors, is even. However, for each room, the number of sides pointing to it is even. Since even subtracted from even gives even, the number of sides pointing to the outside is also even.
+
+2. 
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_8976f0c694062698a29fg-1.jpg?height=203&width=206&top_left_y=549&top_left_x=949)
+
+As shown in the diagram above, the vertices of a regular decagon are colored alternately black and white. We would like to draw colored line segments between the points in such a way that
+
+(a) Every black point is connected to every white point by a line segment.
+
+(b) No two line segments of the same color intersect, except at their endpoints.
+
+Determine, with proof, the minimum number of colors needed.
+
+Solution. Note that the five main diagonals of the decagon must all be different colors since they all intersect at the decagon's center. Thus, at least 5 colors are needed. To see that 5 colors are also sufficient, we can simply assign each black point a color and use that color to connect it with all the white points.
+
+3. Two congruent line segments $A B$ and $C D$ intersect at a point $E$. The perpendicular bisectors of $A C$ and $B D$ intersect at a point $F$ in the interior of $\angle A E C$. Prove that $E F$ bisects $\angle A E C$.
+
+Solution. Since $F$ is on the perpendicular bisectors of $A C$ and $B D, F A=F C$ and $F B=F D$. Also $A B=C D$ is given. Thus $\triangle F A B \cong \triangle F C D$ by SSS. Since corresponding heights of congruent triangles and equal, $F$ is equidistant from $A B$ and $C D$, implying that $F$ is on the bisector of $\angle A E C$.
+
+Remark. The problem was intended to read that $F$ lies in the interior of $\angle A E D$. Some contestants pointed out that the condition given - that $F$ lies in the interior of $\angle A E C$-implies a stronger result, namely that $A E=E C$ and therefore the perpendicular bisectors of $A C$ and $B D$ coincide. This does not, however, impact the truth of the problem as stated.
+
+4. We have a row of boxes that is infinite in one direction, as shown.
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_8976f0c694062698a29fg-1.jpg?height=92&width=669&top_left_y=1729&top_left_x=717)
+
+Determine if it is possible to fill each box with a positive integer such that the number in every box (except the leftmost one) is greater than the average of the numbers in the two neighboring boxes.
+
+Solution. The answer is no.
+
+Denote the numbers in the boxes by $a_{0}, a_{1}, a_{2}, \ldots$ Then, if the conditions are satisfied, we have for all $n \geq 1$
+
+$$
+\begin{aligned}
+a_{n} & >\frac{a_{n-1}+a_{n+1}}{2} \\
+2 a_{n} & >a_{n-1}+a_{n+1} \\
+a_{n}-a_{n-1} & >a_{n+1}-a_{n} .
+\end{aligned}
+$$
+
+This says that the differences $a_{n}-a_{n-1}$ form a strictly decreasing sequences. Since the differences are all integers, they must eventually become negative.
+
+At this point the sequence $a_{n}$ is itself strictly decreasing, so by the same token, eventually the $a_{n}$ 's become negative, contradicting the condition that they are all positive.
+
+5. Given a triangle $A B C$, we draw three circles with respective diameters $A B, B C$, and $C A$. Prove that there exists a point that is inside all three circles.
+
+Solution. We claim that the center $I$ of the triangle's inscribed circle is such a point. To see this, note that
+
+$$
+\angle B I C=180-\angle I C B+\angle C B I=180-\frac{\angle C+\angle B}{2}=180-\frac{180-A}{2}=90+\frac{A}{2}>90 .
+$$
+
+Thus $\triangle B I C$ is obtuse and if we drop a perpendicular $B D$ from $B$ to $C I$, then $D$ lies on the extension of ray $C I$. The circle with diameter $B C$ passes through $D$ since $\angle B D C=90$, and thus $I$, an interior point of chord $C D$, lies inside the circle. By symmetry, we can conclude that $I$ lies inside the other two circles as well.
+
+6. At a certain school, there are 6 subjects offered, and a student can take any combination of them. It is noticed that for any two subjects, there are fewer than 5 students taking both of them and fewer than 5 students taking neither. Determine the maximum possible number of students at the school.
+
+Solution. Let us count in two ways the number of ordered pairs $\left(s,\left\{c_{1}, c_{2}\right\}\right)$, where $s$ is a student, $\left\{c_{1}, c_{2}\right\}$ is an unordered pair of two distinct courses, and $s$ is either taking both $c_{1}$ and $c_{2}$ or neither $c_{1}$ nor $c_{2}$.
+
+First, we consider any of the $\left(\begin{array}{l}6 \\ 2\end{array}\right)=15$ pairs $\left\{c_{1}, c_{2}\right\}$ of distinct courses. By the given conditions, at most 4 students are taking neither $c_{1}$ nor $c_{2}$ and at most 4 take both, implying that $\left\{c_{1}, c_{2}\right\}$ gives rise to at most 8 pairs $\left(s,\left\{c_{1}, c_{2}\right\}\right)$, so the total number of such pairs is at most 120 .
+
+Second, let $N$ be the number of students. Consider any student $s$. If $k$ denotes the number of courses $s$ is taking, then the number of pairs of courses $\left\{c_{1}, c_{2}\right\}$ such that $s$ is taking both or neither of them is
+
+$$
+\left(\begin{array}{l}
+k \\
+2
+\end{array}\right)+\left(\begin{array}{c}
+6-k \\
+2
+\end{array}\right)
+$$
+
+Plugging in $k=0,1, \ldots, 6$ shows that the minimum value of this expression is 6 , attained at $k=3$. Thus there are at least $6 \mathrm{~N}$ pairs $\left(s,\left\{c_{1}, c_{2}\right\}\right)$.
+
+Putting this together, we see that $6 N \leq 120$, so $N$ is at most 20 . A total of 20 students indeed possible, as can be seen by letting each student take a distinct triplet of courses from the $\left(\begin{array}{l}6 \\ 3\end{array}\right)=20$ possible triplets.
+
+7. Let $a$ and $b$ be real numbers. Prove that the polynomial
+
+$$
+P(x)=x^{3}+(2 a+1) x^{2}+\left(2 a^{2}+2 a-3\right) x+b
+$$
+
+does not have three distinct rational roots.
+
+Solution. Assume the contrary, and suppose that the polynomial has rational roots $u, v, w$. Then by Vi'ete's formulas,
+
+$$
+\begin{aligned}
+u+v+w & =-(2 a+1) \\
+u v+v w+w u & =2 a^{2}-2 a-3
+\end{aligned}
+$$
+
+Squaring the first equation and subtracting twice the second yields
+
+$$
+\begin{aligned}
+(u+v+w)^{2}-2 u v-2 v w-2 w u & =(2 a+1)^{2}-\left(4 a^{2}+4 a-6\right) \\
+u^{2}+v^{2}+w^{2} & =4 a^{2}+4 a+1-4 a^{2}-4 a+6=7 .
+\end{aligned}
+$$
+
+We claim that there are no rational numbers $u, v, w$ satisfying $u^{2}+v^{2}+w^{2}=7$. Assuming that there were, let $s$ be their least common denominator, so $u=p / s, v=q / s, w=r / s$ for integers $p, q, r, s$ with GCD 1 . Then
+
+$$
+\begin{aligned}
+& \frac{p^{2}+q^{2}+r^{2}}{s^{2}}=7 \\
+& p^{2}+q^{2}+r^{2}=7 s^{2}
+\end{aligned}
+$$
+
+If we work mod 8 , this equation assumes the symmetrical form
+
+$$
+\begin{aligned}
+p^{2}+q^{2}+r^{2} & \equiv(-1) s^{2} \\
+p^{2}+q^{2}+r^{2}+s^{2} & \equiv 0
+\end{aligned}
+$$
+
+Using the fact that the squares $\bmod 8$ are 0,1 , and 4 , it is not hard to conclude that $p, q, r$, and $s$ must all be even, contradicting the assumption that their GCD is 1.
+

BayArea/md/en-monthly/en-1011-comp6s.md

@@ -0,0 +1,179 @@
+# Berkeley Math Circle Monthly Contest 6 - Solutions 
+
+1. Two thousand and eleven positive integers are chosen, all different and less than or equal to 4020. Prove that two of them have no common factors except 1.
+
+Solution. Split the numbers from 1 to 4020 into the 2010 pairs $\{1,2\},\{3,4\}, \ldots,\{4019,4020\}$. Since 2011 numbers are chosen but there are only 2010 pairs, two numbers have to lie in the same pair. These numbers cannot have any common factors except 1 , because any divisor of both numbers would have to divide their difference, which is 1 .
+
+2. A $3 \times 3 \times 3$ cube is made out of 27 subcubes. On every face shared by two subcubes, there is a door allowing you to move from one cube to the other. Is it possible to visit every subcube exactly once if
+
+(a) You may start and end wherever you like
+
+(b) You must start at the center subcube?
+
+Solution. (a) It is possible. Here is one of many possible routes.
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_63cea5b6ef1be6d7a886g-1.jpg?height=244&width=832&top_left_y=789&top_left_x=674)
+
+Level 1
+
+Level 2
+
+Level 3
+
+(b) It is impossible. Color the subcubes black and white alternately as shown:
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_63cea5b6ef1be6d7a886g-1.jpg?height=217&width=219&top_left_y=1168&top_left_x=671)
+
+Level 1
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_63cea5b6ef1be6d7a886g-1.jpg?height=215&width=209&top_left_y=1166&top_left_x=988)
+
+Level 2
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_63cea5b6ef1be6d7a886g-1.jpg?height=212&width=211&top_left_y=1168&top_left_x=1296)
+
+Level 3
+
+Every door connects a black subcube to a white subcube. Since the central subcube is white, the route must begin
+
+$$
+\text { White } \rightarrow \text { Black } \rightarrow \text { White } \rightarrow \text { Black } \rightarrow \cdots \text {. }
+$$
+
+Examining the first 27 subcubes visited, we see that 14 are white and 13 are black, a contradiction since the actual cube has 14 black and 13 white subcubes.
+
+3. In this fragment of a computer keyboard, the keys are congruent squares touching along their edges, and each letter refers to the point at the center of the corresponding key. Prove that triangles $Q A Z$ and $E S Z$ have the same area.
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_63cea5b6ef1be6d7a886g-1.jpg?height=398&width=398&top_left_y=1731&top_left_x=1362)
+
+Solution. Let us use measuring units in which the side length of each key is 1 . We express the area of quadrilateral $Q A Z E$ in two ways:
+
+(a) By dividing into triangles $Q A Z$ and $Q Z E$. Since $\triangle Q Z E$ has base $Q E=2$ and height 2 , we get
+
+$$
+\text { Area } Q A Z E=\text { Area } Q A Z+\frac{1}{2} \cdot 2 \cdot 2=\operatorname{Area} Q A Z+2
+$$
+
+(b) By dividing to triangles $E S Z, Q W A, W A S, W E S$, and $A S Z$. The four latter triangles all have base 1, height 1, and area $1 / 2$, so
+
+$$
+\text { Area } Q A Z E=\text { Area } E S Z+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\operatorname{Area} E S Z+2
+$$
+
+Since quadrilateral $Q A Z E$ must have the same area in both computations, we deduce that Area $Q A Z=$ Area $E S Z$.
+
+4. Let $x, y, z$, and $u$ be real numbers satisfying the equation
+
+$$
+\frac{x-y}{x+y}+\frac{y-z}{y+z}+\frac{z-u}{z+u}+\frac{u-x}{u+x}=0
+$$
+
+Suppose that $x, y$, and $z$ are rational (i.e. each is the quotient of two integers) and distinct. Prove that $u$ is rational as well. Solution. Let us begin by combining the fractions in pairs.
+
+$$
+\begin{aligned}
+0 & =\left(\frac{x-y}{x+y}+\frac{y-z}{y+z}\right)+\left(\frac{z-u}{z+u}+\frac{u-x}{u+x}\right) \\
+& =\frac{(x-y)(y+z)+(x+y)(y-z)}{(x+y)(y+z)}+\frac{(z-u)(u+x)+(z+u)(u-x)}{(z+u)(u+x)}
+\end{aligned}
+$$
+
+After expanding the numerators and collecting like terms, we get
+
+$$
+\begin{aligned}
+0 & =\frac{2 x y-2 y z}{(x+y)(y+z)}+\frac{2 z u-2 u x}{(z+u)(u+x)} \\
+& =2(x-z)\left[\frac{y}{(x+y)(y+z)}-\frac{u}{(z+u)(u+x)}\right] .
+\end{aligned}
+$$
+
+Because $x \neq z$, we can divide out $2(x-z)$ to get
+
+$$
+\begin{aligned}
+0 & =\frac{y}{(x+y)(y+z)}-\frac{u}{(x+u)(u+z)} \\
+& =\frac{y(x+u)(u+z)-u(x+y)(y+z)}{(x+y)(y+z)(x+u)(u+z)} .
+\end{aligned}
+$$
+
+We now expand the numerator and get, after canceling some terms,
+
+$$
+\begin{aligned}
+0 & =u^{2} y-u y^{2}+x y z-u x z \\
+& =u y(u-y)+x z(y-u) .
+\end{aligned}
+$$
+
+If $u=y$, then we are done since $y$ is rational. Assuming $u \neq y$, we divide through by $(u-y)$ to get $0=u y-x z$. Solving for $u$, we get $u=x z / y$ which is rational given that $x, y$, and $z$ are.
+
+5. Let $a_{1}, a_{2}, a_{3}, \ldots$ be an infinite sequence of positive real numbers such that for all $n \geq 1$,
+
+$$
+a_{n} \leq a_{2 n}+a_{2 n+1}
+$$
+
+Prove that there exists an $N \geq 1$ such that
+
+$$
+\sum_{n=1}^{N} a_{n}>1
+$$
+
+Solution. Let us prove by induction that for any positive integer $k$, there is an $N$ for which
+
+$$
+\sum_{n=1}^{N} a_{n} \geq k a_{1}
+$$
+
+The base case, $k=1$ and $N=1$, is trivial. Given that (1) is true for a given $k$, we have
+
+$$
+k a_{1} \leq \sum_{n=1}^{N} a_{n} \leq \sum_{n=1}^{N}\left(a_{2 n}+a_{2 n+1}\right)=\sum_{\substack{2 \leq n \leq 2 N \\ n \text { even }}} a_{n}+\sum_{\substack{3 \leq n \leq 2 N+1 \\ n \text { odd }}} a_{n}=\sum_{n=2}^{2 N+1} a_{n}
+$$
+
+Adding $a_{1}$, we get
+
+$$
+\sum_{n=1}^{2 N+1} a_{n} \geq(k+1) a_{1}
+$$
+
+as desired.
+
+The problem now follows by taking $k$ large enough so that $k>1 / a_{1}$, so $k a_{1}>1$.
+
+6. Determine whether there exists a $2011 \times 2011$ matrix with the following properties:
+
+- Every cell is filled with an integer from 1 to 4021.
+- For every integer $i(1 \leq i \leq 2011)$, the $i$ th row and the $i$ th column together contain every integer from 1 to 4021 .
+
+Solution. Answer: no. Fix an integer $k$ from 1 to 4021. Let us say that an index $i$ "hits" a cell containing the number $k$ if the cell is in either the $i$ th row or the $i$ th column. The conditions stipulate that each index hits exactly one instance of $k$, so the total number of hits is 4021 . On the other hand, every cell not lying on the main diagonal is hit by exactly two indices (its row number and its column number), while those on the diagonal are hit only once. In particular, to create an odd total number of hits, $k$ must appear on the diagonal. This is a contradiction since there are 4021 permissible values of $k$ and only 2011 spots on the diagonal.
+
+7. A lattice point is a point in the coordinate plane both of whose coordinates are integers. In $\triangle A B C$, all three vertices are lattice points and the area of the triangle is $1 / 2$. Prove that the orthocenter of $\triangle A B C$ is also a lattice point.
+
+Solution. Let us position our coordinate system so that $A$ is the origin. Let $B=(a, b)$ and $C=(c, d)$. Then the formula for the area of a triangle with given vertex coordinates gives $\frac{1}{2}=\frac{1}{2}|a d-b c|$, i.e. $a d-b c= \pm 1$.
+
+Let $H=(x, y)$ be the orthocenter of $\triangle A B C$. The condition $C H \perp A B$ can be expressed in vector form as
+
+$$
+0=\overrightarrow{A B} \cdot \overrightarrow{C H}=\langle a, b\rangle \cdot\langle x-c, y-d\rangle
+$$
+
+or
+
+$$
+a x+b y=a c+b d .
+$$
+
+Similarly the condition $B H \perp A C$ can be expressed as
+
+$$
+c x+d y=a c+b d .
+$$
+
+Adding $d$ times equation (2) to $-b$ times equation (3) causes $y$ to cancel, leaving
+
+$$
+(a d-b c) x=d(a c+b d)-b(a c+b d) .
+$$
+
+Since $a d-b c= \pm 1$, it follows that $x$ is an integer. Similarly, we can prove that $y$ is an integer as well.
+

BayArea/md/en-monthly/en-1011-comp7s.md

@@ -0,0 +1,153 @@
+# Berkeley Math Circle Monthly Contest 7 - Solutions 
+
+1. Let $x$ and $y$ be integers such that $\frac{3 x+4 y}{5}$ is an integer. Prove that $\frac{4 x-3 y}{5}$ is an integer.
+
+Solution. The basic strategy is to combine the facts that $x, y$, and $(3 x+4 y) / 5$ are all integers. Here is one solution:
+
+$$
+2(x)+1(y)-2\left(\frac{3 x+4 y}{5}\right)=\frac{5(2 x+y)-2(3 x+4 y)}{5}=\frac{10 x+5 y-6 x-8 y}{5}=\frac{4 x-3 y}{5}
+$$
+
+Since the left side is clearly an integer, so is the right side.
+
+2. Six rooks are placed on a $6 \times 6$ chessboard, at the locations marked + , so that each rook "attacks" the five squares in the same row and the five squares in the same column. Determine if it is possible to label each empty square with a digit (0 through 9 ) so that for each rook, the ten squares which it attacks are all labeled with different digits.
+
+Solution. The answer is yes. One of many solutions is shown.
+
+3. Let $f(n)$ be the number of digits of a positive integer $n$ (in base 10). Prove that
+
+| + | 5 | 6 | 7 | 8 | 9 |
+| :---: | :---: | :---: | :---: | :---: | :---: |
+| 0 | + | 7 | 8 | 9 | 6 |
+| 1 | 2 | + | 9 | 5 | 8 |
+| 2 | 3 | 4 | + | 6 | 5 |
+| 3 | 4 | 0 | 1 | + | 7 |
+| 4 | 1 | 3 | 0 | 2 | + |
+
+$$
+f\left(2^{n}\right)+f\left(5^{n}\right)=n+1
+$$
+
+Solution. Let $f\left(2^{n}\right)=x$. Then since the smallest number with $x$ digits is $10^{x-1}$ and the largest is $10^{x}-1$, we have
+
+$$
+10^{x-1} \leq 2^{n}<10^{x}
+$$
+
+However, a power of 2 (other than 1) cannot also be a power of 10 , so the inequality is strict:
+
+$$
+10^{x-1}<2^{n}<10^{x} .
+$$
+
+Similarly, if $f\left(5^{n}\right)=y$, then
+
+$$
+10^{y-1}<5^{n}<10^{y} .
+$$
+
+Multiplying (1) by [2],
+
+$$
+\begin{gathered}
+10^{x-1} \cdot 10^{y-1}<2^{n} \cdot 5^{n}<10^{x} \cdot 10^{y} \\
+10^{x+y-2}<10^{n}<10^{x+y} \\
+x+y-2<n<x+y .
+\end{gathered}
+$$
+
+Since the only integer between $x+y-2$ and $x+y$ is $x+y-1$,
+
+$$
+n=x+y-1
+$$
+
+so $x+y=n+1$ as desired.
+
+4. Let $A B C$ be a triangle with incenter $I$. A line through $I$ parallel to $B C$ intersects sides $A B$ and $A C$ at $D$ and $E$ respectively. Prove that the perimeter of $\triangle A D E$ is equal to $A B+A C$.
+
+Solution. Because $D E \| B C, \angle B I D=\angle I B C$ which is the same as $\angle D B I$ since $I$ is on the bisector of $\angle A B C$. Thus $\triangle B D I$ is isosceles, implying $B D=D I$. Similarly $C E=E I$. Thus the perimeter of $\triangle A D E$ is
+
+$$
+A D+A E+D E=A D+A E+D I+E I=A D+A E+B D+C E=A B+A C
+$$
+
+5. Two dice are loaded so that the numbers 1 through 6 come up with various (possibly different) probabilities on each die. Is it possible that, when both dice are rolled, each of the possible totals 2 through 12 has an equal probability of occurring?
+
+Solution. The answer is no. Suppose that each of the totals from 2 to 12 has an equal probability, which must be $1 / 11$ since the sum of all probabilities is 1 . Let $a$ and $b$ be the probabilities of a 1 and a 6 , respectively, on the first die, and let $c$ and $d$ be the corresponding probabilities on the second die.
+
+Since $1 / 11$ is the probability of rolling a total of $2, a c=1 / 11$ so $c=1 /(11 a)$; since $1 / 11$ is the probability of rolling 12 , $b d=1 / 11$ so $d=1 /(11 b)$. Since the probability of rolling a 7 through the combination $1+6$ or $6+1$ is at most $1 / 11$,
+
+$$
+\begin{aligned}
+\frac{1}{11} & \geq a d+b c \\
+\frac{1}{11} & \geq \frac{a}{11 b}+\frac{b}{11 a} \\
+1 & \geq \frac{a}{b}+\frac{b}{a}
+\end{aligned}
+$$
+
+Since $a / b$ and $b / a$ are reciprocals, one of them is at least 1 , so this inequality cannot hold.
+
+6. Let $A B C$ be a triangle with $\angle A=120^{\circ}$. The bisector of $\angle A$ meets side $B C$ at $D$. Prove that
+
+$$
+\frac{1}{A D}=\frac{1}{A B}+\frac{1}{A C}
+$$
+
+Solution. The area of $\triangle A B C$ is the sum of the areas of triangles $A B D$ and $A D C$, so
+
+$$
+\begin{aligned}
+\frac{1}{2} A B \cdot A C \cdot \sin 120^{\circ} & =\frac{1}{2} A B \cdot A D \cdot \sin 60^{\circ}+\frac{1}{2} A D \cdot A C \cdot \sin 60^{\circ} \\
+\frac{1}{2} A B \cdot A C \cdot \frac{\sqrt{3}}{2} & =\frac{1}{2} A B \cdot A D \cdot \frac{\sqrt{3}}{2}+\frac{1}{2} A D \cdot A C \cdot \frac{\sqrt{3}}{2} \\
+A B \cdot A C & =A B \cdot A D+A D \cdot A C .
+\end{aligned}
+$$
+
+Dividing through by $A B \cdot A C \cdot A D$ gives the desired result.
+
+7. Let $n$ and $k$ be positive integers with $n<\sqrt{(k-1) 2^{k}}$. Prove that it is possible to color each element of the set $\{1,2, \ldots, n\}$ red or green such that no $k$-term arithmetic progression is monochromatic.
+
+Solution. Let $A$ be the number of $k$-term arithmetic progressions in $\{1,2, \ldots, n\}$. For any common difference $d$, an arithmetic progression of difference $d$ fits in $\{1,2, \ldots, n\}$ iff its initial term $a$ satisfies
+
+$$
+1 \leq a<a+(k-1) d \leq n
+$$
+
+which is equivalent to
+
+$$
+1 \leq a \leq n-(k-1) d ;
+$$
+
+this inequality has $n-(k-1) d$ solutions, as long as $n-(k-1) d \geq 0$. Thus the total number of arithmetic progressions is
+
+$$
+A=\sum_{d=1}^{\left\lfloor\frac{n}{k-1}\right\rfloor}(n-(k-1) d)
+$$
+
+As shown in the figure, the terms of this sum can be seen as the areas of nonoverlapping rectangles lying under the graph of $y=n-(k-1) x$. Their sum therefore does not exceed the area of the triangle enclosed by this line and the axes:
+
+$$
+A \leq \frac{1}{2} \cdot \frac{n}{k-1} \cdot n=\frac{n^{2}}{2(k-1)}
+$$
+
+Now consider any $k$-term arithmetic progression. The number of colorings in which it is monochromatic is
+
+$$
+2^{n} \cdot \frac{2}{2^{k}}
+$$
+
+![](https://cdn.mathpix.com/cropped/2024_04_17_e48c180e94f881d24d80g-2.jpg?height=480&width=523&top_left_y=1928&top_left_x=1381)
+since, of the $2^{k}$ ways that its terms might be colored, only 2 are monochromatic. Therefore the number of colorings that make no $k$-term arithmetic progression monochromatic is at least
+
+$$
+\begin{aligned}
+& 2^{n}-A\left(2^{n} \cdot \frac{2}{2^{k}}\right) \\
+& \geq 2^{n}-\frac{n^{2}}{2(k-1)} \cdot 2^{n} \cdot \frac{2}{2^{k}} \\
+& =2^{n}\left(1-\frac{n^{2}}{(k-1) 2^{k}}\right)
+\end{aligned}
+$$
+
+If $n^{2}<(k-1) 2^{k}$ (equivalently, $n<\sqrt{(k-1) 2^{k}}$ ) then this lower bound will be positive, implying that there is at least one coloring with the desired property.
+