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Problem 1. Find all prime numbers $p$ for which there exist positive integers $x, y$ and $z$ such that the number
$$
x^{p}+y^{p}+z^{p}-x-y-z
$$
is a product of exactly three distinct prime numbers.
|
Solution. Let $A=x^{p}+y^{p}+z^{p}-x-y-z$. For $p=2$, we take $x=y=4$ and $z=3$. Then $A=30=2 \cdot 3 \cdot 5$. For $p=3$ we can take $x=3$ and $y=2$ and $z=1$. Then again $A=30=2 \cdot 3 \cdot 5$. For $p=5$ we can take $x=2$ and $y=1$ and $z=1$. Again $A=30=2 \cdot 3 \cdot 5$.
Assume now that $p \geqslant 7$. Workin... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 0 |
Problem 3. Triangle $A B C$ is such that $A B<A C$. The perpendicular bisector of side $B C$ intersects lines $A B$ and $A C$ at points $P$ and $Q$, respectively. Let $H$ be the orthocentre of triangle $A B C$, and let $M$ and $N$ be the midpoints of segments $B C$ and $P Q$, respectively. Prove that lines $H M$ and $... |
Solution. We have
$$
\angle A P Q=\angle B P M=90^{\circ}-\angle M B P=90^{\circ}-\angle C B A=\angle H C B
$$
and
$$
\angle A Q P=\angle M Q C=90^{\circ}-\angle Q C M=90^{\circ}-\angle A C B=\angle C B H
$$
From these two equalities, we see that the triangles $A P Q$ and $H C B$ are similar. Moreover, since $M$ a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 2 |
Problem 4. A $5 \times 100$ table is divided into 500 unit square cells, where $n$ of them are coloured black and the rest are coloured white. Two unit square cells are called adjacent if they share a common side. Each of the unit square cells has at most two adjacent black unit square cells. Find the largest possible... |
Solution. If we colour all the cells along all edges of the board together with the entire middle row except the second and the last-but-one cell, the condition is satisfied and there are 302 black cells. The figure below exhibits this colouring for the $5 \times 8$ case.
$ is circumscribed. The incircle of the triangle $A B C$ touches the lines $A B$ and $A C$ at the points $M$ and $N$, respectively. Prove that the incenter of the trapezoid $A B C D$ lies on the line $M N$.
=90^{\circ}-\frac{1}{2} m(\widehat{M A N}) \quad \text { and } \quad m(\widehat{B I C})=90^{\circ}+\frac{1}{2} m(\widehat{M A N})
$$
the quadrilateral $I R N C$ is ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 4 |
Problem 2. Let $a, b$ and $c$ be positive real numbers. Prove that
$$
\frac{8}{(a+b)^{2}+4 a b c}+\frac{8}{(b+c)^{2}+4 a b c}+\frac{8}{(c+a)^{2}+4 a b c}+a^{2}+b^{2}+c^{2} \geq \frac{8}{a+3}+\frac{8}{b+3}+\frac{8}{c+3}
$$
|
Solution. Since $2 a b \leq a^{2}+b^{2}$, it follows that $(a+b)^{2} \leq 2\left(a^{2}+b^{2}\right)$
and $4 a b c \leq 2 c\left(a^{2}+b^{2}\right)$, for any positive reals $a, b, c$.
Adding these inequalities, we find
$$
(a+b)^{2}+4 a b c \leq 2\left(a^{2}+b^{2}\right)(c+1)
$$
so that
$$
\frac{8}{(a+b)^{2}+4 a b ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 5 |
Problem 3. Find all the triples of integers $(a, b, c)$ such that the number
$$
N=\frac{(a-b)(b-c)(c-a)}{2}+2
$$
is a power of 2016 .
(A power of 2016 is an integer of the form $2016^{n}$, where $n$ is a non-negative integer.)
|
Solution. Let $a, b, c$ be integers and $n$ be a positive integer such that
$$
(a-b)(b-c)(c-a)+4=2 \cdot 2016^{n}
$$
We set $a-b=-x, b-c=-y$ and we rewrite the equation as
$$
x y(x+y)+4=2 \cdot 2016^{n}
$$
If $n>0$, then the right hand side is divisible by 7 , so we have that
$$
x y(x+y)+4 \equiv 0 \quad(\bmod 7)... | (,b,)=(k+2,k+1,k),k\in\mathbb{Z} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6 |
Problem 2. Let $n$ three-digit numbers satisfy the following properties:
(1) No number contains the digit 0 .
(2) The sum of the digits of each number is 9 .
(3) The units digits of any two numbers are different.
(4) The tens digits of any two numbers are different.
(5) The hundreds digits of any two numbers are ... |
Solution. Let $S$ denote the set of three-digit numbers that have digit sum equal to 9 and no digit equal to 0 . We will first find the cardinality of $S$. We start from the number 111 and each element of $S$ can be obtained from 111 by a string of 6 A's (which means that we add 1 to the current digit) and $2 G$ 's (w... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7 |
Problem 4. Let $A B C$ be an acute triangle, $A^{\prime}, B^{\prime}$ and $C^{\prime}$ be the reflections of the vertices $A, B$ and $C$ with respect to $B C, C A$, and $A B$, respectively, and let the circumcircles of triangles $A B B^{\prime}$ and $A C C^{\prime}$ meet again at $A_{1}$. Points $B_{1}$ and $C_{1}$ ar... |
Solution. Let $O_{1}, O_{2}$ and $O$ be the circumcenters of triangles $A B B^{\prime}, A C C^{\prime}$ and $A B C$ respectively. As $A B$ is the perpendicular bisector of the line segment $C C^{\prime}, O_{2}$ is the intersection of the perpendicular bisector of $A C$ with $A B$. Similarly, $O_{1}$ is the intersectio... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8 |
# Problem 1
Let $a, b$ and $c$ be positive real numbers such that $a+b+c=1$. Prove that
$$
\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}+6 \geq 2 \sqrt{2}\left(\sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}}\right)
$$
When does equality hold?
| ## Solution
Replacing $1-a, 1-b, 1-c$ with $b+c, c+a, a+b$ respectively on the right hand side, the given inequality becomes
and equivalently
$$
\left(\frac{b+c}{a}-2 \sqrt{2} \sqrt{\frac{... | =b==\frac{1}{3} | Inequalities | proof | Yes | Yes | olympiads | false | 9 |
## Problem 2
Let the circles $k_{1}$ and $k_{2}$ intersect at two distinct points $A$ and $B$, and let $t$ be a common tangent of $k_{1}$ and $k_{2}$, that touches $k_{1}$ and $k_{2}$ at $M$ and $N$, respectively. If $t \perp A M$ and $M N=2 A M$, evaluate $\angle N M B$.
| ## Solution 1
Let $P$ be the symmetric of $A$ with respect to $M$ (Figure 1). Then $A M=M P$ and $t \perp A P$, hence the triangle $A P N$ is isosceles with $A P$ as its base, so $\angle N A P=\angle N P A$. We have $\angle B A P=\angle B A M=\angle B M N$ and $\angle B A N=\angle B N M$.
Thus we have
$$
180^{\circ}... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10 |
## Problem 3
On a board there are $n$ nails each two connected by a string. Each string is colored in one of $n$ given distinct colors. For each three distinct colors, there exist three nails connected with strings in these three colors. Can $n$ be
a) 6 ?
b) 7 ?
|
Solution. (a) The answer is no.
Suppose it is possible. Consider some color, say blue. Each blue string is the side of 4 triangles formed with vertices on the given points. As there exist $\binom{5}{2}=\frac{5 \cdot 4}{2}=10$ pairs of colors other than blue, and for any such pair of colors together with the blue colo... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11 |
## Problem 4
Find all positive integers $x, y, z$ and $t$ such that
$$
2^{x} \cdot 3^{y}+5^{z}=7^{t}
$$
| ## Solution
Reducing modulo 3 we get $5^{z} \equiv 1$, therefore $z$ is even, $z=2 c, c \in \mathbb{N}$.
Next we prove that $t$ is even:
Obviously, $t \geq 2$. Let us suppose that $t$ is odd, say $t=2 d+1, d \in \mathbb{N}$. The equation becomes $2^{x} \cdot 3^{y}+25^{c}=7 \cdot 49^{d}$. If $x \geq 2$, reducing modu... | 3,1,=2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12 |
Problem 1. The real numbers $a, b, c, d$ satisfy simultaneously the equations
$$
a b c-d=1, b c d-a=2, c d a-b=3, d a b-c=-6
$$
Prove that $a+b+c+d \neq 0$.
|
Solution. Suppose that $a+b+c+d=0$. Then
$$
a b c+b c d+c d a+d a b=0
$$
If $a b c d=0$, then one of numbers, say $d$, must be 0 . In this case $a b c=0$, and so at least two of the numbers $a, b, c, d$ will be equal to 0 , making one of the given equations impossible. Hence $a b c d \neq 0$ and, from (1),
$$
\frac... | proof | Algebra | proof | Yes | Yes | olympiads | false | 13 |
Problem 3. Let $A L$ and $B K$ be angle bisectors in the non-isosceles triangle $A B C$ ( $L$ lies on the side $B C, K$ lies on the side $A C$ ). The perpendicular bisector of $B K$ intersects the line $A L$ at point $M$. Point $N$ lies on the line $B K$ such that $L N$ is parallel to $M K$. Prove that $L N=N A$.
|
Solution. The point $M$ lies on the circumcircle of $\triangle A B K$ (since both $A L$ and the perpendicular bisector of $B K$ bisect the arc $B K$ of this circle). Then $\angle C B K=$ $\angle A B K=\angle A M K=\angle N L A$. Thus $A B L N$ is cyclic, whence $\angle N A L=\angle N B L=$ $\angle C B K=\angle N L A$.... | LN=NA | Geometry | proof | Yes | Yes | olympiads | false | 14 |
Problem 1. Let $n(n \geq 1)$ be an integer. Consider the equation
$$
2 \cdot\left\lfloor\frac{1}{2 x}\right\rfloor-n+1=(n+1)(1-n x)
$$
where $x$ is the unknown real variable.
(a) Solve the equation for $n=8$.
(b) Prove that there exists an integer $n$ for which the equation has at least 2021 solutions. (For any re... |
Solution. Let $k=\left[\frac{1}{2 x}\right], k \in \mathbb{Z}$.
(a) For $n=8$, the equation becomes
$$
k=\left[\frac{1}{2 x}\right]=8-36 x \Rightarrow x \neq 0 \text { and } x=\frac{8-k}{36}
$$
Since $x \neq 0$, we have $k \neq 8$, and the last relation implies $k=\left[\frac{1}{2 x}\right]=\left[\frac{18}{8-k}\rig... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16 | |
Problem 2. For any set $A=\left\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right\}$ of five distinct positive integers denote by $S_{A}$ the sum of its elements, and denote by $T_{A}$ the number of triples $(i, j, k)$ with $1 \leqslant i<j<k \leqslant 5$ for which $x_{i}+x_{j}+x_{k}$ divides $S_{A}$.
Find the largest possibl... |
Solution. We will prove that the maximum value that $T_{A}$ can attain is 4 . Let $A=$ $\left\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right\}$ be a set of five positive integers such that $x_{1}x_{4}$ and $x_{3}>x_{2}$. Analogously we can show that any triple of form $(x, y, 5)$ where $y>2$ isn't good.
By above, the numbe... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17 |
Problem 3. Let $A B C$ be an acute scalene triangle with circumcenter $O$. Let $D$ be the foot of the altitude from $A$ to the side $B C$. The lines $B C$ and $A O$ intersect at $E$. Let $s$ be the line through $E$ perpendicular to $A O$. The line $s$ intersects $A B$ and $A C$ at $K$ and $L$, respectively. Denote by ... | ## Solution.
Let us denote angles of triangle $A B C$ with $\alpha, \beta, \gamma$ in a standard way. By basic anglechasing we have
$$
\angle B A D=90^{\circ}-\beta=\angle O A C \text { and ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18 |
Problem 4. Let $M$ be a subset of the set of 2021 integers $\{1,2,3, \ldots, 2021\}$ such that for any three elements (not necessarily distinct) $a, b, c$ of $M$ we have $|a+b-c|>10$. Determine the largest possible number of elements of $M$.
|
Solution. The set $M=\{1016,1017, \ldots, 2021\}$ has 1006 elements and satisfies the required property, since $a, b, c \in M$ implies that $a+b-c \geqslant 1016+1016-2021=11$. We will show that this is optimal.
Suppose $M$ satisfies the condition in the problem. Let $k$ be the minimal element of $M$. Then $k=|k+k-k|... | 1006 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19 |
Problem 1. Find all distinct prime numbers $p, q$ and $r$ such that
$$
3 p^{4}-5 q^{4}-4 r^{2}=26
$$
|
Solution. First notice that if both primes $q$ and $r$ differ from 3 , then $q^{2} \equiv r^{2} \equiv 1(\bmod 3)$, hence the left hand side of the given equation is congruent to zero modulo 3 , which is impossible since 26 is not divisible by 3 . Thus, $q=3$ or $r=3$. We consider two cases.
Case 1. $q=3$.
The equat... | p=5,q=3,r=19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 20 |
Problem 2. Consider an acute triangle $A B C$ with area S. Let $C D \perp A B \quad(D \in A B)$, $D M \perp A C \quad(M \in A C)$ and $\quad D N \perp B C \quad(N \in B C)$. Denote by $H_{1}$ and $H_{2}$ the orthocentres of the triangles $M N C$ and $M N D$ respectively. Find the area of the quadrilateral $\mathrm{AH}... |
Solution 1. Let $O, P, K, R$ and $T$ be the mid-points of the segments $C D, M N$, $C N, C H_{1}$ and $M H_{1}$, respectively. From $\triangle M N C$ we have that $\overline{P K}=\frac{1}{2} \overline{M C}$ and $P K \| M C$. Analogously, from $\Delta M H_{1} C$ we have that $\overline{T R}=\frac{1}{2} \overline{M C}$ ... | S | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21 |
Problem 3. Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that
$$
\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2} \geq 3(a+b+c+1)
$$
When does equality hold?
|
Solution 1. By using AM-GM $\left(x^{2}+y^{2}+z^{2} \geq x y+y z+z x\right)$ we have
$$
\begin{aligned}
\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2} & \geq\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)+\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)+\left(c+\... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 22 |
Problem 4. For a positive integer $n$, two players A and B play the following game: Given a pile of $s$ stones, the players take turn alternatively with A going first. On each turn the player is allowed to take either one stone, or a prime number of stones, or a multiple of $n$ stones. The winner is the one who takes ... |
Solution. Denote by $k$ the sought number and let $\left\{s_{1}, \mathrm{~s}_{2}, \ldots, \mathrm{s}_{k}\right\}$ be the corresponding values for $s$. We call each $s_{i}$ a losing number and every other nonnegative integer a winning numbers.
## Clearly every multiple of $n$ is a winning number.
Suppose there are tw... | n-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 23 |
Problem 1. Find all pairs $(a, b)$ of positive integers such that
$$
11 a b \leq a^{3}-b^{3} \leq 12 a b
$$
|
Solution 1. Let $a-b=t$. Due to $a^{3}-b^{3} \geq 11 a b$ we conclude that $a>b$ so $t$ is a positive integer and the condition can be written as
$$
11 b(b+t) \leq t\left[b^{2}+b(b+t)+(b+t)^{2}\right] \leq 12 b(b+t)
$$
Since
$$
t\left[b^{2}+b(b+t)+(b+t)^{2}\right]=t\left(b^{2}+b^{2}+b t+b^{2}+2 b t+t^{2}\right)=3 t... | (5,2) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 24 |
Problem 2. Let $A B C$ be an acute triangle such that $A H=H D$, where $H$ is the orthocenter of $A B C$ and $D \in B C$ is the foot of the altitude from the vertex $A$. Let $\ell$ denote the line through $H$ which is tangent to the circumcircle of the triangle $B H C$. Let $S$ and $T$ be the intersection points of $\... |
Solution 1. In order to prove that $S M$ and $T N$ are parallel, it suffices to prove that both of them are perpendicular to $S T$. Due to symmetry, we will provide a detailed proof of $S M \perp S T$, whereas the proof of $T N \perp S T$ is analogous. In this solution we will use the following notation: $\angle B A C... | proof | Geometry | proof | Yes | Yes | olympiads | false | 25 |
Problem 4. We call an even positive integer $n$ nice if the set $\{1,2, \ldots, n\}$ can be partitioned into $\frac{n}{2}$ two-element subsets, such that the sum of the elements in each subset is a power of 3 . For example, 6 is nice, because the set $\{1,2,3,4,5,6\}$ can be partitioned into subsets $\{1,2\},\{3,6\},\... |
Solution. For a nice number $n$ and a given partition of the set $\{1,2, \ldots, n\}$ into twoelement subsets such that the sum of the elements in each subset is a power of 3 , we say that $a, b \in\{1,2, \ldots, n\}$ are paired if both of them belong to the same subset.
Let $x$ be a nice number and $k$ be a (unique)... | 2^{2022}-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 26 |
## Problem 1.
Find all pairs $(a, b)$ of positive integers such that $a!+b$ and $b!+a$ are both powers of 5 .
|
Solution. The condition is symmetric so we can assume that $b \leq a$.
The first case is when $a=b$. In this case, $a!+a=5^{m}$ for some positive integer $m$. We can rewrite this as $a \cdot((a-1)!+1)=5^{m}$. This means that $a=5^{k}$ for some integer $k \geq 0$. It is clear that $k$ cannot be 0 . If $k \geq 2$, then... | (1,4),(4,1),(5,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 27 |
## Problem 2.
Prove that for all non-negative real numbers $x, y, z$, not all equal to 0 , the following inequality holds
$$
\frac{2 x^{2}-x+y+z}{x+y^{2}+z^{2}}+\frac{2 y^{2}+x-y+z}{x^{2}+y+z^{2}}+\frac{2 z^{2}+x+y-z}{x^{2}+y^{2}+z} \geqslant 3
$$
Determine all the triples $(x, y, z)$ for which the equality holds.
|
Solution. Let us first write the expression $L$ on the left hand side in the following way
$$
\begin{aligned}
L & =\left(\frac{2 x^{2}-x+y+z}{x+y^{2}+z^{2}}+2\right)+\left(\frac{2 y^{2}+x-y+z}{x^{2}+y+z^{2}}+2\right)+\left(\frac{2 z^{2}+x+y-z}{x^{2}+y^{2}+z}+2\right)-6 \\
& =\left(2 x^{2}+2 y^{2}+2 z^{2}+x+y+z\right)... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 28 |
## Problem 3.
Alice and Bob play the following game on a $100 \times 100$ grid, taking turns, with Alice starting first. Initially the grid is empty. At their turn, they choose an integer from 1 to $100^{2}$ that is not written yet in any of the cells and choose an empty cell, and place it in the chosen cell. When the... |
Solution. We denote by $(i, j)$ the cell in the $i$-th line and in the $j$-th column for every $1 \leq i, j \leq n$. Bob associates the following pair of cells : $(i, 2 k+1),(i, 2 k+2)$ for $1 \leq i \leq 100$ and $0 \leq k \leq 49$ except for $(i, k)=(100,0)$ and $(100,1)$, and the pairs $(100,1),(100,3)$ and $(100,2... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 29 |
Problem 1. Find all triples $(a, b, c)$ of real numbers such that the following system holds:
$$
\left\{\begin{array}{l}
a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \\
a^{2}+b^{2}+c^{2}=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}
\end{array}\right.
$$
|
Solution. First of all if $(a, b, c)$ is a solution of the system then also $(-a,-b,-c)$ is a solution. Hence we can suppose that $a b c>0$. From the first condition we have
$$
a+b+c=\frac{a b+b c+c a}{a b c}
$$
Now, from the first condition and the second condition we get
$$
(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\rig... | (,b,)=(,\frac{1}{},1),(,\frac{1}{},-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 31 |
Problem 2. Let $\triangle A B C$ be a right-angled triangle with $\angle B A C=90^{\circ}$ and let $E$ be the foot of the perpendicular from $A$ on $B C$. Let $Z \neq A$ be a point on the line $A B$ with $A B=B Z$. Let (c) be the circumcircle of the triangle $\triangle A E Z$. Let $D$ be the second point of intersecti... |
Solution. We will first show that $P A$ is tangent to $(c)$ at $A$.
Since $E, D, Z, A$ are concyclic, then $\angle E D C=\angle E A Z=\angle E A B$. Since also the triangles $\triangle A B C$ and $\triangle E B A$ are similar, then $\angle E A B=\angle B C A$, therefore $\angle E D C=\angle B C A$.
Since $\angle F E... | proof | Geometry | proof | Yes | Yes | olympiads | false | 32 |
Problem 3. Alice and Bob play the following game: Alice picks a set $A=\{1,2, \ldots, n\}$ for some natural number $n \geqslant 2$. Then starting with Bob, they alternatively choose one number from the set $A$, according to the following conditions: initially Bob chooses any number he wants, afterwards the number chos... |
Solution. To say that Alice has a winning strategy means that she can find a number $n$ to form the set A, so that she can respond appropriately to all choices of Bob and always get at the end a composite number for the sum of her choices. If such $n$ does not exist, this would mean that Bob has a winning strategy ins... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 33 |
Problem 4. Find all pairs $(p, q)$ of prime numbers such that
$$
1+\frac{p^{q}-q^{p}}{p+q}
$$
is a prime number.
|
Solution. It is clear that $p \neq q$. We set
$$
1+\frac{p^{q}-q^{p}}{p+q}=r
$$
and we have that
$$
p^{q}-q^{p}=(r-1)(p+q)
$$
From Fermat's Little Theorem we have
$$
p^{q}-q^{p} \equiv-q \quad(\bmod p)
$$
Since we also have that
$$
(r-1)(p+q) \equiv-r q-q \quad(\bmod p)
$$
from (3) we get that
$$
r q \equiv 0... | (2,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 34 |
Problem 2. Let $x, y, z$ be positive integers such that $x \neq y \neq z \neq x$. Prove that
$$
(x+y+z)(x y+y z+z x-2) \geq 9 x y z
$$
When does the equality hold?
|
Solution. Since $x, y, z$ are distinct positive integers, the required inequality is symmetric and WLOG we can suppose that $x \geq y+1 \geq z+2$. We consider 2 possible cases:
Case 1. $y \geq z+2$. Since $x \geq y+1 \geq z+3$ it follows that
$$
(x-y)^{2} \geq 1, \quad(y-z)^{2} \geq 4, \quad(x-z)^{2} \geq 9
$$
whic... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 35 |
Problem 3. Let $A B C$ be an acute triangle such that $A B \neq A C$, with circumcircle $\Gamma$ and circumcenter $O$. Let $M$ be the midpoint of $B C$ and $D$ be a point on $\Gamma$ such that $A D \perp B C$. Let $T$ be a point such that $B D C T$ is a parallelogram and $Q$ a point on the same side of $B C$ as $A$ su... |
Solution. Let $X^{\prime}$ be symmetric point to $Q$ in line $B C$. Now since $\angle C B A=\angle C Q M=\angle C X^{\prime} M$, $\angle B C A=\angle B Q M=\angle B X^{\prime} M$, we have
$$
\angle B X^{\prime} C=\angle B X^{\prime} M+\angle C X^{\prime} M=\angle C B A+\angle B C A=180^{\circ}-\angle B A C
$$
we hav... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36 |
Problem 4. Consider a regular $2 n$-gon $P, A_{1} A_{2} \ldots A_{2 n}$ in the plane, where $n$ is a positive integer. We say that a point $S$ on one of the sides of $P$ can be seen from a point $E$ that is external to $P$, if the line segment $S E$ contains no other points that lie on the sides of $P$ except $S$. We ... |
Solution Answer: For $n=2$, the answer is 36 ; for $n=3$, the answer is 30 and for $n \geq 4$, the answer is $6 n$.
Lemma 1. Given a regular $2 n$-gon in the plane and a sequence of $n$ consecutive sides $s_{1}, s_{2}, \ldots, s_{n}$ there is an external point $Q$ in the plane, such that the color of each $s_{i}$ can... | 6n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 37 |
Problem 1. Find all prime numbers $a, b, c$ and positive integers $k$ which satisfy the equation
$$
a^{2}+b^{2}+16 \cdot c^{2}=9 \cdot k^{2}+1
$$
| # Solution:
The relation $9 \cdot k^{2}+1 \equiv 1(\bmod 3)$ implies
$$
a^{2}+b^{2}+16 \cdot c^{2} \equiv 1(\bmod 3) \Leftrightarrow a^{2}+b^{2}+c^{2} \equiv 1(\bmod 3)
$$
Since $a^{2} \equiv 0,1(\bmod 3), \quad b^{2} \equiv 0,1(\bmod 3), c^{2} \equiv 0,1(\bmod 3)$, we have:
| $a^{2}$ | 0 | 0 | 0 | 0 | 1 | 1 | 1 |... | (37,3,3,13),(17,3,3,7),(3,37,3,13),(3,17,3,7),(3,3,2,3) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38 |
Problem 2. Let $a, b, c$ be positive real numbers such that $a+b+c=3$. Find the minimum value of the expression
$$
A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}
$$
$19^{\text {th ... | ## Solution:
We can rewrite $A$ as follows:
$$
\begin{aligned}
& A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-a^{2}-b^{2}-c^{2}= \\
& 2\left(\frac{a b+b c+c a}{a b c}\right)-\left(a^{2}+b^{2}+c^{2}\right)=2\left(\frac{a b+b c+c a}{a b c}\right)-\left((a+b+c... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 39 |
Problem 3. Let $\triangle A B C$ be an acute triangle. The lines $l_{1}, l_{2}$ are perpendicular to $A B$ at the points $A, B$ respectively. The perpendicular lines from the midpoint $M$ of $A B$ to the sides of the triangle $A C, B C$ intersect the lines $l_{1}, l_{2}$ at the points $E, F$, respectively. If $D$ is t... | ## Solution:
Let $H, G$ be the points of intersection of $M E, M F$ with $A C, B C$ respectively. From the similarity of triangles $\triangle M H A$ and $\triangle M A E$ we get $\frac{M H}{M A}=\frac{M A}{M E}$, thus
$$
M A^{2}=M H \cdot M E
$$
Similarly, from the similarity of triangles $\triangle M B G$ and $\tri... | proof | Geometry | proof | Yes | Yes | olympiads | false | 40 |
A1. Let $a, b, c$ be positive real numbers such that $a+b+c+a b+b c+c a+a b c=7$. Prove that
$$
\sqrt{a^{2}+b^{2}+2}+\sqrt{b^{2}+c^{2}+2}+\sqrt{c^{2}+a^{2}+2} \geq 6
$$
|
Solution. First we see that $x^{2}+y^{2}+1 \geq x y+x+y$. Indeed, this is equivalent to
$$
(x-y)^{2}+(x-1)^{2}+(y-1)^{2} \geq 0
$$
Therefore
$$
\begin{aligned}
& \sqrt{a^{2}+b^{2}+2}+\sqrt{b^{2}+c^{2}+2}+\sqrt{c^{2}+a^{2}+2} \\
\geq & \sqrt{a b+a+b+1}+\sqrt{b c+b+c+1}+\sqrt{c a+c+a+1} \\
= & \sqrt{(a+1)(b+1)}+\sqrt... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 42 |
A2. Let $a$ and $b$ be positive real numbers such that $3 a^{2}+2 b^{2}=3 a+2 b$. Find the minimum value of
$$
A=\sqrt{\frac{a}{b(3 a+2)}}+\sqrt{\frac{b}{a(2 b+3)}}
$$
|
Solution. By the Cauchy-Schwarz inequality we have that
$$
5\left(3 a^{2}+2 b^{2}\right)=5\left(a^{2}+a^{2}+a^{2}+b^{2}+b^{2}\right) \geq(3 a+2 b)^{2}
$$
(or use that the last inequality is equivalent to $(a-b)^{2} \geq 0$ ).
So, with the help of the given condition we get that $3 a+2 b \leq 5$. Now, by the AM-GM i... | \frac{2}{\sqrt{5}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 43 |
A3. Let $a, b, c, d$ be real numbers such that $0 \leq a \leq b \leq c \leq d$. Prove the inequality
$$
a b^{3}+b c^{3}+c d^{3}+d a^{3} \geq a^{2} b^{2}+b^{2} c^{2}+c^{2} d^{2}+d^{2} a^{2}
$$
|
Solution. The inequality is equivalent to
$$
\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\right)^{2} \geq\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} d^{2}+d^{2} a^{2}\right)^{2}
$$
By the Cauchy-Schwarz inequality,
$$
\left(a b^{3}+b c^{3}+c d^{3}+d a^{3}\right)\left(a^{3} b+b^{3} c+c^{3} d+d^{3} a\right) \geq\left(a^{2} b^{2}+b^... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 44 |
A4. Let $x, y, z$ be three distinct positive integers. Prove that
$$
(x+y+z)(x y+y z+z x-2) \geq 9 x y z
$$
When does the equality hold?
|
Solution. Since $x, y, z$ are distinct positive integers, the required inequality is symmetric and WLOG we can suppose that $x \geq y+1 \geq z+2$. We consider 2 possible cases:
Case 1. $y \geq z+2$. Since $x \geq y+1 \geq z+3$ it follows that
$$
(x-y)^{2} \geq 1, \quad(y-z)^{2} \geq 4, \quad(x-z)^{2} \geq 9
$$
whic... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 45 |
C1. Consider a regular $2 n+1$-gon $P$ in the plane, where $n$ is a positive integer. We say that a point $S$ on one of the sides of $P$ can be seen from a point $E$ that is external to $P$, if the line segment $S E$ contains no other points that lie on the sides of $P$ except $S$. We want to color the sides of $P$ in... |
Solution. Answer: $n=1$ is clearly a solution, we can just color each side of the equilateral triangle in a different color, and the conditions are satisfied. We prove there is no larger $n$ that fulfills the requirements.
Lemma 1. Given a regular $2 n+1$-gon in the plane, and a sequence of $n+1$ consecutive sides $s... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 46 |
C2. Consider a regular $2 n$-gon $P$ in the plane, where $n$ is a positive integer. We say that a point $S$ on one of the sides of $P$ can be seen from a point $E$ that is external to $P$, if the line segment $S E$ contains no other points that lie on the sides of $P$ except $S$. We want to color the sides of $P$ in 3... |
Solution. Answer: For $n=2$, the answer is 36 ; for $n=3$, the answer is 30 and for $n \geq 4$, the answer is $6 n$.
Lemma 1. Given a regular $2 n$-gon in the plane and a sequence of $n$ consecutive sides $s_{1}, s_{2}, \ldots, s_{n}$ there is an external point $Q$ in the plane, such that the color of each $s_{i}$ ca... | 6n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 47 |
C3. We have two piles with 2000 and 2017 coins respectively. Ann and Bob take alternate turns making the following moves: The player whose turn is to move picks a pile with at least two coins, removes from that pile $t$ coins for some $2 \leqslant t \leqslant 4$, and adds to the other pile 1 coin. The players can choo... |
Solution. Denote the number of coins in the two piles by $X$ and $Y$. We say that the pair $(X, Y)$ is losing if the player who begins the game loses and that the pair $(X, Y)$ is winning otherwise. We shall prove that $(X, Y)$ is loosing if $X-Y \equiv 0,1,7 \bmod 8$, and winning if $X-Y \equiv 2,3,4,5,6 \bmod 8$.
L... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 48 |
G1. Given a parallelogram $A B C D$. The line perpendicular to $A C$ passing through $C$ and the line perpendicular to $B D$ passing through $A$ intersect at point $P$. The circle centered at point $P$ and radius $P C$ intersects the line $B C$ at point $X,(X \neq C)$ and the line $D C$ at point $Y$, $(Y \neq C)$. Pro... |
Solution. Denote the feet of the perpendiculars from $P$ to the lines $B C$ and $D C$ by $M$ and $N$ respectively and let $O=A C \cap B D$. Since the points $O, M$ and $N$ are midpoints of $C A, C X$ and $C Y$ respectively it suffices to prove that $M, N$ and $O$ are collinear. According to Menelaus's theorem for $\tr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 49 |
G2. Let $A B C$ be an acute triangle such that $A B$ is the shortest side of the triangle. Let $D$ be the midpoint of the side $A B$ and $P$ be an interior point of the triangle such that
$$
\Varangle C A P=\Varangle C B P=\Varangle A C B
$$
Denote by $M$ and $N$ the feet of the perpendiculars from $P$ to $B C$ and ... |
Solution. If $\gamma=\Varangle A C B$ then $\Varangle C A P=\Varangle C B P=\Varangle A C B=\gamma$. Let $E=K N \cap A P$ and $F=K M \cap B P$. We show that points $E$ and $F$ are midpoints of $A P$ and $B P$, respectively.
$ such that $2^{x}+3^{y}$ is a perfect square.
|
Solution. In order for the expression $2^{x}+3^{y}$ to be a perfect square, a positive integer $t$ such that $2^{x}+3^{y}=t^{2}$ should exist.
Case 1. If $x$ is even, then there exists a positive integer $z$ such that $x=2 z$. Then
$$
\left(t-2^{z}\right)\left(t+2^{z}\right)=3^{y}
$$
Since $t+2^{z}-\left(t-2^{z}\ri... | (4,2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55 |
NT4. Solve in nonnegative integers the equation $5^{t}+3^{x} 4^{y}=z^{2}$.
|
Solution. If $x=0$ we have
$$
z^{2}-2^{2 y}=5^{t} \Longleftrightarrow\left(z+2^{y}\right)\left(z-2^{y}\right)=5^{t}
$$
Putting $z+2^{y}=5^{a}$ and $z-2^{y}=5^{b}$ with $a+b=t$ we get $5^{a}-5^{b}=2^{y+1}$. This gives us $b=0$ and now we have $5^{t}-1=2^{y+1}$. If $y \geq 2$ then consideration by modulo 8 gives $2 \m... | (,x,y,z)=(1,0,1,3),(0,1,0,2),(2,2,2,13),(0,1,2,7) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 56 |
NT5. Find all positive integers $n$ such that there exists a prime number $p$, such that
$$
p^{n}-(p-1)^{n}
$$
is a power of 3 .
Note. A power of 3 is a number of the form $3^{a}$ where $a$ is a positive integer.
|
Solution. Suppose that the positive integer $n$ is such that
$$
p^{n}-(p-1)^{n}=3^{a}
$$
for some prime $p$ and positive integer $a$.
If $p=2$, then $2^{n}-1=3^{a}$ by $(1)$, whence $(-1)^{n}-1 \equiv 0(\bmod 3)$, so $n$ should be even. Setting $n=2 s$ we obtain $\left(2^{s}-1\right)\left(2^{s}+1\right)=3^{a}$. It ... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 57 |
A2. Let $a, b, c$ be positive real numbers such that abc $=1$. Show that
$$
\frac{1}{a^{3}+b c}+\frac{1}{b^{3}+c a}+\frac{1}{c^{3}+a b} \leq \frac{(a b+b c+c a)^{2}}{6}
$$
so
|
Solution. By the AM-GM inequality we have $a^{3}+b c \geq 2 \sqrt{a^{3} b c}=2 \sqrt{a^{2}(a b c)}=2 a$ and
$$
\frac{1}{a^{3}+b c} \leq \frac{1}{2 a}
$$
Similarly; $\frac{1}{b^{3}+c a} \leq \frac{1}{2 b} \cdot \frac{1}{c^{3}+a b} \leq \frac{1}{2 c}$ and then
$$
\frac{1}{a^{3}+b c}+\frac{1}{b^{3}+c a}+\frac{1}{c^{3}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 58 |
A3. Let $a . b$ c ce positue real numbers such that $a+b+c=a^{2}+b^{2}+c^{2}$. Shou that
$$
\frac{a^{2}}{a^{2}+a b}+\frac{b^{2}}{b^{2}+b c}+\frac{c^{2}}{c^{2}+c a} \geq \frac{a+b+c}{2}
$$
|
Solution. By the Cauchy-Schwarz inequality it is
$$
\begin{aligned}
& \left(\frac{a^{2}}{a^{2}+a b}+\frac{b^{2}}{b^{2}+b c}+\frac{c^{2}}{c^{2}+c a}\right)\left(\left(a^{2}+a b\right)+\left(b^{2}+b c\right)+\left(c^{2}+c a\right)\right) \geq(a+b+c)^{2} \\
\Rightarrow & \frac{a^{2}}{a^{2}+a b}+\frac{b^{2}}{b^{2}+b c}+\... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 59 |
A4. Solve the following equation for $x, y, z \in \mathbb{N}$
$$
\left(1+\frac{x}{y+z}\right)^{2}+\left(1+\frac{y}{z+x}\right)^{2}+\left(1+\frac{z}{x+y}\right)^{2}=\frac{27}{4}
$$
|
Solution 1. Call $a=1+\frac{x}{y+z}, b=1+\frac{y}{z+x}, c=1+\frac{z}{x+y}$ to get
$$
a^{2}+b^{2}+c^{2}=\frac{27}{4}
$$
Since it is also true that
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2
$$
the quadratic-harmonic means inequality implies
$$
\frac{3}{2}=\sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}} \geq \frac{3}{\frac{1}{a}+... | z | Algebra | math-word-problem | Yes | Yes | olympiads | false | 60 |
A5. Find the largest positive integer $n$ for which the inequality
$$
\frac{a+b+c}{a b c+1}+\sqrt[n]{a b c} \leq \frac{5}{2}
$$
holds for all $a, b, c \in[0,1]$. Here $\sqrt[1]{a b c}=a b c$.
|
Solution. Let $n_{\max }$ be the sought largest value of $n$, and let $E_{a, b, c}(n)=\frac{a+b+c}{a b c+1}+\sqrt[n]{a b c}$. Then $E_{a, b, c}(m)-E_{a, b, c}(n)=\sqrt[m]{a b c}-\sqrt[n]{a b c}$ and since $a . b c \leq 1$ we clearly have $E_{a, b, c}(m) \geq$ $E_{a, b, c}(n)$ for $m \geq n$. So if $E_{a, b, c}(n) \geq... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 61 |
G1. Let $A B C$ be an equilateral triangle, and $P$ a point on the circumcircle of the triangle $A B C$ and distinct from $A, B$ and $C$. If the lines through $P$ and parallel to $B C, C A, A B$ intersect the lines $C A, A B, B C$ at $M, N$ and $Q$ respectively, prove that $M, N$ and $Q$ are collinear.
|
Solution. Without any loss of generality, let $P$ be in the minor arc of the chord $A C$ as in Figure 1. Since $\angle P N A=\angle N P M=60^{\circ}$ and $\angle N A M=\angle P M A=120^{\circ}$, it follows that the points $A, M, P$ and $N$ are concyclic. This yields
$$
\angle N M P=\angle N A P
$$
, then $\angle B O A=$ $2 y$, and since $O A=O B$, it is $\angle O A B=\angle O B A=90^{\circ}-y$. Also since $A H \perp B C$, it is
$\angle H A C=90^{\circ}-y=\angle O A B$ and the claim follows.
Since $A . A^{\prime}$ bisect... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 63 |
C1. Along a round table are arranged 11 cards with the names (all distinct) of the 11 members of the $16^{\text {th }}$ JBMO Problem Selection Committee. The distances between each two consecutive cards are equal. Assume that in the first meeting of the Committee none of its 11 members sits in front of the card with h... |
Solution. Yes it is: Rotating the table by the angles $\frac{360^{\circ}}{11}, 2 \cdot \frac{360^{\circ}}{11}, 3 \cdot \frac{360^{\circ}}{11}, \ldots, 10 \cdot \frac{360^{\circ}}{11}$, we obtain 10 new positions of the table. By the assumption, it is obvious that every one of the 11 members of the Committee will be se... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 64 |
C2. $n$ nails nailed on a board are connected by two via a string. Each string is colored in one of $n$ given colors. For any three colors there exist three nails conne.cted by two with strings in these three colors. Can $n$ be: (a) 6, (b) 7?
|
Solution. (a) The answer is no:
Suppose it is possible. Consider some color, say blue. Each blue string is the side of 4 triangles formed with vertices on the given points. As there exist $\binom{5}{2}=\frac{5 \cdot 4}{2}=10$ pairs of colors other than blue, and for any such pair of colors together with the blue colo... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 65 |
C3. In a circle of diameter 1 consider 65 points no three of which are collinear. Prove that there exist 3 among these points which form a triangle with area less then or equal to $\frac{1}{72}$.
|
Solution. Lemma: If a triangle $A B C$ lies in a rectangle $K L M N$. with sides $K L=a$ and $L M=b$, then the area of the triangle is less then or equal to $\frac{a b}{2}$.
Proof of the lemma: Writhout any loss of generality assume that among the distance of $A, B, C$ from $K L$, that of $A$ is between the other two... | proof | Geometry | proof | Yes | Yes | olympiads | false | 66 |
NT5. Find all the positive integers $x, y, z, t$ such that $2^{x} \cdot 3^{y}+5^{z}=7^{t}$.
|
Solution. Reducing modulo 3 we get $5^{z} \equiv 1$, therefore $z$ is even, $z=2 c, c \in \mathbb{N}$. .
Next we prove that $t$ is even. Obviously, $t \geq 2$. Let us suppose that $t$ is odd, $t=2 d+1$, $d \in \mathbb{N}$. The equation becomes $2^{x} \cdot 3^{y}+25^{c}=7 \cdot 49^{d}$.
If $x \geq 2$, reducing modulo... | 3,1,=2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 67 |
A1. Find all ordered triples $(x, y, z)$ of real numbers satisfying the following system of equations:
$$
\begin{aligned}
x^{3} & =\frac{z}{y}-2 \frac{y}{z} \\
y^{3} & =\frac{x}{z}-2 \frac{z}{x} \\
z^{3} & =\frac{y}{x}-2 \frac{x}{y}
\end{aligned}
$$
|
Solution. We have
$$
\begin{aligned}
& x^{3} y z=z^{2}-2 y^{2} \\
& y^{3} z x=x^{2}-2 z^{2} \\
& z^{3} x y=y^{2}-2 x^{2}
\end{aligned}
$$
with $x y z \neq 0$.
Adding these up we obtain $\left(x^{2}+y^{2}+z^{2}\right)(x y z+1)=0$. Hence $x y z=-1$. Now the system of equations becomes:
$$
\begin{aligned}
& x^{2}=2 y... | (1,1,-1),(1,-1,1),(-1,1,1),(-1,-1,-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 68 |
A2. Find the largest possible value of the expression $\left|\sqrt{x^{2}+4 x+8}-\sqrt{x^{2}+8 x+17}\right|$ where $x$ is a real number.
|
Solution. We observe that
$$
\left|\sqrt{x^{2}+4 x+8}-\sqrt{x^{2}+8 x+17}\right|=\left|\sqrt{(x-(-2))^{2}+(0-2)^{2}}-\sqrt{\left.(x-(-4))^{2}+(0-1)^{2}\right)}\right|
$$
is the absolute difference of the distances from the point $P(x, 0)$ in the $x y$-plane to the points $A(-2,2)$ and $B(-4,1)$.
By the Triangle Ine... | \sqrt{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 69 |
A3. Show that
$$
\left(a+2 b+\frac{2}{a+1}\right)\left(b+2 a+\frac{2}{b+1}\right) \geq 16
$$
for all positive real numbers $a, b$ satisfying $a b \geq 1$.
|
Solution 1. By the AM-GM Inequality we have:
$$
\frac{a+1}{2}+\frac{2}{a+1} \geq 2
$$
Therefore
$$
a+2 b+\frac{2}{a+1} \geq \frac{a+3}{2}+2 b
$$
and, similarly,
$$
b+2 a+\frac{2}{b+1} \geq 2 a+\frac{b+3}{2}
$$
On the other hand,
$$
(a+4 b+3)(b+4 a+3) \geq(\sqrt{a b}+4 \sqrt{a b}+3)^{2} \geq 64
$$
by the Cauchy... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 70 |
C3. All possible pairs of $n$ apples are weighed and the results are given to us in an arbitrary order. Can we determine the weights of the apples if a. $n=4$, b. $n=5$, c. $n=6$ ?
|
Solution. a. No. Four apples with weights $1,5,7,9$ and with weights $2,4,6,10$ both give the results $6,8,10,12,14,16$ when weighed in pairs.
b. Yes. Let $a \leq b \leq c \leq d \leq e$ be the weights of the apples. As each apple is weighed 4 times, by adding all 10 pairwise weights and dividing the sum by 4 , we ob... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 71 |
G1. Let $A B$ be a diameter of a circle $\omega$ with center $O$ and $O C$ be a radius of $\omega$ which is perpendicular to $A B$. Let $M$ be a point on the line segment $O C$. Let $N$ be the second point of intersection of the line $A M$ with $\omega$, and let $P$ be the point of intersection of the lines tangent to... |
Solution. Since the lines $P N$ and $B P$ are tangent to $\omega, N P=P B$ and $O P$ is the bisector of $\angle N O B$. Therefore the lines $O P$ and $N B$ are perpendicular. Since $\angle A N B=90^{\circ}$, it follows that the lines $A N$ and $O P$ are parallel. As $M O$ and $P B$ are also parallel and $A O=O B$, the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 72 |
G3. Let $D$ be a point on the side $B C$ of an acute triangle $A B C$ such that $\angle B A D=\angle C A O$ where $O$ is the center of the circumcircle $\omega$ of the triangle $A B C$. Let $E$ be the second point of intersection of $\omega$ and the line $A D$. Let $M, N, P$ be the midpoints of the line segments $B E,... |
Solution. We will show that $M O P D$ is a parallelogram. From this it follows that $M, N$, $P$ are collinear.
Since $\angle B A D=\angle C A O=90^{\circ}-\angle A B C, D$ is the foot of the perpendicular from $A$ to side $B C$. Since $M$ is the midpoint of the line segment $B E$, we have $B M=M E=M D$ and hence $\an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 73 |
G4. Let $I$ be the incenter and $A B$ the shortest side of a triangle $A B C$. The circle with center $I$ and passing through $C$ intersects the ray $A B$ at the point $P$ and the ray $B A$ at the point $Q$. Let $D$ be the point where the excircle of the triangle $A B C$ belonging to angle $A$ touches the side $B C$, ... |
Solution. First we will show that points $P$ and $Q$ are not on the line segment $A B$.
Assume that $Q$ is on the line segment $A B$. Since $C I=Q I$ and $\angle I B Q=\angle I B C$, either the triangles $C B I$ and $Q B I$ are congruent or $\angle I C B+\angle I Q B=180^{\circ}$. In the first case, we have $B C=B Q$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 74 |
G5. A circle passing through the midpoint $M$ of the side $B C$ and the vertex $A$ of a triangle $A B C$ intersects the sides $A B$ and $A C$ for the second time at the points $P$ and $Q$, respectively. Show that if $\angle B A C=60^{\circ}$ then
$$
A P+A Q+P Q<A B+A C+\frac{1}{2} B C
$$
|
Solution. Since the quadrilateral $A P M Q$ is cyclic, we have $\angle P M Q=180^{\circ}-\angle P A Q=$ $180^{\circ}-\angle B A C=120^{\circ}$. Therefore $\angle P M B+\angle Q M C=180^{\circ}-\angle P M Q=60^{\circ}$.
Let the point $B^{\prime}$ be the symmetric of the point $B$ with respect to the line $P M$ and the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 75 |
N3. Find all ordered pairs $(a, b)$ of positive integers for which the numbers $\frac{a^{3} b-1}{a+1}$ and $\frac{b^{3} a+1}{b-1}$ are positive integers.
|
Solution. As $a^{3} b-1=b\left(a^{3}+1\right)-(b+1)$ and $a+1 \mid a^{3}+1$, we have $a+1 \mid b+1$.
As $b^{3} a+1=a\left(b^{3}-1\right)+(a+1)$ and $b-1 \mid b^{3}-1$, we have $b-1 \mid a+1$.
So $b-1 \mid b+1$ and hence $b-1 \mid 2$.
- If $b=2$, then $a+1 \mid b+1=3$ gives $a=2$. Hence $(a, b)=(2,2)$ is the only so... | (1,3),(2,2),(3,3) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 77 |
N5. Find all ordered triples $(x, y, z)$ of positive integers satisfying the equation
$$
\frac{1}{x^{2}}+\frac{y}{x z}+\frac{1}{z^{2}}=\frac{1}{2013}
$$
|
Solution. We have $x^{2} z^{2}=2013\left(x^{2}+x y z+z^{2}\right)$. Let $d=\operatorname{gcd}(x, z)$ and $x=d a, z=d b$. Then $a^{2} b^{2} d^{2}=2013\left(a^{2}+a b y+b^{2}\right)$.
As $\operatorname{gcd}(a, b)=1$, we also have $\operatorname{gcd}\left(a^{2}, a^{2}+a b y+b^{2}\right)=1$ and $\operatorname{gcd}\left(b... | (x,y,z)=(2013n,2013n^{2}-2,2013n) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 78 |
N6. Find all ordered triples $(x, y, z)$ of integers satisfying the following system of equations:
$$
\begin{aligned}
x^{2}-y^{2} & =z \\
3 x y+(x-y) z & =z^{2}
\end{aligned}
$$
|
Solution. If $z=0$, then $x=0$ and $y=0$, and $(x, y, z)=(0,0,0)$.
Let us assume that $z \neq 0$, and $x+y=a$ and $x-y=b$ where $a$ and $b$ are nonzero integers such that $z=a b$. Then $x=(a+b) / 2$ and $y=(a-b) / 2$, and the second equations gives $3 a^{2}-3 b^{2}+4 a b^{2}=4 a^{2} b^{2}$.
Hence
$$
b^{2}=\frac{3 a... | (0,0,0),(1,0,1),(0,1,-1),(1,2,-3),(2,1,3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 79 |
## A1 MLD
Let $x, y, z$ be real numbers, satisfying the relations
$$
\left\{\begin{array}{l}
x \geq 20 \\
y \geq 40 \\
z \geq 1675 \\
x+y+z=2015
\end{array}\right.
$$
Find the greatest value of the product $P=x \cdot y \cdot z$.
| ## Solution 1:
By virtue of $z \geq 1675$ we have
$$
y+z<2015 \Leftrightarrow y<2015-z \leq 2015-1675<1675
$$
It follows that $(1675-y) \cdot(1675-z) \leq 0 \Leftrightarrow y \cdot z \leq 1675 \cdot(y+z-1675)$.
By using the inequality $u \cdot v \leq\left(\frac{u+v}{2}\right)^{2}$ for all real numbers $u, v$ we obt... | 48407500 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 80 |
## A2 ALB
3) If $x^{3}-3 \sqrt{3} x^{2}+9 x-3 \sqrt{3}-64=0$, find the value of $x^{6}-8 x^{5}+13 x^{4}-5 x^{3}+49 x^{2}-137 x+2015$.
|
Solution
$x^{3}-3 \sqrt{3} x^{2}+9 x-3 \sqrt{3}-64=0 \Leftrightarrow(x-\sqrt{3})^{3}=64 \Leftrightarrow(x-\sqrt{3})=4 \Leftrightarrow x-4=\sqrt{3} \Leftrightarrow x^{2}-8 x+16=3 \Leftrightarrow$ $x^{2}-8 x+13=0$
$x^{6}-8 x^{5}+13 x^{4}-5 x^{3}+49 x^{2}-137 x+2015=\left(x^{2}-8 x+13\right)\left(x^{4}-5 x+9\right)+189... | 1898 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 81 |
## A3 MNE
Let $a, b, c$ be positive real numbers. Prove that
$$
\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}>2
$$
| ## Solution:
Starting from the double expression on the left-hand side of given inequality, and applying twice the Arithmetic-Geometric mean inequality, we find that
$$
\begin{aligned}
2 \frac{a}{b}+2 \sqrt{\frac{b}{c}}+2 \sqrt[3]{\frac{c}{a}} & =\frac{a}{b}+\left(\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt{\frac{b}{c}}\rig... | 4 | Inequalities | proof | Yes | Yes | olympiads | false | 82 |
## A5 MKCD
Let $x, y, z$ be positive real numbers that satisfy the equality $x^{2}+y^{2}+z^{2}=3$. Prove that
$$
\frac{x^{2}+y z}{x^{2}+y z+1}+\frac{y^{2}+z x}{y^{2}+z x+1}+\frac{z^{2}+x y}{z^{2}+x y+1} \leq 2
$$
| ## Solution:
We have
$$
\begin{aligned}
& \frac{x^{2}+y z}{x^{2}+y z+1}+\frac{y^{2}+z x}{y^{2}+z x+1}+\frac{z^{2}+x y}{z^{2}+x y+1} \leq 2 \Leftrightarrow \\
& \frac{x^{2}+y z+1}{x^{2}+y z+1}+\frac{y^{2}+z x+1}{y^{2}+z x+1}+\frac{z^{2}+x y+1}{z^{2}+x y+1} \leq 2+\frac{1}{x^{2}+y z+1}+\frac{1}{y^{2}+z x+1}+\frac{1}{z^... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 83 |
## G1 MNE
Around the triangle $A B C$ the circle is circumscribed, and at the vertex $C$ tangent $t$ to this circle is drawn. The line $p$ which is parallel to this tangent intersects the lines $B C$ and $A C$ at the points $D$ and $E$, respectively. Prove that the points $A, B, D, E$ belong to the same circle.
| ## Solution:
Let $O$ be the center of a circumscribed circle $k$ of the triangle $A B C$, and let $F$ and $G$ be the points of intersection of the line $C O$ with the line $p$ and the circle $k$, respectively (see Figure). From $p \| t$ it follows that $p \perp C O$. Furthermore, $\angle A B C=\angle A G C$, because t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 84 |
## G2 MLD
The point $P$ is outside of the circle $\Omega$. Two tangent lines, passing from the point $P$, touch the circle $\Omega$ at the points $A$ and $B$. The median $A M, M \in(B P)$, intersects the circle $\Omega$ at the point $C$ and the line $P C$ intersects again the circle $\Omega$ at the point $D$. Prove th... | ## Solution:
Since $\angle B A C=\angle B A M=\angle M B C$, we have $\triangle M A B \cong \triangle M B C$.
We obtain $\frac{M A}{M B}=\frac{M B}{M C}=\frac{A B}{B C}$. The equality $\qua... | proof | Geometry | proof | Yes | Yes | olympiads | false | 85 |
## G3 GRE
Let $c \equiv c(O, K)$ be a circle with center $O$ and radius $R$ and $A, B$ be two points on it, not belonging to the same diameter. The bisector of the angle $A \hat{B} O$ intersects the circle $c$ at point $C$, the circumcircle of the triangle $A O B$, say ${ }^{c_{1}}$ at point $K$ and the circumcircle o... | ## Solution:
The segments $O B, O C$ are equal, as radii of the circle ${ }^{c}$. Hence $O B C$ is an isosceles triangle and
$$
\hat{B}_{1}=\hat{C}_{1}=\hat{x}
$$
The chord $B C$ is the b... | proof | Geometry | proof | Yes | Yes | olympiads | false | 86 |
G4
CYP
Let $\triangle A B C$ be an acute triangle. The lines $\left(\varepsilon_{1}\right),\left(\varepsilon_{2}\right)$ are perpendicular to $A B$ at the points $A$, $B$, respectively. The perpendicular lines from the midpoint $M$ of $A B$ to the sides of the triangle $A C_{;} B C$ intersect the lines $\left(\varep... | ## Solution:
Let $H, G$ be the points of intersection of $M E, M F$, with $A C, B C$ respectively. From the similarity of triangles $\triangle M H A$ and $\triangle M A E$ we get
$$
\frac{M H}{M A}=\frac{M A}{M E}
$$
thus, $M A^{2}=M H \cdot M E$
Similarly, from the similarity of triangles $\triangle M B G$ and $\t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 87 |
G5 ROU
Let $A B C$ be an acute triangle with $A B \neq A C$. The incircle $\omega$ of the triangle touches the sides $B C, C A$ and $A B$ at $D, E$ and $F$, respectively. The perpendicular line erected at $C$ onto $B C$ meets $E F$ at $M$, and similarly, the perpendicular line erected at $B$ onto $B C$ meets $E F$ at... | ## Solution:
## Proof 1.1.
Let $\{T\}=E F \cap B C$. Applying Menelaus' theorem to the triangle $A B C$ and the transversal line $E-F-T$ we obtain $\frac{T B}{T C} \cdot \frac{E C}{E A} \cdot \frac{F A}{F B}=1$, i.e. $\frac{T B}{T C} \cdot \frac{s-c}{s-a} \cdot \frac{s-a}{s-b}=1$, or $\frac{T B}{T C}=\frac{s-b}{s-c}$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 88 |
NT1 SAU
What is the greatest number of integers that can be selected from a set of 2015 consecutive numbers so that no sum of any two selected numbers is divisible by their difference?
| ## Solution:
We take any two chosen numbers. If their difference is 1 , it is clear that their sum is divisible by their difference. If their difference is 2 , they will be of the same parity, and their sum is divisible by their difference. Therefore, the difference between any chosen numbers will be at least 3 . In o... | 672 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 89 |
NT2 BUL
A positive integer is called a repunit, if it is written only by ones. The repunit with $n$ digits will be denoted by $\underbrace{11 \ldots 1}_{n}$. Prove that:
a) the repunit $\underbrace{11 \ldots 1}_{n}$ is divisible by 37 if and only if $n$ is divisible by 3 ;
b) there exists a positive integer $k$ suc... | ## Solution:
a) Let $n=3 m+r$, where $m$ and $r$ are non-negative integers and $r<3$.
Denote by $\underbrace{00 \ldots 0}_{p}$ a recording with $p$ zeroes and $\underbrace{a b c a b c \ldots a b c}_{p x a b c c}$ recording with $p$ times $a b c$. We
have: $\quad \underbrace{11 \ldots 1}_{n}=\underbrace{11 \ldots 1}_... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 90 |
## NT3 ALB
a) Show that the product of all differences of possible couples of six given positive integers is divisible by 960 (original from Albania).
b) Show that the product of all differences of possible couples of six given positive integers. is divisible by 34560 (modified by problem selecting committee).
| ## Solution:
a) Since we have six numbers then at least two of them have a same residue when divided by 3 , so at least one of the differences in our product is divisible by 3 .
Since we have six numbers then at least two of them have a same residuc when divided by 5 , so at least one of the differences in our produc... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 91 |
## NT5 MNE
Does there exist positive integers $a, b$ and a prime $p$ such that
$$
a^{3}-b^{3}=4 p^{2} ?
$$
| ## Solution:
The given equality may be written as
$$
(a-b)\left(a^{2}+a b+b^{2}\right)=4 p^{2}
$$
Since $a-b3$ and $3^{2} \equiv 0(\bmod 3)$, it follows that the congruence (5) is not satisfied for any odd prime $p$.
If $a-b=4$, then substituting $a=b+4$ in (1) we obtain
$$
3 b^{2}+12 b+16=p^{2}
$$
whence it foll... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 93 |
## C1 BUL
A board $n \times n(n \geq 3)$ is divided into $n^{2}$ unit squares. Integers from 0 to $n$ included are written down: one integer in each unit square, in such a way that the sums of integers in each $2 \times 2$ square of the board are different. Find all $n$ for which such boards exist.
| ## Solution:
The number of the $2 \times 2$ squares in a board $n \times n$ is equal to $(n-1)^{2}$. All possible sums of the numbers in such squares are $0,1, \ldots, 4 n$. A necessary condition for the existence of a board with the required property is $4 n+1 \geq(n-1)^{2}$ and consequently $n(n-6) \leq 0$. Thus $n ... | 3\leqn\leq6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 94 |
## C3 ALB
Positive integers are put into the following table
| 1 | 3 | 6 | 10 | 15 | 21 | 28 | 36 | | |
| ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: |
| 2 | 5 | 9 | 14 | 20 | 27 | 35 | 44 | | |
| 4 | 8 | 13 | 19 | 26 | 34 | 43 | 53 | | |
| 7 | 12 | 18 | 25 | 33 | 42 | | | | |
| 11 |... | ## Solution 1:
We shall observe straights lines as on the next picture. We can call these lines diagonals.
| 1 | $\sqrt{3}$ | 6 | 10 | 15 | 21 | 28 | 36 | |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 2 | 5 | 9 | 14 | 20 | 27 | 35 | 44 | |
| 4 | 8 | 13 | 19 | 26 | 34 | 43 | 53 | |
|... | 2015 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 95 |
C4
GRE
Let $n \geq 1$ be a positive integer. A square of side length $n$ is divided by lines parallel to each side into $n^{2}$ squares of side length 1 . Find the number of parallelograms which have vertices among the vertices of the $n^{2}$ squares of side length 1 , with both sides smaller or equal to 2 , and whi... |
Solution:
We can divide all these parallelograms into 7 classes (types I-VII), according to Figure.
Type 1: There are $n$ ways to choose the strip for the horizontal (shorter) side of th... | 7n^{2}-12n+1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 96 |
A1 Let $a$ be a real positive number such that $a^{3}=6(a+1)$. Prove that the equation $x^{2}+a x+a^{2}-6=0$ has no solution in the set of the real number.
| ## Solution
The discriminant of the equation is $\Delta=3\left(8-a^{2}\right)$. If we accept that $\Delta \geq 0$, then $a \leq 2 \sqrt{2}$ and $\frac{1}{a} \geq \frac{\sqrt{2}}{4}$, from where $a^{2} \geq 6+6 \cdot \frac{\sqrt{2}}{4}=6+\frac{6}{a} \geq 6+\frac{3 \sqrt{2}}{2}>8$ (contradiction).
| proof | Algebra | proof | Yes | Yes | olympiads | false | 97 |
A2 Prove that $\frac{a^{2}-b c}{2 a^{2}+b c}+\frac{b^{2}-c a}{2 b^{2}+c a}+\frac{c^{2}-a b}{2 c^{2}+a b} \leq 0$ for any real positive numbers $a, b, c$.
| ## Solution
The inequality rewrites as $\sum \frac{2 a^{2}+b c-3 b c}{2 a^{2}+b c} \leq 0$, or $3-3 \sum \frac{b c}{2 a^{2}+b c} \leq 0$ in other words $\sum \frac{b c}{2 a^{2}+b c} \geq 1$.
Using Cauchy-Schwarz inequality we have
$$
\sum \frac{b c}{2 a^{2}+b c}=\sum \frac{b^{2} c^{2}}{2 a^{2} b c+b^{2} c^{2}} \geq ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 98 |
A3 Let $A$ be a set of positive integers containing the number 1 and at least one more element. Given that for any two different elements $m, n$ of $A$ the number $\frac{m+1}{(m+1, n+1)}$ is also an element of $A$, prove that $A$ coincides with the set of positive integers.
| ## Solution
Let $a>1$ be lowest number in $A \backslash\{1\}$. For $m=a, n=1$ one gets $y=\frac{a+1}{(2, a+1)} \in A$. Since $(2, a+1)$ is either 1 or 2 , then $y=a+1$ or $y=\frac{a+1}{2}$.
But $1<\frac{a+1}{2}<a$, hence $y=a+1$. Applying the given property for $m=a+1, n=a$ one has $\frac{a+2}{(a+2, a+1)}=a+2 \in A$,... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 99 |
A4 Let $a$ and $b$ be positive integers bigger than 2. Prove that there exists a positive integer $k$ and a sequence $n_{1}, n_{2}, \ldots, n_{k}$ consisting of positive integers, such that $n_{1}=a$, $n_{k}=b$, and $\left(n_{i}+n_{i+1}\right) \mid n_{i} n_{i+1}$ for all $i=1,2, \ldots, k-1$.
| ## Solution
We write $a \Leftrightarrow b$ if the required sequence exists. It is clear that $\Leftrightarrow$ is equivalence relation, i.e. $a \Leftrightarrow a,(a \Leftrightarrow b$ implies $b \Rightarrow a)$ and $(a \Leftrightarrow b, b \Leftrightarrow c$ imply $a \Leftrightarrow c$ ).
We shall prove that for ever... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 100 |
A5 The real numbers $x, y, z, m, n$ are positive, such that $m+n \geq 2$. Prove that
$$
\begin{gathered}
x \sqrt{y z(x+m y)(x+n z)}+y \sqrt{x z(y+m x)(y+n z)}+z \sqrt{x y(z+m x)(x+n y)} \leq \\
\frac{3(m+n)}{8}(x+y)(y+z)(z+x) .
\end{gathered}
$$
| ## Solution
Using the AM-GM inequality we have
$$
\begin{aligned}
& \sqrt{y z(x+m y)(x+n z)}=\sqrt{(x z+m y z)(x y+n y z)} \leq \frac{x y+x z+(m+n) y z}{2} \\
& \sqrt{x z(y+m x)(y+n z)}=\sqrt{(y z+m x z)(x y+n x z)} \leq \frac{x y+y z+(m+n) x z}{2} \\
& \sqrt{x y(z+m x)(z+n y)}=\sqrt{(y z+m x y)(x z+n x y)} \leq \fra... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 101 |
C1 We call a tiling of an $m \times n$ rectangle with corners (see figure below) "regular" if there is no sub-rectangle which is tiled with corners. Prove that if for some $m$ and $n$ there exists a "regular" tiling of the $m \times n$ rectangular then there exists a "regular" tiling also for the $2 m \times 2 n$ rect... | ## Solution
A corner-shaped tile consists of 3 squares. Let us call "center of the tile" the square that has two neighboring squares. Notice that in a "regular" tiling, the squares situated in the corners of the rectangle have to be covered by the "center" of a tile, otherwise a $2 \times 3$ (or $3 \times 2$ ) rectang... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 102 |
C2 Consider 50 points in the plane, no three of them belonging to the same line. The points have been colored into four colors. Prove that there are at least 130 scalene triangles whose vertices are colored in the same color.
| ## Solution
Since $50=4 \cdot 12+2$, according to the pigeonhole principle we will have at least 13 points colored in the same color. We start with the:
Lemma. Given $n>8$ points in the plane, no three of them collinear, then there are at least $\frac{n(n-1)(n-8)}{6}$ scalene triangles with vertices among the given p... | 130 | Combinatorics | proof | Yes | Yes | olympiads | false | 103 |
C3 The nonnegative integer $n$ and $(2 n+1) \times(2 n+1)$ chessboard with squares colored alternatively black and white are given. For every natural number $m$ with $1<m<2 n+1$, an $m \times m$ square of the given chessboard that has more than half of its area colored in black, is called a $B$-square. If the given ch... | ## Solution
Every square with even side length will have an equal number of black and white $1 \times 1$ squares, so it isn't a $B$-square. In a square with odd side length, there is one more $1 \times 1$ black square than white squares, if it has black corner squares. So, a square with odd side length is a $B$-square... | \frac{(n+1)(2n^{2}+4n+3)}{3} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 104 |
G1 Let $M$ be an interior point of the triangle $A B C$ with angles $\varangle B A C=70^{\circ}$ and $\varangle A B C=80^{\circ}$. If $\varangle A C M=10^{\circ}$ and $\varangle C B M=20^{\circ}$, prove that $A B=M C$.
| ## Solution
Let $O$ be the circumcenter of the triangle $A B C$. Because the triangle $A B C$ is acute, $O$ is in the interior of $\triangle A B C$. Now we have that $\varangle A O C=2 \varangle A B C=160^{\circ}$, so $\varangle A C O=10^{\circ}$ and $\varangle B O C=2 \varangle B A C=140^{\circ}$, so $\varangle C B O... | proof | Geometry | proof | Yes | Yes | olympiads | false | 105 |
G2 Let $A B C D$ be a convex quadrilateral with $\varangle D A C=\varangle B D C=36^{\circ}, \varangle C B D=18^{\circ}$ and $\varangle B A C=72^{\circ}$. If $P$ is the point of intersection of the diagonals $A C$ and $B D$, find the measure of $\varangle A P D$.
| ## Solution
On the rays ( $D A$ and ( $B A$ we take points $E$ and $Z$, respectively, such that $A C=A E=$ $A Z$. Since $\varangle D E C=\frac{\varangle D A C}{2}=18^{\circ}=\varangle C B D$, the quadrilateral $D E B C$ is cyclic.
Similarly, the quadrilateral $C B Z D$ is cyclic, because $\varangle A Z C=\frac{\varan... | 108 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 106 |
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