problem stringlengths 10 5.15k | answer stringlengths 0 1.22k | solution stringlengths 0 11.1k | difficulty float64 0.75 2.02k | difficulty_raw listlengths 3 8 |
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Let $\triangle ABC$ be an acute scalene triangle with circumcircle $\omega$. The tangents to $\omega$ at $B$ and $C$ intersect at $T$. Let $X$ and $Y$ be the projections of $T$ onto lines $AB$ and $AC$, respectively. Suppose $BT = CT = 16$, $BC = 22$, and $TX^2 + TY^2 + XY^2 = 1143$. Find $XY^2$. | 717 | Using the Claim (below) we get $\triangle ABC \sim \triangle XTM \sim \triangle YMT.$
Corresponding sides of similar $\triangle XTM \sim \triangle YMT$ is $MT,$ so
$\triangle XTM = \triangle YMT \implies MY = XT, MX = TY \implies XMYT$ – parallelogram.
\[4 TD^2 = MT^2 = \sqrt{BT^2 - BM^2} =\sqrt{153}.\] The formula fo... | 6.625 | [
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Zou and Chou are practicing their $100$-meter sprints by running $6$ races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is $\frac23$ if they won the previous race but only $\frac13$ if they lost the previous race. The probability that Zou will win exactly $5$... | 97 | Note that Zou wins one race. The probability that he wins the last race is $\left(\frac{2}{3}\right)^4\left(\frac{1}{3}\right)=\frac{16}{243}.$ Now, if he doesn't win the last race, then there must be two races where the winner of the previous race loses. We can choose any $4$ of the middle races for Zou to win. So the... | 5.75 | [
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In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$, and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. [asy] p... | 109 | Let the intersection of $AE$ and $BC$ be $G$, and let $BG=x$, so $CG=11-x$.
By the Pythagorean theorem, ${AG}^2={AB}^2+{BG}^2$, so $AG=\sqrt{x^2+9}$, and thus $EG=9-\sqrt{x^2+9}$.
By the Pythagorean theorem again, ${CG}^2={EG}^2+{CE}^2$: \[11-x=\sqrt{7^2+(9-\sqrt{x^2+9})^2}.\]
Solving, we get $x=\frac{9}{4}$, so the... | 5.5 | [
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Find the number of positive integers less than $1000$ that can be expressed as the difference of two integral powers of $2.$ | 50 | We look for all positive integers of the form $2^a-2^b<1000,$ where $0\leq b<a.$ Performing casework on $a,$ we can enumerate all possibilities in the table below: \[\begin{array}{c|c} & \\ [-2.25ex] \boldsymbol{a} & \boldsymbol{b} \\ \hline & \\ [-2ex] 1 & 0 \\ 2 & 0,1 \\ 3 & 0,1,2 \\ 4 & 0,1,2,3 \\ 5 & 0,1,2,3,4 \\ 6... | 4.5 | [
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Find the number of ways $66$ identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile. | 331 | Suppose we have $1$ coin in the first pile. Then $(1, 2, 63), (1, 3, 62), \ldots, (1, 32, 33)$ all work for a total of $31$ piles. Suppose we have $2$ coins in the first pile, then $(2, 3, 61), (2, 4, 60), \ldots, (2, 31, 33)$ all work, for a total of $29$. Continuing this pattern until $21$ coins in the first pile, we... | 4.375 | [
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Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences. | 31 | Proceed as in solution 2, until we reach \[3x^2+2d^2=xd^2,\] Write
$d^2=\frac{3x^2}{x-2}$, it follows that $x-2=3k^2$ for some (positive) integer k and $k \mid x$.
Taking both sides modulo $k$, $-2 \equiv 0 \pmod{k}$, so $k \mid 2 \rightarrow k=1,2$.
When $k=1$, we have $x=5$ and $d=5$. When $k=2$, we have $x=14$ and... | 6.125 | [
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Segments $\overline{AB}, \overline{AC},$ and $\overline{AD}$ are edges of a cube and $\overline{AG}$ is a diagonal through the center of the cube. Point $P$ satisfies $BP=60\sqrt{10}$, $CP=60\sqrt{5}$, $DP=120\sqrt{2}$, and $GP=36\sqrt{7}$. Find $AP.$ | 192 | Let $E$ be the vertex of the cube such that $ABED$ is a square. By the British Flag Theorem, we can easily we can show that \[PA^2 + PE^2 = PB^2 + PD^2\] and \[PA^2 + PG^2 = PC^2 + PE^2\] Hence, adding the two equations together, we get $2PA^2 + PG^2 = PB^2 + PC^2 + PD^2$. Substituting in the values we know, we get $2P... | 7 | [
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Find the number of pairs $(m,n)$ of positive integers with $1\le m<n\le 30$ such that there exists a real number $x$ satisfying \[\sin(mx)+\sin(nx)=2.\] | 63 | We know that the range of $sin$ is between $-1$ and $1$.
Thus, the only way for the sum to be $2$ is for $sin$ of $mx$ and $nx$ to both be $1$.
The $sin$ of $(90+360k)$ is equal to 1.
Assuming $mx$ and $nm$ are both positive, m and n could be $1,5,9,13,17,21,25,29$. There are $8$ ways, so $\dbinom{8}{2}$.
If both a... | 4.5 | [
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Find the number of integers $c$ such that the equation \[\left||20|x|-x^2|-c\right|=21\]has $12$ distinct real solutions. | 57 | Graph $y=|20|x|-x^2|$ (If you are having trouble, look at the description in the next two lines and/or the diagram in Solution 1). Notice that we want this to be equal to $c-21$ and $c+21$.
We see that from left to right, the graph first dips from very positive to $0$ at $x=-20$, then rebounds up to $100$ at $x=-10$, ... | 6 | [
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Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$
Diagram
[asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, C, D, E, F, G, H; A = (-45sqrt(2)/8,... | 567 | Draw the distances in terms of $B$, as shown in the diagram. By similar triangles, $\triangle{AEC}\sim\triangle{BIC}$. As a result, let $AB=u$, then $BC=AD=\frac{6}{5}u$ and $2AC=3BC$. The triangle $ABC$ is $6-5-9$ which $\cos(\angle{ABC})=-\frac{1}{3}$. By angle subtraction, $\cos(180-\theta)=-\cos\theta$. Therefore, ... | 6.125 | [
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Consider the sequence $(a_k)_{k\ge 1}$ of positive rational numbers defined by $a_1 = \frac{2020}{2021}$ and for $k\ge 1$, if $a_k = \frac{m}{n}$ for relatively prime positive integers $m$ and $n$, then
\[a_{k+1} = \frac{m + 18}{n+19}.\]Determine the sum of all positive integers $j$ such that the rational number $a_j$ ... | 59 | We know that $a_{1}=\tfrac{t}{t+1}$ when $t=2020$ so $1$ is a possible value of $j$. Note also that $a_{2}=\tfrac{2038}{2040}=\tfrac{1019}{1020}=\tfrac{t}{t+1}$ for $t=1019$. Then $a_{2+q}=\tfrac{1019+18q}{1020+19q}$ unless $1019+18q$ and $1020+19q$ are not relatively prime which happens when $q+1$ divides $18q+1019$ o... | 6.25 | [
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Let $ABCD$ be a cyclic quadrilateral with $AB=4,BC=5,CD=6,$ and $DA=7.$ Let $A_1$ and $C_1$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to line $BD,$ and let $B_1$ and $D_1$ be the feet of the perpendiculars from $B$ and $D,$ respectively, to line $AC.$ The perimeter of $A_1B_1C_1D_1$ is $\frac mn... | 301 | The angle $\theta$ between diagonals satisfies \[\tan{\frac{\theta}{2}}=\sqrt{\frac{(s-b)(s-d)}{(s-a)(s-c)}}\] (see https://en.wikipedia.org/wiki/Cyclic_quadrilateral#Angle_formulas). Thus, \[\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-4)(11-6)}{(11-5)(11-7)}}\text{ or }\tan{\frac{\theta}{2}}=\sqrt{\frac{(11-5)(11-7)}{(11-4... | 6.875 | [
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Let $A_1A_2A_3\ldots A_{12}$ be a dodecagon ($12$-gon). Three frogs initially sit at $A_4,A_8,$ and $A_{12}$. At the end of each minute, simultaneously, each of the three frogs jumps to one of the two vertices adjacent to its current position, chosen randomly and independently with both choices being equally likely. Al... | 19 | We can solve the problem by removing $1$ frog, and calculate the expected time for the remaining $2$ frogs. In the original problem, when the movement stops, $2$ of the $3$ frogs meet. Because the $3$ frogs cannot meet at one vertex, the probability that those two specific frogs meet is $\frac13$. If the expected time ... | 7.25 | [
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Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$, respectively, intersect at distinct points $A$ and $B$. A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$. Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120... | 672 | Suppose we label the points as shown here. By radical axis, the tangents to $\omega$ at $D$ and $E$ intersect on $AB$. Thus $PDQE$ is harmonic, so the tangents to $\omega$ at $P$ and $Q$ intersect at $X \in DE$. Moreover, $OX \parallel O_1O_2$ because both $OX$ and $O_1O_2$ are perpendicular to $AB$, and $OX = 2OP$ bec... | 6.375 | [
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For any positive integer $a, \sigma(a)$ denotes the sum of the positive integer divisors of $a$. Let $n$ be the least positive integer such that $\sigma(a^n)-1$ is divisible by $2021$ for all positive integers $a$. Find the sum of the prime factors in the prime factorization of $n$. | 125 | Since the problem works for all positive integers $a$, let's plug in $a=2$ and see what we get. Since $\sigma({2^n}) = 2^{n+1}-1,$ we have $2^{n+1} \equiv 2 \pmod{2021}.$ Simplifying using CRT and Fermat's Little Theorem, we get that $2^n \equiv 0 \pmod{42}$ and $2^n \equiv 0 \pmod{46}.$ Then, we can look at $a=2022$ j... | 7.125 | [
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Let $S$ be the set of positive integers $k$ such that the two parabolas\[y=x^2-k~~\text{and}~~x=2(y-20)^2-k\]intersect in four distinct points, and these four points lie on a circle with radius at most $21$. Find the sum of the least element of $S$ and the greatest element of $S$.
Diagram
Graph in Desmos: https://www.... | 285 | Make the translation $y \rightarrow y+20$ to obtain $20+y=x^2-k$ and $x=2y^2-k$. Multiply the first equation by $2$ and sum, we see that $2(x^2+y^2)=3k+40+2y+x$. Completing the square gives us $\left(y- \frac{1}{2}\right)^2+\left(x - \frac{1}{4}\right)^2 = \frac{325+24k}{16}$; this explains why the two parabolas inters... | 7.125 | [
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Find the arithmetic mean of all the three-digit palindromes. (Recall that a palindrome is a number that reads the same forward and backward, such as $777$ or $383$.) | 550 | Recall that the arithmetic mean of all the $n$ digit palindromes is just the average of the largest and smallest $n$ digit palindromes, and in this case the $2$ palindromes are $101$ and $999$ and $\frac{101+999}{2}=\boxed{550},$ which is the final answer.
~ math31415926535 | 2.125 | [
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Equilateral triangle $ABC$ has side length $840$. Point $D$ lies on the same side of line $BC$ as $A$ such that $\overline{BD} \perp \overline{BC}$. The line $\ell$ through $D$ parallel to line $BC$ intersects sides $\overline{AB}$ and $\overline{AC}$ at points $E$ and $F$, respectively. Point $G$ lies on $\ell$ such t... | 336 | Since $\triangle AFG$ is isosceles, $AF = FG$, and since $\triangle AEF$ is equilateral, $AF = EF$. Thus, $EF = FG$, and since these triangles share an altitude, they must have the same area.
Drop perpendiculars from $E$ and $F$ to line $BC$; call the meeting points $P$ and $Q$, respectively. $\triangle BEP$ is clearl... | 6.125 | [
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Find the number of permutations $x_1, x_2, x_3, x_4, x_5$ of numbers $1, 2, 3, 4, 5$ such that the sum of five products \[x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2\] is divisible by $3$. | 80 | Since $3$ is one of the numbers, a product with a $3$ in it is automatically divisible by $3,$ so WLOG $x_3=3,$ we will multiply by $5$ afterward since any of $x_1, x_2, \ldots, x_5$ would be $3,$ after some cancelation we see that now all we need to find is the number of ways that $x_5x_1(x_4+x_2)$ is divisible by $3,... | 5.625 | [
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There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$ | 330 | We plug -20 into the equation obtaining $(-20)^3-20a+b$, likewise, plugging -21 into the second equation gets $(-21)^3+441c+d$.
Both equations must have 3 solutions exactly, so the other two solutions must be $m + \sqrt{n} \cdot i$ and $m - \sqrt{n} \cdot i$.
By Vieta's, the sum of the roots in the first equation is ... | 6.375 | [
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For positive real numbers $s$, let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$. The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$. Find $a^2+b^2$. | 736 | Note: Archimedes15 Solution which I added an answer here are two cases. Either the $4$ and $10$ are around an obtuse angle or the $4$ and $10$ are around an acute triangle. If they are around the obtuse angle, the area of that triangle is $<20$ as we have $\frac{1}{2} \cdot 40 \cdot \sin{\alpha}$ and $\sin$ is at most ... | 6.75 | [
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For any finite set $S$, let $|S|$ denote the number of elements in $S$. Find the number of ordered pairs $(A,B)$ such that $A$ and $B$ are (not necessarily distinct) subsets of $\{1,2,3,4,5\}$ that satisfy \[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\] | 454 | The answer is \begin{align*} \sum_{k=0}^{5}\left[2\binom{5}{k}2^{5-k}-\binom{5}{k}\right] &= 2\sum_{k=0}^{5}\binom{5}{k}2^{5-k}-\sum_{k=0}^{5}\binom{5}{k} \\ &=2(2+1)^5-(1+1)^5 \\ &=2(243)-32 \\ &=\boxed{454}. \end{align*} ~MRENTHUSIASM | 5.75 | [
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Let $a, b, c,$ and $d$ be real numbers that satisfy the system of equations \begin{align*} a + b &= -3, \\ ab + bc + ca &= -4, \\ abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{align*} There exist relatively prime positive integers $m$ and $n$ such that \[a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.\]Find $m + n$. | 145 | From the fourth equation we get $d=\frac{30}{abc}.$ substitute this into the third equation and you get $abc + \frac{30(ab + bc + ca)}{abc} = abc - \frac{120}{abc} = 14$. Hence $(abc)^2 - 14(abc)-120 = 0$. Solving we get $abc = -6$ or $abc = 20$. From the first and second equation we get $ab + bc + ca = ab-3c = -4 \Lon... | 6.5 | [
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An ant makes a sequence of moves on a cube where a move consists of walking from one vertex to an adjacent vertex along an edge of the cube. Initially the ant is at a vertex of the bottom face of the cube and chooses one of the three adjacent vertices to move to as its first move. For all moves after the first move, th... | 49 | Note that we don't care which exact vertex the ant is located at, just which level (either top face or bottom face). Consider the ant to be on any of the two levels and having moved at least one move. Define $p_i$ to be the probability that after $i$ moves, the ant ends up on the level it started on.
On the first move... | 6.125 | [
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Find the number of ordered pairs $(m, n)$ such that $m$ and $n$ are positive integers in the set $\{1, 2, ..., 30\}$ and the greatest common divisor of $2^m + 1$ and $2^n - 1$ is not $1$. | 295 | This solution refers to the Remarks section.
By the Euclidean Algorithm, we have \[\gcd\left(2^m+1,2^m-1\right)=\gcd\left(2,2^m-1\right)=1.\] We are given that $\gcd\left(2^m+1,2^n-1\right)>1.$ Multiplying both sides by $\gcd\left(2^m-1,2^n-1\right)$ gives \begin{align*} \gcd\left(2^m+1,2^n-1\right)\cdot\gcd\left(2^m-... | 6.25 | [
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Two spheres with radii $36$ and one sphere with radius $13$ are each externally tangent to the other two spheres and to two different planes $\mathcal{P}$ and $\mathcal{Q}$. The intersection of planes $\mathcal{P}$ and $\mathcal{Q}$ is the line $\ell$. The distance from line $\ell$ to the point where the sphere with ra... | 335 | The isosceles triangle of centers $O_1 O_2 O$ ($O$ is the center of sphere of radii $13$) has sides $O_1 O = O_2 O = 36 + 13 = 49,$ and $O_1 O_2 = 36 + 36 = 72.$
Let $N$ be the midpoint $O_1 O_2$.
The isosceles triangle of points of tangency $T_1 T_2 T$ has sides $T_1 T = T_2 T = 2 \sqrt{13 \cdot 36} = 12 \sqrt{13}$ ... | 7 | [
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A teacher was leading a class of four perfectly logical students. The teacher chose a set $S$ of four integers and gave a different number in $S$ to each student. Then the teacher announced to the class that the numbers in $S$ were four consecutive two-digit positive integers, that some number in $S$ was divisible by $... | 258 | We know right away that $42\not\in S$ and $84\not\in S$ as stated in Solution 1.
To get a feel for the problem, let’s write out some possible values of $S$ based on the teacher’s remarks. The first multiple of 7 that is two-digit is 14. The closest multiple of six from 14 is 12, and therefore there are two possible se... | 7 | [
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A convex quadrilateral has area $30$ and side lengths $5, 6, 9,$ and $7,$ in that order. Denote by $\theta$ the measure of the acute angle formed by the diagonals of the quadrilateral. Then $\tan \theta$ can be written in the form $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 47 | By Bretschneider's Formula, \[30=\tfrac{1}{4}\sqrt{4u^2v^2-(b^2+d^2-a^2-c^2)^2}=\tfrac{1}{4}\sqrt{4u^2v^2-441}.\] Thus, $uv=3\sqrt{1649}$. Also, \[[ABCD]=\tfrac 12 \cdot uv\sin{\theta};\] solving for $\sin{\theta}$ yields $\sin{\theta}=\tfrac{40}{\sqrt{1649}}$. Since $\theta$ is acute, $\cos{\theta}$ is positive, from ... | 6.25 | [
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Find the least positive integer $n$ for which $2^n + 5^n - n$ is a multiple of $1000$. | 797 | \[5^n \equiv n \pmod{8}\] Note that $5^n \equiv 5,1,5,1,...$ and $n \equiv 0,..,7$ so $n$ is periodic every $[2,8]=8$. Easy to check that only $n\equiv 5 \pmod{8}$ satisfy. Let $n=8a+5$. Note that by binomial theorem, \[2^{8a+5}=32\cdot 2^{8a} \equiv 7(1+15)^{2a} \equiv 7(1+30a)\pmod{25}\] So we have $7(1+30a) \equiv 8... | 6.875 | [
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Let $\Delta ABC$ be an acute triangle with circumcenter $O$ and centroid $G$. Let $X$ be the intersection of the line tangent to the circumcircle of $\Delta ABC$ at $A$ and the line perpendicular to $GO$ at $G$. Let $Y$ be the intersection of lines $XG$ and $BC$. Given that the measures of $\angle ABC, \angle BCA,$ and... | 592 | Notice that $\triangle ABC$ looks isosceles, so we assume it's isosceles. Then, let $\angle BAC = \angle ABC = 13x$ and $\angle BCA = 2x.$ Taking the sum of the angles in the triangle gives $28x=180,$ so $13x = \frac{13}{28} \cdot 180 = \frac{585}{7}$ so the answer is $\boxed{592}.$ | 6.5 | [
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Let $f(n)$ and $g(n)$ be functions satisfying \[f(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 1 + f(n+1) & \text{ otherwise} \end{cases}\]and \[g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 2 + g(n+2) & \text{ otherwise} \end{cases}\]for positive integers $n$.... | 258 | Since $n$ isn't a perfect square, let $n=m^2+k$ with $0<k<2m+1$. If $m$ is odd, then $f(n)=g(n)$. If $m$ is even, then \begin{align*} f(n)&=(m+1)^2-(m^2+k)+(m+1)=3m+2-k, \\ g(n)&=(m+2)^2-(m^2+k)+(m+2)=5m+6-k, \end{align*} from which \begin{align*} 7(3m+2-k)&=4(5m+6-k) \\ m&=3k+10. \end{align*} Since $m$ is even, $k$ is... | 6.75 | [
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Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$ | 116 | Let $R(x)=P(x)+Q(x).$ Since the $x^2$-terms of $P(x)$ and $Q(x)$ cancel, we conclude that $R(x)$ is a linear polynomial.
Note that \begin{alignat*}{8} R(16) &= P(16)+Q(16) &&= 54+54 &&= 108, \\ R(20) &= P(20)+Q(20) &&= 53+53 &&= 106, \end{alignat*} so the slope of $R(x)$ is $\frac{106-108}{20-16}=-\frac12.$
It follow... | 4.375 | [
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Find the three-digit positive integer $\underline{a}\,\underline{b}\,\underline{c}$ whose representation in base nine is $\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},$ where $a,$ $b,$ and $c$ are (not necessarily distinct) digits. | 227 | As shown in Solution 1, we get $99a = 71b+8c$.
We can see that $99$ is $28$ larger than $71$, and we have an $8c$. We can clearly see that $56$ is a multiple of $8$, and any larger than $56$ would result in $c$ being larger than $9$. Therefore, our only solution is $a = 2, b = 2, c = 7$. Our answer is $\underline{a}\,... | 4.125 | [
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In isosceles trapezoid $ABCD$, parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$, respectively, and $AD=BC=333$. The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$, and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$. Find $PQ$.
Diagram
[asy] /* Made by MRENTHUSI... | 242 | This will be my first solution on AoPS. My apologies in advance for any errors.
Angle bisectors can be thought of as the locus of all points equidistant from the lines whose angle they bisect. It can thus be seen that $P$ is equidistant from $AB, AD,$ and $CD$ and $Q$ is equidistant from $AB, BC,$ and $CD.$ If we let ... | 6.5 | [
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Let $w = \dfrac{\sqrt{3} + i}{2}$ and $z = \dfrac{-1 + i\sqrt{3}}{2},$ where $i = \sqrt{-1}.$ Find the number of ordered pairs $(r,s)$ of positive integers not exceeding $100$ that satisfy the equation $i \cdot w^r = z^s.$ | 834 | We rewrite $w$ and $z$ in polar form: \begin{align*} w &= e^{i\cdot\frac{\pi}{6}}, \\ z &= e^{i\cdot\frac{2\pi}{3}}. \end{align*} The equation $i \cdot w^r = z^s$ becomes \begin{align*} e^{i\cdot\frac{\pi}{2}} \cdot \left(e^{i\cdot\frac{\pi}{6}}\right)^r &= \left(e^{i\cdot\frac{2\pi}{3}}\right)^s \\ e^{i\left(\frac{\pi... | 7 | [
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A straight river that is $264$ meters wide flows from west to east at a rate of $14$ meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of $D$ meters downstream from Sherry. Relative to the water, Melanie swims at $80$ meters per minute, and Sherry swims at $60$ meters per ... | 550 | Claim
Median $AM$ and altitude $AH$ are drawn in triangle $ABC$. $AB = c, AC = b < c, BC = a$ are known. Let's denote $MH = x$.
Prove that \begin{align*}2ax = c^{2} – b^{2}\end{align*}
Proof \[BH + CH = a,\] \begin{align*} BH^{2} – CH^{2} = c^{2} – b^{2}\implies BH - CH &= \frac{c^{2} – b^{2}} {a},\end{align*} \[BH =... | 6.625 | [
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Find the number of ordered pairs of integers $(a, b)$ such that the sequence\[3, 4, 5, a, b, 30, 40, 50\]is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression. | 228 | divide cases into $7\leq a<20; 21\leq a\leq28$.(Notice that $a$ can't be equal to $6,20$, that's why I divide them into two parts. There are three cases that arithmetic sequence forms: $3,12,21,30;4,16,28,40;3,5,7,9$.(NOTICE that $5,20,35,50$ IS NOT A VALID SEQUENCE!) So when $7\leq a<20$, there are $10+11+12+...+22-3-... | 6.125 | [
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Let $a,b,c,d,e,f,g,h,i$ be distinct integers from $1$ to $9.$ The minimum possible positive value of \[\dfrac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}\] can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | 289 | To minimize a positive fraction, we minimize its numerator and maximize its denominator. It is clear that $\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i} \geq \frac{1}{7\cdot8\cdot9}.$
If we minimize the numerator, then $a \cdot b \cdot c - d \cdot e \cdot f = 1.$ Note that $a \cdot b \cdot c \cdot d ... | 5.625 | [
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Equilateral triangle $\triangle ABC$ is inscribed in circle $\omega$ with radius $18.$ Circle $\omega_A$ is tangent to sides $\overline{AB}$ and $\overline{AC}$ and is internally tangent to $\omega.$ Circles $\omega_B$ and $\omega_C$ are defined analogously. Circles $\omega_A,$ $\omega_B,$ and $\omega_C$ meet in six po... | 378 | Let $O$ be the center, $R = 18$ be the radius, and $CC'$ be the diameter of $\omega.$ Let $r$ be the radius, $E,D,F$ are the centers of $\omega_A, \omega_B,\omega_C.$ Let $KGH$ be the desired triangle with side $x.$ We find $r$ using \[CC' = 2R = C'K + KC = r + \frac{r}{\sin 30^\circ} = 3r.\] \[r = \frac{2R}{3} = 12.\]... | 7 | [
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Ellina has twelve blocks, two each of red ($\textbf{R}$), blue ($\textbf{B}$), yellow ($\textbf{Y}$), green ($\textbf{G}$), orange ($\textbf{O}$), and purple ($\textbf{P}$). Call an arrangement of blocks $\textit{even}$ if there is an even number of blocks between each pair of blocks of the same color. For example, the... | 247 | We can simply use constructive counting. First, let us place the red balls; choose the first slot in $12$ ways, and the second in $6$ ways, because the number is cut in half due to the condition in the problem. This gives $12 \cdot 6$ ways to place the blue balls. Similarly, there are $10 \cdot 5$ ways to place the blu... | 5.625 | [
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Three spheres with radii $11$, $13$, and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A$, $B$, and $C$, respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560$. Find $AC^2$.
Diagrams
[asy] size(500); p... | 756 | Let the distance between the center of the sphere to the center of those circular intersections as $a,b,c$ separately. $a-11,b-13,c-19$. According to the problem, we have $a^2-11^2=b^2-13^2=c^2-19^2;(11+13)^2-(b-a)^2=560$. After solving we have $b-a=4$, plug this back to $11^2-a^2=13^2-b^2; a=4,b=8,c=16$
The desired v... | 6.625 | [
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Let $ABCD$ be a parallelogram with $\angle BAD < 90^{\circ}$. A circle tangent to sides $\overline{DA}$, $\overline{AB}$, and $\overline{BC}$ intersects diagonal $\overline{AC}$ at points $P$ and $Q$ with $AP < AQ$, as shown. Suppose that $AP = 3$, $PQ = 9$, and $QC = 16$. Then the area of $ABCD$ can be expressed in th... | 150 | Let $\omega$ be the circle, let $r$ be the radius of $\omega$, and let the points at which $\omega$ is tangent to $AB$, $BC$, and $AD$ be $H$, $K$, and $T$, respectively. PoP on $A$ and $C$ with respect to $\omega$ yields \[AT=6, CK=20.\]
Let $TG = AC, CG||AT.$
In $\triangle KGT$ $KT \perp BC,$ $KT = \sqrt{GT^2 – (KC... | 7 | [
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For any finite set $X$, let $| X |$ denote the number of elements in $X$. Define \[ S_n = \sum | A \cap B | , \] where the sum is taken over all ordered pairs $(A, B)$ such that $A$ and $B$ are subsets of $\left\{ 1 , 2 , 3, \cdots , n \right\}$ with $|A| = |B|$. For example, $S_2 = 4$ because the sum is taken over the... | 245 | We take cases based on the number of values in each of the subsets in the pair. Suppose we have $k$ elements in each of the subsets in a pair (for a total of n elements in the set). The expected number of elements in any random pair will be $n \cdot \frac{k}{n} \cdot \frac{k}{n}$ by linearity of expectation because for... | 7.25 | [
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Let $S$ be the set of all rational numbers that can be expressed as a repeating decimal in the form $0.\overline{abcd},$ where at least one of the digits $a,$ $b,$ $c,$ or $d$ is nonzero. Let $N$ be the number of distinct numerators obtained when numbers in $S$ are written as fractions in lowest terms. For example, bot... | 392 | $0.abcd=\frac{\overline{abcd}}{9999}$, $9999=9\times 11\times 101$.
Then we need to find the number of positive integers less than $10000$ that can meet the requirement. Suppose the number is $x$.
Case $1$: $(9999, x)=1$. Clearly $x$ satisfies. \[\varphi \left( 9999 \right) =9999\times \left( 1-\frac{1}{3} \right) \t... | 6.875 | [
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Given $\triangle ABC$ and a point $P$ on one of its sides, call line $\ell$ the $\textit{splitting line}$ of $\triangle ABC$ through $P$ if $\ell$ passes through $P$ and divides $\triangle ABC$ into two polygons of equal perimeter. Let $\triangle ABC$ be a triangle where $BC = 219$ and $AB$ and $AC$ are positive intege... | 459 | We wish to solve the Diophantine equation $a^2+ab+b^2=3^2 \cdot 73^2$. It can be shown that $3|a$ and $3|b$, so we make the substitution $a=3x$ and $b=3y$ to obtain $x^2+xy+y^2=73^2$ as our new equation to solve for.
Notice that $r^2+r+1=(r-\omega)(r-{\omega}^2)$, where $\omega=e^{i\frac{2\pi}{3}}$. Thus, \[x^2+xy+y^2... | 6.625 | [
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Let $x,$ $y,$ and $z$ be positive real numbers satisfying the system of equations: \begin{align*} \sqrt{2x-xy} + \sqrt{2y-xy} &= 1 \\ \sqrt{2y-yz} + \sqrt{2z-yz} &= \sqrt2 \\ \sqrt{2z-zx} + \sqrt{2x-zx} &= \sqrt3. \end{align*} Then $\left[ (1-x)(1-y)(1-z) \right]^2$ can be written as $\frac{m}{n},$ where $m$ and $n$ ar... | 33 | Let $1-x=a;1-y=b;1-z=c$, rewrite those equations
$\sqrt{(1-a)(1+b)}+\sqrt{(1+a)(1-b)}=1$;
$\sqrt{(1-b)(1+c)}+\sqrt{(1+b)(1-c)}=\sqrt{2}$
$\sqrt{(1-a)(1+c)}+\sqrt{(1-c)(1+a)}=\sqrt{3}$
square both sides, get three equations:
$2ab-1=2\sqrt{(1-a^2)(1-b^2)}$
$2bc=2\sqrt{(1-b^2)(1-c^2)}$
$2ac+1=2\sqrt{(1-c^2)(1-a^2)}$
Get... | 6.875 | [
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Adults made up $\frac5{12}$ of the crowd of people at a concert. After a bus carrying $50$ more people arrived, adults made up $\frac{11}{25}$ of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived. | 154 | Let $x$ be the number of people at the party before the bus arrives. We know that $x\equiv 0\pmod {12}$, as $\frac{5}{12}$ of people at the party before the bus arrives are adults. Similarly, we know that $x + 50 \equiv 0 \pmod{25}$, as $\frac{11}{25}$ of the people at the party are adults after the bus arrives. $x + 5... | 5.125 | [
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Azar, Carl, Jon, and Sergey are the four players left in a singles tennis tournament. They are randomly assigned opponents in the semifinal matches, and the winners of those matches play each other in the final match to determine the winner of the tournament. When Azar plays Carl, Azar will win the match with probabili... | 125 | Let $A$ be Azar, $C$ be Carl, $J$ be Jon, and $S$ be Sergey. The $4$ circles represent the $4$ players, and the arrow is from the winner to the loser with the winning probability as the label.
This problem can be solved by using $2$ cases.
$\textbf{Case 1:}$ $C$'s opponent for the semifinal is $A$
The probability $C$... | 6.75 | [
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A right square pyramid with volume $54$ has a base with side length $6.$ The five vertices of the pyramid all lie on a sphere with radius $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 21 | Although I can't draw the exact picture of this problem, but it is quite easy to imagine that four vertices of the base of this pyramid is on a circle (Radius $\frac{6}{\sqrt{2}} = 3\sqrt{2}$). Since all five vertices are on the sphere, the distances of the spherical center and the vertices are the same: $l$. Because o... | 5.375 | [
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There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that\[\log_{20x} (22x)=\log_{2x} (202x).\]The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 112 | By the change of base rule, we have $\frac{\log 22x}{\log 20x}=\frac{\log 202x}{\log 2x}$, or $\frac{\log 22 +\log x}{\log 20 +\log x}=\frac{\log 202 +\log x}{\log 2 +\log x}=k$. We also know that if $a/b=c/d$, then this also equals $\frac{a-b}{c-d}$. We use this identity and find that $k=\frac{\log 202 -\log 22}{\log ... | 5.875 | [
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Twenty distinct points are marked on a circle and labeled $1$ through $20$ in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original $20$ points. | 72 | Let $a$, $b$, and $c$ be the vertex of a triangle that satisfies this problem, where $a > b > c$. \[a - b = p_1\] \[b - c = p_2\] \[a - c = p_3\]
$p_3 = a - c = a - b + b - c = p_1 + p_2$. Because $p_3$ is the sum of two primes, $p_1$ and $p_2$, $p_1$ or $p_2$ must be $2$. Let $p_1 = 2$, then $p_3 = p_2 + 2$. There are... | 6.625 | [
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Let $x_1\leq x_2\leq \cdots\leq x_{100}$ be real numbers such that $|x_1| + |x_2| + \cdots + |x_{100}| = 1$ and $x_1 + x_2 + \cdots + x_{100} = 0$. Among all such $100$-tuples of numbers, the greatest value that $x_{76} - x_{16}$ can achieve is $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find... | 841 | To find the greatest value of $x_{76} - x_{16}$, $x_{76}$ must be as large as possible, and $x_{16}$ must be as small as possible. If $x_{76}$ is as large as possible, $x_{76} = x_{77} = x_{78} = \dots = x_{100} > 0$. If $x_{16}$ is as small as possible, $x_{16} = x_{15} = x_{14} = \dots = x_{1} < 0$. The other numbers... | 6.25 | [
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A circle with radius $6$ is externally tangent to a circle with radius $24$. Find the area of the triangular region bounded by the three common tangent lines of these two circles. | 192 | [asy] //Created by isabelchen size(12cm, 12cm); draw(circle((0,0),24)); draw(circle((30,0),6)); draw((72/5, 96/5) -- (40,0)); draw((72/5, -96/5) -- (40,0)); draw((24, 12) -- (24, -12)); draw((0, 0) -- (40, 0)); draw((72/5, 96/5) -- (0,0)); draw((168/5, 24/5) -- (30,0)); draw((54/5, 72/5) -- (30,0)); dot((72/5, 96/5)); ... | 6.25 | [
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Find the number of positive integers $n \le 600$ whose value can be uniquely determined when the values of $\left\lfloor \frac n4\right\rfloor$, $\left\lfloor\frac n5\right\rfloor$, and $\left\lfloor\frac n6\right\rfloor$ are given, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to the real n... | 80 | 1. For $n$ to be uniquely determined, $n$ AND $n + 1$ both need to be a multiple of $4, 5,$ or $6.$ Since either $n$ or $n + 1$ is odd, we know that either $n$ or $n + 1$ has to be a multiple of $5.$ We can state the following cases:
1. $n$ is a multiple of $4$ and $n+1$ is a multiple of $5$
2. $n$ is a multiple of $... | 5.625 | [
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Let $\ell_A$ and $\ell_B$ be two distinct perpendicular lines. For positive integers $m$ and $n$, distinct points $A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m$ lie on $\ell_A$, and distinct points $B_1, B_2, B_3, \ldots, B_n$ lie on $\ell_B$. Additionally, when segments $\overline{A_iB_j}$ are drawn ... | 244 | We want to derive a general function $f(m,n)$ that indicates the number of bounded regions. Observing symmetry, we know this is a symmetric function about $m$ and $n$. Now let's focus on $f(m+1, n)-f(m, n)$, which is the difference caused by adding one point to the existing $m$ points of line $\ell_A$. This new point, ... | 6.375 | [
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Find the remainder when\[\binom{\binom{3}{2}}{2} + \binom{\binom{4}{2}}{2} + \dots + \binom{\binom{40}{2}}{2}\]is divided by $1000$.
~ pi_is_3.14 | 4 | Doing simple algebra calculation will give the following equation: \begin{align*} \binom{\binom{n}{2}}{2} = \frac{\frac{n(n-1)}{2} \cdot (\frac{n(n-1)}{2}-1)}{2} \\ = \frac{n(n-1)(n^2-n-2)}{8} \\ = \frac{(n+1)n(n-1)(n-2)}{8} \\ = \frac{(n+1)!}{8\cdot (n-3)!} = 3 \cdot \frac{(n+1)!}{4!\cdot (n-3)!} \\ = 3 \binom{n+1}{4}... | 6.375 | [
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Let $ABCD$ be a convex quadrilateral with $AB=2$, $AD=7$, and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}$. Find the square of the area of $ABCD$. | 180 | 2022 AIME II Q11(Hand-drawn picture)
According to the problem, we have $AB=AB'=2$, $DC=DC'=3$, $MB=MB'$, $MC=MC'$, and $B'C'=7-2-3=2$
Because $M$ is the midpoint of $BC$, we have $BM=MC$, so: \[MB=MB'=MC'=MC.\]
Then, we can see that $\bigtriangleup{MB'C'}$ is an isosceles triangle with $MB'=MC'$
Therefore, we could... | 6.375 | [
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Let $a, b, x,$ and $y$ be real numbers with $a>4$ and $b>1$ such that\[\frac{x^2}{a^2}+\frac{y^2}{a^2-16}=\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2}=1.\]Find the least possible value of $a+b.$ | 23 | Denote $P = \left( x , y \right)$.
Because $\frac{x^2}{a^2}+\frac{y^2}{a^2-16} = 1$, $P$ is on an ellipse whose center is $\left( 0 , 0 \right)$ and foci are $\left( - 4 , 0 \right)$ and $\left( 4 , 0 \right)$.
Hence, the sum of distance from $P$ to $\left( - 4 , 0 \right)$ and $\left( 4 , 0 \right)$ is equal to twic... | 6.625 | [
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There is a polynomial $P(x)$ with integer coefficients such that\[P(x)=\frac{(x^{2310}-1)^6}{(x^{105}-1)(x^{70}-1)(x^{42}-1)(x^{30}-1)}\]holds for every $0<x<1.$ Find the coefficient of $x^{2022}$ in $P(x)$. | 220 | Note that $2022 = 210\cdot 9 +132$. Since the only way to express $132$ in terms of $105$, $70$, $42$, or $30$ is $135 = 30+30+30+42$, we are essentially just counting the number of ways to express $210*9$ in terms of these numbers. Since $210 = 2*105=3*70=5*42=7*30$, it can only be expressed as a sum in terms of only ... | 6.875 | [
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For positive integers $a$, $b$, and $c$ with $a < b < c$, consider collections of postage stamps in denominations $a$, $b$, and $c$ cents that contain at least one stamp of each denomination. If there exists such a collection that contains sub-collections worth every whole number of cents up to $1000$ cents, let $f(a, ... | 188 | Notice that we must have $a = 1$, otherwise $1$ cent stamp cannot be represented. At least $b-1$ numbers of $1$ cent stamps are needed to represent the values less than $b$. Using at most $c-1$ stamps of value $1$ and $b$, it can have all the values from $1$ to $c-1$ cents. Plus $\lfloor \frac{999}{c} \rfloor$ stamps o... | 6.75 | [
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Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2$, respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D$, as shown. Suppose that $AB = 2$, $O_1O_2 = 15$, $CD = 16$, and $ABO_1CDO_2$ is a convex hexagon... | 140 | Let points $A'$ and $B'$ be the reflections of $A$ and $B,$ respectively, about the perpendicular bisector of $O_1 O_2.$ \[B'O_2 = BO_1 = O_1 P = O_1 C,\] \[A'O_1 = AO_2 = O_2 P = O_2 D.\] We establish the equality of the arcs and conclude that the corresponding chords are equal \[\overset{\Large\frown} {CO_1} + \overs... | 7.75 | [
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Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | 191 | We get around the condition that each man can't be opposite to another man by simply considering all $7$ diagonals, and choosing $5$ where there will be a single man. For each diagonal, the man can go on either side, and there are $\binom{14}{5}$ ways to arrange the men and the women in total. Thus our answer is $\frac... | 5.375 | [
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Positive real numbers $b \not= 1$ and $n$ satisfy the equations \[\sqrt{\log_b n} = \log_b \sqrt{n} \qquad \text{and} \qquad b \cdot \log_b n = \log_b (bn).\] The value of $n$ is $\frac{j}{k},$ where $j$ and $k$ are relatively prime positive integers. Find $j+k.$ | 881 | Denote $x = \log_b n$. Hence, the system of equations given in the problem can be rewritten as \begin{align*} \sqrt{x} & = \frac{1}{2} x , \\ bx & = 1 + x . \end{align*} Solving the system gives $x = 4$ and $b = \frac{5}{4}$. Therefore, \[n = b^x = \frac{625}{256}.\] Therefore, the answer is $625 + 256 = \boxed{881}$.
... | 5.5 | [
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A plane contains $40$ lines, no $2$ of which are parallel. Suppose that there are $3$ points where exactly $3$ lines intersect, $4$ points where exactly $4$ lines intersect, $5$ points where exactly $5$ lines intersect, $6$ points where exactly $6$ lines intersect, and no points where more than $6$ lines intersect. Fin... | 607 | In this solution, let $\boldsymbol{n}$-line points be the points where exactly $n$ lines intersect. We wish to find the number of $2$-line points.
There are $\binom{40}{2}=780$ pairs of lines. Among them:
The $3$-line points account for $3\cdot\binom32=9$ pairs of lines.
The $4$-line points account for $4\cdot\binom42... | 6.125 | [
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The sum of all positive integers $m$ such that $\frac{13!}{m}$ is a perfect square can be written as $2^a3^b5^c7^d11^e13^f,$ where $a,b,c,d,e,$ and $f$ are positive integers. Find $a+b+c+d+e+f.$ | 012 | We first rewrite $13!$ as a prime factorization, which is $2^{10}\cdot3^5\cdot5^2\cdot7\cdot11\cdot13.$
For the fraction to be a square, it needs each prime to be an even power. This means $m$ must contain $7\cdot11\cdot13$. Also, $m$ can contain any even power of $2$ up to $2^{10}$, any odd power of $3$ up to $3^{5}$... | 6.625 | [
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Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$ | 106 | WLOG, let $P$ be on minor arc $\overarc {AB}$. Let $r$ and $O$ be the radius and center of the circumcircle respectively, and let $\theta = \angle AOP$.
By the Pythagorean Theorem, the area of the square is $2r^2$. We can use the Law of Cosines on isosceles triangles $\triangle AOP, \, \triangle COP, \, \triangle BOP,... | 6.125 | [
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Alice knows that $3$ red cards and $3$ black cards will be revealed to her one at a time in random order. Before each card is revealed, Alice must guess its color. If Alice plays optimally, the expected number of cards she will guess correctly is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. ... | 051 | Denote by $N_{i,j}$ the optimal expected number of cards that Alice guesses correctly, where the number of cards are $i$ and $j \ge i.$
If $i = 0$ then Alice guesses correctly all cards, so $N_{0,j} = j.$
If $j = i$ then Alice guesses next card with probability $\frac {1}{2} \implies N_{i,i} = \frac {1}{2} + N_{i-1,i... | 6.5 | [
7,
6,
7,
6,
6,
7,
7,
6
] |
Call a positive integer $n$ extra-distinct if the remainders when $n$ is divided by $2, 3, 4, 5,$ and $6$ are distinct. Find the number of extra-distinct positive integers less than $1000$. | 049 | Because the LCM of all of the numbers we are dividing by is $60$, we know that all of the remainders are $0$ again at $60$, meaning that we have a cycle that repeats itself every $60$ numbers.
After listing all of the remainders up to $60$, we find that $35$, $58$, and $59$ are extra-distinct. So, we have $3$ numbers ... | 4.875 | [
4,
4,
4,
5,
5,
6,
6,
5
] |
Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$
Diagram
[asy] /* Made by MRENTHUSIASM; inspired by Math Jams. */ size(300); pair A, B, C, D,... | 125 | Notation is shown on diagram, $RT \perp AD, FG \perp AB, E = AD \cap \omega, E' = FG \cap AD.$ $RT = 9 + 16 = 25 = FG$ as hights of rhombus. \[RP = QT = 9, PQ = 16 - 9 = 7, GE' = PF = 5,\] \[PE' = 25 - 5 - 5 = 15, RE = \sqrt{RP \cdot RQ} = \sqrt{9 \cdot 16} = 12.\] \[PE = \sqrt{RP^2 + RE^2} = 15 \implies E = E'.\] \[\s... | 6.5 | [
6,
7,
6,
7,
7,
6,
6,
7
] |
Find the number of cubic polynomials $p(x) = x^3 + ax^2 + bx + c,$ where $a, b,$ and $c$ are integers in $\{-20,-19,-18,\ldots,18,19,20\},$ such that there is a unique integer $m \not= 2$ with $p(m) = p(2).$ | 738 | $p(x)-p(2)$ is a cubic with at least two integral real roots, therefore it has three real roots, which are all integers.
There are exactly two distinct roots, so either $p(x)=p(2)+(x-2)^2(x-m)$ or $p(x)=p(2)+(x-2)(x-m)^2$, with $m\neq 2$.
In the first case $p(x)=x^3-(4+m)x^2+(4+4m)x-4m+p(2)$, with $|4+4m|\leq 20$ (wh... | 7 | [
7,
7,
7,
7,
7,
7,
7,
7
] |
There exists a unique positive integer $a$ for which the sum \[U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor\] is an integer strictly between $-1000$ and $1000$. For that unique $a$, find $a+U$.
(Note that $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$.) | 944 | Consider the integral \[\int_{0}^{2023} \dfrac{n^2-na}{5} \, dn.\] We hope this will give a good enough appoximation of $U$ to find $a.$ However, this integral can be easily evaluated(if you know calculus) to be \[\dfrac{1}{15}2023^3-\dfrac{a}{10}2023^2=2023^2\left(\dfrac{2023}{15}-\dfrac{a}{10}\right).\] Because we wa... | 7.125 | [
8,
7,
7,
7,
7,
7,
7,
7
] |
Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$ | 235 | The problem is the same as laying out a line of polynomoes to cover spots $0,1,...10$: 1 triomino ($RGG$), $n$ dominoes ($RG$), and $8-2n$ monominoes ($R$). The $G$ spots cover the members of the subset. The total number spots is 11, because one $R$ spot always covers the 0, and the other spots cover 1 through 10.
The... | 5.125 | [
4,
5,
5,
5,
5,
6,
5,
6
] |
Let $\triangle ABC$ be an equilateral triangle with side length $55.$ Points $D,$ $E,$ and $F$ lie on $\overline{BC},$ $\overline{CA},$ and $\overline{AB},$ respectively, with $BD = 7,$ $CE=30,$ and $AF=40.$ Point $P$ inside $\triangle ABC$ has the property that \[\angle AEP = \angle BFP = \angle CDP.\] Find $\tan^2(\a... | 075 | We begin by using the fact stated in Solution 3 that, for any point in an equilateral triangle, the lengths of the three perpendicular lines dropped to the sides of the triangle add up to the altitude of that triangle. To make things simple, let's assign $\angle AEP = \angle BFP = \angle CDP = \alpha$. We can label the... | 7 | [
7,
6,
8,
7,
8,
6,
7,
7
] |
Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths $\sqrt{21}$ and $\sqrt{31}$. The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. A parallelepiped is a sol... | 125 | Let us inscribe a tetrahedron $ACB'D'$ in given parallelepiped so that its edges coincide with the diagonals of the faces of the parallelepiped. Note that the three edges outgoing from the vertex $B'$ have the same length $b$, and the three edges at the base have a different length $a.$
The volume of the tetrahedron $... | 7.125 | [
7,
7,
7,
7,
6,
8,
8,
7
] |
The following analog clock has two hands that can move independently of each other. [asy] unitsize(2cm); draw(unitcircle,black+linewidth(2)); for (int i = 0; i < 12; ++i) { draw(0.9*dir(30*i)--dir(30*i)); } for (int i = 0; i < 4; ++i) { draw(0.85*dir(90*i)--dir(90*i),black+linewidth(2)); } for (int i = 1; i < 13; ++i) ... | 608 | This is more of a solution sketch and lacks rigorous proof for interim steps, but illustrates some key observations that lead to a simple solution.
Note that one can visualize this problem as walking on a $N \times N$ grid where the edges warp. Your goal is to have a single path across all nodes on the grid leading ba... | 7.25 | [
8,
7,
7,
7,
7,
7,
8,
7
] |
Find the largest prime number $p<1000$ for which there exists a complex number $z$ satisfying
the real and imaginary part of $z$ are both integers;
$|z|=\sqrt{p},$ and
there exists a triangle whose three side lengths are $p,$ the real part of $z^{3},$ and the imaginary part of $z^{3}.$ | 349 | Assume that $z=a+bi$. Then, \[z^3=(a^3-3ab^2)+(3a^2b-b^3)i\]Note that by the Triangle Inequality, \[|(a^3-3ab^2)-(3a^2b-b^3)|<p\implies |a^3+b^3-3ab^2-3a^2b|<a^2+b^2\]Thus, we know \[|a+b||a^2+b^2-4ab|<a^2+b^2\]Without loss of generality, assume $a>b$ (as otherwise, consider $i^3\overline z=b+ai$). If $|a/b|\geq 4$, th... | 7.25 | [
7,
7,
7,
8,
8,
7,
7,
7
] |
The numbers of apples growing on each of six apple trees form an arithmetic sequence where the greatest number of apples growing on any of the six trees is double the least number of apples growing on any of the six trees. The total number of apples growing on all six trees is $990.$ Find the greatest number of apples ... | 220 | Let the terms in the sequence be defined as \[a_1, a_2, ..., a_6.\]
Since this is an arithmetic sequence, we have $a_1+a_6=a_2+a_5=a_3+a_4.$ So, \[\sum_{i=1}^6 a_i=3(a_1+a_6)=990.\] Hence, $(a_1+a_6)=330.$ And, since we are given that $a_6=2a_1,$ we get $3a_1=330\implies a_1=110$ and $a_6=\boxed{220}.$
~Kiran | 3.75 | [
4,
4,
4,
3,
3,
3,
4,
5
] |
Recall that a palindrome is a number that reads the same forward and backward. Find the greatest integer less than $1000$ that is a palindrome both when written in base ten and when written in base eight, such as $292 = 444_{\text{eight}}.$ | 585 | Assuming that such palindrome is greater than $777_8 = 511,$ we conclude that the palindrome has four digits when written in base $8.$ Let such palindrome be \[(\underline{ABBA})_8 = 512A + 64B + 8B + A = 513A + 72B.\]
It is clear that $A=1,$ so we repeatedly add $72$ to $513$ until we get palindromes less than $1000:... | 5.625 | [
6,
6,
6,
5,
6,
6,
5,
5
] |
Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$
Diagram
[asy] /* Made by MRENTHUSIASM */ size(200); pair A, B, C, P; A = origin; B = (0,10*sqrt(5)); ... | 250 | Denote the area of $X$ by $[X].$ As in previous solutions, we see that $\angle APC = 90 ^\circ, \triangle BPC \sim \triangle APB$ with ratio $k = \sqrt{2}\implies$ \[\frac {PC}{PB} = \frac {PB}{PA} = k \implies PC = k^2 \cdot AP = 20 \implies [APC] = \frac {AP \cdot PC}{2} = 100.\] \[[BPC] = k^2 [APB] = 2 [APB].\] \[AB... | 6.75 | [
7,
7,
7,
7,
6,
7,
7,
6
] |
Let $x,y,$ and $z$ be real numbers satisfying the system of equations \begin{align*} xy + 4z &= 60 \\ yz + 4x &= 60 \\ zx + 4y &= 60. \end{align*} Let $S$ be the set of possible values of $x.$ Find the sum of the squares of the elements of $S.$ | 273 | We first subtract the second equation from the first, noting that they both equal $60$. \begin{align*} xy+4z-yz-4x&=0 \\ 4(z-x)-y(z-x)&=0 \\ (z-x)(4-y)&=0 \end{align*}
Case 1: Let $y=4$.
The first and third equations simplify to: \begin{align*} x+z&=15 \\ xz&=44 \end{align*} from which it is apparent that $x=4$ and $... | 5.625 | [
5,
6,
6,
5,
6,
5,
5,
7
] |
Let $S$ be the set of all positive rational numbers $r$ such that when the two numbers $r$ and $55r$ are written as fractions in lowest terms, the sum of the numerator and denominator of one fraction is the same as the sum of the numerator and denominator of the other fraction. The sum of all the elements of $S$ can be... | 719 | Denote $r = \frac{a}{b}$, where $\left( a, b \right) = 1$. We have $55 r = \frac{55a}{b}$. Suppose $\left( 55, b \right) = 1$, then the sum of the numerator and the denominator of $55r$ is $55a + b$. This cannot be equal to the sum of the numerator and the denominator of $r$, $a + b$. Therefore, $\left( 55, b \right) \... | 6.875 | [
7,
7,
7,
6,
7,
7,
7,
7
] |
Consider the L-shaped region formed by three unit squares joined at their sides, as shown below. Two points $A$ and $B$ are chosen independently and uniformly at random from inside the region. The probability that the midpoint of $\overline{AB}$ also lies inside this L-shaped region can be expressed as $\frac{m}{n},$ w... | 035 | Consider this diagram:
First, the one of points must be in the uppermost box and the other in the rightmost box. This happens with probability 2/3*1/3=2/9.
We need the midpoints of the $x$ coordinates to be greater than $1$ but less than $2.$
We need the midpoints of the $y$ coordinates to be greater than $1$ but le... | 5 | [
5,
6,
4,
5,
4,
5,
6,
5
] |
Each vertex of a regular dodecagon ($12$-gon) is to be colored either red or blue, and thus there are $2^{12}$ possible colorings. Find the number of these colorings with the property that no four vertices colored the same color are the four vertices of a rectangle. | 928 | Note that the condition is equivalent to stating that there are no 2 pairs of oppositely spaced vertices with the same color.
Case 1: There are no pairs. This yields $2$ options for each vertices 1-6, and the remaining vertices 7-12 are set, yielding $2^6=64$ cases.
Case 2: There is one pair. Again start with 2 optio... | 6.375 | [
6,
6,
6,
7,
6,
6,
7,
7
] |
Let $\omega = \cos\frac{2\pi}{7} + i \cdot \sin\frac{2\pi}{7},$ where $i = \sqrt{-1}.$ Find the value of the product\[\prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right).\] | 024 | The product can be factored into $-(r-1)(s-1)(t-1)(r-w)(s-w)(t-w)(r-w^2)(s-w^2)(t-w^2)....(r-w^6)(s-w^6)(t-w^6)$,
where $r,s,t$ are the roots of the polynomial $x^3+x+1=0$.
This is then $-(r^7-1)(s^7-1)(t^7-1)$ because $(r^7-1)$ and $(r-1)(r-w)(r-w^2)...(r-w^6)$ share the same roots.
To find $-(r^7-1)(s^7-1)(t^7-1)$... | 7.875 | [
8,
8,
8,
8,
8,
8,
7,
8
] |
Circles $\omega_1$ and $\omega_2$ intersect at two points $P$ and $Q,$ and their common tangent line closer to $P$ intersects $\omega_1$ and $\omega_2$ at points $A$ and $B,$ respectively. The line parallel to $AB$ that passes through $P$ intersects $\omega_1$ and $\omega_2$ for the second time at points $X$ and $Y,$ r... | 033 | Notice that line $\overline{PQ}$ is the radical axis of circles $\omega_1$ and $\omega_2$. By the radical axis theorem, we know that the tangents of any point on line $\overline{PQ}$ to circles $\omega_1$ and $\omega_2$ are equal. Therefore, line $\overline{PQ}$ must pass through the midpoint of $\overline{AB}$, call t... | 6.625 | [
7,
6,
7,
6,
6,
7,
7,
7
] |
Let $N$ be the number of ways to place the integers $1$ through $12$ in the $12$ cells of a $2 \times 6$ grid so that for any two cells sharing a side, the difference between the numbers in those cells is not divisible by $3.$ One way to do this is shown below. Find the number of positive integer divisors of $N.$ \[\be... | 144 | We replace the numbers which are 0 mod(3) to 0, 1 mod(3) to 1, and 2 mod(3) to 2.
Then, the problem is equivalent to arranging 4 0's,4 1's, and 4 2's into the grid (and then multiplying by $4!^3$ to account for replacing the remainder numbers with actual numbers) such that no 2 of the same numbers are adjacent.
Then,... | 6.75 | [
7,
7,
6,
7,
6,
7,
7,
7
] |
Find the number of collections of $16$ distinct subsets of $\{1,2,3,4,5\}$ with the property that for any two subsets $X$ and $Y$ in the collection, $X \cap Y \not= \emptyset.$ | 081 | Denote the $A$ as $\{ 1,2,3,4,5 \}$ and the collection of subsets as $S$.
Case 1: There are only sets of size $3$ or higher in $S$: Any two sets in $S$ must have at least one element common to both of them (since $3+3>5$). Since there are $16$ subsets of $A$ that have size $3$ or higher, there is only one possibility f... | 6.375 | [
6,
7,
7,
6,
7,
6,
6,
6
] |
In $\triangle ABC$ with side lengths $AB = 13,$ $BC = 14,$ and $CA = 15,$ let $M$ be the midpoint of $\overline{BC}.$ Let $P$ be the point on the circumcircle of $\triangle ABC$ such that $M$ is on $\overline{AP}.$ There exists a unique point $Q$ on segment $\overline{AM}$ such that $\angle PBQ = \angle PCQ.$ Then $AQ$... | 247 | We use the law of Cosine and get \[AB^2 = AM^2 + BM^2 - 2 AM \cdot BM \cos \angle AMB,\] \[AC^2 = AM^2 + CM^2 + 2 AM \cdot CM \cos \angle AMB \implies\] \[AM^2 = \frac {AB^2 + AC^2}{2}- BM^2 = \sqrt{148} \approx 12.\] We use the power of point $M$ with respect circumcircle $\triangle ABC$ and get \[AM \cdot MP = BM \cd... | 6.625 | [
7,
7,
6,
7,
7,
6,
7,
6
] |
Let $A$ be an acute angle such that $\tan A = 2 \cos A.$ Find the number of positive integers $n$ less than or equal to $1000$ such that $\sec^n A + \tan^n A$ is a positive integer whose units digit is $9.$ | 167 | \[\tan A = 2 \cos A \implies \sin A = 2 \cos^2 A \implies \sin^2 A + \cos^2 A = 4 \cos^4 A + \cos^2 A = 1\] \[\implies \cos^2 A = \frac {\sqrt {17} - 1}{8}.\] \[c_n = \sec^n A + \tan^n A = \frac {1}{\cos^n A} + 2^n \cos^n A = (4\cos^2 A +1)^{\frac {n}{2}}+(4 \cos^2 A)^{\frac {n}{2}} =\] \[= \left(\frac {\sqrt {17} + 1}... | 6.875 | [
6,
7,
7,
7,
7,
7,
7,
7
] |
A cube-shaped container has vertices $A,$ $B,$ $C,$ and $D,$ where $\overline{AB}$ and $\overline{CD}$ are parallel edges of the cube, and $\overline{AC}$ and $\overline{BD}$ are diagonals of faces of the cube, as shown. Vertex $A$ of the cube is set on a horizontal plane $\mathcal{P}$ so that the plane of the rectangl... | 751 | Denote $h(X)$ the distance from point $X$ to $\mathcal{P}, h(A) = 0, h(B) = 2,$ $h(C) = 8, h(D) = 10, h(G) = h(I) = h(H) = 7, AB = a, AC = a \sqrt{2}.$
Let slope $AB$ to $\mathcal{P}$ be $\alpha.$ Notation is shown in the diagram. \[\tan \alpha = \frac {\sin \alpha}{\cos \alpha} = \frac {h(B)}{AB}\cdot \frac {AC}{h(C)... | 6.875 | [
7,
7,
7,
7,
7,
7,
7,
6
] |
For each positive integer $n$ let $a_n$ be the least positive integer multiple of $23$ such that $a_n \equiv 1 \pmod{2^n}.$ Find the number of positive integers $n$ less than or equal to $1000$ that satisfy $a_n = a_{n+1}.$ | 363 | Observe that if $a_{n-1} - 1$ is divisible by $2^n$, $a_n = a_{n-1}$. If not, $a_n = a_{n-1} + 23 \cdot 2^{n-1}$.
This encourages us to let $b_n = \frac{a_n - 1}{2^n}$. Rewriting the above equations, we have \[b_n = \begin{cases} \frac{b_{n-1}}{2} & \text{if } 2 \text{ } \vert \text{ } b_{n-1} \\ \frac{b_{n-1}+23}{2} ... | 6.625 | [
7,
6,
6,
8,
6,
6,
7,
7
] |
The school now introduces a new color, silver, for the flag design. Crestview's school colors are now purple, gold, and silver. The students are designing a flag using three solid-colored horizontal stripes. Using one, two, or all three of the school colors, how many different flags are possible if adjacent stripes may... | 27 | 3.875 | [
5,
5,
3,
4,
5,
3,
3,
3
] | |
Determine the sum of the real numbers \( x \) for which \(\frac{2 x}{x^{2}+5 x+3}+\frac{3 x}{x^{2}+x+3}=1\). | -4 | 4.5 | [
6,
5,
5,
4,
4,
4,
4,
4
] | |
The blue parabola shown is the graph of the equation \( x = ay^2 + by + c \). The vertex of the parabola is at \( (5, 3) \), and it passes through the point \( (3, 5) \). Find \( c \). | \frac{1}{2} | 4.375 | [
4,
5,
5,
4,
4,
5,
4,
4
] | |
Observe the following equations:
1. $\cos 2\alpha = 2\cos^2\alpha - 1$;
2. $\cos 4\alpha = 8\cos^4\alpha - 8\cos^2\alpha + 1$;
3. $\cos 6\alpha = 32\cos^6\alpha - 48\cos^4\alpha + 18\cos^2\alpha - 1$;
4. $\cos 8\alpha = 128\cos^8\alpha - 256\cos^6\alpha + 160\cos^4\alpha - 32\cos^2\alpha + 1$;
5. $\cos 10\alpha = m\co... | 962 | 6.5 | [
7,
7,
7,
5,
6,
6,
7,
7
] | |
Given 4 distinct books that are to be distributed evenly between two students, find the probability that the Chinese language book and the Mathematics book are given to the same student. | \frac{1}{3} | 4 | [
4,
4,
4,
3,
5,
4,
4,
4
] | |
For real numbers $s,$ the intersection points of the lines $2x - 3y = 4s + 6$ and $2x + y = 3s + 1$ are plotted. All these points lie on a particular line. Determine the slope of this line. | -\frac{2}{13} | 4.5 | [
4,
6,
4,
4,
6,
4,
4,
4
] | |
An elastic ball falls from a height of 128m, where each bounce reaches half of its previous height before falling again. Calculate the total distance traveled by the ball when it hits the ground for the 8th time. | 382 | 5.25 | [
5,
5,
4,
6,
5,
6,
5,
6
] | |
Leah and Jackson run for 45 minutes on a circular track. Leah runs clockwise at 200 m/min in a lane with a radius of 40 meters, while Jackson runs counterclockwise at 280 m/min in a lane with a radius of 55 meters, starting on the same radial line as Leah. Calculate how many times they pass each other after the start. | 72 | 4.625 | [
4,
4,
5,
6,
4,
4,
6,
4
] | |
2011 warehouses are connected by roads in such a way that from any warehouse you can reach any other, possibly passing through several roads. There are $x_{1}, \ldots, x_{2011}$ kg of cement in the warehouses, respectively. In one trip, an arbitrary amount of cement can be transported from one warehouse to another alo... | 2010 | 6 | [
6,
6,
5,
6,
7,
6,
6,
6
] |
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