MathDB

2023 AIME

Part of AIME Problems

Subcontests

(15)
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Time is ticking

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 = 0; i < 12; ++i) { label("\small" + (string) i, dir(90 - i * 30) * 0.75); } draw((0,0)--0.6*dir(90),black+linewidth(2),Arrow(TeXHead,2bp)); draw((0,0)--0.4*dir(90),black+linewidth(2),Arrow(TeXHead,2bp)); [/asy] Initially, both hands point to the number 12. The clock performs a sequence of hand movements so that on each movement, one of the two hands moves clockwise to the next number on the clock while the other hand does not move.
Let NN be the number of sequences of 144 hand movements such that during the sequence, every possible positioning of the hands appears exactly once, and at the end of the 144 movements, the hands have returned to their initial position. Find the remainder when NN is divided by 1000.

Cube on a Plane

A cube-shaped container has vertices AA, BB, CC, and DD where AB‾\overline{AB} and CD‾\overline{CD} are parallel edges of the cube, and AC‾\overline{AC} and BD‾\overline{BD} are diagonals of the faces of the cube. Vertex AA of the cube is set on a horizontal plane P\mathcal P so that the plane of the rectangle ABCDABCD is perpendicular to P\mathcal P, vertex BB is 22 meters above P\mathcal P, vertex CC is 88 meters above P\mathcal P, and vertex DD is 1010 meters above P\mathcal P. The cube contains water whose surface is 77 meters above P\mathcal P. The volume of the water is mn\tfrac mn cubic meters, where mm and nn are relatively prime positive integers. Find m+nm+n. [asy] size(250); defaultpen(linewidth(0.6)); pair A = origin, B = (6,3), X = rotate(40)*B, Y = rotate(70)*X, C = X+Y, Z = X+B, D = B+C, W = B+Y; pair P1 = 0.8*C+0.2*Y, P2 = 2/3*C+1/3*X, P3 = 0.2*D+0.8*Z, P4 = 0.63*D+0.37*W; pair E = (-20,6), F = (-6,-5), G = (18,-2), H = (9,8); filldraw(E--F--G--H--cycle,rgb(0.98,0.98,0.2)); fill(A--Y--P1--P4--P3--Z--B--cycle,rgb(0.35,0.7,0.9)); draw(A--B--Z--X--A--Y--C--X^^C--D--Z); draw(P1--P2--P3--P4--cycle^^D--P4); dot("AA",A,S); dot("BB",B,S); dot("CC",C,N); dot("DD",D,N); label("P\mathcal P",(-13,4.5)); [/asy]
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6
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Alice's Cards

Alice knows that 33 red cards and 33 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 mn\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+nm+n.

double integration on the AIME

Consider the L-shaped region formed by three unit squares joined at their sides, as shown below. Two points AA and BB are chosen independently and uniformly at random from inside this region. The probability that the midpoint of AB‾\overline{AB} also lies inside this L-shaped region can be expressed as mn\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+nm+n. [asy] size(2.5cm); draw((0,0)--(0,2)--(1,2)--(1,1)--(2,1)--(2,0)--cycle); draw((0,1)--(1,1)--(1,0), dotted); [/asy]
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maa REALLY loves rhombi

Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths 21\sqrt{21} and 31\sqrt{31}. The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is mn\frac mn, where mm and nn are relatively prime positive integers. Find m+nm+n. A parallelepiped is a solid with six parallelogram faces such as the one shown below. [asy] unitsize(2cm); pair o = (0, 0), u = (1, 0), v = 0.8*dir(40), w = dir(70); draw(o--u--(u+v)); draw(o--v--(u+v), dotted); draw(shift(w)*(o--u--(u+v)--v--cycle)); draw(o--w); draw(u--(u+w)); draw(v--(v+w), dotted); draw((u+v)--(u+v+w)); [/asy]

i ran out of interesting titles

Let AA be an acute angle such that tan⁡A=2cos⁡A\tan A = 2\cos A. Find the number of positive integers nn less than or equal to 10001000 such that sec⁡nA+tan⁡nA\sec^n A + \tan^n A is a positive integer whose units digit is 99.
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9
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*Fun* Polynomial Counting!

Find the number of cubic polynomials p(x)=x3+ax2+bx+cp(x) = x^3 + ax^2 + bx + c, where aa, bb, and cc are integers in {−20,−19,−18,…,18,19,20}\{-20, -19,-18, \dots , 18, 19, 20\}, such that there is a unique integer m≠2m \neq 2 with p(m)=p(2)p(m) = p(2).

Geo Geo Geo

Circles ω1\omega_1 and ω2\omega_2 intersect at two points PP and QQ, and their common tangent line closer to PP intersects ω1\omega_1 and ω2\omega_2 at points AA and BB, respectively. The line parallel to line ABAB that passes through PP intersects ω1\omega_1 and ω2\omega_2 for the second time at points XX and YY, respectively. Suppose PX=10,PY=14,PX = 10, PY = 14, and PQ=5PQ = 5. Then the area of trapezoid XABYXABY is mnm\sqrt{n} where mm and nn are positive integers and nn is not divisible by the square of any prime. Find m+nm + n.
3
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8
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15
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7
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10
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did U silly this?

There exists a unique positive integer aa for which the sum U=∑n=12023⌊n2−na5⌋U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor is an integer strictly between −1000-1000 and 10001000. For that unique aa, find a+Ua+U.
(Note that ⌊x⌋\lfloor x\rfloor denotes the greatest integer that is less than or equal to xx.)

Integers in a Grid

Let NN be the number of ways to place the integers 11 through 1212 in the 1212 cells of a 2×62\times 6 grid so that for any two cells sharing a side, the difference between the numbers in those cells is not divisible by 33. One way to do this is shown below. Find the number of positive integer divisors of NN.
[asy] size(160); defaultpen(linewidth(0.6)); for(int j=0;j<=6;j=j+1) { draw((j,0)--(j,2)); } for(int i=0;i<=2;i=i+1) { draw((0,i)--(6,i)); } for(int k=1;k<=12;k=k+1) { label(""+((string)k)+""+((string) k)+"",(floor((k-1)/2)+0.5,k%2+0.5)); } [/asy]
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Miquel Point

Let △ABC\triangle ABC be an equilateral triangle with side length 5555. Points DD, EE, and FF lie on sides BC‾\overline{BC}, CA‾\overline{CA}, and AB‾\overline{AB}, respectively, with BD=7BD=7, CE=30CE=30, and AF=40AF=40. A unique point PP inside △ABC\triangle ABC has the property that ∡AEP=∡BFP=∡CDP.\measuredangle AEP=\measuredangle BFP=\measuredangle CDP. Find tan⁡2(∡AEP)\tan^{2}\left(\measuredangle AEP\right).

why maa why

In △ABC\triangle ABC with side lengths AB=13AB=13, BC=14BC=14, and CA=15CA=15, let MM be the midpoint of BC‾\overline{BC}. Let PP be the point on the circumcircle of △ABC\triangle ABC such that MM is on AP‾\overline{AP}. There exists a unique point QQ on segment AM‾\overline{AM} such that ∠PBQ=∠PCQ\angle PBQ = \angle PCQ. Then AQAQ can be written as mn\frac{m}{\sqrt{n}}, where mm and nn are relatively prime positive integers. Find m+nm+n.
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