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Difference between revisions of "2023 AIME II Problems"

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==Problem 10==
 
==Problem 10==
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These problems will not be available until the 2023 AIME II is released in February 2023.
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Let <math>N</math> be the number of ways to place the integers <math>1</math> through <math>12</math> in the <math>12</math> cells of a <math>2 \times 6</math> grid so that for any two cells sharing a side, the difference between the numbers in those cells is not divisible by <math>3.</math> One way to do this is shown below. Find the number of positive integer divisors of <math>N.</math>
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<cmath>\begin{array}{|c|c|c|c|c|c|} \hline
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\,1\, & \,3\, & \,5\, & \,7\, & \,9\, & 11 \\ \hline
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\,2\, & \,4\, & \,6\, & \,8\, & 10 & 12 \\ \hline
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\end{array}</cmath>
  
 
[[2023 AIME II Problems/Problem 10|Solution]]
 
[[2023 AIME II Problems/Problem 10|Solution]]

Revision as of 16:02, 16 February 2023

2023 AIME II (Answer Key)
Printable version | AoPS Contest Collections • PDF

Instructions

  1. This is a 15-question, 3-hour examination. All answers are integers ranging from $000$ to $999$, inclusive. Your score will be the number of correct answers; i.e., there is neither partial credit nor a penalty for wrong answers.
  2. No aids other than scratch paper, graph paper, ruler, compass, and protractor are permitted. In particular, calculators and computers are not permitted.
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Problem 1

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 growing on any of the six trees.

Solution

Problem 2

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 $292 = 444_{\text{eight}}.$

Solution

Problem 3

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 $\triangle ABC.$

Solution

Problem 4

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 $S.$

Solution

Problem 5

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 expressed in the form $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$

Solution

Problem 6

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 $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ [asy] unitsize(2cm); draw((0,0)--(2,0)--(2,1)--(1,1)--(1,2)--(0,2)--cycle); draw((0,1)--(1,1)--(1,0),dashed); [/asy]

Solution

Problem 7

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.

Solution

Problem 8

Let $\omega = \cos\frac{2\pi}{7} + i \cdot \sin\frac{2\pi}{7},$ where $i = \sqrt{-1}.$ Find the value of the product \[\prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right).\]

Solution

Problem 9

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 $Y,$ respectively. Suppose $PX=10,$ $PY=14,$ and $PQ=5.$ Then the area of trapezoid $XABY$ is $m\sqrt{n},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n.$

Solution

Problem 10

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 $N.$ \[\begin{array}{|c|c|c|c|c|c|} \hline \,1\, & \,3\, & \,5\, & \,7\, & \,9\, & 11 \\ \hline \,2\, & \,4\, & \,6\, & \,8\, & 10 & 12 \\ \hline \end{array}\]

Solution

Problem 11

These problems will not be available until the 2023 AIME II is released in February 2023.

Solution

Problem 12

These problems will not be available until the 2023 AIME II is released in February 2023.

Solution

Problem 13

These problems will not be available until the 2023 AIME II is released in February 2023.

Solution

Problem 14

These problems will not be available until the 2023 AIME II is released in February 2023.

Solution

Problem 15

These problems will not be available until the 2023 AIME II is released in February 2023.

Solution

See also

2023 AIME II (Problems • Answer Key • Resources)
Preceded by
2023 AIME I 		Followed by
2024 AIME I
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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