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2022 SSMO Team Round Problems

Revision as of 21:43, 31 May 2023 by Pinkpig (talk | contribs) (Created page with "==Problem 1== In triangle <math>ABC</math>, circumcircle <math>\omega</math> is drawn. Let <math>I</math> be the incenter of <math>\triangle{ABC}</math>. Let <math>H_A</math>...")
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Problem 1

In triangle $ABC$, circumcircle $\omega$ is drawn. Let $I$ be the incenter of $\triangle{ABC}$. Let $H_A$ be the intersection of the $A$-altitude and $\omega.$ Given that $AB=13,AC=15,$ and $BC=14,$ the area of triangle $AIH_A$ can be expressed as $\frac{m}{n},$ for relatively prime positive integers $m$ and $n.$ Find $m+n.$

Solution

Problem 2

Consider $8$ marbles in a line, where the color of each marble is either black or white and is randomly chosen. Define the period of a lineup of 8 marbles to be the length of the smallest lineup of marbles such that if we consider the infinite repeating sequence of marbles formed by repeating that lineup, the original lineup of 8 marbles can be found within that sequence.

A good ordering of these marbles is defined to be an ordering such that the period of the ordering is at most $6$. For example, $bwwbbbww$ is a good ordering because we may consider the lineup $bwwbb$, which has a length equal to $5.$ If the probability that the marbles form a good ordering can be expressed as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

Solution

Problem 3

Let $ABCD$ be an isosceles trapezoid such that $AB\parallel CD.$ Let $E$ be a point on $CD$ such that $AB=CE.$ Let the midpoint of $DE$ be $M$ such that $BD$ intersects $AM$ at $G$ and $AE$ at $F.$ If $DC=36, AB=24,$ and $AD=10,$ then $[AGF]$ can be expressed as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

[asy] unitsize(2mm); fill((6,8/5)--(10,8/3)--(6,8)--cycle,lightgray); draw((0,0)--(36,0)--(30,8)--(6,8)--cycle); draw((6,8)--(12,0)--(0,0)); draw((0,0)--(30,8)); draw((6,8)--(6,0)); label("A", (6,8), NW); dot((6,8)); label("B", (30,8), NE); dot((30,8)); label("C", (36,0), SE); dot((36,0)); label("D", (0,0), SW); dot((0,0)); label("E",(12,0),S); dot((12,0)); label("M",(6,0),S); dot((6,0)); dot((6,8/5)); dot((10,8/3)); [/asy]

Solution

Problem 4

Solution

Problem 5

Solution

Problem 6

Solution

Problem 7

Solution

Problem 8

Solution

Problem 9

Solution

Problem 10

Solution

Problem 11

Solution

Problem 12

Solution

Problem 13

Solution

Problem 14

Solution

Problem 15

Solution

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