Difference between revisions of "2024 SSMO Relay Round 1 Problems"

(Created page with "==Problem 1== Let <math>AD</math> be an altitude of triangle <math>ABC</math> and let <math>DE</math> be an altitude of triangle <math>ACD.</math> If <math>AB=29,CE=9,</math>...")
 
 
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Let <math>AD</math> be an altitude of triangle <math>ABC</math> and let <math>DE</math> be an altitude of triangle <math>ACD.</math> If <math>AB=29,CE=9,</math> and <math>DE=12,</math> what is the area of triangle <math>ABC?</math>
 
Let <math>AD</math> be an altitude of triangle <math>ABC</math> and let <math>DE</math> be an altitude of triangle <math>ACD.</math> If <math>AB=29,CE=9,</math> and <math>DE=12,</math> what is the area of triangle <math>ABC?</math>
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<asy>
 
<asy>
 
size(200);
 
size(200);

Latest revision as of 16:01, 2 May 2025

Problem 1

Let $AD$ be an altitude of triangle $ABC$ and let $DE$ be an altitude of triangle $ACD.$ If $AB=29,CE=9,$ and $DE=12,$ what is the area of triangle $ABC?$

[asy] size(200); pair A,B,C,D,EE; A = (0,20); B = (-21,0); C = (15,0); D = (0,0); EE = 0.36*A+0.64*C; draw(A--B--C--cycle); draw(A--D--EE); dot("\(A\)", A, N); dot("\(B\)", B, W); dot("\(C\)", C, E); dot("\(D\)", D, S); dot("\(E\)", EE, dir(EE)); [/asy]

Solution

Problem 2

Let $T = TNYWR.$ A circular necklace is called $interesting$ if it has $T$ black beads and $T$ white beads. A move consists of cutting out a segment of consecutive beads and reattaching it in reverse. It is possible to change any $interesting$ necklace into any other $interesting$ necklace using at most $x$ moves. Find $x$. (Note: Rotations and reflections of a necklace are considered the same necklace).

Solution

Problem 3

Let $T = TNYWR.$ In a circle, there are $T$ people. $T-2$ of them have red shoes, and two of them have blue shoes. First, they will randomly eliminate somebody from the circle. Then, they will randomly eliminate somebody with red shoes from the circle, and the cycle repeats until there is only one person left. If the probability this person has blue shoes is $\frac{m}{n},$ for relatively prime positive integers $m$ and $n,$ find $m+n.$

Solution