2020 CIME II Problems/Problem 7

Problem 7

Let $ABC$ be a triangle with $AB=340$ , $BC=146$ , and $CA=390$ . If $M$ is a point on the interior of segment $BC$ such that the length $AM$ is an integer, then the average of all distinct possible values of $AM$ can be expressed in the form $\tfrac pq$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ .

$[asy] size(3.5cm); defaultpen(fontsize(10pt)); pair A,B,C,M; A=dir(95); B=dir(-117); C=dir(-63); M=(2B+3C)/5; draw(A--B--C--A); draw(A--M,dashed); dot("$A$",A,N); dot("$B$",B,SW); dot("$C$",C,SE); dot("$M$",M,NW); label("$340$",A--B,W); label("$390$",A--C,E); label("$146$",B--C,S); [/asy]$

Solution

Given that the length $AM$ is an integer and that it lies on the interior of segment $BC$ , the shortest possible length of $AM$ is the length of the altitude dropped straight down from vertex $A$ . This can be calculated as $\frac{2[\triangle ABC]}{BC}$ , which is equal to $\frac{2[\triangle ABC]}{146}$ or $\frac{[\triangle ABC]}{73}.$ The area of triangle $ABC$ can be found using Heron's formula. It is just $\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{438 \cdot 98 \cdot 292 \cdot 48}=24528.$ The shortest possible length of $AM$ is $\frac{24528}{73}=336.$ $AM$ can be anything greater than or equal to $336$ , but the condition that point $M$ lies in the interior of segment $BC$ limits the values that we can reach. Starting at $B$ and heading east (we cannot get $340$ because $M$ is strictly between $B$ and $C$ ), we reach the integers $AM=339, 338, 337, 336,$ and then as we move further east the length of $AM$ will start to increase. We then reach $AM=337, 338, 339, 340,..., 386, 387, 388, 389.$ We cannot get $390$ because then $M=C$ which is not allowed. The distinct possible values of $AM$ are $336, 337, 338,..., 388, 389.$ The average is $\frac{336+389}{2}=\frac{725}{2}.$ The answer is $725+2=\boxed{727}.$

2020 CIME II (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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2020 CIME II Problems/Problem 7

Problem 7

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