# 2020 CIME II Problems/Problem 7

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## 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 $725+2=\boxed{727}.$