2020 CIME II Problems/Problem 7

Revision as of 20:41, 5 September 2020 by Jbala (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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]


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

See also

2020 CIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All CIME Problems and Solutions

The problems on this page are copyrighted by the MAC's Christmas Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS