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Difference between revisions of "2020 CIME II Problems/Problem 7"
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Let <math>ABC</math> be a triangle with <math>AB=340</math>, <math>BC=146</math>, and <math>CA=390</math>. If <math>M</math> is a point on the interior of segment <math>BC</math> such that the length <math>AM</math> is an integer, then the average of all distinct possible values of <math>AM</math> can be expressed in the form <math>\tfrac pq</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.
 
Let <math>ABC</math> be a triangle with <math>AB=340</math>, <math>BC=146</math>, and <math>CA=390</math>. If <math>M</math> is a point on the interior of segment <math>BC</math> such that the length <math>AM</math> is an integer, then the average of all distinct possible values of <math>AM</math> can be expressed in the form <math>\tfrac pq</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.
  
−
[[Image:2020_CIME_II_-7.png]]
+
[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("<math>A</math>",A,N);
 +
dot("<math>B</math>",B,SW);
 +
dot("<math>C</math>",C,SE);
 +
dot("<math>M</math>",M,NW);
 +
label("<math>340</math>",A--B,W);
 +
label("<math>390</math>",A--C,E);
 +
label("<math>146</math>",B--C,S);
 +
[/asy]
  
 
==Solution==
 
==Solution==

Revision as of 18:43, 5 September 2020

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

See also

2020 CIME II (Problems • Answer Key • Resources)
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

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