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Difference between revisions of "2020 CIME II Problems/Problem 7"

(Created page with "==Problem 7== 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 segmen...")
 
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==Solution==
 
==Solution==
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Given that the length <math>AM</math> is an integer and that it lies on the interior of segment <math>BC</math>, the shortest possible length of <math>AM</math> is the length of the altitude dropped straight down from vertex <math>A</math>. This can be calculated as <math>\frac{2[\triangle ABC]}{BC}</math>, which is equal to <cmath>\frac{2[\triangle ABC]}{146}</cmath> or <cmath>\frac{[\triangle ABC]}{73}</cmath>. The area of triangle <math>ABC</math> can be found using Heron's formula. It is just <cmath>\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{438 \cdot 98 \cdot 292 \cdot 48}=24528.</cmath> The shortest possible length of <math>AM</math> is <cmath>\frac{24528}{73}=336.</cmath> <math>AM</math> can be anything greater than or equal to <math>336</math>, but the condition that point <math>M</math> lies in the interior of segment <math>BC</math> limits the values that we can reach. Starting at <math>B</math> and heading east (we cannot get <math>340</math> because <math>M</math> is strictly between <math>B</math> and <math>C</math>), we reach the integers <cmath>AM=339, 338, 337, 336</cmath>, and then as we move further east the length of <math>AM</math> will start to increase. We then reach <cmath>AM=337, 338, 339, 340,..., 386, 387, 388, 389</cmath>. We cannot get <math>390</math> because then <math>M=C</math> which is not allowed. The distinct possible values of <math>AM</math> are <cmath>336, 337, 338,..., 388, 389.</cmath> The average is <cmath>\frac{336+389}{2}=\frac{725}{2}</cmath>. The answer is <math>725+2=\boxed{727}.</math>
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Given that the length <math>AM</math> is an integer and that it lies on the interior of segment <math>BC</math>, the shortest possible length of <math>AM</math> is the length of the altitude dropped straight down from vertex <math>A</math>. This can be calculated as <math>\frac{2[\triangle ABC]}{BC}</math>, which is equal to <cmath>\frac{2[\triangle ABC]}{146}</cmath> or <cmath>\frac{[\triangle ABC]}{73}.</cmath> The area of triangle <math>ABC</math> can be found using Heron's formula. It is just <cmath>\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{438 \cdot 98 \cdot 292 \cdot 48}=24528.</cmath> The shortest possible length of <math>AM</math> is <cmath>\frac{24528}{73}=336.</cmath> <math>AM</math> can be anything greater than or equal to <math>336</math>, but the condition that point <math>M</math> lies in the interior of segment <math>BC</math> limits the values that we can reach. Starting at <math>B</math> and heading east (we cannot get <math>340</math> because <math>M</math> is strictly between <math>B</math> and <math>C</math>), we reach the integers <cmath>AM=339, 338, 337, 336,</cmath> and then as we move further east the length of <math>AM</math> will start to increase. We then reach <cmath>AM=337, 338, 339, 340,..., 386, 387, 388, 389.</cmath> We cannot get <math>390</math> because then <math>M=C</math> which is not allowed. The distinct possible values of <math>AM</math> are <cmath>336, 337, 338,..., 388, 389.</cmath> The average is <cmath>\frac{336+389}{2}=\frac{725}{2}.</cmath> The answer is <math>725+2=\boxed{727}.</math>
  
 
==See also==
 
==See also==

Revision as of 17:09, 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 $p+q$.

2020 CIME II -7.png

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
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