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>. | ||
| − | + | <asy> | |
size(3.5cm); defaultpen(fontsize(10pt)); | size(3.5cm); defaultpen(fontsize(10pt)); | ||
pair A,B,C,M; | pair A,B,C,M; | ||
| Line 12: | Line 12: | ||
draw(A--B--C--A); | draw(A--B--C--A); | ||
draw(A--M,dashed); | draw(A--M,dashed); | ||
| − | dot(" | + | dot("$A$",A,N); |
| − | dot(" | + | dot("$B$",B,SW); |
| − | dot(" | + | dot("$C$",C,SE); |
| − | dot(" | + | dot("$M$",M,NW); |
| − | label(" | + | label("$340$",A--B,W); |
| − | label(" | + | label("$390$",A--C,E); |
| − | label(" | + | label("$146$",B--C,S); |
| − | + | </asy> | |
==Solution== | ==Solution== | ||
Latest revision as of 20:41, 5 September 2020
Problem 7
Let 
be a triangle with 
, 
, and 
. If 
is a point on the interior of segment 
such that the length 
is an integer, then the average of all distinct possible values of can be expressed in the form 
, where 
and 
are relatively prime positive integers. Find 
.
![[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]](http://latex.artofproblemsolving.com/6/7/b/67b384cbecaa595bc5afa5a24f722a8887611cc7.png)
Solution
Given that the length is an integer and that it lies on the interior of segment , the shortest possible length of is the length of the altitude dropped straight down from vertex 
. This can be calculated as ![$\frac{2[\triangle ABC]}{BC}$](http://latex.artofproblemsolving.com/e/d/c/edc54a3a69e46d191bf7226cd1a996f3b9c5a3f5.png)
, which is equal to ![\[\frac{2[\triangle ABC]}{146}\]](http://latex.artofproblemsolving.com/b/5/b/b5b8e2a396e6068718549901c58451f6a39ad8e3.png)
or ![\[\frac{[\triangle ABC]}{73}.\]](http://latex.artofproblemsolving.com/7/6/5/765bdd4c99cf7ccb5d661dfb4afe099fb09e5d17.png)
The area of triangle can be found using Heron's formula. It is just ![\[\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{438 \cdot 98 \cdot 292 \cdot 48}=24528.\]](http://latex.artofproblemsolving.com/6/2/3/623ccfce0dfafe401f04cef9c752a9b3c2e9448d.png)
The shortest possible length of is ![\[\frac{24528}{73}=336.\]](http://latex.artofproblemsolving.com/8/8/2/8829b98774f1597d278cc095f02f3ad8985bee65.png)
can be anything greater than or equal to 
, but the condition that point lies in the interior of segment limits the values that we can reach. Starting at 
and heading east (we cannot get 
because is strictly between and 
), we reach the integers ![\[AM=339, 338, 337, 336,\]](http://latex.artofproblemsolving.com/7/a/8/7a8e01538310ba9bf6e53b4dd562d0122fec6e69.png)
and then as we move further east the length of will start to increase. We then reach ![\[AM=337, 338, 339, 340,..., 386, 387, 388, 389.\]](http://latex.artofproblemsolving.com/e/5/0/e50e90aac731ddd0cc999e57cc726bf5490e740f.png)
We cannot get 
because then 
which is not allowed. The distinct possible values of are ![\[336, 337, 338,..., 388, 389.\]](http://latex.artofproblemsolving.com/6/e/7/6e7ba3739a38c01a597ad1837f3adc8eccf2c213.png)
The average is ![\[\frac{336+389}{2}=\frac{725}{2}.\]](http://latex.artofproblemsolving.com/6/3/0/630f7efe28c48c60cdb1e5725a3a4a085cec5031.png)
The answer is 
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 | ||
The problems on this page are copyrighted by the MAC's Christmas Mathematics Competitions. 
