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)); | ||
+ | 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== | ==Solution== |
Latest revision as of 21:41, 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
.
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
, which is equal to
or
The area of triangle
can be found using Heron's formula. It is just
The shortest possible length of
is
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
and then as we move further east the length of
will start to increase. We then reach
We cannot get
because then
which is not allowed. The distinct possible values of
are
The average is
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.