Difference between revisions of "2013 AIME I Problems/Problem 9"
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A paper equilateral triangle <math>ABC</math> has side length <math>12</math>. The paper triangle is folded so that vertex <math>A</math> touches a point on side <math>\overline{BC}</math> a distance <math>9</math> from point <math>B</math>. The length of the line segment along which the triangle is folded can be written as <math>\frac{m\sqrt{p}}{n}</math>, where <math>m</math>, <math>n</math>, and <math>p</math> are positive integers, <math>m</math> and <math>n</math> are relatively prime, and <math>p</math> is not divisible by the square of any prime. Find <math>m+n+p</math>. | A paper equilateral triangle <math>ABC</math> has side length <math>12</math>. The paper triangle is folded so that vertex <math>A</math> touches a point on side <math>\overline{BC}</math> a distance <math>9</math> from point <math>B</math>. The length of the line segment along which the triangle is folded can be written as <math>\frac{m\sqrt{p}}{n}</math>, where <math>m</math>, <math>n</math>, and <math>p</math> are positive integers, <math>m</math> and <math>n</math> are relatively prime, and <math>p</math> is not divisible by the square of any prime. Find <math>m+n+p</math>. | ||
Revision as of 20:08, 24 January 2021
Contents
Problem
A paper equilateral triangle has side length
. The paper triangle is folded so that vertex
touches a point on side
a distance
from point
. The length of the line segment along which the triangle is folded can be written as
, where
,
, and
are positive integers,
and
are relatively prime, and
is not divisible by the square of any prime. Find
.
Solution 1
Let and
be the points on
and
, respectively, where the paper is folded.
Let be the point on
where the folded
touches it.
Let ,
, and
be the lengths
,
, and
, respectively.
We have ,
,
,
,
, and
.
Using the Law of Cosines on :
Using the Law of Cosines on :
Using the Law of Cosines on :
The solution is .
Solution 2
Proceed with the same labeling as in Solution 1.
Therefore, .
Similarly, .
Now, and
are similar triangles, so
.
Solving this system of equations yields and
.
Using the Law of Cosines on :
The solution is .
Note
Once you find and
, you can scale down the triangle by a factor of
so that all sides are integers. Applying Law of cosines becomes easier, you just need to remember to scale back up.
Solution 3 (Coordinate Bash)
We let the original position of be
, and the position of
after folding be
. Also, we put the triangle on the coordinate plane such that
,
,
, and
.
Note that since is reflected over the fold line to
, the fold line is the perpendicular bisector of
. We know
and
. The midpoint of
(which is a point on the fold line) is
. Also, the slope of
is
, so the slope of the fold line (which is perpendicular), is the negative of the reciprocal of the slope of
, or
. Then, using point slope form, the equation of the fold line is
Note that the equations of lines
and
are
and
, respectively. We will first find the intersection of
and the fold line by substituting for
:
Therefore, the point of intersection is
. Now, lets find the intersection with
. Substituting for
yields
Therefore, the point of intersection is
. Now, we just need to use the distance formula to find the distance between
and
.
The number 39 is in all of the terms, so let's factor it out:
Therefore, our answer is
, and we are done.
Solution by nosaj.
Video Solution
https://www.youtube.com/watch?v=581ZtcQFCaE&t=98s
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.