2013 AIME I Problems/Problem 9
Problem 9
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 .
Solution 3 (Coordinate Bash)
e 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
.
[asy]
import cse5;
size(10cm);
pen tpen = defaultpen + 1.337;
real a = 39/5.0;
real b = 39/7.0;
pair B = MP("B", (0,0), dir(200));
pair A = (9,0);
pair C = MP("C", (12,0), dir(-20));
pair K = (6,10.392);
pair M = (a*B+(12-a)*K) / 12;
pair N = (b*C+(12-b)*K) / 12;
draw(B--M--N--C--cycle);
draw(M--A--N--cycle);
label("", A, S);
pair X = (6,6*sqrt(3));
draw(B--X--C);
label("
",X,dir(90));
draw(A--X);
[/asy]
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.
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 |
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