Difference between revisions of "2013 AIME I Problems/Problem 9"
(→Solution 3 (Coordinate Bash)) |
(→Solution 3 (Coordinate Bash)) |
||
Line 89: | Line 89: | ||
e let the original position of <math>A</math> be <math>A</math>, and the position of <math>A</math> after folding be <math>D</math>. Also, we put the triangle on the coordinate plane such that <math>A=(0,0)</math>, <math>B=(-6,-6\sqrt3)</math>, <math>C=(6,-6\sqrt3)</math>, and <math>D=(3,-6\sqrt3)</math>. | e let the original position of <math>A</math> be <math>A</math>, and the position of <math>A</math> after folding be <math>D</math>. Also, we put the triangle on the coordinate plane such that <math>A=(0,0)</math>, <math>B=(-6,-6\sqrt3)</math>, <math>C=(6,-6\sqrt3)</math>, and <math>D=(3,-6\sqrt3)</math>. | ||
− | + | <asy> | |
− | |||
size(10cm); | size(10cm); | ||
pen tpen = defaultpen + 1.337; | pen tpen = defaultpen + 1.337; | ||
Line 103: | Line 102: | ||
draw(B--M--N--C--cycle); | draw(B--M--N--C--cycle); | ||
draw(M--A--N--cycle); | draw(M--A--N--cycle); | ||
− | label(" | + | label("$D$", A, S); |
pair X = (6,6*sqrt(3)); | pair X = (6,6*sqrt(3)); | ||
draw(B--X--C); | draw(B--X--C); | ||
− | label(" | + | label("$A$",X,dir(90)); |
draw(A--X); | draw(A--X); | ||
− | + | </asy> | |
Note that since <math>A</math> is reflected over the fold line to <math>D</math>, the fold line is the perpendicular bisector of <math>AD</math>. We know <math>A=(0,0)</math> and <math>D=(3,-6\sqrt3)</math>. The midpoint of <math>AD</math> (which is a point on the fold line) is <math>(\tfrac32, -3\sqrt3)</math>. Also, the slope of <math>AD</math> is <math>\frac{-6\sqrt3}{3}=-2\sqrt3</math>, so the slope of the fold line (which is perpendicular), is the negative of the reciprocal of the slope of <math>AD</math>, or <math>\frac{1}{2\sqrt3}=\frac{\sqrt3}{6}</math>. Then, using point slope form, the equation of the fold line is | Note that since <math>A</math> is reflected over the fold line to <math>D</math>, the fold line is the perpendicular bisector of <math>AD</math>. We know <math>A=(0,0)</math> and <math>D=(3,-6\sqrt3)</math>. The midpoint of <math>AD</math> (which is a point on the fold line) is <math>(\tfrac32, -3\sqrt3)</math>. Also, the slope of <math>AD</math> is <math>\frac{-6\sqrt3}{3}=-2\sqrt3</math>, so the slope of the fold line (which is perpendicular), is the negative of the reciprocal of the slope of <math>AD</math>, or <math>\frac{1}{2\sqrt3}=\frac{\sqrt3}{6}</math>. Then, using point slope form, the equation of the fold line is |
Revision as of 14:53, 26 May 2017
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
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.