Difference between revisions of "2013 AIME I Problems/Problem 9"
Alexlikemath (talk | contribs) |
|||
Line 84: | Line 84: | ||
The solution is <math>39 + 39 + 35 = \boxed{113}</math>. | The solution is <math>39 + 39 + 35 = \boxed{113}</math>. | ||
+ | |||
+ | ===Note=== | ||
+ | Once you find <math>DP</math> and <math>DQ</math>, you can scale down the triangle by a factor of <math>\frac{39}{35}</math> 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) == | == Solution 3 (Coordinate Bash) == | ||
Line 125: | Line 128: | ||
==Video Solution== | ==Video Solution== | ||
https://www.youtube.com/watch?v=581ZtcQFCaE&t=98s | https://www.youtube.com/watch?v=581ZtcQFCaE&t=98s | ||
+ | |||
+ | |||
+ | |||
== See also == | == See also == | ||
{{AIME box|year=2013|n=I|num-b=8|num-a=10}} | {{AIME box|year=2013|n=I|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 13:09, 23 December 2020
Contents
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 .
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.