Difference between revisions of "2013 AIME I Problems/Problem 9"
(→Problem 9) |
Zephyr38688 (talk | contribs) m (→Solution 1) |
||
Line 54: | Line 54: | ||
<math>x^{2} = (\frac{39}{5})^2 + (\frac{39}{7})^2 - (\frac{39}{5} \times \frac{39}{7})</math> | <math>x^{2} = (\frac{39}{5})^2 + (\frac{39}{7})^2 - (\frac{39}{5} \times \frac{39}{7})</math> | ||
− | <math>x | + | <math>x = \frac{39 \sqrt{39}}{35}</math> |
The solution is <math>39 + 39 + 35 = \boxed{113}</math>. | The solution is <math>39 + 39 + 35 = \boxed{113}</math>. |
Revision as of 18:14, 17 February 2014
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 .
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.