Difference between revisions of "2015 AIME I Problems/Problem 11"
Line 10: | Line 10: | ||
Then <math>\mathrm{cos}{(\theta)} = \dfrac{x}{8}</math> | Then <math>\mathrm{cos}{(\theta)} = \dfrac{x}{8}</math> | ||
− | and <math>\mathrm{cos}{(2\theta)} = \dfrac{x}{y} = 2\mathrm{cos^2}{(\theta)} - 1 = \dfrac{x^2-32}{32} | + | and <math>\mathrm{cos}{(2\theta)} = \dfrac{x}{y} = 2\mathrm{cos^2}{(\theta)} - 1 = \dfrac{x^2-32}{32}</math>. |
+ | |||
+ | Cross-multiplying yields <math>32x = y(x^2-32)</math>. | ||
Since <math>x,y>0</math>, <math>x^2-32</math> must be positive, so <math>x > 5.5</math>. | Since <math>x,y>0</math>, <math>x^2-32</math> must be positive, so <math>x > 5.5</math>. |
Revision as of 07:19, 21 March 2015
Problem
Triangle has positive integer side lengths with
. Let
be the intersection of the bisectors of
and
. Suppose
. Find the smallest possible perimeter of
.
Solution
Let be the midpoint of
. Then by SAS Congruence,
, so
.
Now let ,
, and
.
Then
and .
Cross-multiplying yields .
Since ,
must be positive, so
.
Additionally, since is a right triangle with hypotenuse
of length
,
.
Therefore, given that is an integer, the only possible values for
are
,
,
, and
.
However, only one of these values, yields an integral value for
, so we conclude that
and
.
Thus the perimeter of must be
.
See Also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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.