Difference between revisions of "2015 AIME I Problems/Problem 11"
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Therefore, given that <math>BC=2x</math> is an integer, the only possible values for <math>x</math> are <math>6</math>, <math>6.5</math>, <math>7</math>, and <math>7.5</math>. | Therefore, given that <math>BC=2x</math> is an integer, the only possible values for <math>x</math> are <math>6</math>, <math>6.5</math>, <math>7</math>, and <math>7.5</math>. | ||
− | However, only one of these values, <math>x=6</math> yields an integral value for <math>AB=y</math>, so we conclude that <math>x=6</math> and <math>y=\dfrac{32(6)}{(6)^2-32}=48</math>. | + | However, only one of these values, <math>x=6</math>, yields an integral value for <math>AB=y</math>, so we conclude that <math>x=6</math> and <math>y=\dfrac{32(6)}{(6)^2-32}=48</math>. |
Thus the perimeter of <math>\triangle ABC</math> must be <math>2(x+y) = \boxed{108}</math>. | Thus the perimeter of <math>\triangle ABC</math> must be <math>2(x+y) = \boxed{108}</math>. |
Revision as of 07:20, 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 has 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.