Difference between revisions of "2015 AIME I Problems/Problem 11"
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==Solution 1 (No Trig)== | ==Solution 1 (No Trig)== | ||
− | Let <math>AB=x</math> and the foot of the altitude from <math>A</math> to <math>BC</math> be point <math>E</math> and <math>BE=y</math>. Since ABC is isosceles, <math>I</math> is on <math>AE</math>. By pythagorean theorem, <math>AE=\sqrt{x^2-y^2}</math>. Let <math>IE=a</math> and <math>IA=b</math>. By angle bisector theorem, <math>\frac{y}{a}=\frac{x}{b}</math>. Also, <math>a+b=\sqrt{x^2-y^2}</math>. Solving for <math>a</math>, we get <math>a=\frac{\sqrt{x^2-y^2}}{1+\frac{x}{y}}</math>. Then, using pythagorean on <math>BEI</math> we have <math>y^2+\left(\frac{\sqrt{x^2-y^2}}{1+\frac{x}{y}}\right)^2=8^2=64</math>. Simplifying, we have <math>y^2+y^2\frac{x^2-y^2}{(x+y)^2}=64</math>. Factoring out the <math>y^2</math>, we have <math>y^2\left(1+\frac{x^2-y^2}{(x+y)^2}\right)=64</math>. Adding 1 to the fraction and simplifying, we have <math>\frac{y^2x(x+y)}{(x+y)^2}=32</math>. Crossing out the <math>x+y</math>, and solving for <math>x</math> yields <math>32y = x(y^2-32)</math>. Then, we continue as | + | Let <math>AB=x</math> and the foot of the altitude from <math>A</math> to <math>BC</math> be point <math>E</math> and <math>BE=y</math>. Since ABC is isosceles, <math>I</math> is on <math>AE</math>. By pythagorean theorem, <math>AE=\sqrt{x^2-y^2}</math>. Let <math>IE=a</math> and <math>IA=b</math>. By angle bisector theorem, <math>\frac{y}{a}=\frac{x}{b}</math>. Also, <math>a+b=\sqrt{x^2-y^2}</math>. Solving for <math>a</math>, we get <math>a=\frac{\sqrt{x^2-y^2}}{1+\frac{x}{y}}</math>. Then, using pythagorean on <math>BEI</math> we have <math>y^2+\left(\frac{\sqrt{x^2-y^2}}{1+\frac{x}{y}}\right)^2=8^2=64</math>. Simplifying, we have <math>y^2+y^2\frac{x^2-y^2}{(x+y)^2}=64</math>. Factoring out the <math>y^2</math>, we have <math>y^2\left(1+\frac{x^2-y^2}{(x+y)^2}\right)=64</math>. Adding 1 to the fraction and simplifying, we have <math>\frac{y^2x(x+y)}{(x+y)^2}=32</math>. Crossing out the <math>x+y</math>, and solving for <math>x</math> yields <math>32y = x(y^2-32)</math>. Then, we continue as Solution 2 does. |
Revision as of 12:42, 8 June 2017
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 1 (No Trig)
Let and the foot of the altitude from
to
be point
and
. Since ABC is isosceles,
is on
. By pythagorean theorem,
. Let
and
. By angle bisector theorem,
. Also,
. Solving for
, we get
. Then, using pythagorean on
we have
. Simplifying, we have
. Factoring out the
, we have
. Adding 1 to the fraction and simplifying, we have
. Crossing out the
, and solving for
yields
. Then, we continue as Solution 2 does.
Solution 2 (Trig)
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.