Difference between revisions of "2015 AIME I Problems/Problem 11"
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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>. | ||
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+ | ==Solution 3== | ||
+ | Let <math>AB=x</math>, call the midpoint of <math>BC</math> point <math>E</math>, call the point where the incircle meets <math>AB</math> point <math>D</math>, and let <math>BE=y</math>. We are looking for the minimum value of <math>2(x+y)</math>. <math>AE</math> is an altitude because the triangle is isosceles. By Pythagoras on <math>BEI</math>, the inradius is <math>\sqrt{64-y^2}</math> and by Pythagoras on <math>ABE</math>, <math>AE</math> is <math>\sqrt{x^2-y^2}</math>. By equal tangents, <math>BE=BD=y</math>, so <math>AD=x-y</math>. Since <math>ID</math> is an inradius, <math>ID=IE</math> and using pythagoras on <math>ADI</math> yields <math>AI=</math><math>\sqrt{x^2-2xy+64}</math>. <math>ADI</math> is similar to <math>AEB</math> by <math>AA</math>, so we can write <math>\frac{x-y}{\sqrt{x^2-2xy+64}}=\frac{\sqrt{x^2-y^2}}{x}</math>. Simplifying, <math>\frac{x}{\sqrt{x^2-2xy+64}}=\sqrt{\frac{x+y}{x-y}}</math>. Squaring, subtracting 1 from both sides, and multiplying everything out, we get <math>yx^2-2xy^2+64y=yx^2 -32x+32y-xy^2</math>, which turns into <math>32y=x(y^2-32)</math>. Finish as above. | ||
==See Also== | ==See Also== |
Revision as of 23:39, 26 February 2018
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 Theorem 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
.
Solution 3
Let , call the midpoint of
point
, call the point where the incircle meets
point
, and let
. We are looking for the minimum value of
.
is an altitude because the triangle is isosceles. By Pythagoras on
, the inradius is
and by Pythagoras on
,
is
. By equal tangents,
, so
. Since
is an inradius,
and using pythagoras on
yields
.
is similar to
by
, so we can write
. Simplifying,
. Squaring, subtracting 1 from both sides, and multiplying everything out, we get
, which turns into
. Finish as above.
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 |
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