Difference between revisions of "2015 AIME I Problems/Problem 11"
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Define angle <math>IBC = x</math>. Foot of perpendicular from <math>I</math> to <math>BC</math> is point <math>P</math>. | Define angle <math>IBC = x</math>. Foot of perpendicular from <math>I</math> to <math>BC</math> is point <math>P</math>. | ||
<math>\overline{BC} = 2\overline{BP} = 2(8\cos(x)) = N</math>, where <math>N</math> is an integer. Thus, <math>\cos(x) = \frac{N}{16}</math>. Via double angle, we calculate <math>\overline{AB}</math> to be <math>\frac{8\cos(x)}{2\cos(x)^2 - 1} = \frac{64N}{N^2 - 128}</math>. This is to be an integer. We can bound <math>N</math> now, as <math>N > 11</math> to avoid negative values and <math>N < 16</math> due to triangle inequality. Testing, <math>N = 12</math> works, giving <math>\overline{AB} = 48, \overline{BC} = 12</math>. | <math>\overline{BC} = 2\overline{BP} = 2(8\cos(x)) = N</math>, where <math>N</math> is an integer. Thus, <math>\cos(x) = \frac{N}{16}</math>. Via double angle, we calculate <math>\overline{AB}</math> to be <math>\frac{8\cos(x)}{2\cos(x)^2 - 1} = \frac{64N}{N^2 - 128}</math>. This is to be an integer. We can bound <math>N</math> now, as <math>N > 11</math> to avoid negative values and <math>N < 16</math> due to triangle inequality. Testing, <math>N = 12</math> works, giving <math>\overline{AB} = 48, \overline{BC} = 12</math>. | ||
− | Our answer is | + | Our answer is <math>2 * 48 + 12 = \boxed{108}</math>. |
==See Also== | ==See Also== |
Revision as of 16:07, 28 September 2019
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
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 2 (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 1 does.
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 in Solution 1.
Solution 4
Angle bisectors motivate trig bash.
Define angle . Foot of perpendicular from
to
is point
.
, where
is an integer. Thus,
. Via double angle, we calculate
to be
. This is to be an integer. We can bound
now, as
to avoid negative values and
due to triangle inequality. Testing,
works, giving
.
Our answer is
.
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.