Difference between revisions of "2009 AIME I Problems/Problem 12"
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In right <math>\triangle ABC</math> with hypotenuse <math>\overline{AB}</math>, <math>AC = 12</math>, <math>BC = 35</math>, and <math>\overline{CD}</math> is the altitude to <math>\overline{AB}</math>. Let <math>\omega</math> be the circle having <math>\overline{CD}</math> as a diameter. Let <math>I</math> be a point outside <math>\triangle ABC</math> such that <math>\overline{AI}</math> and <math>\overline{BI}</math> are both tangent to circle <math>\omega</math>. The ratio of the perimeter of <math>\triangle ABI</math> to the length <math>AB</math> can be expressed in the form <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. | In right <math>\triangle ABC</math> with hypotenuse <math>\overline{AB}</math>, <math>AC = 12</math>, <math>BC = 35</math>, and <math>\overline{CD}</math> is the altitude to <math>\overline{AB}</math>. Let <math>\omega</math> be the circle having <math>\overline{CD}</math> as a diameter. Let <math>I</math> be a point outside <math>\triangle ABC</math> such that <math>\overline{AI}</math> and <math>\overline{BI}</math> are both tangent to circle <math>\omega</math>. The ratio of the perimeter of <math>\triangle ABI</math> to the length <math>AB</math> can be expressed in the form <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. | ||
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==Solution 2== | ==Solution 2== |
Revision as of 05:26, 30 September 2022
Contents
Problem
In right with hypotenuse
,
,
, and
is the altitude to
. Let
be the circle having
as a diameter. Let
be a point outside
such that
and
are both tangent to circle
. The ratio of the perimeter of
to the length
can be expressed in the form
, where
and
are relatively prime positive integers. Find
.
Solution 2
As in Solution , let
and
be the intersections of
with
and
respectively.
First, by pythagorean theorem, . Now the area of
is
, so
and the inradius of
is
.
Now from we find that
so
and similarly,
.
Note ,
, and
. So we have
,
. Now we can compute the area of
in two ways: by heron's formula and by inradius times semiperimeter, which yields
The quadratic formula now yields . Plugging this back in, the perimeter of
is
so the ratio of the perimeter to
is
and our answer is
Solution 3
As in Solution , let
and
be the intersections of
with
and
respectively.
Recall that the distance from a point outside a circle to that circle is the same along both tangent lines to the circle drawn from the point.
Recall also that the length of the altitude to the hypotenuse of a right-angle triangle is the geometric mean of the two segments into which it cuts the hypotenuse.
Let . Let
. Let
. The semi-perimeter of
is
.
Since the lengths of the sides of
are
,
and
, the square of its area by Heron's formula is
.
The radius of
is
. Therefore
. As
is the in-circle of
, the area of
is also
, and so the square area is
.
Therefore Dividing both sides by
we get:
and so
. The semi-perimeter of
is therefore
and the whole perimeter is
. Now
, so the ratio of the perimeter of
to the hypotenuse
is
and our answer is
and .
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.