2012 AIME I Problems/Problem 12
Contents
Problem 12
Let be a right triangle with right angle at
Let
and
be points on
with
between
and
such that
and
trisect
If
then
can be written as
where
and
are relatively prime positive integers, and
is a positive integer not divisible by the square of any prime. Find
Solutions
Solution 1
Let . Using angle bisector theorem on
,
, so
. Then, drop the altitude from
to
and call the foot
. Thus,
,
, and
. Finally,
. Our answer is
.
Solution 2
Without loss of generality, set . Then, by the Angle Bisector Theorem on triangle
, we have
. We apply the Law of Cosines to triangle
to get
, which we can simplify to get
.
Now, we have by another application of the Law of Cosines to triangle
, so
. In addition,
, so
.
Our final answer is .
Solution 3
(This solution does not use the Angle Bisector Theorem or the Law of Cosines, but it uses the Law of Sines and more trig)
Find values for all angles in terms of .
,
,
,
, and
.
Use the law of sines on and
:
In ,
. This simplifies to
.
In ,
. This simplifies to
.
Solve for and equate them so that you get
.
From this, .
Use a trig identity on the denominator on the right to obtain:
This simplifies to
This gives
Dividing by
, we have
. Our final answer is
.
Solution 4
(This solution avoids advanced trigonometry)
Let be the foot of the perpendicular from
to
, and let
be the foot of the perpendicular from
to
.
Now let . Clearly, triangles
and
are similar with
, so
.
Since triangles and
are 30-60-90 right triangles, we can easily find other lengths in terms of
. For example, we see that
and
. Therefore
.
Again using the fact that triangles and
are similar, we see that
, so
.
Thus , and our answer is
.
Solution 5
(Another solution without trigonometry)
Extend to point
such that
. It is then clear that
is similar to
.
Let ,
. Then
.
With the Angle Bisector Theorem, we get that . From 30-60-90
, we get that
and
.
From , we have that
. Simplifying yields
, and
, so our answer is
.
Solution 6
Let , and let the feet of the altitudes from
and
to
be
and
, respectively. Also, let
and
. We see that
and
by right triangles
and
. From this we have that
. With the same triangles we have
and
. From 30-60-90 triangles
and
, we see that
and
, so
. From our two values of
we get:
Our answer is then
.
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012aimei/352
~ dolphin7
Solution 7
WLOG, let and
. First, by the Law of Sines on
, we find that
Now, we will find
. Consider the following diagram:
We have constructed equilateral triangle , and its circumcircle. Since
,
lies on
as well. Let
be the point diametrically opposite
on
, and let
be the foot of
on
(this is the midpoint of
). It is easy to compute that
and
. Therefore, by the Pythagorean Theorem,
. Now, by Power of a Point, we know that
, which means that
From before, we know that
. It's now easy to compute
as well using the Pythagorean identity; we find that
, and thus
for an answer of
.
See also
2012 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 |
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