Mock Geometry AIME 2011 Problems/Problem 12
Problem
A triangle has the property that its sides form an arithmetic progression, and that the angle opposite the longest side is three times the angle opposite the shortest side. The ratio of the longest side to the shortest side can be expressed as , where
are positive integers,
is not divisible by the square of any prime, and
and
are relatively prime. Find
.
Solution 1
Let the triangle be , such that
, and
, so that
. Construct two points
and
on
such that
.
Since ,
. Therefore,
, and
. Now
.
Let . From Angle Bisector Theorem on
with angle bisector
,
, so
.
Therefore, , and so
. Now we have
, and
.
Now we use Angle Bisector Theorem on with angle bisector
. We have
, so
.
Finally, notice that , so
.
Now, , so from the Quadratic Formula,
(we neglect the negative root), and so
.
Therefore, , and the answer is
.
Solution 2 (Trig)
Let the sides of the triangle be ,
, and
, since they form arithmetic progression. Let the their corresponding opposite angles be
and
respectively. Using Law of Sines, we find that:
.
Using the identity , we find that
. Applying double angle, we find
.
Then we use Law of Sines again, and use the substitution for , given that
.
.
Then factoring out and canceling it we get:
.
Cross multiplying yields:
Given that , and the earlier identity for
we get:
We can safely cancel since the solution,
would go out of range for the triangle sum of angles.
Expanding in the ensuing expression,
Cancelling and expanding
yields a cubic equal to 0 with GCD
. Since
would not work either, we are left with a quadratic. Solving for
, and neglecting the negative root since it would go out of bounds too, we find
.
The problem asks for the ratio of to
, using Law of Sines, this is
. This equals
. Substituting value of
, we find the ratio as
, or
.