2010 AIME I Problems/Problem 15
Problem
In with
,
, and
, let
be a point on
such that the incircles of
and
have equal radii. Let
and
be positive relatively prime integers such that
. Find
.
Solution
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Solution 1
Let , then
. Also let
Clearly,
. We can also express each area by the rs formula. Then
. Equating and cross-multiplying yields
or
Note that for d to be positive, we must have
.
By Stewart's Theorem, we have or
Brute forcing by plugging in our previous result for
, we have
Clearing the fraction and gathering like terms, we get
Aside: Since must be rational in order for our answer to be in the desired form, we can use the Rational Root Theorem to reveal that
is an integer. The only such
in the above-stated range is
.
Legitimately solving that quartic, note that and
should clearly be solutions, corresponding to the sides of the triangle and thus degenerate cevians. Factoring those out, we get
The only solution in the desired range is thus
. Then
, and our desired ratio
, giving us an answer of
.
Solution 2
Let and
so
. Let the incenters of
and
be
and
respectively, and their equal inradii be
. From
, we find that
Let the incircle of meet
at
and the incircle of
meet
at
. Then note that
is a rectangle. Also,
is right because
and
are the angle bisectors of
and
respectively and
. By properties of tangents to circles
and
. Now notice that the altitude of
to
is of length
, so by similar triangles we find that
(3). Equating (3) with (1) and (2) separately yields
and adding these we have
Solution 3
Let the incircle of hit
,
,
at
, and let the incircle of
hit
,
,
at
. Draw the incircle of
, and let it be tangent to
at
. Observe that we have a homothety centered at A sending the incircle of
to that of
, and one centered at
taking the incircle of
to that of
. These have the same power, since they take incircles of the same size to the same circle. Also, the power of the homothety is
.
By standard computations, and
. Now, let
and
. We will now go around and chase lengths. Observe that
. Then,
. We also have
, so
and
.
Observe now that . Also,
. Solving, we get
and
(as a side note, note that
, a result that I actually believe appears in Mandelbrot 1995-2003, or some book in that time-range).
Now, we get . To finish, we will compute area ratios.
. Also, since their inradii are equal, we get
. Equating and cross multiplying yields the quadratic
, so
. However, observe that
, so we take
. Our ratio is therefore
, giving the answer
.
See Also
- <url>viewtopic.php?t=338911 Discussion</url>
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