2010 AIME II Problems/Problem 15
Contents
Problem 15
In triangle ,
,
, and
. Points
and
lie on
with
and
. Points
and
lie on
with
and
. Let
be the point, other than
, of intersection of the circumcircles of
and
. Ray
meets
at
. The ratio
can be written in the form
, where
and
are relatively prime positive integers. Find
.
Solution
Let .
since
. Since quadrilateral
is cyclic,
and
, yielding
and
. Multiplying these together yields *
.
.
Now we claim that . To prove this, we can use cyclic quadrilaterals.
From ,
and
. So,
and
.
From ,
and
. Thus,
and
.
Thus, from AA similarity, .
Therefore, , which can easily be computed by the angle bisector theorem to be
. It follows that *
, giving us an answer of
.
- These two ratios are the same thing and can also be derived from the Ratio Lemma.
Ratio Lemma :, for any cevian AD of a triangle ABC.
Source: [1] by Zhero
Extension
The work done in this problem leads to a nice extension of this problem:
Given a and points
,
,
,
,
,
, such that
,
,
,
, and
,
, then let
be the circumcircle of
and
be the circumcircle of
. Let
be the intersection point of
and
distinct from
. Define
and
similarly. Then
,
, and
concur.
This can be proven using Ceva's theorem and the work done in this problem, which effectively allows us to compute the ratio that line divides the opposite side
into and similarly for the other two sides.
See Also
2010 AIME II (Problems • Answer Key • Resources) | ||
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