Difference between revisions of "2016 AIME I Problems/Problem 6"
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WLOG assume <math>\triangle ABC</math> is isosceles. Then, <math>L</math> is the midpoint of <math>AB</math>, and <math>\angle CLB=\angle CLA=90^\circ</math>. Draw the perpendicular from <math>I</math> to <math>CB</math>, and let it meet <math>CB</math> at <math>E</math>. Since <math>IL=2</math>, <math>IE</math> is also <math>2</math> (they are both inradii). Set <math>BD</math> as <math>x</math>. Then, triangles <math>BLD</math> and <math>CEI</math> are similar, and <math>\tfrac{2}{3}=\tfrac{CI}{x}</math>. Thus, <math>CI=\tfrac{2x}{3}</math>. <math>\triangle CBD~\triangle CEI</math>, so <math>\tfrac{IE}{DB}=\tfrac{CI}{CD}</math>. Thus <math>\tfrac{2}{x}=\tfrac{(2x/3)}{(2x/3+5)}</math>. Solving for <math>x</math>, we have: | WLOG assume <math>\triangle ABC</math> is isosceles. Then, <math>L</math> is the midpoint of <math>AB</math>, and <math>\angle CLB=\angle CLA=90^\circ</math>. Draw the perpendicular from <math>I</math> to <math>CB</math>, and let it meet <math>CB</math> at <math>E</math>. Since <math>IL=2</math>, <math>IE</math> is also <math>2</math> (they are both inradii). Set <math>BD</math> as <math>x</math>. Then, triangles <math>BLD</math> and <math>CEI</math> are similar, and <math>\tfrac{2}{3}=\tfrac{CI}{x}</math>. Thus, <math>CI=\tfrac{2x}{3}</math>. <math>\triangle CBD~\triangle CEI</math>, so <math>\tfrac{IE}{DB}=\tfrac{CI}{CD}</math>. Thus <math>\tfrac{2}{x}=\tfrac{(2x/3)}{(2x/3+5)}</math>. Solving for <math>x</math>, we have: |
Revision as of 10:57, 13 March 2016
Problem
In let
be the center of the inscribed circle, and let the bisector of
intersect
at
. The line through
and
intersects the circumscribed circle of
at the two points
and
. If
and
, then
, where
and
are relatively prime positive integers. Find
.
Solution
Solution 1
Suppose we label the angles as shown below.
As
and
intercept the same arc, we know that
. Similarly,
. Also, using
, we find
. Therefore,
. Therefore,
, so
must be isosceles with
. Similarly,
. Then
, hence
. Also,
bisects
, so by the Angle Bisector Theorem
. Thus
, and the answer is
.
Solution 2
WLOG assume is isosceles. Then,
is the midpoint of
, and
. Draw the perpendicular from
to
, and let it meet
at
. Since
,
is also
(they are both inradii). Set
as
. Then, triangles
and
are similar, and
. Thus,
.
, so
. Thus
. Solving for
, we have:
, or
.
is positive, so
. As a result,
and the answer is
Solution 3
WLOG assume is isosceles (with vertex
). Let
be the center of the circumcircle,
the circumradius, and
the inradius. A simple sketch will reveal that
must be obtuse (as an acute triangle will result in
being greater than
) and that
and
are collinear. Next, if
,
and
. Euler gives us that
, and in this case,
. Thus,
. Solving for
, we have
, then
, yielding
. Next,
so
. Finally,
gives us
, and
. Our answer is then
.
See also
2016 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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.