2016 AIME I Problems/Problem 6
Contents
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
.
Solution 4
Since and
,
. Also,
and
so
. Now we can call
,
and
,
. By angle bisector theorem,
. So let
and
for some value of
. Now call
. By the similar triangles we found earlier,
and
. We can simplify this to
and
. So we can plug the
into the first equation and get
. We can now draw a line through
and
that intersects
at
. By mass points, we can assign a mass of
to
,
to
, and
to
. We can also assign a mass of
to
by angle bisector theorem. So the ratio of
. So since
, we can plug this back into the original equation to get
. This means that
which has roots -2 and
which means our
and our answer is
.
Solution 5
Since and
both intercept arc
, it follows that
. Note that
by the external angle theorem. It follows that
, so we must have that
is isosceles, yielding
. Note that
, so
. This yields
. It follows that
, giving a final answer of
.
Solution 6
Let be the excenter opposite to
in
. By the incenter-excenter lemma
. Its well known that
.
~Pluto1708
Alternate solution: "We can use the angle bisector theorem on and bisector
to get that
. Since
, we get
. Thus,
and
."
(https://artofproblemsolving.com/community/c759169h1918283_geometry_problem)
Solution 7
We can just say that quadrilateral is a right kite with right angles at
and
. Let us construct another similar right kite with the points of tangency on
and
called
and
respectively, point
, and point
. Note that we only have to look at one half of the circle since the diagram is not symmetrical. Let us call
for simplicity's sake. Based on the fact that
is similar to
we can use triangle proportionality to say that
is
. Using geometric mean theorem we can show that
must be
. With Pythagorean Theorem we can say that
. Multiplying both sides by
and moving everything to LHS will give you
Since
must be in the form
we can assume that
is most likely a positive fraction in the form
where
is a factor of
. Testing the factors in synthetic division would lead
, giving us our desired answer
.
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 | ||
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