Difference between revisions of "2016 AIME I Problems/Problem 6"
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Since <math>\angle BCD</math> and <math>\angle BAD</math> both intercept arc <math>BD</math>, it follows that <math>\angle BAD=\gamma</math>. Note that <math>\angle AID=\alpha+\gamma</math> by the external angle theorem. It follows that <math>\angle DAI=\angle AID=\alpha+\gamma</math>, so we must have that <math>\triangle AID</math> is isosceles, yielding <math>AD=ID=5</math>. Note that <math>\triangle DLA \sim \triangle DAC</math>, so <math>\frac{DA}{DL} = \frac{DC}{DA}</math>. This yields <math>DC = \frac{25}{3}</math>. It follows that <math>CI = DC - DI = \frac{10}{3}</math>, giving a final answer of <math>\boxed{013}</math>. | Since <math>\angle BCD</math> and <math>\angle BAD</math> both intercept arc <math>BD</math>, it follows that <math>\angle BAD=\gamma</math>. Note that <math>\angle AID=\alpha+\gamma</math> by the external angle theorem. It follows that <math>\angle DAI=\angle AID=\alpha+\gamma</math>, so we must have that <math>\triangle AID</math> is isosceles, yielding <math>AD=ID=5</math>. Note that <math>\triangle DLA \sim \triangle DAC</math>, so <math>\frac{DA}{DL} = \frac{DC}{DA}</math>. This yields <math>DC = \frac{25}{3}</math>. It follows that <math>CI = DC - DI = \frac{10}{3}</math>, giving a final answer of <math>\boxed{013}</math>. | ||
− | == | + | Let <math>I_C</math> be the excenter opposite to <math>C</math> in <math>ABC</math>.By Fact 5 <math>DI=DC \therefore</math> <math>LI_C=8,LI=2,II_C=10</math>. Its well known that <math>(I_C,I,L,C)=-1 \implies \dfrac{LI_C}{LI}.\dfrac{CI}{CI_C}=-1 \implies \dfrac{CI}{CI+10}=\dfrac{1}{4} \implies \boxed{CI=\dfrac{10}{3}}</math>.<math>\blacksquare</math> |
− | {{ | + | ~Pluto1708 |
− | {{ |
Revision as of 16:42, 16 March 2019
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
.
Let be the excenter opposite to
in
.By Fact 5
. Its well known that
.
~Pluto1708