Difference between revisions of "2016 AIME I Problems/Problem 6"
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==Solution 3== | ==Solution 3== | ||
− | WLOG assume <math>\triangle ABC</math> is isosceles (with vertex <math>C</math>). Let <math>O</math> be the center of the circumcircle, <math>r</math> the inradius, and <math>R</math> the circumradius. A simple sketch will reveal that <math>\triangle ABC</math> must also be obtuse (as an acute triangle will result in <math>LI</math> being greater than <math>DL</math>) and that <math>O</math> and <math>I</math> are collinear. Next, if <math>OI=d</math>, <math>DO+OI=R+d</math> and <math>R+d=DL+DI=5</math>. Euler gives us that <math>d^{2}=R(R-2r)</math>, and in this case, <math>r=LI=2</math>. Thus, <math>d=\sqrt{R^{2}-4R}</math>. Solving for <math>d</math>, we have <math>R+\sqrt{R^{2}-4R}=5</math>, then <math>R^{2}-4R=25-10R+R^{2}</math>, yielding <math>R=\frac{25}{6}</math>. Next, <math>R+d=5</math> so <math>d=\frac{5}{6}</math>. Finally, <math>OC=OI+ | + | WLOG assume <math>\triangle ABC</math> is isosceles (with vertex <math>C</math>). Let <math>O</math> be the center of the circumcircle, <math>r</math> the inradius, and <math>R</math> the circumradius. A simple sketch will reveal that <math>\triangle ABC</math> must also be obtuse (as an acute triangle will result in <math>LI</math> being greater than <math>DL</math>) and that <math>O</math> and <math>I</math> are collinear. Next, if <math>OI=d</math>, <math>DO+OI=R+d</math> and <math>R+d=DL+DI=5</math>. Euler gives us that <math>d^{2}=R(R-2r)</math>, and in this case, <math>r=LI=2</math>. Thus, <math>d=\sqrt{R^{2}-4R}</math>. Solving for <math>d</math>, we have <math>R+\sqrt{R^{2}-4R}=5</math>, then <math>R^{2}-4R=25-10R+R^{2}</math>, yielding <math>R=\frac{25}{6}</math>. Next, <math>R+d=5</math> so <math>d=\frac{5}{6}</math>. Finally, <math>OC=OI+CI</math> gives us <math>R=d+CI</math>, and <math>CI=\frac{25}{6}-\frac{5}{6}=\frac{10}{3}</math>. Our answer is then <math>\boxed{013}</math>. |
== See also == | == See also == | ||
{{AIME box|year=2016|n=I|num-b=5|num-a=7}} | {{AIME box|year=2016|n=I|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:40, 4 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
It is well known that and so we have
. Then
and so
and from the angle bisector theorem
so
and our answer is
Solution 2
This is a cheap solution.
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 inradius, and
the circumradius. A simple sketch will reveal that
must also 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 |
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