2011 AMC 12A Problems/Problem 25
Problem
Triangle has
,
,
, and
. Let
,
, and
be the orthocenter, incenter, and circumcenter of
, respectively. Assume that the area of pentagon
is the maximum possible. What is
?
Solution
1) Let the circumcircle of have a center
. Since
,
is a chord that intercept an arc of
2) Define the length of as a unit length.
3) Draw the diameter to
. Let's call the interception of the diameter with
(because it is the midpoint) and interception with the circle
.
4) Since OMB and XMC are fixed, the area is a constant. Thus, also achieved maximum area.
Lemma:
For , we fixed it to
when we drew the diagram.
Let ,
Now, let's isolate the points ,
,
, and
.
,
$m\angle BIC = 180^{\circ} - \frac{\beta}{2} - \frac{\gamma}{2} = 180^{\circ} - ({180^{\circ} - 120^{\circ})= 120 ^{\circ}$ (Error compiling LaTeX. Unknown error_msg)
Now, lets isolate the points ,
,
, and
.
,
Lemma proven. The lemma yields that BOIHC is a cyclic pentagon.
Since XOIHC also achieved maximum area,
Let ,
,
, and the radius is
(which will drop out.)
Then the area = , where
So we want to maximize , Note that
.
Let's do some multivariable calculus.
,
If the partial derivatives with respect to and
are zero, then
, and it is very easy to show that
is the maximum with the second derivative test (left for the reader).
Now, we need to verify that such a situation exists and find the angle for this situation.
Let's extend to the direction of
, since
is the angle bisector, so that
intersects the midpoint of the arc
. Hence, if such a case exists,
, so
.
If the angle is , it is clear that since
and
are on the second circle (follows from the lemma).
will be at the right place.
can be easily verified too.
Hence, the answer is .
See also
2011 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
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All AMC 12 Problems and Solutions |
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