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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