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 (MAA)
By the Inscribed Angle Theorem, Let and be the feet of the altitudes of from and , respectively. In we get , and as exterior angle Because the lines and are bisectors of and , respectively, it follows thatThus the points , and are all on a circle. Further, since we have .
Because , it is sufficient to maximize the area of quadrilateral . If , are two points in an arc of circle with , then the maximum area of occurs when . Indeed, if , then replacing by the point located halfway in the arc of the circle yields a triangle with larger area than , and the area of remains the same. Similarly, if .
Therefore the maximum is achieved when , that is, when Thus and .
Solution 2
Let , , for convenience.
It's well-known that , , and (verifiable by angle chasing). Then, as , it follows that and consequently pentagon is cyclic. Observe that is fixed, hence the circumcircle of cyclic pentagon is also fixed. Similarly, as (both are radii), it follows that and also is fixed. Since is maximal, it suffices to maximize .
Verify that , by angle chasing; it follows that since by Triangle Angle Sum. Similarly, (isosceles base angles are equal), hence Since , by Inscribed Angles.
There are two ways to proceed.
Letting and be the circumcenter and circumradius, respectively, of cyclic pentagon , the most straightforward is to write , whence and, using the fact that is fixed, maximize with Jensen's Inequality.
A more elegant way is shown below.
Lemma: is maximized only if .
Proof by contradiction: Suppose is maximized when . Let be the midpoint of minor arc be and the midpoint of minor arc . Then since the altitude from to is greater than that from to ; similarly . Taking , to be the new orthocenter, incenter, respectively, this contradicts the maximality of , so our claim follows.
With our lemma() and from above, along with the fact that inscribed angles that intersect the same length chords are equal,
Video Solution by Osman Nal
https://www.youtube.com/watch?v=O3amRG9zEHE&ab_channel=OsmanNal
See also
2011 AMC 12A (Problems • Answer Key • Resources) | |
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