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
Let ,
,
for convenience.
It's well-known that ,
, and
(indeed, all are verifiable by angle chasing). Then, as
, it follows that
and consequently pentagon
is cyclic. Observe that
is fixed, whence the circumcircle of cyclic pentagon
is also fixed. Similarly, as
, it follows that
is the midpoint of minor arc
, so it's fixed as well. This implies that
is fixed, and since
is maximal, it suffices to maximize
.
Verify that ,
by angle chasing; it follows that
since
by Triangle Angle Sum. Similarly,
, whence
and consequently
by Inscribed Angles.
There are several 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 much more elegant way is shown below.
Lemma: is maximized only if
.
Proof: Suppose for the sake of contradiction that 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
, whence the claim follows.
It's necessary to show the existence of a maximum (although the wording of the problem gives it to you for free), which is not hard. Either way, since
by our lemma and
from above, it follows that
-Solution by thecmd999
See also
2011 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.