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 that
Thus 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) | |
Preceded by Problem 24 |
Followed by Last Problem |
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All AMC 12 Problems and Solutions |
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