2009 AIME II Problems/Problem 15
Contents
Problem
Let be a diameter of a circle with diameter 1. Let
and
be points on one of the semicircular arcs determined by
such that
is the midpoint of the semicircle and
. Point
lies on the other semicircular arc. Let
be the length of the line segment whose endpoints are the intersections of diameter
with chords
and
. The largest possible value of
can be written in the form
, where
and
are positive integers and
is not divisible by the square of any prime. Find
.
Solutions
Solution 1 (Quick)
Let and
. Further more let
and
. Angle chasing reveals
and
. Additionally
and
by the Pythagorean Theorem.
By the Angle Bisector Formula,
As we compute
and
, and finally
. Taking the derivative of
with respect to
, we arrive at
Clearly the maximum occurs when
. Plugging this back in, using the fact that
and
, we get
with
Solution 2
Let be the center of the circle. Define
,
, and let
and
intersect
at points
and
, respectively. We will express the length of
as a function of
and maximize that function in the interval
.
Let be the foot of the perpendicular from
to
. We compute
as follows.
(a) By the Extended Law of Sines in triangle , we have
(b) Note that and
. Since
and
are similar right triangles, we have
, and hence,
(c) We have and
, and hence by the Law of Sines,
(d) Multiplying (a), (b), and (c), we have
,
which is a function of (and the constant
). Differentiating this with respect to
yields
,
and the numerator of this is
,
which vanishes when . Therefore, the length of
is maximized when
, where
is the value in
that satisfies
.
Note that
,
so . We compute
,
so the maximum length of is
, and the answer is
.
Solution 3
Suppose and
intersect
at
and
, respectively, and let
and
. Since
is the midpoint of arc
,
bisects
, and we get
To find
, we note that
and
, so
Writing
, we can substitute known values and multiply the equations to get
The value we wish to maximize is
By the AM-GM inequality,
, so
giving the answer of
. Equality is achieved when
subject to the condition
, which occurs for
and
.
Solution 4 (Projective)
By Pythagoras in we get
Since cross ratios are preserved upon projecting, note that By definition of a cross ratio, this becomes
Let
such that
We know that
so the LHS becomes
In the RHS, we are given every value except for However, Ptolemy's Theorem on
gives
Substituting, we get
where we use
Again using we have
Then
Since this is a function in
we differentiate WRT
to find its maximum. By quotient rule, it suffices to solve
Substituting back yields
so
is the answer.
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See Also
2009 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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