Difference between revisions of "2009 AIME II Problems/Problem 15"
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Suppose <math>\overline{AC}</math> and <math>\overline{BC}</math> intersect <math>\overline{MN}</math> at <math>D</math> and <math>E</math>, respectively, and let <math>MC = x</math> and <math>NC = y</math>. Since <math>A</math> is the midpoint of arc <math>MN</math>, <math>\overline{CA}</math> bisects <math>\angle MCN</math>, and we get | Suppose <math>\overline{AC}</math> and <math>\overline{BC}</math> intersect <math>\overline{MN}</math> at <math>D</math> and <math>E</math>, respectively, and let <math>MC = x</math> and <math>NC = y</math>. Since <math>A</math> is the midpoint of arc <math>MN</math>, <math>\overline{CA}</math> bisects <math>\angle MCN</math>, and we get | ||
<cmath>\frac{MC}{MD} = \frac{NC}{ND}\Rightarrow MD = \frac{x}{x + y}.</cmath> | <cmath>\frac{MC}{MD} = \frac{NC}{ND}\Rightarrow MD = \frac{x}{x + y}.</cmath> | ||
− | To find <math>ME</math>, we note that <math>\triangle | + | To find <math>ME</math>, we note that <math>\triangle BNE\sim\triangle MCE</math> and <math>\triangle BME\sim\triangle NCE</math>, so |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\frac{BN}{NE} &= \frac{MC}{CE} \\ | \frac{BN}{NE} &= \frac{MC}{CE} \\ |
Revision as of 01:34, 28 August 2015
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
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 2
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
.
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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.