Difference between revisions of "2009 AIME II Problems/Problem 15"
5849206328x (talk | contribs) |
5849206328x (talk | contribs) m (→Solution 2) |
||
Line 119: | Line 119: | ||
By the AM-GM inequality, <math>3(x/y) + 4(y/x)\geq 2\sqrt{12} = 4\sqrt{3}</math>, so | By the AM-GM inequality, <math>3(x/y) + 4(y/x)\geq 2\sqrt{12} = 4\sqrt{3}</math>, so | ||
<cmath>DE\leq \frac{1}{4\sqrt{3} + 7} = 7 - 4\sqrt{3},</cmath> | <cmath>DE\leq \frac{1}{4\sqrt{3} + 7} = 7 - 4\sqrt{3},</cmath> | ||
− | giving the answer of <math>7 + 4 + 3 = \boxed{014}</math>. Equality is achieved when <math>3(x/y) = 4(y/x)</math> subject to the | + | giving the answer of <math>7 + 4 + 3 = \boxed{014}</math>. Equality is achieved when <math>3(x/y) = 4(y/x)</math> subject to the condition <math>x^2 + y^2 = 1</math>, which occurs for <math>x = \frac{2\sqrt{7}}{7}</math> and <math>y = \frac{\sqrt{21}}{7}</math> |
==See Also== | ==See Also== | ||
{{AIME box|year=2009|n=II|num-b=14|after=Last Problem}} | {{AIME box|year=2009|n=II|num-b=14|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:36, 7 April 2014
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.