Difference between revisions of "2009 AIME II Problems/Problem 15"
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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 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> | + | 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 23:37, 6 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.