Difference between revisions of "2009 AIME II Problems/Problem 15"
5849206328x (talk | contribs) (→Solution 2) |
m (→Solution 2) |
||
Line 103: | Line 103: | ||
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 00:34, 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.