Difference between revisions of "2009 AIME II Problems/Problem 15"
(→Solutions) |
(→Solutions) |
||
Line 4: | Line 4: | ||
== Solutions == | == Solutions == | ||
− | ===Solution 1=== | + | ===Solution 1 (Quick)=== |
+ | Let <math>V = \overline{NM} \cap \overline{AC}</math> and <math>W = \overline{NM} \cap \overline{BC}</math>. Further more let <math>\angle MNC = \alpha</math> and <math>\angle NMC = 90^\circ - \alpha</math>. Angle chasing reveals <math>\angle NBC = \angle NAC = \alpha</math> and <math>\angle MBC = \angle MAC = 90^\circ - \alpha</math>. Additionally <math>NB = \frac{4}{5}</math> and <math>AN = AM</math> by the Pythagorean Theorem. | ||
+ | |||
+ | By the Angle Bisector Formula, | ||
+ | <cmath>\frac{NV}{MV} = \frac{\sin (\alpha)}{\sin (90^\circ - \alpha)} = \tan (\alpha)</cmath> | ||
+ | <cmath>\frac{MW}{NW} = \frac{3\sin (90^\circ - \alpha)}{4\sin (\alpha)} = \frac{3}{4} \cot (\alpha)</cmath> | ||
+ | |||
+ | As <math>NV + MV =MW + NW = 1</math> we compute <math>NW = \frac{1}{1+\frac{3}{4}\cot(\alpha)}</math> and <math>MV = \frac{1}{1+\tan (\alpha)}</math>, and finally <math>VW = NW + MV - 1 = \frac{1}{1+\frac{3}{4}\cot(\alpha)} + \frac{1}{1+\tan (\alpha)} - 1</math>. Taking the derivative of <math>VW</math> with respect to <math>\alpha</math>, we arrive at | ||
+ | <cmath>VW' = \frac{7\cos^2 (\alpha) - 4}{(\sin(\alpha) + \cos(\alpha))^2(4\sin(\alpha)+3\cos(\alpha))^2}</cmath> | ||
+ | Clearly the maximum occurs when <math>\alpha = \cos^{-1}\left(\frac{2}{\sqrt{7}}\right)</math>. Plugging this back in, using the fact that <math>\tan(\cos^{-1}(x)) = \frac{\sqrt{1-x^2}}{x}</math> and <math>\cot(\cos^{-1}(x)) = \frac{x}{\sqrt{1-x^2}}</math>, we get | ||
+ | |||
+ | <math>VW = 7 - 4\sqrt{3}</math> | ||
+ | with <math>7 + 4 + 3 = \boxed{014}</math> | ||
+ | ===Solution 2=== | ||
Let <math>O</math> be the center of the circle. Define <math>\angle{MOC}=t</math>, <math>\angle{BOA}=2a</math>, and let <math>BC</math> and <math>AC</math> intersect <math>MN</math> at points <math>X</math> and <math>Y</math>, respectively. We will express the length of <math>XY</math> as a function of <math>t</math> and maximize that function in the interval <math>[0, \pi]</math>. | Let <math>O</math> be the center of the circle. Define <math>\angle{MOC}=t</math>, <math>\angle{BOA}=2a</math>, and let <math>BC</math> and <math>AC</math> intersect <math>MN</math> at points <math>X</math> and <math>Y</math>, respectively. We will express the length of <math>XY</math> as a function of <math>t</math> and maximize that function in the interval <math>[0, \pi]</math>. | ||
Line 86: | Line 99: | ||
so the maximum length of <math>XY</math> is <math>\sin(a)\times\frac{\sin(t')}{\sin(t' + a) + \cos(a)} = 7 - 4\sqrt{3}</math>, and the answer is <math>7 + 4 + 3 = \boxed{014}</math>. | so the maximum length of <math>XY</math> is <math>\sin(a)\times\frac{\sin(t')}{\sin(t' + a) + \cos(a)} = 7 - 4\sqrt{3}</math>, and the answer is <math>7 + 4 + 3 = \boxed{014}</math>. | ||
− | ===Solution | + | ===Solution 3=== |
<asy> | <asy> | ||
unitsize(144); | unitsize(144); | ||
Line 121: | Line 134: | ||
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>. | ||
− | ===Solution | + | ===Solution 4 (Projective)=== |
By Pythagoras in <math>\triangle BMN,</math> we get <math>BN=\dfrac{4}{5}.</math> | By Pythagoras in <math>\triangle BMN,</math> we get <math>BN=\dfrac{4}{5}.</math> | ||
Revision as of 12:29, 31 July 2017
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.
~Generic_Username
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.