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Difference between revisions of "2009 AIME II Problems/Problem 15"

(→‎See Also)
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Let <math>\overline{MN}</math> be a diameter of a circle with diameter 1. Let <math>A</math> and <math>B</math> be points on one of the semicircular arcs determined by <math>\overline{MN}</math> such that <math>A</math> is the midpoint of the semicircle and <math>MB=\frac{3}5</math>. Point <math>C</math> lies on the other semicircular arc. Let <math>d</math> be the length of the line segment whose endpoints are the intersections of diameter <math>\overline{MN}</math> with chords <math>\overline{AC}</math> and <math>\overline{BC}</math>. The largest possible value of <math>d</math> can be written in the form <math> r-s\sqrt{t} </math>, where <math>r, s</math> and <math>t</math> are positive integers and <math>t</math> is not divisible by the square of any prime. Find <math>r+s+t</math>.
 
Let <math>\overline{MN}</math> be a diameter of a circle with diameter 1. Let <math>A</math> and <math>B</math> be points on one of the semicircular arcs determined by <math>\overline{MN}</math> such that <math>A</math> is the midpoint of the semicircle and <math>MB=\frac{3}5</math>. Point <math>C</math> lies on the other semicircular arc. Let <math>d</math> be the length of the line segment whose endpoints are the intersections of diameter <math>\overline{MN}</math> with chords <math>\overline{AC}</math> and <math>\overline{BC}</math>. The largest possible value of <math>d</math> can be written in the form <math> r-s\sqrt{t} </math>, where <math>r, s</math> and <math>t</math> are positive integers and <math>t</math> is not divisible by the square of any prime. Find <math>r+s+t</math>.
  
−
== Solution ==
+
== Solutions ==
  
 +
===Solution 1===
 
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>.
  
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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 2===
 +
<asy>
 +
unitsize(144);
 +
pair A, B, C, M, n;
 +
A = (0,1); B = (-7/25, 24/25); C=(1/7,-4*sqrt(3)/7); M = (-1,0); n = (1,0);
 +
pair [] D = intersectionpoints(A--C,M--n); pair [] e = intersectionpoints(B--C,M--n);
 +
 +
draw(circle((0,0),1));
 +
draw(M--n--B--M--A--n--C--A--B--C--cycle);
 +
 +
label("$A$",A,N); label("$B$",B,NNW); label("$M$",M,W); label("$C$",C,SSE); label("$N$",n,E);
 +
label("$D$",D[0],SE); label("$E$",e[0],SW);
 +
label("$x$",(M+C)/2,SW); label("$y$",(n+C)/2,SE);
 +
</asy>
 +
 +
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>
 +
To find <math>ME</math>, we note that <math>\triangle BND\sim\triangle MCD</math> and <math>\triangle BMD\sim\triangle NCD</math>, so
 +
<cmath>\begin{align*}
 +
\frac{BN}{NE} &= \frac{MC}{CE} \\
 +
\frac{ME}{BM} &= \frac{CE}{NC}.
 +
\end{align*}</cmath>
 +
Writing <math>NE = 1 - ME</math>, we can substitute known values and multiply the equations to get
 +
<cmath>\frac{4(ME)}{3 - 3(ME)} = \frac{x}{y}\Rightarrow ME = \frac{3x}{3x + 4y}.</cmath>
 +
The value we wish to maximize is
 +
<cmath>\begin{align*}
 +
DE &= MD - ME \\
 +
&= \frac{x}{x + y} - \frac{3x}{3x + 4y} \\
 +
&= \frac{xy}{3x^2 + 7xy + 4y^2} \\
 +
&= \frac{1}{3(x/y) + 4(y/x) + 7}.
 +
\end{align*}</cmath>
 +
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>
 +
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 conditions <math>x^2 + y^2 = 1</math> and <math>x > 0 > y</math>, which occurs for <math>(x,y) = \left(\frac{2\sqrt{7}}{7}, -\frac{\sqrt{21}}{7}\right)</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 11:04, 4 April 2014

Problem

Let $\overline{MN}$ be a diameter of a circle with diameter 1. Let $A$ and $B$ be points on one of the semicircular arcs determined by $\overline{MN}$ such that $A$ is the midpoint of the semicircle and $MB=\frac{3}5$. Point $C$ lies on the other semicircular arc. Let $d$ be the length of the line segment whose endpoints are the intersections of diameter $\overline{MN}$ with chords $\overline{AC}$ and $\overline{BC}$. The largest possible value of $d$ can be written in the form $r-s\sqrt{t}$, where $r, s$ and $t$ are positive integers and $t$ is not divisible by the square of any prime. Find $r+s+t$.

Solutions

Solution 1

Let $O$ be the center of the circle. Define $\angle{MOC}=t$, $\angle{BOA}=2a$, and let $BC$ and $AC$ intersect $MN$ at points $X$ and $Y$, respectively. We will express the length of $XY$ as a function of $t$ and maximize that function in the interval $[0, \pi]$.

Let $C'$ be the foot of the perpendicular from $C$ to $MN$. We compute $XY$ as follows.

(a) By the Extended Law of Sines in triangle $ABC$, we have

\[CA\]

\[= \sin\angle{ABC}\]

\[= \sin\left(\frac{\widehat{AN} + \widehat{NC}}{2}\right)\]

\[= \sin\left(\frac{\frac{\pi}{2} + (\pi-t)}{2}\right)\]

\[= \sin\left(\frac{3\pi}{4} - \frac{t}{2}\right)\]

\[= \sin\left(\frac{\pi}{4} + \frac{t}{2}\right)\]

(b) Note that $CC' = CO\sin(t) = \left(\frac{1}{2}\right)\sin(t)$ and $AO = \frac{1}{2}$. Since $CC'Y$ and $AOY$ are similar right triangles, we have $CY/AY = CC'/AO = \sin(t)$, and hence,

\[CY/CA\]

\[= \frac{CY}{CY + AY}\]

\[= \frac{\sin(t)}{1 + \sin(t)}\]

\[= \frac{\sin(t)}{\sin\left(\frac{\pi}{2}\right) + \sin(t)}\]

\[= \frac{\sin(t)}{2\sin\left(\frac{\pi}{4} + \frac{t}{2}\right)\cos\left(\frac{\pi}{4} - \frac{t}{2}\right)}\]

(c) We have $\angle{XCY} = \frac{\widehat{AB}}{2}=a$ and $\angle{CXY} = \frac{\widehat{MB}+\widehat{CN}}{2} = \frac{\left(\frac{\pi}{2} - 2a\right) + (\pi - t)}{2} = \frac{3\pi}{4} - a - \frac{t}{2}$, and hence by the Law of Sines,

\[XY/CY\]

\[= \frac{\sin\angle{XCY}}{\sin\angle{CXY}}\]

\[= \frac{\sin(a)}{\sin\left(\frac{3\pi}{4} - a - \frac{t}{2}\right)}\]

\[= \frac{\sin(a)}{\sin\left(\frac{\pi}{4} + a + \frac{t}{2}\right)}\]

(d) Multiplying (a), (b), and (c), we have

\[XY\]

\[= CA * (CY/CA) * (XY/CY)\]

\[= \frac{\sin(t)\sin(a)}{2\cos\left(\frac{\pi}{4} - \frac{t}{2}\right)\sin\left(\frac{\pi}{4} + a + \frac{t}{2}\right)}\]

\[= \frac{\sin(t)\sin(a)}{\sin\left(\frac{\pi}{2} + a\right) + \sin(a + t)}\]

\[= \sin(a)\times\frac{\sin(t)}{\sin(t + a) + \cos(a)}\],

which is a function of $t$ (and the constant $a$). Differentiating this with respect to $t$ yields

\[\sin(a)\times\frac{\cos(t)(\sin(t + a) + \cos(a)) - \sin(t)\cos(t + a)}{(\sin(t + a) + \cos(a))^2}\],

and the numerator of this is

\[\sin(a) \times(\sin(t + a)\cos(t) - \cos(t + a)\sin(t) + \cos(a)\cos(t))\] \[= \sin(a) \times (\sin(a) + \cos(a)\cos(t))\],

which vanishes when $\sin(a) + \cos(a)\cos(t) = 0$. Therefore, the length of $XY$ is maximized when $t=t'$, where $t'$ is the value in $[0, \pi]$ that satisfies $\cos(t') = -\tan(a)$.

Note that

\[\frac{1 - \tan(a)}{1 + \tan(a)} = \tan\left(\frac{\pi}{4} - a\right) = \tan((\widehat{MB})/2) = \tan\angle{MNB} = \frac{3}{4}\],

so $\tan(a) = \frac{1}{7}$. We compute

\[\sin(a) = \frac{\sqrt{2}}{10}\]

\[\cos(a) = \frac{7\sqrt{2}}{10}\]

\[\cos(t') = -\tan(a) = -\frac{1}{7}\]

\[\sin(t') = \frac{4\sqrt{3}}{7}\]

\[\sin(t' + a)=\sin(t')\cos(a) + \cos(t')\sin(a) = \frac{28\sqrt{6} - \sqrt{2}}{70}\],

so the maximum length of $XY$ is $\sin(a)\times\frac{\sin(t')}{\sin(t' + a) + \cos(a)} = 7 - 4\sqrt{3}$, and the answer is $7 + 4 + 3 = \boxed{014}$.

Solution 2

[asy] unitsize(144); pair A, B, C, M, n; A = (0,1); B = (-7/25, 24/25); C=(1/7,-4*sqrt(3)/7); M = (-1,0); n = (1,0); pair [] D = intersectionpoints(A--C,M--n); pair [] e = intersectionpoints(B--C,M--n); draw(circle((0,0),1)); draw(M--n--B--M--A--n--C--A--B--C--cycle); label("$A$",A,N); label("$B$",B,NNW); label("$M$",M,W); label("$C$",C,SSE); label("$N$",n,E); label("$D$",D[0],SE); label("$E$",e[0],SW); label("$x$",(M+C)/2,SW); label("$y$",(n+C)/2,SE); [/asy]

Suppose $\overline{AC}$ and $\overline{BC}$ intersect $\overline{MN}$ at $D$ and $E$, respectively, and let $MC = x$ and $NC = y$. Since $A$ is the midpoint of arc $MN$, $\overline{CA}$ bisects $\angle MCN$, and we get \[\frac{MC}{MD} = \frac{NC}{ND}\Rightarrow MD = \frac{x}{x + y}.\] To find $ME$, we note that $\triangle BND\sim\triangle MCD$ and $\triangle BMD\sim\triangle NCD$, so \begin{align*} \frac{BN}{NE} &= \frac{MC}{CE} \\ \frac{ME}{BM} &= \frac{CE}{NC}. \end{align*} Writing $NE = 1 - ME$, we can substitute known values and multiply the equations to get \[\frac{4(ME)}{3 - 3(ME)} = \frac{x}{y}\Rightarrow ME = \frac{3x}{3x + 4y}.\] The value we wish to maximize is \begin{align*} DE &= MD - ME \\ &= \frac{x}{x + y} - \frac{3x}{3x + 4y} \\ &= \frac{xy}{3x^2 + 7xy + 4y^2} \\ &= \frac{1}{3(x/y) + 4(y/x) + 7}. \end{align*} By the AM-GM inequality, $3(x/y) + 4(y/x)\geq 2\sqrt{12} = 4\sqrt{3}$, so \[DE\leq \frac{1}{4\sqrt{3} + 7} = 7 - 4\sqrt{3},\] giving the answer of $7 + 4 + 3 = \boxed{014}$. Equality is achieved when $3(x/y) = 4(y/x)$ subject to the conditions $x^2 + y^2 = 1$ and $x > 0 > y$, which occurs for $(x,y) = \left(\frac{2\sqrt{7}}{7}, -\frac{\sqrt{21}}{7}\right)$.

See Also

2009 AIME II (Problems • Answer Key • Resources)
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