Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2009 AIME II Problems/Problem 15"
(→‎Solution)
Line 5: Line 5:
 
(For some reason, I can't submit LaTeX for this page.)
 
(For some reason, I can't submit LaTeX for this page.)
  
−
Let O be the center of the circle. Define <MOC = t, <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 <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 C' be the foot of the perpendicular from C to MN. We compute XY as follows.
+
Let <math>C'</math> be the foot of the perpendicular from <math>C</math> to <math>MN</math>. We compute <math>XY</math> as follows.
  
−
(a) By the Extended Law of Sines in triangle ABC, we have
+
(a) By the Extended Law of Sines in triangle <math>ABC</math>, we have
  
−
CA
+
<cmath>CA</cmath>
  
−
= sin<ABC
+
<cmath>= \sin\angle{ABC}</cmath>
  
−
= sin((arc AN + arc NC)/2)
+
<cmath>= \sin\left(\frac{\widehat{AN} + \widehat{NC}}{2}\right)</cmath>
  
−
= sin((pi/2 + (pi-t))/2)
+
<cmath>= \sin\left(\frac{\frac{\pi}{2} + (\pi-t)}{2}\right)</cmath>
  
−
= sin(3pi/4 - t/2)
+
<cmath>= \sin\left(\frac{3\pi}{4} - \frac{t}{2}\right)</cmath>
  
−
= sin(pi/4 + t/2)
+
<cmath>= \sin\left(\frac{\pi}{4} + \frac{t}{2}\right)</cmath>
  
−
(b) Note that CC' = COsin(t) = (1/2)sin(t) and AO = 1/2. Since CC'Y and AOY are similar right triangles, we have CY/AY = CC'/AO = sin(t), and hence,
+
(b) Note that <math>CC' = CO\sin(t) = \left(\frac{1}{2}\right)\sin(t)</math> and <math>AO = \frac{1}{2}</math>. Since <math>CC'Y</math> and <math>AOY</math> are similar right triangles, we have <math>CY/AY = CC'/AO = \sin(t)</math>, and hence,
  
−
CY/CA
+
<cmath>CY/CA</cmath>
  
−
= CY/(CY + AY)
+
<cmath>= \frac{CY}{CY + AY}</cmath>
  
−
= sin(t) / (1 + sin(t))
+
<cmath>= \frac{\sin(t)}{1 + \sin(t)}</cmath>
  
−
= sin(t) / (sin(pi/2) + sin(t))
+
<cmath>= \frac{\sin(t)}{\sin\left(\frac{\pi}{2}\right) + \sin(t)}</cmath>
  
−
= sin(t) / (2sin(pi/4 + t/2)cos(pi/4 - t/2))
+
<cmath>= \frac{\sin(t)}{2\sin\left(\frac{\pi}{4} + \frac{t}{2}\right)\cos\left(\frac{\pi}{4} - \frac{t}{2}\right)}</cmath>
  
−
(c) We have <XCY = (arc AB)/2 = a and <CXY = (arc MB + arc CN)/2 = ((pi/2 - 2a) + (pi - t))/2 = 3pi/4 - a - t/2, and hence by the Law of Sines,
+
(c) We have <math>\angle{XCY} = \frac{\widehat{AB}}{2}=a</math> and <math>\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}</math>, and hence by the Law of Sines,
  
−
XY/CY
+
<cmath>XY/CY</cmath>
  
−
= sin<XCY / sin<CXY
+
<cmath>= \frac{\sin\angle{XCY}}{\sin\angle{CXY}}</cmath>
  
−
= sin(a) / sin(3pi/4 - a - t/2)
+
<cmath>= \frac{\sin(a)}{\sin\left(\frac{3\pi}{4} - a - \frac{t}{2}\right)}</cmath>
  
−
= sin(a) / sin(pi/4 + a + t/2).
+
<cmath>= \frac{\sin(a)}{\sin\left(\frac{\pi}{4} + a + \frac{t}{2}\right)}</cmath>
  
 
(d) Multiplying (a), (b), and (c), we have
 
(d) Multiplying (a), (b), and (c), we have
  
−
XY
+
<cmath>XY</cmath>
  
−
= CA * (CY/CA) * (XY/CY)
+
<cmath>= CA * (CY/CA) * (XY/CY)</cmath>
  
−
= sin(t)sin(a) / (2cos(pi/4 - t/2)sin(pi/4 + a + t/2))
+
<cmath>= \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)}</cmath>
  
−
= sin(t)sin(a) / (sin(pi/2 + a) + sin(a + t))
+
<cmath>= \frac{\sin(t)\sin(a)}{\sin\left(\frac{\pi}{2} + a\right) + \sin(a + t)}</cmath>
  
−
= sin(a) * sin(t) / (sin(t + a) + cos(a)),
+
<cmath>= \sin(a)\times\frac{\sin(t)}{\sin(t + a) + \cos(a)}</cmath>,
  
−
which is a function of t (and the constant a). Differentiating this with respect to t yields
+
which is a function of <math>t</math> (and the constant <math>a</math>). Differentiating this with respect to <math>t</math> yields
  
−
sin(a) * (cos(t)(sin(t + a) + cos(a)) - sin(t)cos(t + a)) / (sin(t + a) + cos(a))^2,
+
<cmath>\sin(a)\times\frac{\cos(t)(\sin(t + a) + \cos(a)) - \sin(t)\cos(t + a)}{(\sin(t + a) + \cos(a))^2}</cmath>,
  
 
and the numerator of this is
 
and the numerator of this is
  
−
sin(a) * (sin(t + a)cos(t) - cos(t + a)sin(t) + cos(a)cos(t)) = sin(a) * (sin(a) + cos(a)cos(t)),
+
<cmath>\sin(a) \times(\sin(t + a)\cos(t) - \cos(t + a)\sin(t) + \cos(a)\cos(t))</cmath>
 +
<cmath>= \sin(a) \times (\sin(a) + \cos(a)\cos(t))</cmath>,
  
−
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).
+
which vanishes when <math>\sin(a) + \cos(a)\cos(t) = 0</math>. Therefore, the length of <math>XY</math> is maximized when <math>t=t'</math>, where <math>t'</math> is the value in <math>[0, \pi]</math> that satisfies <math>\cos(t') = -\tan(a)</math>.
  
 
Note that
 
Note that
  
−
(1 - tan(a)) / (1 + tan(a)) = tan(pi/4 - a) = tan((arc MB)/2) = tan<MNB = 3/4,
+
<cmath>\frac{1 - \tan(a)}{1 + \tan(a)} = \tan\left(\frac{\pi}{4} - a\right) = \tan((\widehat{MB})/2) = \tan\angle{MNB} = \frac{3}{4}</cmath>,
  
−
so tan(a) = 1/7. We compute
+
so <math>\tan(a) = \frac{1}{7}</math>. We compute
  
−
sin(a) = sqrt(2)/10
+
<cmath>\sin(a) = \frac{\sqrt{2}}{10}</cmath>
  
−
cos(a) = 7sqrt(2)/10
+
<cmath>\cos(a) = \frac{7\sqrt{2}}{10}</cmath>
  
−
cos(t') = -tan(a) = -1/7
+
<cmath>\cos(t') = -\tan(a) = -\frac{1}{7}</cmath>
  
−
sin(t') = 4sqrt(3)/7
+
<cmath>\sin(t') = \frac{4\sqrt{3}}{7}</cmath>
  
−
sin(t' + a) = sin(t')cos(a) + cos(t')sin(a) = (28sqrt(6) - sqrt(2))/70,
+
<cmath>\sin(t' + a)=\sin(t')\cos(a) + \cos(t')\sin(a) = \frac{28\sqrt{6} - \sqrt{2}}{70}</cmath>,
  
−
so the maximum length of XY is sin(a) * sin(t') / (sin(t' + a) + cos(a)) = 7 - 4sqrt(3), and the answer is 7 + 4 + 3 = 014.
+
so the maximum length of <math>XY</math> is <math>\sin(a)\times\frac{\sin(t')}{\sin(t' + a) + \cos(a)} = 7 - 4sqrt(3)</math>, and the answer is <math>7 + 4 + 3 = \boxed{014}</math>.

Revision as of 21:29, 26 March 2011

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 $r+s+t$.

Solution

(For some reason, I can't submit LaTeX for this page.)

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_II_Problems/Problem_15&oldid=37789"