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 "1996 AHSME Problems/Problem 30"

m (→‎Solution 5 (Ptolemy's theorem))
(→‎Solution 1)
 
(3 intermediate revisions by one other user not shown)
Line 4: Line 4:
 
<math>\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389  \qquad \textbf{(E)}\ 409 </math>
 
<math>\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389  \qquad \textbf{(E)}\ 409 </math>
  
−
==Solution 1==
+
==Solution 1 (Alcumus)==
 
In hexagon <math>ABCDEF</math>, let <math>AB=BC=CD=3</math> and let <math>DE=EF=FA=5</math>. Since arc <math>BAF</math> is one third of the circumference of the circle, it follows that <math>\angle BCF = \angle BEF=60^{\circ}</math>. Similarly, <math>\angle CBE =\angle CFE=60^{\circ}</math>. Let <math>P</math> be the intersection of <math>\overline{BE}</math> and <math>\overline{CF}</math>, <math>Q</math> that of <math>\overline{BE}</math> and <math>\overline{AD}</math>, and <math>R</math> that of <math>\overline{CF}</math> and <math>\overline{AD}</math>. Triangles <math>EFP</math> and  <math>BCP</math> are equilateral, and by symmetry, triangle <math>PQR</math> is isosceles and thus also equilateral.  
 
In hexagon <math>ABCDEF</math>, let <math>AB=BC=CD=3</math> and let <math>DE=EF=FA=5</math>. Since arc <math>BAF</math> is one third of the circumference of the circle, it follows that <math>\angle BCF = \angle BEF=60^{\circ}</math>. Similarly, <math>\angle CBE =\angle CFE=60^{\circ}</math>. Let <math>P</math> be the intersection of <math>\overline{BE}</math> and <math>\overline{CF}</math>, <math>Q</math> that of <math>\overline{BE}</math> and <math>\overline{AD}</math>, and <math>R</math> that of <math>\overline{CF}</math> and <math>\overline{AD}</math>. Triangles <math>EFP</math> and  <math>BCP</math> are equilateral, and by symmetry, triangle <math>PQR</math> is isosceles and thus also equilateral.  
 
<asy>
 
<asy>
Line 108: Line 108:
 
[[File:1996AHSMEP305.png|500px|center]]
 
[[File:1996AHSMEP305.png|500px|center]]
  
−
Note that minor arc <math>\overarc{AB}</math> is a third of the circumference, therefore, <math>\angle AOB = 120^{\circ}</math>.
+
Note that major arc <math>\overarc{AE}</math> is two thirds of the circumference, therefore, <math>\angle AFE = 120^{\circ}</math>.
  
−
By the Law of Cosine, <math>AB = \sqrt{ 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cdot \cos 120^{\circ}} = \sqrt{ 9 + 25 + 15 } = 7</math>
+
By the Law of Cosine, <math>AE= \sqrt{ 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cdot \cos 120^{\circ}} = \sqrt{ 9 + 25 + 15 } = 7</math>
  
−
By the Ptolemy's theorem of quadrilateral <math>ABDE</math>, <math>AD \cdot BE = AB \cdot DE + BD \cdot AE</math>, <math>AD^2= 3 \cdot 5 + 7^2 = 64</math>, <math>AD = 8</math>
+
By the Ptolemy's theorem of quadrilateral <math>ABDE</math>, <math>AD \cdot BE = AB \cdot DE + BD \cdot AE</math>, <math>AD = BE</math>, <math>AD^2= 3 \cdot 5 + 7^2 = 64</math>, <math>AD = 8</math>
  
 
By the Ptolemy's theorem of quadrilateral <math>ABCD</math>, <math>AC \cdot BD = BC \cdot AD + AB \cdot CD</math>, <math>7AC = 3 \cdot 8 + 3 \cdot 5 = 39</math>, <math>AC = \frac{39}{7}</math>
 
By the Ptolemy's theorem of quadrilateral <math>ABCD</math>, <math>AC \cdot BD = BC \cdot AD + AB \cdot CD</math>, <math>7AC = 3 \cdot 8 + 3 \cdot 5 = 39</math>, <math>AC = \frac{39}{7}</math>
  
−
By the Ptolemy's theorem of quadrilateral <math>ABCF</math>, <math>AC \cdot BF = AB \cdot CF + BC \cdot AF</math>, <math>(\frac{39}{7})^2 = 3 \cdot CF + 3 \cdot 3</math>, <math>3CF = (\frac{39}{7})^2 - 9 = \frac{1521-441}{49} = \frac{1080}{49}</math>, <math>CF = \frac{360}{49}</math>
+
By the Ptolemy's theorem of quadrilateral <math>ABCF</math>, <math>AC \cdot BF = AB \cdot CF + BC \cdot AF</math>, <math>AC = BF</math>, <math>(\frac{39}{7})^2 = 3 \cdot CF + 3 \cdot 3</math>, <math>CF = \frac{360}{49}</math>
  
 
Therefore, the answer is <math>\boxed{\textbf{(E) } 409}</math>.
 
Therefore, the answer is <math>\boxed{\textbf{(E) } 409}</math>.

Latest revision as of 12:09, 20 December 2023

Problem

A hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5. The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length 3 and the other with three sides each of length 5, has length equal to $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

$\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389 \qquad \textbf{(E)}\ 409$

Solution 1 (Alcumus)

In hexagon $ABCDEF$, let $AB=BC=CD=3$ and let $DE=EF=FA=5$. Since arc $BAF$ is one third of the circumference of the circle, it follows that $\angle BCF = \angle BEF=60^{\circ}$. Similarly, $\angle CBE =\angle CFE=60^{\circ}$. Let $P$ be the intersection of $\overline{BE}$ and $\overline{CF}$, $Q$ that of $\overline{BE}$ and $\overline{AD}$, and $R$ that of $\overline{CF}$ and $\overline{AD}$. Triangles $EFP$ and $BCP$ are equilateral, and by symmetry, triangle $PQR$ is isosceles and thus also equilateral. [asy] import olympiad; import geometry; size(150); defaultpen(linewidth(0.8)); real angleUnit = 15; draw(Circle(origin,1)); pair D = dir(22.5); pair C = dir(3*angleUnit + degrees(D)); pair B = dir(3*angleUnit + degrees(C)); pair A = dir(3*angleUnit + degrees(B)); pair F = dir(5*angleUnit + degrees(A)); pair E = dir(5*angleUnit + degrees(F)); draw(A--B--C--D--E--F--cycle); dot("$A$",A,A); dot("$B$",B,B); dot("$C$",C,C); dot("$D$",D,D); dot("$E$",E,E); dot("$F$",F,F); draw(A--D^^B--E^^C--F); label("$3$",D--C,SW); label("$3$",B--C,S); label("$3$",A--B,SE); label("$5$",A--F,NE); label("$5$",F--E,N); label("$5$",D--E,NW); [/asy]

Furthermore, $\angle BAD$ and $\angle BED$ subtend the same arc, as do $\angle ABE$ and $\angle ADE$. Hence triangles $ABQ$ and $EDQ$ are similar. Therefore, \[\frac{AQ}{EQ}=\frac{BQ}{DQ}=\frac{AB}{ED}=\frac{3}{5}.\] It follows that \[\frac{\frac{AD-PQ}{2}}{PQ+5} =\frac{3}{5}\quad \mbox {and}\quad \frac{3-PQ}{\frac{AD+PQ}{2}}=\frac{3}{5}.\] Solving the two equations simultaneously yields $AD=360/49,$ so $m+n=\boxed{409}. \blacksquare$

Solution 2

All angle measures are in degrees. Let the first trapezoid be $ABCD$, where $AB=BC=CD=3$. Then the second trapezoid is $AFED$, where $AF=FE=ED=5$. We look for $AD$.

Since $ABCD$ is an isosceles trapezoid, we know that $\angle BAD=\angle CDA$ and, since $AB=BC$, if we drew $AC$, we would see $\angle BCA=\angle BAC$. Anyway, $\widehat{AB}=\widehat{BC}=\widehat{CD}$ ($\widehat{AB}$ means arc AB). Using similar reasoning, $\widehat{AF}=\widehat{FE}=\widehat{ED}$.

Let $\widehat{AB}=2\phi$ and $\widehat{AF}=2\theta$. Since $6\theta+6\phi=360$ (add up the angles), $2\theta+2\phi=120$ and thus $\widehat{AB}+\widehat{AF}=\widehat{BF}=120$. Therefore, $\angle FAB=\frac{1}{2}\widehat{BDF}=\frac{1}{2}(240)=120$. $\angle CDE=120$ as well.

Now I focus on triangle $FAB$. By the Law of Cosines, $BF^2=3^2+5^2-30\cos{120}=9+25+15=49$, so $BF=7$. Seeing $\angle ABF=\theta$ and $\angle AFB=\phi$, we can now use the Law of Sines to get: \[\sin{\phi}=\frac{3\sqrt{3}}{14}\;\text{and}\;\sin{\theta}=\frac{5\sqrt{3}}{14}.\]

Now I focus on triangle $AFD$. $\angle AFD=3\phi$ and $\angle ADF=\theta$, and we are given that $AF=5$, so \[\frac{\sin{\theta}}{5}=\frac{\sin{3\phi}}{AD}.\] We know $\sin{\theta}=\frac{5\sqrt{3}}{14}$, but we need to find $\sin{3\phi}$. Using various identities, we see \begin{align*}\sin{3\phi}&=\sin{(\phi+2\phi)}=\sin{\phi}\cos{2\phi}+\cos{\phi}\sin{2\phi}\\ &=\sin{\phi}(1-2\sin^2{\phi})+2\sin{\phi}\cos^2{\phi}\\ &=\sin{\phi}\left(1-2\sin^2{\phi}+2(1-\sin^2{\phi})\right)\\ &=\sin{\phi}(3-4\sin^2{\phi})\\ &=\frac{3\sqrt{3}}{14}\left(3-\frac{27}{49}\right)=\frac{3\sqrt{3}}{14}\left(\frac{120}{49}\right)=\frac{180\sqrt{3}}{343} \end{align*} Returning to finding $AD$, we remember \[\frac{\sin{\theta}}{5}=\frac{\sin{3\phi}}{AD}\;\text{so}\;AD=\frac{5\sin{3\phi}}{\sin{\theta}}.\] Plugging in and solving, we see $AD=\frac{360}{49}$. Thus, the answer is $360 + 49 = 409$, which is answer choice $\boxed{\textbf{(E)}}$.

Solution 3

Let $x$ be the desired length. One can use Parameshvara's circumradius formula, which states that for a cyclic quadrilateral with sides $a, b, c, d$ the circumradius $R$ satisfies \[R^2=\frac{1}{16}\cdot\frac{(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)},\] where $s$ is the semiperimeter. Applying this to the trapezoid with sides $3, 3, 3, x$, we see that many terms cancel and we are left with \[R^2=\frac{27}{9-x}\] Similar canceling occurs for the trapezoid with sides $5, 5, 5, x$, and since the two quadrilaterals share the same circumradius, we can equate: \[\frac{27}{9-x}=\frac{125}{15-x}\] Solving for $x$ gives $x=\frac{360}{49}$, so the answer is $\fbox{(E) 409}$.

Solution 4

1996AHSMEP30.png

Note that minor arc $\overarc{AB}$ is a third of the circumference, therefore, $\angle AOB = 120^{\circ}$. Major arc $\overarc{AB}$$=240^{\circ}$, $\angle ACB = 120^{\circ}$

By the Law of Cosine, $AB = \sqrt{ 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cdot \cos 120^{\circ}} = \sqrt{ 9 + 25 + 15 } = 7$

$\frac{\angle AOB}{2} = 60^{\circ}$, therefore, $r = \frac{\frac{AB}{2}}{\sin 60^{\circ}} = \frac{\frac{7}{2}}{ \frac{\sqrt{3}}{2} } = \frac{7\sqrt{3}}{3}$

$\sin \frac{\theta}{2} = \frac{\frac32}{r} = \frac{3}{2r} = \frac{3}{2 \cdot \frac{7\sqrt{3}}{3}} = \frac{3 \sqrt{3}}{14}$

Let $x$ be the length of the chord, $\sin \frac{3 \theta}{2} = \frac{\frac{x}{2}}{r}$

By the triple angle formula, $\sin \frac{3 \theta}{2} = 3 \cdot \sin \frac{\theta}{2} - 4 \cdot \sin(\frac{ \theta}{2})^3 = 3 \cdot \frac{3 \sqrt{3}}{14} - 4 \cdot (\frac{3 \sqrt{3}}{14})^3$

$x = 2 \cdot \frac{7\sqrt{3}}{3} \cdot [3 \cdot \frac{3 \sqrt{3}}{14} - 4 \cdot (\frac{3 \sqrt{3}}{14})^3] = 2 \cdot \frac{7\sqrt{3}}{3} \cdot (\frac{9\sqrt{3}}{14} - \frac{82\sqrt{3}}{2 \cdot 7^3}) = 9 - \frac{81}{49} = \frac{360}{49}$

Therefore, the answer is $\boxed{\textbf{(E) } 409}$.

~isabelchen

Solution 5

1996AHSMEP302.png

Note that minor arc $\overarc{AB}$ is a third of the circumference, therefore, $\angle AOB = 120^{\circ}$.

$\sin \frac{\alpha}{2} = \frac{\frac32}{r}$, $\sin \frac{\alpha}{2} = \frac{3}{2r}$

$\sin \frac{120^{\circ}-\alpha}{2} = \frac{\frac52}{r}$, $\sin (60^{\circ} - \frac{\alpha}{2}) = \frac{5}{2r}$

$\frac{\sin \frac{\alpha}{2}}{\sin (60^{\circ} - \frac{\alpha}{2}) } = \frac{\frac{3}{2r}}{\frac{5}{2r}} = \frac35$, $5 \cdot \sin \frac{\alpha}{2} = 3 \cdot \sin (60^{\circ} - \frac{\alpha}{2})$

$5 \cdot \sin \frac{\alpha}{2} = 3 ( \sin 60^{\circ} \cos \frac{\alpha}{2} - \sin \frac{\alpha}{2} \cos 60^{\circ}) = 3 ( \frac{\sqrt{3}}{2} \cdot \cos \frac{\alpha}{2} - \frac12 \cdot \sin \frac{\alpha}{2})$

$13 \cdot \sin \frac{\alpha}{2} = \frac{3\sqrt{3}}{2} \cdot \cos \frac{\alpha}{2}$

Let $\sin \frac{\alpha}{2} = a$, $\cos \frac{\alpha}{2} = \sqrt{1-a^2}$, $13a = \frac{3\sqrt{3}}{2} \cdot \sqrt{1-a^2}$

$169a^2 = 27-27a^2$, $196a^2=27$, $\sin \frac{\alpha}{2} = a = \sqrt{\frac{27}{196}} = \frac{3 \sqrt{3}}{14}$

Let $x$ be the length of the chord, $\sin \frac{3 \alpha}{2} = \frac{\frac{x}{2}}{r}$

By the triple angle formula, $\sin \frac{3 \alpha}{2} = 3 \cdot \sin \frac{\alpha}{2} - 4 \cdot \sin(\frac{ \alpha}{2})^3 = 3 \cdot \frac{3 \sqrt{3}}{14} - 4 \cdot (\frac{3 \sqrt{3}}{14})^3$

$x = 2 \cdot \frac{7\sqrt{3}}{3} \cdot [3 \cdot \frac{3 \sqrt{3}}{14} - 4 \cdot (\frac{3 \sqrt{3}}{14})^3] = 2 \cdot \frac{7\sqrt{3}}{3} \cdot (\frac{9\sqrt{3}}{14} - \frac{82\sqrt{3}}{2 \cdot 7^3}) = 9 - \frac{81}{49} = \frac{360}{49}$

Therefore, the answer is $\boxed{\textbf{(E) } 409}$.

~isabelchen

Solution 6 (Ptolemy's theorem)

1996AHSMEP305.png

Note that major arc $\overarc{AE}$ is two thirds of the circumference, therefore, $\angle AFE = 120^{\circ}$.

By the Law of Cosine, $AE= \sqrt{ 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cdot \cos 120^{\circ}} = \sqrt{ 9 + 25 + 15 } = 7$

By the Ptolemy's theorem of quadrilateral $ABDE$, $AD \cdot BE = AB \cdot DE + BD \cdot AE$, $AD = BE$, $AD^2= 3 \cdot 5 + 7^2 = 64$, $AD = 8$

By the Ptolemy's theorem of quadrilateral $ABCD$, $AC \cdot BD = BC \cdot AD + AB \cdot CD$, $7AC = 3 \cdot 8 + 3 \cdot 5 = 39$, $AC = \frac{39}{7}$

By the Ptolemy's theorem of quadrilateral $ABCF$, $AC \cdot BF = AB \cdot CF + BC \cdot AF$, $AC = BF$, $(\frac{39}{7})^2 = 3 \cdot CF + 3 \cdot 3$, $CF = \frac{360}{49}$

Therefore, the answer is $\boxed{\textbf{(E) } 409}$.

~isabelchen

See also

1996 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 29 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1996_AHSME_Problems/Problem_30&oldid=207994"