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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 11"

(combine solutions 2 and 3)
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==Solution 2 (Pythagorean Theorem)==
 
==Solution 2 (Pythagorean Theorem)==
  
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Label the center of both circles <math>O</math>. Label the chord in the larger circle <math>\overline{AD}</math>; let this chord intersect the smaller circle at <math>B</math> and <math>C</math>, where <math>B</math> is between <math>A</math> and <math>C</math>. Construct the radius perpendicular to the chord, and label its intersection with the chord as <math>M</math>. <math>M</math> is the midpoint of <math>AB</math> because a radius that is perpendicular to a chord bisects the chord.  
+
Label the center of both circles <math>O</math>. Label the chord in the larger circle <math>\overline{ABCD}</math> where <math>A</math> and <math>D</math> are on the larger circle and <math>B</math> and <math>C</math> are on the smaller circle. Construct the radius perpendicular to the chord and label their intersection as <math>M</math>. <math>M</math> is the midpoint of the chord because a radius that is perpendicular to a chord bisects the chord.  
  
 
Construct segments <math>\overline{AO}</math> and <math>\overline{BO}</math>. These are radii with lengths 17 and 19 respectively.  
 
Construct segments <math>\overline{AO}</math> and <math>\overline{BO}</math>. These are radii with lengths 17 and 19 respectively.  
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</cmath>
 
</cmath>
  
−
In <math>\triangle OMC</math>, we have
+
In <math>\triangle OMB</math>, we have
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
−
OM^2 & = OC^2 - CM^2 \\
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OM^2 & = OB^2 - BM^2 \\
 
& = OC^2 - \left( \frac{BC}{2} \right)^2 \\
 
& = OC^2 - \left( \frac{BC}{2} \right)^2 \\
 
& = OC^2 - \left( \frac{AD}{4} \right)^2 \\
 
& = OC^2 - \left( \frac{AD}{4} \right)^2 \\
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</cmath>
 
</cmath>
  
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Equating these two expressions, we get <cmath>19^2 - \frac{AB^2}{4} = 17^2 - \frac{AB^2}{16}</cmath> and <math>AD=\boxed{\textbf{(E) }8 \sqrt{6}}</math>.
+
Equating these two expressions, we get <cmath>19^2 - \frac{AD^2}{4} = 17^2 - \frac{AD^2}{16}</cmath> and <math>AD=\boxed{\textbf{(E) }8 \sqrt{6}}</math>.
  
 
~eisthefifthletter and [https://www.professorchenedu.com Steven Chen]
 
~eisthefifthletter and [https://www.professorchenedu.com Steven Chen]

Revision as of 17:42, 26 November 2021

Problem

Consider two concentric circles of radius $17$ and $19.$ The larger circle has a chord, half of which lies inside the smaller circle. What is the length of the chord in the larger circle?

$\textbf{(A)}\ 12\sqrt{2} \qquad\textbf{(B)}\ 10\sqrt{3} \qquad\textbf{(C)}\ \sqrt{17 \cdot 19} \qquad\textbf{(D)}\ 18 \qquad\textbf{(E)}\ 8\sqrt{6}$

Solution 1 (Power of a Point)

Draw the diameter perpendicular to the chord. Call the intersection between that diameter and the chord $A$. In the circle of radius $17$, let the shorter piece of the diameter cut by the chord would be of length $x$, making the longer piece $34-x.$ In that same circle, let the $y$ be the length of the portion of the chord in the smaller circle that is cut by the diameter we drew. Thus, in the circle of radius $7$, the shorter piece of the diameter cut by the chord would be of length $x+2$, making the longer piece $36-x,$ and length of the piece of the chord cut by the diameter would be $2y$ (as given in the problem). By Power of a Point, we can construct the system of equations \[x(34-x) = y^2\]\[(x+2)(36-x) = (y+2)^2\]Expanding both equations, we get $34x-x^2 = y^2$ and $36x-x^2+72-2x = 4y^2,$ in which the $34x$ and $-x^2$ terms magically cancel when we subtract the first equation from the second equation. Thus, now we have $72 = 3y^2 \implies y = 2\sqrt{6} \implies 4y = \boxed{\textbf{(E)} \: 8\sqrt{6}}$.

-fidgetboss_4000

Solution 2 (Pythagorean Theorem)

Label the center of both circles $O$. Label the chord in the larger circle $\overline{ABCD}$ where $A$ and $D$ are on the larger circle and $B$ and $C$ are on the smaller circle. Construct the radius perpendicular to the chord and label their intersection as $M$. $M$ is the midpoint of the chord because a radius that is perpendicular to a chord bisects the chord.

Construct segments $\overline{AO}$ and $\overline{BO}$. These are radii with lengths 17 and 19 respectively.

Then, use the Pythagorean Theorem. In $\triangle OMA$, we have \begin{align*} OM^2 & = OA^2 - AM^2 \\ & = OA^2 - \left( \frac{AD}{2} \right)^2 \\ & = 19^2 - \frac{AD^2}{4} . \end{align*}

In $\triangle OMB$, we have \begin{align*} OM^2 & = OB^2 - BM^2 \\ & = OC^2 - \left( \frac{BC}{2} \right)^2 \\ & = OC^2 - \left( \frac{AD}{4} \right)^2 \\ & = 17^2 - \frac{AD^2}{16} . \end{align*}

Equating these two expressions, we get \[19^2 - \frac{AD^2}{4} = 17^2 - \frac{AD^2}{16}\] and $AD=\boxed{\textbf{(E) }8 \sqrt{6}}$.

~eisthefifthletter and Steven Chen

See Also

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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
All AMC 12 Problems and Solutions

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