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

m (→‎Solution 3)
(combine solutions 2 and 3)
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==Solution 2 (Pythagorean Theorem)==
 
==Solution 2 (Pythagorean Theorem)==
  
−
Label the intersection of the chord with the smaller and larger circle as A and B respectively.  
+
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.  
  
−
Construct the radius perpendicular to the chord and label its intersection with the chord as M. Because a radius that is perpendicular to a chord also bisects the chord and because half of the chord in the larger circle lies in the smaller circle, <math>BM = 2AM</math>.
+
Construct segments <math>\overline{AO}</math> and <math>\overline{BO}</math>. These are radii with lengths 17 and 19 respectively.  
  
−
Construct segments AO and BO. These are radii with lengths 17 and 19 respectively.
+
Then, use the [[Pythagorean Theorem]]. In <math>\triangle OMA</math>, we have
−
 
 
−
Then, use the [[Pythagorean Theorem]] on right triangles <math>AMO</math> and <math>BMO</math> to get the following system of equations:
 
−
<cmath>OM^2+AM^2=17^2</cmath>
 
−
<cmath>OM^2+(2AM)^2=19^2</cmath>
 
−
 
 
−
So, <math>AM=2\sqrt{6}</math>, and the length of the chord in the larger circle <math> = 4AM = \boxed{\textbf{(E)} \: 8\sqrt{6}}</math>.
 
−
 
 
−
== Solution 3 ==
 
−
Denote by <math>O</math> the center of both circles.
 
−
Denote by <math>AB</math> this chord.
 
−
Let this chord intersect the smaller circle at <math>C</math> and <math>D</math>, where <math>C</math> is between <math>A</math> and <math>D</math>.
 
−
Denote by <math>M</math> the midpoint of <math>AB</math>.
 
−
 
 
−
In <math>\triangle OMA</math>, following from the Pythagorean theorem, we have
 
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
 
OM^2 & = OA^2 - AM^2 \\
 
OM^2 & = OA^2 - AM^2 \\
−
& = OA^2 - \left( \frac{AB}{2} \right)^2 \\
+
& = OA^2 - \left( \frac{AD}{2} \right)^2 \\
−
& = 19^2 - \frac{AB^2}{4} .
+
& = 19^2 - \frac{AD^2}{4} .
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
−
In <math>\triangle OMC</math>, following from the Pythagorean theorem, we have
+
 
 +
In <math>\triangle OMC</math>, we have
 
<cmath>
 
<cmath>
 
\begin{align*}
 
\begin{align*}
 
OM^2 & = OC^2 - CM^2 \\
 
OM^2 & = OC^2 - CM^2 \\
−
& = OC^2 - \left( \frac{CD}{2} \right)^2 \\
+
& = OC^2 - \left( \frac{BC}{2} \right)^2 \\
−
& = OC^2 - \left( \frac{AB}{4} \right)^2 \\
+
& = OC^2 - \left( \frac{AD}{4} \right)^2 \\
−
& = 17^2 - \frac{AB^2}{16} .
+
& = 17^2 - \frac{AD^2}{16} .
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
−
Equating these two expressions, we get
 
−
<cmath>
 
−
\[
 
−
19^2 - \frac{AB^2}{4} = 17^2 - \frac{AB^2}{16} .
 
−
\]
 
−
</cmath>
 
−
Therefore, the answer is <math>\boxed{\textbf{(E) }8 \sqrt{6}}</math>.
 
  
−
~Steven Chen (www.professorchenedu.com)
+
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>.
 +
 
 +
~eisthefifthletter and [https://www.professorchenedu.com Steven Chen]
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2021 Fall|ab=A|num-b=10|num-a=12}}
 
{{AMC12 box|year=2021 Fall|ab=A|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:35, 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 $\overline{AD}$; let this chord intersect the smaller circle at $B$ and $C$, where $B$ is between $A$ and $C$. Construct the radius perpendicular to the chord, and label its intersection with the chord as $M$. $M$ is the midpoint of $AB$ 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 $\triangle OMC$, we have \begin{align*} OM^2 & = OC^2 - CM^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 \[19^2 - \frac{AB^2}{4} = 17^2 - \frac{AB^2}{16}\] 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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