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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 11"
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<math>\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}</math>
 
<math>\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}</math>
  
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==Solution 1 (Power of a Point)==
+
==Solution 1 (Pythagorean Theorem)==
  
−
Draw the diameter perpendicular to the chord. Call the intersection between that diameter and the chord <math>A</math>. In the circle of radius <math>17</math>, let the shorter piece of the diameter cut by the chord would be of length <math>x</math>, making the longer piece <math>34-x.</math> In that same circle, let the <math>y</math> 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 <math>7</math>, the shorter piece of the diameter cut by the chord would be of length <math>x+2</math>, making the longer piece <math>36-x,</math> and length of the piece of the chord cut by the diameter would be <math>2y</math> (as given in the problem statement). By [[Power of a Point Theorem|Power of a Point]], we can construct the system of equations <cmath>x(34-x) = y^2</cmath><cmath>(x+2)(36-x) = (y+2)^2</cmath>Expanding both equations, we get <math>34x-x^2 = y^2</math> and <math>36x-x^2+72-2x = 4y^2,</math> in which the <math>34x</math> and <math>-x^2</math> terms magically cancel when we subtract the first equation from the second equation. Thus, now we have <math>72 = 3y^2 \implies y = 2\sqrt{6} \implies 4y = \boxed{ 8\sqrt{6}}</math>.
+
Label the center of both circles <math>O</math>. Label the chord in the larger circle as <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>. Because a radius that is perpendicular to a chord bisects the chord, <math>M</math> is the midpoint of the chord.
 +
 
 +
Construct segments <math>\overline{AO}</math> and <math>\overline{BO}</math>. These are radii with lengths 17 and 19 respectively.
 +
 
 +
Then, use the [[Pythagorean Theorem]]. In <math>\triangle OMA</math>, we have
 +
<cmath>
 +
\begin{align*}
 +
OM^2 & = OA^2 - AM^2 \\
 +
& = OA^2 - \left( \frac{AD}{2} \right)^2 \\
 +
& = 19^2 - \frac{AD^2}{4} .
 +
\end{align*}
 +
</cmath>
 +
 
 +
In <math>\triangle OMB</math>, we have
 +
<cmath>
 +
\begin{align*}
 +
OM^2 & = OB^2 - BM^2 \\
 +
& = OB^2 - \left( \frac{BC}{2} \right)^2 \\
 +
& = OB^2 - \left( \frac{AD}{4} \right)^2 \\
 +
& = 17^2 - \frac{AD^2}{16} .
 +
\end{align*}
 +
</cmath>
 +
 
 +
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]
 +
 
 +
==Solution 2 (Power of a Point)==
 +
 
 +
Draw the diameter perpendicular to the chord. Notice that by symmetry this diameter bisects the chord.
 +
 
 +
Call the intersection between that diameter and the chord <math>A</math>. In the smaller circle, let the shorter piece of the diameter cut by the chord be <math>x</math>, making the longer piece <math>34-x.</math> In that same circle, let the <math>y</math> be the length of the portion of the chord that is cut by the diameter.  
 +
 
 +
The radius of the larger circle is <math>2</math> more than the radius of the small circle. So, in the larger circle, the shorter piece of the diameter cut by the chord is of length <math>x+2</math> and the longer piece is <math>36-x.</math> As given in the problem, the bisected length of the chord in the larger circle is twice as much, so it is of length <math>2y</math>. By [[Power of a Point Theorem|Power of a Point]], we can construct a system of equations
 +
<cmath>x(34-x) = y^2</cmath>
 +
<cmath>(x+2)(36-x) = (2y)^2.</cmath>
 +
Expanding both equations, we get
 +
<cmath>34x-x^2 = y^2</cmath>
 +
<cmath>36x-x^2+72-2x = 4y^2,</cmath>
 +
in which the <math>34x</math> and <math>-x^2</math> terms magically cancel when we subtract the first equation from the second equation. Thus, now we have <cmath>72 = 3y^2 \implies y = 2\sqrt{6} \implies 4y = \boxed{\textbf{(E)} \: 8\sqrt{6}}.</cmath>
  
 
-fidgetboss_4000
 
-fidgetboss_4000
  
−
==Solution 2 (Pythagorean Theorem)==
+
==Solution 3 (Coordinate Geometry)==
 +
Represent the circles as <math>x^{2}+y^{2}=17^{2}</math>, and <math>x^{2}+y^{2}=19^{2}</math>. Solving for <math>x</math> in these equations we obtain <math>x=\sqrt{17^{2}-y^{2}}</math> and <math>x=\sqrt{19^{2}-y^{2}}</math>. Because half of the chord is in the smaller circle, the larger circle should have an <math>x</math> value that is twice as big as the smaller circle's <math>x</math> value. We set up the equation:
 +
<cmath>2\sqrt{289-y^{2}}=\sqrt{361-y^{2}}</cmath>
 +
<cmath>4(289-y^{2})=361-y^{2}</cmath>
 +
<cmath>y^{2}=265</cmath>
 +
Substituting <math>y^{2}</math> into <math>x^{2}+y^{2}=19^{2}</math>, we obtain <math>x=\sqrt{96}=4\sqrt{6}</math>. However, this is only half of the chord length, so we must double it to obtain <math>\boxed{\textbf{(E)} \: 8\sqrt{6}}.</math>
 +
 
 +
-helpmebro
 +
 
  
−
Label an intersection of the chord with the smaller and larger circle as A and B respectively. Draw the radius perpendicular to the chord and label its intersection with the chord M. Note that <math>BM = 2AM</math> because half of the chord lies in the smaller circle and because if a radius is perpendicular to a chord, then the radius also bisects the chord.
+
==Video Solution by TheBeautyofMath==
 +
https://youtu.be/ToiOlqWz3LY
  
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Construct segments AO and BO. Note that these are radii with lengths 17 and 19 respectively.
+
~IceMatrix
  
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We have two right triangles. Use the [[Pythagorean Theorem]] to get the following system of equations:
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==Video Solution (Logic and Geometry)==
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<cmath>MO^2+AM^2=17^2</cmath>
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https://youtu.be/iG1vVXeTv58
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<cmath>MO^2+(2AM)^2=19^2</cmath>
 
  
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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>.
+
~Education, the Study of Everything
  
 
==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}}

Latest revision as of 20:42, 22 November 2024

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 (Pythagorean Theorem)

Label the center of both circles $O$. Label the chord in the larger circle as $\overline{ABCD}$, 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 $M$. Because a radius that is perpendicular to a chord bisects the chord, $M$ is the midpoint of 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 \\ & = OB^2 - \left( \frac{BC}{2} \right)^2 \\ & = OB^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

Solution 2 (Power of a Point)

Draw the diameter perpendicular to the chord. Notice that by symmetry this diameter bisects the chord.

Call the intersection between that diameter and the chord $A$. In the smaller circle, let the shorter piece of the diameter cut by the chord be $x$, making the longer piece $34-x.$ In that same circle, let the $y$ be the length of the portion of the chord that is cut by the diameter.

The radius of the larger circle is $2$ more than the radius of the small circle. So, in the larger circle, the shorter piece of the diameter cut by the chord is of length $x+2$ and the longer piece is $36-x.$ As given in the problem, the bisected length of the chord in the larger circle is twice as much, so it is of length $2y$. By Power of a Point, we can construct a system of equations \[x(34-x) = y^2\] \[(x+2)(36-x) = (2y)^2.\] Expanding both equations, we get \[34x-x^2 = y^2\] \[36x-x^2+72-2x = 4y^2,\] 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 3 (Coordinate Geometry)

Represent the circles as $x^{2}+y^{2}=17^{2}$, and $x^{2}+y^{2}=19^{2}$. Solving for $x$ in these equations we obtain $x=\sqrt{17^{2}-y^{2}}$ and $x=\sqrt{19^{2}-y^{2}}$. Because half of the chord is in the smaller circle, the larger circle should have an $x$ value that is twice as big as the smaller circle's $x$ value. We set up the equation: \[2\sqrt{289-y^{2}}=\sqrt{361-y^{2}}\] \[4(289-y^{2})=361-y^{2}\] \[y^{2}=265\] Substituting $y^{2}$ into $x^{2}+y^{2}=19^{2}$, we obtain $x=\sqrt{96}=4\sqrt{6}$. However, this is only half of the chord length, so we must double it to obtain $\boxed{\textbf{(E)} \: 8\sqrt{6}}.$

-helpmebro


Video Solution by TheBeautyofMath

https://youtu.be/ToiOlqWz3LY

~IceMatrix

Video Solution (Logic and Geometry)

https://youtu.be/iG1vVXeTv58

~Education, the Study of Everything

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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