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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 (Power of a Point)==
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==Solution 1 (Power of a Point)==
  
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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, 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>.
+
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>.
  
 
-fidgetboss_4000
 
-fidgetboss_4000
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We have two right triangles. Use the [[Pythagorean Theorem]] to get the following system of equations:
 
We have two right triangles. Use the [[Pythagorean Theorem]] to get the following system of equations:
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<cmath>MO^2+AM^2=17^2</cmath><cmath>MO^2+(2AM)^2=19^2</cmath>
+
<cmath>MO^2+AM^2=17^2</cmath>
 +
<cmath>MO^2+(2AM)^2=19^2</cmath>
  
−
Solve to get <math>AM=2\sqrt{6}</math>. If a radius is perpendicular to a chord, then the radius also bisects the chord. So, the length of the chord in the larger circle is <math>4AM</math> or <math>\boxed{ 8\sqrt{6}}</math>.
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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>.
  
 
==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 21:53, 23 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 statement). 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 $72 = 3y^2 \implies y = 2\sqrt{6} \implies 4y = \boxed{ 8\sqrt{6}}$.

-fidgetboss_4000

Solution 2 (Pythagorean Theorem)

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 $BM = 2AM$ 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.

Construct segments AO and BO. Note that these are radii with lengths 17 and 19 respectively.

We have two right triangles. Use the Pythagorean Theorem to get the following system of equations: \[MO^2+AM^2=17^2\] \[MO^2+(2AM)^2=19^2\]

So, $AM=2\sqrt{6}$, and the length of the chord in the larger circle $= 4AM = \boxed{\textbf{(E)} \: 8\sqrt{6}}$.

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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