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

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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>.
 
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>
 +
\begin{align*}
 +
OM^2 & = OA^2 - AM^2 \\
 +
& = OA^2 - \left( \frac{AB}{2} \right)^2 \\
 +
& = 19^2 - \frac{AB^2}{4} .
 +
\end{align*}
 +
</cmath>
 +
 +
In <math>\triangle OMC</math>, following from the Pythagorean theorem, we have
 +
<cmath>
 +
\begin{align*}
 +
OM^2 & = OC^2 - CM^2 \\
 +
& = OC^2 - \left( \frac{CD}{2} \right)^2 \\
 +
& = OC^2 - \left( \frac{AB}{4} \right)^2 \\
 +
& = 17^2 - \frac{AB^2}{16} .
 +
\end{align*}
 +
</cmath>
 +
 +
Equating these two equations, 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)
 +
  
 
==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 20:42, 25 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 intersection of the chord with the smaller and larger circle as A and B respectively.

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, $BM = 2AM$.

Construct segments AO and BO. These are radii with lengths 17 and 19 respectively.

Then, use the Pythagorean Theorem on right triangles $AMO$ and $BMO$ to get the following system of equations: \[OM^2+AM^2=17^2\] \[OM^2+(2AM)^2=19^2\]

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

Solution 3

Denote by $O$ the center of both circles. Denote by $AB$ this chord. Let this chord intersect the smaller circle at $C$ and $D$, where $C$ is between $A$ and $D$. Denote by $M$ the midpoint of $AB$.

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

In $\triangle OMC$, following from the Pythagorean theorem, we have \begin{align*} OM^2 & = OC^2 - CM^2 \\ & = OC^2 - \left( \frac{CD}{2} \right)^2 \\ & = OC^2 - \left( \frac{AB}{4} \right)^2 \\ & = 17^2 - \frac{AB^2}{16} . \end{align*}

Equating these two equations, we get \[ 19^2 - \frac{AB^2}{4} = 17^2 - \frac{AB^2}{16} . \]

Therefore, the answer is $\boxed{\textbf{(E) }8 \sqrt{6}}$.

~Steven Chen (www.professorchenedu.com)


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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