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

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

@@ Line 37: / Line 37: @@
 ==Solution 2 (Power of a Point)==
-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>19</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). 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) = (2y)^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{\textbf{(E)} \: 8\sqrt{6}}</math>.
+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
+==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
 ==Video Solution by TheBeautyofMath==

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

Page

Toolbox

Search