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

Revision as of 00:54, 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 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 expressions, 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)

@@ Line 38: / Line 38: @@
 \end{align*}
 </cmath>
 In <math>\triangle OMC</math>, following from the Pythagorean theorem, we have
 <cmath>
@@ Line 48: / Line 47: @@
 \end{align*}
 </cmath>
+Equating these two expressions, we get
-Equating these two equations, we get
 <cmath>
 \[
@@ Line 55: / Line 53: @@
 \]
 </cmath>
 Therefore, the answer is <math>\boxed{\textbf{(E) }8 \sqrt{6}}</math>.
 ~Steven Chen (www.professorchenedu.com)
 ==See Also==
 {{AMC12 box|year=2021 Fall|ab=A|num-b=10|num-a=12}}
 {{MAA Notice}}

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