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

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==Solution 2 (Pythagorean Theorem)==
 
==Solution 2 (Pythagorean Theorem)==
  
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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.
+
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, <math>BM = 2AM</math>.
  
 
Construct segments AO and BO. Note that these are radii with lengths 17 and 19 respectively.  
 
Construct segments AO and BO. Note that these are radii with lengths 17 and 19 respectively.  
  
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We have two right triangles. Use the [[Pythagorean Theorem]] to get the following system of equations:
+
We have the two right triangles <math>AMO</math> and <math>BMO</math>. Using the [[Pythagorean Theorem]] leads to the following system of equations:
 
<cmath>OM^2+AM^2=17^2</cmath>
 
<cmath>OM^2+AM^2=17^2</cmath>
 
<cmath>OM^2+(2AM)^2=19^2</cmath>
 
<cmath>OM^2+(2AM)^2=19^2</cmath>

Revision as of 22:02, 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 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. Note that these are radii with lengths 17 and 19 respectively.

We have the two right triangles $AMO$ and $BMO$. Using the Pythagorean Theorem leads to 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 $= 4AM = \boxed{\textbf{(E)} \: 8\sqrt{6}}$.

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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