Difference between revisions of "2021 AMC 12B Problems/Problem 8"
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Three equally spaced parallel lines intersect a circle, creating three chords of lengths <math>38,38,</math> and <math>34</math>. What is the distance between two adjacent parallel lines? | Three equally spaced parallel lines intersect a circle, creating three chords of lengths <math>38,38,</math> and <math>34</math>. What is the distance between two adjacent parallel lines? | ||
<math>\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12</math> | <math>\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12</math> | ||
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==Solution== | ==Solution== | ||
Revision as of 20:07, 11 February 2021
Problem
Three equally spaced parallel lines intersect a circle, creating three chords of lengths 
and 
. What is the distance between two adjacent parallel lines?
Solution
Since the two chords of length 
have the same length, they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be 
. Thus, the distance from the center of the circle to the chord of length is
![\[2d + d = 3d\]](http://latex.artofproblemsolving.com/1/0/1/101174599b6d666f76b2e275aa60e3187d3116f6.png)
and the distance between each of the chords is just 
. Let the radius of the circle be 
. Drawing radii to the points where the chords touch the circle, we create two different right triangles:
- One with base 
, height , and hypotenuse
- Another with base 
, height 
, and hypotenuse
By the Pythagorean theorem, we can create the following systems of equations:
![\[19^2 + d^2 = r^2\]](http://latex.artofproblemsolving.com/f/9/a/f9a21b320ccc483bd60bb7101a3799e2bd7d2496.png)
![\[17^2 + (3d)^2 = r^2\]](http://latex.artofproblemsolving.com/6/1/a/61a1ea9f1f4bd7a07beb17baa81f6cc87b3a549b.png)
Solving, we find 
, so 
-Solution by Joeya (someone draw a diagram n fix my latex please)
See Also
| 2021 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 7 |
Followed by Problem 9 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
