Difference between revisions of "2021 AMC 12B Problems/Problem 8"

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Solving, we find <math>d = 3</math>, so <math>2d = \boxed{(B) 6}</math>
 
Solving, we find <math>d = 3</math>, so <math>2d = \boxed{(B) 6}</math>
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-Solution by Joeya (someone draw a diagram please)

Revision as of 16:36, 11 February 2021

Problem 8

Three equally spaced parallel lines intersect a circle, creating three chords of lengths $38,38,$ and $34$. What is the distance between two adjacent parallel lines?

$\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12$


Solution

Since the two chords of length $38$ 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 $d$. Thus, the distance from the center to the chord of length $34$ is

\[2d + d = 3d\]

and the distance between each of the chords is just $2d$. Let the radius of the circle be $r$. Drawing radii to the points where the chords touch the circle, we create two different right triangles:

- One with base $\frac{38}{2}= 19$, height $d$, and hypotenuse $r$

- Another with base $\frac{34}{2} = 17$, height $3d$, and hypotenuse $r$

By the Pythagorean theorem, we can create the following systems of equations:

\[19^2 + d^2 = r^2\]

\[17^2 + (3d)^2 = r^2\]

Solving, we find $d = 3$, so $2d = \boxed{(B) 6}$

-Solution by Joeya (someone draw a diagram please)

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