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

## Problem

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 of the circle 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 lines intersect 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 n fix my latex please)