# 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 1

$[asy] unitsize(10); pair O = (0, 4), A = (0, 5), B = (0, 7), R = (3.873, 5), L = (2.645, 7); draw(O--A--B); draw(O--R); draw(O--L); label("A", A, NW); label("B", B, N); label("R", R, E); label("L", L, E); label("O", O, S); label("d", O--A, W); label("2d", A--B, W*2+0.5*N); label("r", O--R, S); label("r", O--L, S*0.5 + 1.5 * E); dot(O); dot(A); dot(B); dot(R); dot(L); draw(circle((0, 4), 4)); draw((-7, 3) -- (7, 3)); draw((-7, 5) -- (7, 5)); draw((-7, 7) -- (7, 7)); [/asy]$

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)

## Solution 2 (Coordinates)

Because we know that the equation of a circle is $(x-a)^2 + (y-b)^2 = r^2$ where the center of the circle is $(a, b)$ and the radius is $r$, we can find the equation of this circle by centering it on the origin. Doing this, we get that the equation is $x^2 + y^2 = r^2$. Now, we can set the distance between the chords as $2d$ so the distance from the chord with length 38 to the diameter is $d$.

Therefore, the following points are on the circle as the y-axis splits the chord in half, that is where we get our x value:

$(19, d)$

$(19, -d)$

$(17, -3d)$

Now, we can plug one of the first two value in as well as the last one to get the following equations:

$$19^2 + d^2 = r^2$$

$$17^2 + (3d)^2 = r^2$$

Subtracting these two equations, we get $19^2 - 17^2 = 8d^2$ - therefore, we get $72 = 8d^2 \rightarrow d^2 = 9 \rightarrow d = 3$. We want to find $2d = 6$ because that's the distance between two chords. So, our answer is $\boxed{B}$.

~Tony_Li2007