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 and diagram by Jamess2022(burntTacos). (Someone fix the diagram if possible.)

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

Video Solution by Punxsutawney Phil

https://youtu.be/yxt8-rUUosI

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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All AMC 12 Problems and Solutions

2021 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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All AMC 10 Problems and Solutions

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2021 AMC 12B Problems/Problem 8

Contents

Problem

Solution 1

Solution 2 (Coordinates)

Video Solution by Punxsutawney Phil

See Also