# 2021 AMC 12B Problems/Problem 8

The following problem is from both the 2021 AMC 10B #14 and 2021 AMC 12B #8, so both problems redirect to this page.

## 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 $[asy] size(8cm); pair O = (0, 0), A = (0, 3), B = (0, 9), R = (19, 3), L = (17, 9); draw(O--A--B); draw(O--R); draw(O--L); label("A", A, NE); label("B", B, N); label("R", R, NE); label("L", L, NE); label("O", O, S); label("d", O--A, W); label("2d", A--B, W); label("r", O--R, S); label("r", O--L, NW); dot(O); dot(A); dot(B); dot(R); dot(L); draw(circle((0, 0), sqrt(370))); draw(-R -- (R.x, -R.y)); draw((-R.x, R.y) -- R); draw((-L.x, L.y) -- L); [/asy]$

Since two parallel chords have the same length ( $38$), 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 $r$ ( $\triangle RAO$ on the diagram)

- Another with base $\frac{34}{2} = 17$, height $2d + d$, and hypotenuse $r$ ( $\triangle LBO$ on the diagram)

By the Pythagorean theorem, we can create the following system of equations: $$19^2 + d^2 = r^2$$ $$17^2 + (2d + d)^2 = r^2$$

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

-Solution by Joeya and diagram by Jamess2022(burntTacos).

## 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

## Solution 3 (Stewart's Theorem) $[asy] real r=sqrt(370); draw(circle((0, 0), r)); pair A = (-19, 3); pair B = (19, 3); draw(A--B); pair C = (-19, -3); pair D = (19, -3); draw(C--D); pair E = (-17, -9); pair F = (17, -9); draw(E--F); pair O = (0, 0); pair P = (0, -3); pair Q = (0, -9); draw(O--Q); draw(O--C); draw(O--D); draw(O--E); draw(O--F); label("O", O, N); label("C", C, SW); label("D", D, SE); label("E", E, SW); label("F", F, SE); label("P", P, SW); label("Q", Q, S); [/asy]$ If $d$ is the requested distance, and $r$ is the radius of the circle, Stewart's Theorem applied to $\triangle OCD$ with cevian $\overleftrightarrow{OP}$ gives $$19\cdot 38\cdot 19 + \tfrac{1}{2}d\cdot 38\cdot\tfrac{1}{2}d=19r^{2}+19r^{2}.$$ This simplifies to $13718+\tfrac{19}{2}d^{2}=38r^{2}$. Similarly, another round of Stewart's Theorem applied to $\triangle OEF$ with cevian $\overleftrightarrow{OQ}$ gives $$17\cdot 34\cdot 17 + \tfrac{3}{2}d\cdot 34\cdot\tfrac{3}{2}d=17r^{2}+17r^{2}.$$ This simplifies to $9826+\tfrac{153}{2}d^{2}=34r^{2}$. Dividing the top equation by $38$ and the bottom equation by $34$ results in the system of equations \begin{align*} 361+\tfrac{1}{4}d^{2} &= r^{2} \\ 289+\tfrac{9}{4}d^{2} &= r^{2} \\ \end{align*} By transitive, $361+\tfrac{1}{4}d^{2}=289+\tfrac{9}{4}d^{2}$. Therefore $(\tfrac{9}{4}-\tfrac{1}{4})d^{2}=361-289\rightarrow 2d^{2}=72\rightarrow d^{2}=36\rightarrow d=\boxed{\textbf{(B)} ~6}.$

~Punxsutawney Phil

## Video Solution by TheBeautyofMath

https://youtu.be/L1iW94Ue3eI?t=1118 (for AMC 10B)

https://youtu.be/kuZXQYHycdk?t=574 (for AMC 12B)

~IceMatrix

## Video Solution by Interstigation

~Interstigation

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 