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.

1 Problem
2 Solution 1 (Pythagorean Theorem)
3 Solution 2 (Coordinates)
4 Solution 3 (Stewart's Theorem)
5 Video Solution (Super Fast. Just 1 min!)
6 Video Solution by Hawk Math
7 Video Solution by Punxsutawney Phil
8 Video Solution by OmegaLearn (Circular Geometry)
9 Video Solution by TheBeautyofMath
10 Video Solution by Interstigation
11 See Also

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 (Pythagorean Theorem)

$[asy] size(8cm); pair O = (0, 0), A = (0, 3), B = (0, 9), R = (19, 3), L = (17, 9); draw(O--A--B); draw(O--R); draw(O--L); label("$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 $3d$ , 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, diagram by Jamess2022(burntTacos), and minor edits by lpieleanu.

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 (Super Fast. Just 1 min!)

https://youtu.be/145UJbG4aCQ

~Education, the Study of Everything

Video Solution by Hawk Math

https://www.youtube.com/watch?v=VzwxbsuSQ80

Video Solution by Punxsutawney Phil

https://youtu.be/yxt8-rUUosI

Video Solution by OmegaLearn (Circular Geometry)

https://youtu.be/XNYq4ZMBtBU

~pi_is_3.14

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

https://youtu.be/lYxKkS252Og

~Interstigation

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

2021 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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