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

(Solution 1)
(Video Solution by TheBeautyofMath)
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~IceMatrix
 
~IceMatrix
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==Video Solution by Interstigation==
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https://youtu.be/lYxKkS252Og
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~Interstigation
  
 
==See Also==
 
==See Also==

Revision as of 02:03, 4 March 2021

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(6cm); 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, NE); label("$L$", L, N); 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((-3.873, 3) -- (3.873, 3)); draw((-3.873, 5) -- (3.873, 5)); draw((-2.645, 7) -- (2.645, 7)); [/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). (Someone fix the diagram if possible. - Done. )

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 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

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

See Also

2021 AMC 12B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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

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