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

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(Solution)
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<math>\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12</math>
 
<math>\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12</math>
  
==Solution==
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==Solution 1==
  
 
Since the two chords of length <math>38</math> 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 <math>d</math>. Thus, the distance from the center of the circle to the chord of length <math>34</math> is  
 
Since the two chords of length <math>38</math> 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 <math>d</math>. Thus, the distance from the center of the circle to the chord of length <math>34</math> is  
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-Solution by Joeya (someone draw a diagram n fix my latex please)
 
-Solution by Joeya (someone draw a diagram n fix my latex please)
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==Solution 2 (Coordinates)==
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Because we know that the equation of a circle is <math>(x-a)^2 + (y-b)^2 = r^2</math> where the center of the circle is <math>(a, b)</math> and the radius is <math>r</math>, we can find the equation of this circle by centering it on the origin. Doing this, we get that the equation is <math>x^2 + y^2 = r^2</math>. Now, we can set the distance between the chords as <math>2d</math> so the distance from the cord with length 38 to the diameter is <math>d</math>.
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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:
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<math>(19, d)</math>
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<math>(19, -d)</math>
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<math>(17, -3d)</math>
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Now, we can plug one of the first two value in as well as the last one to get the following equations:
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<math>19^2 + d^2 = r^2</math>
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<math>17^2 + (3d)^2 = r^2</math>
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Subtracting these two equations, we get <math>19^2 - 17^2 = 8d^2</math> - therefore, we get <math>72 = 8d^2</math> -> <math>d^2 = 9</math> -> <math>d = 3</math>. We want to find <math>2d = 6</math> because that's the distance between two chords. So, our answer is <math>\boxed{B}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2021|ab=B|num-b=7|num-a=9}}
 
{{AMC12 box|year=2021|ab=B|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:49, 11 February 2021

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

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 cord 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 $72 = 8d^2$ -> $d^2 = 9$ -> $d = 3$. We want to find $2d = 6$ because that's the distance between two chords. So, our answer is $\boxed{B}$.

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

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