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

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{{duplicate|[[2021 AMC 10B Problems#Problem 14|2021 AMC 10B #14]] and [[2021 AMC 12B Problems#Problem 8|2021 AMC 12B #8]]}}
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==Problem==
 
==Problem==
 
Three equally spaced parallel lines intersect a circle, creating three chords of lengths <math>38,38,</math> and <math>34</math>. What is the distance between two adjacent parallel lines?
 
Three equally spaced parallel lines intersect a circle, creating three chords of lengths <math>38,38,</math> and <math>34</math>. What is the distance between two adjacent parallel lines?
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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 <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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Since two parallel chords have the same length (<math>38</math>), 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  
  
 
<cmath>2d + d = 3d</cmath>
 
<cmath>2d + d = 3d</cmath>
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- One with base <math>\frac{38}{2}= 19</math>, height <math>d</math>, and hypotenuse <math>r</math>  (<math>\triangle RAO</math> on the diagram)
 
- One with base <math>\frac{38}{2}= 19</math>, height <math>d</math>, and hypotenuse <math>r</math>  (<math>\triangle RAO</math> on the diagram)
  
- Another with base <math>\frac{34}{2} = 17</math>, height <math>3d</math>, and hypotenuse <math>r</math>  (<math>\triangle LBO</math> on the diagram)
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- Another with base <math>\frac{34}{2} = 17</math>, height <math>2d + d</math>, and hypotenuse <math>r</math>  (<math>\triangle LBO</math> on the diagram)
  
 
By the Pythagorean theorem, we can create the following system of equations:
 
By the Pythagorean theorem, we can create the following system of equations:
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<cmath>19^2 + d^2 = r^2</cmath>
 
<cmath>19^2 + d^2 = r^2</cmath>
  
<cmath>17^2 + (3d)^2 = r^2</cmath>
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<cmath>17^2 + (2d + d)^2 = r^2</cmath>
  
 
Solving, we find <math>d = 3</math>, so <math>2d = \boxed{\textbf{(B)}\ 6}</math>.
 
Solving, we find <math>d = 3</math>, so <math>2d = \boxed{\textbf{(B)}\ 6}</math>.
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~Tony_Li2007
 
~Tony_Li2007
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==Solution 3 (Stewart's Theorem)==
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<asy>
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real r=sqrt(370);
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draw(circle((0, 0), r));
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pair A = (-19, 3);
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pair B = (19, 3);
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draw(A--B);
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pair C = (-19, -3);
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pair D = (19, -3);
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draw(C--D);
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pair E = (-17, -9);
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pair F = (17, -9);
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draw(E--F);
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pair O = (0, 0);
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pair P = (0, -3);
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pair Q = (0, -9);
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draw(O--Q);
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draw(O--C);
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draw(O--D);
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draw(O--E);
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draw(O--F);
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label("$O$", O, N);
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label("$C$", C, SW);
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label("$D$", D, SE);
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label("$E$", E, SW);
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label("$F$", F, SE);
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label("$P$", P, SW);
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label("$Q$", Q, S);
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</asy>
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If <math>d</math> is the requested distance, and <math>r</math> is the radius of the circle, Stewart's Theorem applied to <math>\triangle OCD</math> with cevian <math>\overleftrightarrow{OP}</math> gives <cmath>19\cdot 38\cdot 19 + \tfrac{1}{2}d\cdot 38\cdot\tfrac{1}{2}d=19r^{2}+19r^{2}.</cmath> This simplifies to <math>13718+\tfrac{19}{2}d^{2}=38r^{2}</math>. Similarly, another round of Stewart's Theorem applied to <math>\triangle OEF</math> with cevian <math>\overleftrightarrow{OQ}</math> gives <cmath>17\cdot 34\cdot 17 + \tfrac{3}{2}d\cdot 34\cdot\tfrac{3}{2}d=17r^{2}+17r^{2}.</cmath> This simplifies to <math>9826+\tfrac{153}{2}d^{2}=34r^{2}</math>. Dividing the top equation by <math>38</math> and the bottom equation by <math>34</math> results in the system of equations
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<cmath>\begin{align*}
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361+\tfrac{1}{4}d^{2} &= r^{2} \\
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289+\tfrac{9}{4}d^{2} &= r^{2} \\
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\end{align*}</cmath>
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By transitive, <math>361+\tfrac{1}{4}d^{2}=289+\tfrac{9}{4}d^{2}</math>. Therefore <math>(\tfrac{9}{4}-\tfrac{1}{4})d^{2}=361-289\rightarrow 2d^{2}=72\rightarrow d^{2}=36\rightarrow d=\boxed{\textbf{(B)} ~6}.</math>
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~Punxsutawney Phil
  
 
==Video Solution by Hawk Math==
 
==Video Solution by Hawk Math==
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== Video Solution by OmegaLearn (Circular Geometry) ==
 
== Video Solution by OmegaLearn (Circular Geometry) ==
 
https://youtu.be/XNYq4ZMBtBU
 
https://youtu.be/XNYq4ZMBtBU
 +
 +
==Video Solution by TheBeautyofMath==
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https://youtu.be/L1iW94Ue3eI?t=1118 (for AMC 10B)
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https://youtu.be/kuZXQYHycdk?t=574 (for AMC 12B)
 +
 +
~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==

Latest revision as of 09:23, 29 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

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 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

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