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

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==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==
 
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?
  
 
<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>
  
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==Solution 1==
 +
 +
<asy>
 +
size(6cm);
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pair O = (0, 4), A = (0, 5), B = (0, 7), R = (3.873, 5), L = (2.645, 7);
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draw(O--A--B);
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draw(O--R);
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draw(O--L);
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label("$A$", A, NW);
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label("$B$", B, N);
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label("$R$", R, NE);
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label("$L$", L, N);
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label("$O$", O, S);
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label("$d$", O--A, W);
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label("$2d$", A--B, W*2+0.5*N);
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label("$r$", O--R, S);
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label("$r$", O--L, S*0.5 + 1.5 * E);
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dot(O);
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dot(A);
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dot(B);
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dot(R);
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dot(L);
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draw(circle((0, 4), 4));
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draw((-3.873, 3) -- (3.873, 3));
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draw((-3.873, 5) -- (3.873, 5));
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draw((-2.645, 7) -- (2.645, 7));
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</asy>
  
==Solution==
 
  
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 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>
  
and the distance between each of the chords is just <math>2d</math>. Let the radius of the circle be <math>r</math>. Drawing radii to the points where the chords touch the circle, we create two different right triangles:  
+
and the distance between each of the chords is just <math>2d</math>. Let the radius of the circle be <math>r</math>. Drawing radii to the points where the lines intersect the circle, we create two different right triangles:  
 +
 
 +
- 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)
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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:
 +
 
 +
<cmath>19^2 + d^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>.
 +
 
 +
-Solution by Joeya and diagram by Jamess2022(burntTacos).
 +
(Someone fix the diagram if possible. -<i> Done. </i>)
 +
 
 +
==Solution 2 (Coordinates)==
 +
 
 +
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 chord 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:
 +
 
 +
<math>(19, d)</math>
  
- One with base <math>\frac{38}{2}= 19</math>, height <math>d</math>, and hypotenuse <math>r</math>
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<math>(19, -d)</math>
  
- Another with base <math>\frac{34}{2} = 17</math>, height <math>3d</math>, and hypotenuse <math>r</math>  
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<math>(17, -3d)</math>
  
By the Pythagorean theorem, we can create the following systems of equations:
+
 
 +
Now, we can plug one of the first two value in as well as the last one to get the following equations:
  
 
<cmath>19^2 + d^2 = r^2</cmath>
 
<cmath>19^2 + d^2 = r^2</cmath>
Line 23: Line 77:
 
<cmath>17^2 + (3d)^2 = r^2</cmath>
 
<cmath>17^2 + (3d)^2 = r^2</cmath>
  
Solving, we find <math>d = 3</math>, so <math>2d = \boxed{(B) 6}</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 \rightarrow d^2 = 9 \rightarrow 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>.
 +
 
 +
~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
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 +
==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
  
-Solution by Joeya (someone draw a diagram n fix my latex please)
+
~Interstigation
  
 
==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}}
 +
{{AMC10 box|year=2021|ab=B|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

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

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

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