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

(Added in Solution 3.)
(→‎Solution 3 (Stewart's Theorem))
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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 1==
+
==Solution 1 (Pythagorean Theorem)==
  
 
<asy>
 
<asy>
−
size(6cm);
+
size(8cm);
−
pair O = (0, 4), A = (0, 5), B = (0, 7), R = (3.873, 5), L = (2.645, 7);
+
pair O = (0, 0), A = (0, 3), B = (0, 9), R = (19, 3), L = (17, 9);
 
draw(O--A--B);
 
draw(O--A--B);
 
draw(O--R);
 
draw(O--R);
 
draw(O--L);
 
draw(O--L);
−
label("$A$", A, NW);
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label("$A$", A, NE);
 
label("$B$", B, N);
 
label("$B$", B, N);
 
label("$R$", R, NE);
 
label("$R$", R, NE);
−
label("$L$", L, N);
+
label("$L$", L, NE);
 
label("$O$", O, S);
 
label("$O$", O, S);
 
label("$d$", O--A, W);
 
label("$d$", O--A, W);
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label("$2d$", A--B, W*2+0.5*N);
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label("$2d$", A--B, W);
 
label("$r$", O--R, S);
 
label("$r$", O--R, S);
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label("$r$", O--L, S*0.5 + 1.5 * E);
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label("$r$", O--L, NW);
 
dot(O);
 
dot(O);
 
dot(A);
 
dot(A);
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dot(L);
 
dot(L);
  
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draw(circle((0, 4), 4));
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draw(circle((0, 0), sqrt(370)));
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draw((-3.873, 3) -- (3.873, 3));
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draw(-R -- (R.x, -R.y));
−
draw((-3.873, 5) -- (3.873, 5));
+
draw((-R.x, R.y) -- R);
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draw((-2.645, 7) -- (2.645, 7));
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draw((-L.x, L.y) -- L);
 
</asy>
 
</asy>
  
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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>2d + d</math>, and hypotenuse <math>r</math>  (<math>\triangle LBO</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)
  
 
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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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>.
  
−
-Solution by Joeya and diagram by Jamess2022(burntTacos).
+
-Solution by Joeya, diagram by Jamess2022(burntTacos), and minor edits by lpieleanu.
−
(Someone fix the diagram if possible. -<i> Done. </i>)
 
  
 
==Solution 2 (Coordinates)==
 
==Solution 2 (Coordinates)==
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~Tony_Li2007
 
~Tony_Li2007
  
−
==Solution 3==
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==Solution 3 (Stewart's Theorem)==
 
<asy>
 
<asy>
 
real r=sqrt(370);
 
real r=sqrt(370);
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label("$Q$", Q, S);
 
label("$Q$", Q, S);
 
</asy>
 
</asy>
−
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 \begin{align*}
+
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  
 +
<cmath>\begin{align*}
 
361+\tfrac{1}{4}d^{2} &= r^{2} \\
 
361+\tfrac{1}{4}d^{2} &= r^{2} \\
 
289+\tfrac{9}{4}d^{2} &= r^{2} \\
 
289+\tfrac{9}{4}d^{2} &= r^{2} \\
−
\end{align*} 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>
+
\end{align*}</cmath>
 +
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>
  
 
~Punxsutawney Phil
 
~Punxsutawney Phil
 +
 +
 +
==Video Solution (Super Fast. Just 1 min!)==
 +
https://youtu.be/145UJbG4aCQ
 +
 +
<i>~Education, the Study of Everything </i>
  
 
==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
 +
 +
~pi_is_3.14
  
 
==Video Solution by TheBeautyofMath==
 
==Video Solution by TheBeautyofMath==

Revision as of 23:15, 18 July 2023

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 (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

See Also

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

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

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