Difference between revisions of "2021 AMC 12B Problems/Problem 8"
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MRENTHUSIASM (talk | contribs) m (→Solution 3: Fixed the coding and added the title.) |
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~Tony_Li2007 | ~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 <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} \\ | ||
+ | 289+\tfrac{9}{4}d^{2} &= r^{2} \\ | ||
+ | \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 | ||
==Video Solution by Hawk Math== | ==Video Solution by Hawk Math== | ||
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~IceMatrix | ~IceMatrix | ||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/lYxKkS252Og | ||
+ | |||
+ | ~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.
Contents
Problem
Three equally spaced parallel lines intersect a circle, creating three chords of lengths and . What is the distance between two adjacent parallel lines?
Solution 1
Since two parallel chords 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 . Thus, the distance from the center of the circle to the chord of length is
and the distance between each of the chords is just . Let the radius of the circle be . Drawing radii to the points where the lines intersect the circle, we create two different right triangles:
- One with base , height , and hypotenuse ( on the diagram)
- Another with base , height , and hypotenuse ( on the diagram)
By the Pythagorean theorem, we can create the following system of equations:
Solving, we find , so .
-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 where the center of the circle is and the radius is , we can find the equation of this circle by centering it on the origin. Doing this, we get that the equation is . Now, we can set the distance between the chords as so the distance from the chord with length 38 to the diameter is .
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:
Now, we can plug one of the first two value in as well as the last one to get the following equations:
Subtracting these two equations, we get - therefore, we get . We want to find because that's the distance between two chords. So, our answer is .
~Tony_Li2007
Solution 3 (Stewart's Theorem)
If is the requested distance, and is the radius of the circle, Stewart's Theorem applied to with cevian gives This simplifies to . Similarly, another round of Stewart's Theorem applied to with cevian gives This simplifies to . Dividing the top equation by and the bottom equation by results in the system of equations By transitive, . Therefore
~Punxsutawney Phil
Video Solution by Hawk Math
https://www.youtube.com/watch?v=VzwxbsuSQ80
Video Solution by Punxsutawney Phil
Video Solution by OmegaLearn (Circular Geometry)
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
~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.