Difference between revisions of "2021 AMC 12B Problems/Problem 8"
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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== | + | ==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) | ||
+ | |||
+ | ==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 cord with length 38 to the diameter is <math>d</math>. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | <math>(19, -d)</math> | ||
+ | |||
+ | <math>(17, -3d)</math> | ||
+ | |||
+ | |||
+ | Now, we can plug one of the first two value in as well as the last one to get the following equations: | ||
+ | |||
+ | <math>19^2 + d^2 = r^2</math> | ||
+ | <math>17^2 + (3d)^2 = r^2</math> | ||
+ | |||
+ | 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 and . What is the distance between two adjacent parallel lines?
Solution 1
Since the two chords of length 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
- Another with base , height , and hypotenuse
By the Pythagorean theorem, we can create the following systems of equations:
Solving, we find , so
-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 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 cord 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 .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.