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

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Solving, we find <math>d = 3</math>, so <math>2d = \boxed{(B) 6}</math> | Solving, we find <math>d = 3</math>, so <math>2d = \boxed{(B) 6}</math> | ||

− | -Solution by Joeya (someone draw a diagram please) | + | -Solution by Joeya (someone draw a diagram n fix my latex please) |

## Revision as of 16:42, 11 February 2021

## Problem 8

Three equally spaced parallel lines intersect a circle, creating three chords of lengths and . What is the distance between two adjacent parallel lines?

## Solution

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 chords touch 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)