2020 AMC 12B Problems/Problem 12
Let be a diameter in a circle of radius Let be a chord in the circle that intersects at a point such that and What is
Let be the center of the circle, and be the midpoint of . Let and . This implies that . Since , we now want to find . Since is a right angle, by Pythagorean theorem . Thus, our answer is .
Solution 2 (Power of a Point)
Let be the center of the circle, and be the midpoint of . Draw triangle , and median . Because , is isosceles, so is also an altitude of . , and because angle is degrees and triangle is right, . Because triangle is right, . Thus, . We are looking for + which is also . Because . By power of a point, so . Finally, .
Solution 3 (Law of Cosines)
Let be the center of the circle. Notice how , where is the radius of the circle. By applying the law of cosines on triangle , . Similarly, by applying the law of cosines on triangle , . By subtracting these two equations, we get . We can rearrange it to get . Because both and are both positive, we can safely divide both sides by to obtain . Because , . Through power of a point, we can find out that , so .
Solution 4 (Reflections)
Let be the center of the circle. By reflecting across the line to produce , we have that . Since , . Since , by the Pythagorean Theorem, our desired solution is just . Looking next to circle arcs, we know that , so . Since , and , . Thus, . Since , by the Pythagorean Theorem, the desired .
On The Spot STEM: https://www.youtube.com/watch?v=h-hhRa93lK4
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