Difference between revisions of "2020 AMC 12B Problems/Problem 12"
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− | Let <math>O</math> be the center of the circle, and <math>X</math> be the midpoint of <math>\overline{CD}</math>. Let <math>CX=a</math> and <math>EX=b</math>. This implies that <math>DE = a - b</math>. Since <math>CE = CX + EX = a + b</math>, we now want to find <math>(a+b)^2+(a-b)^2=2(a^2+b^2)</math>. Since <math>\angle CXO</math> is a right angle, by Pythagorean theorem <math>a^2 + b^2 = CX^2 + OX^2 = (5\sqrt{2})^2=50</math>. Thus, our answer is <math>2 | + | Let <math>O</math> be the center of the circle, and <math>X</math> be the midpoint of <math>\overline{CD}</math>. Let <math>CX=a</math> and <math>EX=b</math>. This implies that <math>DE = a - b</math>. Since <math>CE = CX + EX = a + b</math>, we now want to find <math>(a+b)^2+(a-b)^2=2(a^2+b^2)</math>. Since <math>\angle CXO</math> is a right angle, by Pythagorean theorem <math>a^2 + b^2 = CX^2 + OX^2 = (5\sqrt{2})^2=50</math>. Thus, our answer is <math>2\times50=\boxed{\textbf{(E) } 100}</math>. |
~JHawk0224 | ~JHawk0224 |
Revision as of 21:45, 16 August 2020
Contents
Problem
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
Solution 1
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 .
~JHawk0224
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 .
~Math_Wiz_3.14
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 .
~sofas103
Video Solution
On The Spot STEM: https://www.youtube.com/watch?v=h-hhRa93lK4
Video Solution 2
~IceMatrix
See Also
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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.