Difference between revisions of "2020 AMC 12B Problems/Problem 12"
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==Solution 2 (Power of a Point)== | ==Solution 2 (Power of a Point)== | ||
− | Let <math>O</math> be the center of the circle, and <math>X</math> be the midpoint of <math>CD</math>. Draw triangle <math>OCD</math>, and median <math>OX</math>. Because <math>OC = OD</math>, <math>OCD</math> is isosceles, so <math>OX</math> is also an altitude of <math>OCD</math>. <math> | + | Let <math>O</math> be the center of the circle, and <math>X</math> be the midpoint of <math>CD</math>. Draw triangle <math>OCD</math>, and median <math>OX</math>. Because <math>OC = OD</math>, <math>OCD</math> is isosceles, so <math>OX</math> is also an altitude of <math>OCD</math>. <math>OE = 5\sqrt2 - 2\sqrt5</math>, and because angle <math>OEC</math> is <math>45</math> degrees and triangle <math>OXE</math> is right, <math>OX = EX = \frac{5\sqrt2 - 2\sqrt5}{\sqrt2} = 5 - \sqrt{10}</math>. Because triangle <math>OXC</math> is right, <math>CX = \sqrt{(5\sqrt2)^2 - (5 - \sqrt{10})^2} = \sqrt{15 + 10\sqrt{10}}</math>. Thus, <math>CD = 2\sqrt{15 + 10\sqrt{10}}</math>. We are looking for <math>CE^2</math> + <math>DE^2</math> which is also <math>(CE + DE)^2 - 2 \cdot CE \cdot DE</math>. Because <math>CE + DE = CD = 2\sqrt{15 + 10\sqrt{10}}, (CE + DE)^2 = 4(15 + 10\sqrt{10}) = 60 + 40\sqrt{10}</math>. By power of a point, <math>CE \cdot DE = AE \cdot BE = 2\sqrt5\cdot(10\sqrt2 - 2\sqrt5) = 20\sqrt{10} - 20</math> so <math>2 \cdot CE \cdot DE = 40\sqrt{10} - 40</math>. Finally, <math>CE^2 + DE^2 = 60 + 40\sqrt{10} - (40\sqrt{10} - 40) = \boxed{(E) 100}</math>. |
==Solution 3 (Law of Cosines)== | ==Solution 3 (Law of Cosines)== |
Latest revision as of 17:36, 23 November 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.