Difference between revisions of "2020 AMC 12B Problems/Problem 12"
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<math>\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100</math> | <math>\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100</math> | ||
− | ==Solution 1== | + | ==Diagram== |
+ | [[File:2020 AMC 12B Problem 12.png|center]] | ||
+ | |||
+ | ~MRENTHUSIASM (by Geometry Expressions) | ||
+ | |||
+ | ==Solution 1 (Pythagorean Theorem) == | ||
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>. | 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>. | ||
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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>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 | + | 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}}</math>, <math>(CE + DE)^2 = CD^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 = (CE+ED)^2-2\cdot CE \cdot DE=(60 + 40\sqrt{10}) - (40\sqrt{10} - 40) = \boxed{(E) 100}</math>. | ||
==Solution 3 (Law of Cosines)== | ==Solution 3 (Law of Cosines)== | ||
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~sofas103 | ~sofas103 | ||
− | ==Video | + | ==Video Solutions== |
− | + | https://www.youtube.com/watch?v=h-hhRa93lK4 | |
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https://youtu.be/0xgTR3UEqbQ | https://youtu.be/0xgTR3UEqbQ | ||
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==See Also== | ==See Also== |
Revision as of 06:34, 22 April 2021
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
Diagram
~MRENTHUSIASM (by Geometry Expressions)
Solution 1 (Pythagorean Theorem)
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 Solutions
https://www.youtube.com/watch?v=h-hhRa93lK4
See Also
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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All AMC 12 Problems and Solutions |
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