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Difference between revisions of "2020 AMC 12B Problems/Problem 12"

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==Solution==
 
==Solution==
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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>XC=a</math> and <math>XE=b</math>. This implies that <math>ED = a - b</math>. Since <math>CE = XC + XE = 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 = XC^2 + XO^2 = (5\sqrt{2})^2=50</math>. Thus, our answer is <math>2(50)=\boxed{\textbf{(E) } 100}</math>.
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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(50)=\boxed{\textbf{(E) } 100}</math>.
  
 
~JHawk0224
 
~JHawk0224

Revision as of 23:11, 7 February 2020

Problem

Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$

$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100$

Solution

Let $O$ be the center of the circle, and $X$ be the midpoint of $\overline{CD}$. Let $CX=a$ and $EX=b$. This implies that $DE = a - b$. Since $CE = CX + EX = a + b$, we now want to find $(a+b)^2+(a-b)^2=2(a^2+b^2)$. Since $\angle CXO$ is a right angle, by Pythagorean theorem $a^2 + b^2 = CX^2 + OX^2 = (5\sqrt{2})^2=50$. Thus, our answer is $2(50)=\boxed{\textbf{(E) } 100}$.

~JHawk0224

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

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