# 2020 AMC 12B Problems/Problem 12

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## 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 $CE^2+DE^2?$

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

## Diagram

~MRENTHUSIASM (by Geometry Expressions)

## Solution 1 (Pythagorean Theorem)

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 $2\times50=\boxed{\textbf{(E) } 100}$.

~JHawk0224

## Solution 2 (Power of a Point)

Let $O$ be the center of the circle, and $X$ be the midpoint of $CD$. Draw triangle $OCD$, and median $OX$. Because $OC = OD$, $OCD$ is isosceles, so $OX$ is also an altitude of $OCD$. $OE = 5\sqrt2 - 2\sqrt5$, and because angle $OEC$ is $45$ degrees and triangle $OXE$ is right, $OX = EX = \frac{5\sqrt2 - 2\sqrt5}{\sqrt2} = 5 - \sqrt{10}$. Because triangle $OXC$ is right, $CX = \sqrt{(5\sqrt2)^2 - (5 - \sqrt{10})^2} = \sqrt{15 + 10\sqrt{10}}$. Thus, $CD = 2\sqrt{15 + 10\sqrt{10}}$.

We are looking for $CE^2$ + $DE^2$ which is also $(CE + DE)^2 - 2 \cdot CE \cdot DE$.

Because $CE + DE = CD = 2\sqrt{15 + 10\sqrt{10}}$, $(CE + DE)^2 = CD^2=4(15 + 10\sqrt{10}) = 60 + 40\sqrt{10}$.

By Power of a Point, $CE \cdot DE = AE \cdot BE = 2\sqrt5\cdot(10\sqrt2 - 2\sqrt5) = 20\sqrt{10} - 20$, so $2 \cdot CE \cdot DE = 40\sqrt{10} - 40$.

Finally, $CE^2 + DE^2 = (CE+ED)^2-2\cdot CE \cdot DE=(60 + 40\sqrt{10}) - (40\sqrt{10} - 40) = \boxed{(E) 100}$.

## Solution 3 (Law of Cosines)

Let $O$ be the center of the circle. Notice how $OC = OD = r$, where $r$ is the radius of the circle. By applying the law of cosines on triangle $OCE$, $r^2=CE^2+OE^2-2(CE)(OE)\cos{45}=CE^2+OE^2-(CE)(OE)\sqrt{2}$. Similarly, by applying the law of cosines on triangle $ODE$, $r^2=DE^2+OE^2-2(DE)(OE)\cos{135}=DE^2+OE^2+(DE)(OE)\sqrt{2}$. By subtracting these two equations, we get $CE^2-DE^2-(CE)(OE)\sqrt{2}-(DE)(OE)\sqrt{2}=0$. We can rearrange it to get $CE^2-DE^2=(CE)(OE)\sqrt{2}+(DE)(OE)\sqrt{2}=(CE+DE)(OE\sqrt{2})$. Because both $CE$ and $DE$ are both positive, we can safely divide both sides by $(CE+DE)$ to obtain $CE-DE=OE\sqrt{2}$. Because $OE = OB - BE = 5\sqrt{2} - 2\sqrt{5}$, $(CE-DE)^2 = CE^2+DE^2 - 2(CE)(DE) = (OE\sqrt{2})^2 =2(5\sqrt{2} - 2\sqrt{5})^2 = 140 - 40\sqrt{10}$. Through power of a point, we can find out that $(CE)(DE)=20\sqrt{10} - 20$, so $CE^2+DE^2 = (CE-DE)^2+ 2(CE)(DE)= (140 - 40\sqrt{10}) + 2(20\sqrt{10} - 20) = \boxed{\textbf{(E) } 100}$.

~Math_Wiz_3.14

## Solution 4 (Reflections)

$[asy] draw(circle((0,0),7.07)); dot((-7.07,0)); label("A",(-7.07,0),W); dot((7.07,0)); label("B",(7.07,0),E); dot((0,0)); label("O",(0,0),N); dot((-2.24,-6.71)); label("C",(-2.24,-6.71),SSW); dot((6.71,2.24)); label("D",(6.71,2.24),NE); draw((-2.24,-6.71)--(6.71,2.24)); dot((6.71,-2.24)); label("D'",(6.71,-2.24),SE); draw((4.51,0)--(6.71,-2.24)); dot((4.51,0)); label("E",(4.51,0),NNW); draw((-7.07,0)--(7.07,0)); draw((0,0)--(-2.24,-6.71)); draw((0,0)--(6.71,-2.24)); draw((-2.24,-6.71)--(6.71,-2.24)); [/asy]$ Let $O$ be the center of the circle. By reflecting $D$ across the line $AB$ to produce $D'$, we have that $\angle BED'=45$. Since $\angle AEC=45$, $\angle CED'=90$. Since $DE=ED'$, by the Pythagorean Theorem, our desired solution is just $CD'^2$. Looking next to circle arcs, we know that $\angle AEC=\frac{\overarc{AC}+\overarc{BD}}{2}=45$, so $\overarc{AC}+\overarc{BD}=90$. Since $\overarc{BD'}=\overarc{BD}$, and $\overarc{AC}+\overarc{BD'}+\overarc{CD'}=180$, $\overarc{CD'}=90$. Thus, $\angle COD'=90$. Since $OC=OD'=5\sqrt{2}$, by the Pythagorean Theorem, the desired $CD'^2= \boxed{\textbf{(E) } 100}$.

~sofas103