2020 AMC 12B Problems/Problem 12

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

$[asy] size(300); defaultpen(linewidth(0.4)+fontsize(10)); pen s = linewidth(0.8)+fontsize(8); pair O,A,B,C,D,X,F; O=origin; A=(-1,0); B=(1,0); real r = 1-0.6324; X = (r,0); pair Cp = X+3*dir(135); pair Dp = X-3*dir(135); path c1 = CR(O,1); C = IP(c1, Cp--Dp); D = OP(c1, Cp--Dp); F = foot(O,C,D); draw(C--D); dot("$A$", A, W); dot("$B$", B, E); dot("$C$", C, C-X); dot("$D$", D, D-X); dot("$O$", O, S); dot("$E$", X, 2*S); MA("45^\circ",C,X,O,0.075); label("$2\sqrt{5}$",B--X,S); label("$5\sqrt{2}$",A--O,S); draw(c1, s); draw(A--B, heavygreen); [/asy]$

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

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 (legibility changes by eagleye)

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

Video Solutions

https://www.youtube.com/watch?v=h-hhRa93lK4

https://youtu.be/0xgTR3UEqbQ

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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