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

(→‎Diagram)
(→‎Diagram: Thank you Shihan especially for converting diagram to Asy. Minor edits on the diagram: 1. lines have the same thickness, 2. O is not labeled, as the question does not mention it 3. Larger labels.)
Line 6: Line 6:
 
==Diagram==
 
==Diagram==
 
<asy>
 
<asy>
−
size(200);
+
/* Made by Shihan; edited by MRENTHUSIASM */
−
defaultpen(linewidth(0.4)+fontsize(10));
+
size(250);
−
pen s = linewidth(0.8)+fontsize(8);
 
  
−
pair O,A,B,C,D,X,F;
+
pair O, A, B, C, D, E;
−
O=origin;
+
O = origin;
−
A=(-1,0);  
+
A = (-5*sqrt(2),0);
−
B=(1,0);
+
B = (5*sqrt(2),0);
−
real r = 1-0.6324;
+
E = (5*sqrt(2)-2*sqrt(5),0);
−
X = (r,0);
+
path p;
−
pair Cp = X+3*dir(135);
+
p = Circle(O,5*sqrt(2));
−
pair Dp = X-3*dir(135);
+
C = intersectionpoint(p,E--E+10*dir(135));
−
path c1 = CR(O,1);
+
D = intersectionpoint(p,E--E+10*dir(-45));
−
C = IP(c1, Cp--Dp);
+
draw(p);
−
D = OP(c1, Cp--Dp);
+
dot(O,linewidth(4));
−
F = foot(O,C,D);
+
dot("$A$",A,1.5*dir(A),linewidth(4));
−
draw(C--D);
+
dot("$B$",B,1.5*dir(B),linewidth(4));
−
dot("$A$", A, W);
+
dot("$E$",E,1.5*dir(135/2),linewidth(4));
−
dot("$B$", B, E);
+
dot("$C$",C,1.5*dir(C),linewidth(4));
−
dot("$C$", C, C-X);
+
dot("$D$",D,1.5*dir(D),linewidth(4));
−
dot("$D$", D, D-X);
+
draw(A--B^^C--D);
−
dot("$O$", O, S);
+
label("$45^\circ$",E,3.5*dir(155.5),red+fontsize(10));
−
dot("$E$", X, 2*S);
+
label("$5\sqrt2$",midpoint(A--O),S);
−
MA("45^\circ",C,X,O,0.075);
+
label("$2\sqrt5$",midpoint(E--B),S);
−
label("$2\sqrt{5}$",B--X,S);
 
−
label("$5\sqrt{2}$",A--O,S);
 
−
draw(c1, s);
 
−
draw(A--B, heavygreen);
 
 
</asy>
 
</asy>
 +
~Shihan ~MRENTHUSIASM
  
 
==Solution 1 (Pythagorean Theorem) ==
 
==Solution 1 (Pythagorean Theorem) ==

Revision as of 12:34, 30 September 2021

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] /* Made by Shihan; edited by MRENTHUSIASM */ size(250); pair O, A, B, C, D, E; O = origin; A = (-5*sqrt(2),0); B = (5*sqrt(2),0); E = (5*sqrt(2)-2*sqrt(5),0); path p; p = Circle(O,5*sqrt(2)); C = intersectionpoint(p,E--E+10*dir(135)); D = intersectionpoint(p,E--E+10*dir(-45)); draw(p); dot(O,linewidth(4)); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$E$",E,1.5*dir(135/2),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); draw(A--B^^C--D); label("$45^\circ$",E,3.5*dir(155.5),red+fontsize(10)); label("$5\sqrt2$",midpoint(A--O),S); label("$2\sqrt5$",midpoint(E--B),S); [/asy] ~Shihan ~MRENTHUSIASM

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

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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