Difference between revisions of "2016 AMC 8 Problems/Problem 23"

Revision as of 10:50, 24 October 2018

Two congruent circles centered at points $A$ and $B$ each pass through the other circle's center. The line containing both $A$ and $B$ is extended to intersect the circles at points $C$ and $D$ . The circles intersect at two points, one of which is $E$ . What is the degree measure of $\angle CED$ ?

$\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150$

Solution 1

Drawing the diagram:

$[asy] pair A, B, C, D, E; A = (0,0); B = (10,0); C = (-10,0); D = (20,0); E = (5, 8.75); draw(Circle(A, 10)); draw(Circle(B, 10)); dot(A); dot(B); dot(C); dot(D); dot(E); draw(C--D); draw(A--E); draw(B--E); draw(C--E); draw(D--E); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, SW); label("$D$", D, SE); label("$E$", E, N); [/asy]$

we see that $\triangle EAB$ is equilateral as each side is the radius of one of the two circles. Therefore, $\overarc{EB}=m\angle EAB=60^\circ$ . Therefore, since it is an inscribed angle, $m\angle ECB=\frac{60^\circ}{2}=30^\circ$ . So, in $\triangle ECD$ , $m\angle ECB=m\angle EDA=30^\circ$ , and $m\angle CED=180^\circ-30^\circ-30^\circ=120^\circ$ . Our answer is $\boxed{\textbf{(C) }\ 120}$ .

Solution 2

As in Solution 1, observe that $\triangle{EAB}$ is equilateral. Therefore, $m\angle{AEB}=m\angle{EAB}=m\angle{EBA} = 60^{\circ}$ . Since $CD$ is a straight line, we conclude that $m\angle{EBD} = 180^{\circ}-60^{\circ}=120^{\circ}$ . Since $BE=BD$ (both are radii of the same circle), $\triangle{BED}$ is isosceles, meaning that $m\angle{BED}=m\angle{BDE}=30^{\circ}$ . Similarly, $m\angle{AEC}=m\angle{ACE}=30^{\circ}$ .

Now, $\angle{CED}=m\angle{AEC}+m\angle{AEB}+m\angle{BED} = 30^{\circ}+60^{\circ}+30^{\circ} = 120^{\circ}$ . Therefore, the answer is $\boxed{\textbf{(C) }\ 120}$ .

2016 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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
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Difference between revisions of "2016 AMC 8 Problems/Problem 23"

Revision as of 10:50, 24 October 2018

Solution 1

Solution 2

@@ Line 6: / Line 6: @@
 Drawing the diagram:
-[asy]
+<asy>
 pair A, B, C, D, E;
 A = (0,0);
@@ Line 25: / Line 25: @@
 draw(C--E);
 draw(D--E);
-label("<math>A</math>", A, SW);
+label("$A$", A, SW);
-label("<math>B</math>", B, SE);
+label("$B$", B, SE);
-label("<math>C</math>", C, SW);
+label("$C$", C, SW);
-label("<math>D</math>", D, SE);
+label("$D$", D, SE);
-label("<math>E</math>", E, N);
+label("$E$", E, N);
-[/asy]
+</asy>
-we see that <math>\triangle EAB</math> is equilateral as each side is the radius of one of the two circles. Therefore, <math>\overarc{EB}=m\angle EAB-60^\circ</math>. Therefore, since it is an inscribed angle, <math>m\angle ECB=\frac{60^\circ}{2}=30^\circ</math>. So, in <math>\triangle ECD</math>, <math>m\angle ECB=m\angle EDA=30^\circ</math>, and <math>m\angle CED=180^\circ-30^\circ-30^\circ=120^\circ</math>. Our answer is <math>\boxed{\textbf{(C) }\ 120}</math>.
+we see that <math>\triangle EAB</math> is equilateral as each side is the radius of one of the two circles. Therefore, <math>\overarc{EB}=m\angle EAB=60^\circ</math>. Therefore, since it is an inscribed angle, <math>m\angle ECB=\frac{60^\circ}{2}=30^\circ</math>. So, in <math>\triangle ECD</math>, <math>m\angle ECB=m\angle EDA=30^\circ</math>, and <math>m\angle CED=180^\circ-30^\circ-30^\circ=120^\circ</math>. Our answer is <math>\boxed{\textbf{(C) }\ 120}</math>.
 ==Solution 2==