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

(Solution 1)
(Solution 1)
Line 6: Line 6:
 
Drawing the diagram:
 
Drawing the diagram:
  
[asy]
+
<asy>
 
pair A, B, C, D, E;
 
pair A, B, C, D, E;
 
A = (0,0);
 
A = (0,0);
Line 25: Line 25:
 
draw(C--E);
 
draw(C--E);
 
draw(D--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==
 
==Solution 2==

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 (ProblemsAnswer KeyResources)
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
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS