Difference between revisions of "2006 AMC 12A Problems/Problem 16"

Latest revision as of 09:16, 19 December 2021

The following problem is from both the 2006 AMC 12A #16 and 2006 AMC 10A #23, so both problems redirect to this page.

Problem

Circles with centers $A$ and $B$ have radius 3 and 8, respectively. A common internal tangent intersects the circles at $C$ and $D$ , respectively. Lines $AB$ and $CD$ intersect at $E$ , and $AE=5$ . What is $CD$ ?

$[asy] unitsize(2.5mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=3; pair A=(0,0), Ep=(5,0), B=(5+40/3,0); pair M=midpoint(A--Ep); pair C=intersectionpoints(Circle(M,2.5),Circle(A,3))[1]; pair D=B+8*dir(180+degrees(C)); dot(A); dot(C); dot(B); dot(D); draw(C--D); draw(A--B); draw(Circle(A,3)); draw(Circle(B,8)); label("$A$",A,W); label("$B$",B,E); label("$C$",C,SE); label("$E$",Ep,SSE); label("$D$",D,NW); [/asy]$

$\textbf{(A) } 13\qquad\textbf{(B) } \frac{44}{3}\qquad\textbf{(C) } \sqrt{221}\qquad\textbf{(D) } \sqrt{255}\qquad\textbf{(E) } \frac{55}{3}\qquad$

Solution

$\angle AED$ and $\angle BEC$ are vertical angles so they are congruent, as are angles $\angle ADE$ and $\angle BCE$ (both are right angles because the radius and tangent line at a point on a circle are always perpendicular). Thus, $\triangle ACE \sim \triangle BDE$ . By the Pythagorean Theorem, line segment $DE = 4$ . The sides are proportional, so $\frac{DE}{AD} = \frac{CE}{BC} \Rightarrow \frac{4}{3} = \frac{CE}{8}$ . This makes $CE = \frac{32}{3}$ and $CD = CE + DE = 4 + \frac{32}{3} = \boxed{\textbf{(B) }\frac{44}{3}}$ .

2006 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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

2006 AMC 10A (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
All AMC 10 Problems and Solutions

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

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Difference between revisions of "2006 AMC 12A Problems/Problem 16"

Latest revision as of 09:16, 19 December 2021

Problem

Solution

See also

@@ Line 1: / Line 1: @@
 {{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #16]] and [[2006 AMC 10A Problems/Problem 23|2006 AMC 10A #23]]}}
 == Problem ==
-[[Circle]]s with [[center]]s <math>A</math> and <math>B</math> have [[radius |radii]] 3 and 8, respectively. A [[common internal tangent line | common internal tangent]] [[intersect]]s the circles at <math>C</math> and <math>D</math>, respectively. [[Line]]s <math>AB</math> and <math>CD</math> intersect at <math>E</math>, and <math>AE=5</math>. What is <math>CD</math>?
+Circles with centers <math>A</math> and <math>B</math> have radius 3 and 8, respectively. A [[common internal tangent line | common internal tangent]] intersects the circles at <math>C</math> and <math>D</math>, respectively. Lines <math>AB</math> and <math>CD</math> intersect at <math>E</math>, and <math>AE=5</math>. What is <math>CD</math>?
+<asy>
+unitsize(2.5mm);
+defaultpen(fontsize(10pt)+linewidth(.8pt));
+dotfactor=3;
+pair A=(0,0), Ep=(5,0), B=(5+40/3,0);
+pair M=midpoint(A--Ep);
+pair C=intersectionpoints(Circle(M,2.5),Circle(A,3))[1];
+pair D=B+8*dir(180+degrees(C));
+dot(A);
+dot(C);
+dot(B);
+dot(D);
+draw(C--D);
+draw(A--B);
+draw(Circle(A,3));
+draw(Circle(B,8));
-<math>\mathrm{(A) \ } 13\qquad\mathrm{(B) \ } \frac{44}{3}\qquad\mathrm{(C) \ } \sqrt{221}\qquad\mathrm{(D) \ } \sqrt{255}\qquad\mathrm{(E) \ } \frac{55}{3}\qquad</math>
+label("$A$",A,W);
+label("$B$",B,E);
+label("$C$",C,SE);
+label("$E$",Ep,SSE);
+label("$D$",D,NW);
+</asy>
-[[Image:2006_AMC12A-16.png]]
+<math>\textbf{(A) } 13\qquad\textbf{(B) } \frac{44}{3}\qquad\textbf{(C) } \sqrt{221}\qquad\textbf{(D) } \sqrt{255}\qquad\textbf{(E) } \frac{55}{3}\qquad</math>
 == Solution ==
 [[Image:2006_AMC12A-16a.png]]
-<math>\angle AED</math> and <math>\angle BEC</math> are vertical angles so they are [[congruent (geometry) | congruent]], as are [[angle]]s <math>\angle ADE</math> and <math>\angle BCE</math> (both are [[right angle]]s because the radius and [[tangent line]] at a point on a circle are always [[perpendicular]]). Thus, <math>\triangle ACE \sim \triangle BDE</math>.
+<math>\angle AED</math> and <math>\angle BEC</math> are vertical angles so they are congruent, as are angles <math>\angle ADE</math> and <math>\angle BCE</math> (both are right angles because the radius and [[tangent line]] at a point on a circle are always perpendicular). Thus, <math>\triangle ACE \sim \triangle BDE</math>.
-By the [[Pythagorean Theorem]], [[line segment]] <math>DE = 4</math>.  The sides are [[proportion]]al, so <math>\frac{DE}{AD} = \frac{CE}{BC} \Rightarrow \frac{4}{3} = \frac{CE}{8}</math>. This makes <math>CE = \frac{32}{3}</math> and <math>CD = CE + DE = 4 + \frac{32}{3} = \frac{44}{3} \Longrightarrow \mathrm{B}</math>.
+By the [[Pythagorean Theorem]], line segment <math>DE = 4</math>.  The sides are proportional, so <math>\frac{DE}{AD} = \frac{CE}{BC} \Rightarrow \frac{4}{3} = \frac{CE}{8}</math>. This makes <math>CE = \frac{32}{3}</math> and <math>CD = CE + DE = 4 + \frac{32}{3} = \boxed{\textbf{(B) }\frac{44}{3}}</math>.
 == See also ==