Difference between revisions of "2006 AMC 12A Problems/Problem 16"
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{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #16]] and [[2006 AMC 10A Problems/Problem 23|2006 AMC 10A #23]]}} | {{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #16]] and [[2006 AMC 10A Problems/Problem 23|2006 AMC 10A #23]]}} | ||
+ | |||
== Problem == | == Problem == | ||
− | + | 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)); | ||
− | + | label("$A$",A,W); | |
+ | label("$B$",B,E); | ||
+ | label("$C$",C,SE); | ||
+ | label("$E$",Ep,SSE); | ||
+ | label("$D$",D,NW); | ||
+ | </asy> | ||
− | + | <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 == | == Solution == | ||
[[Image:2006_AMC12A-16a.png]] | [[Image:2006_AMC12A-16a.png]] | ||
− | <math>\angle AED</math> and <math>\angle BEC</math> are vertical angles so they are | + | <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]], | + | 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 == | == See also == |
Latest revision as of 10:16, 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 and
have radius 3 and 8, respectively. A common internal tangent intersects the circles at
and
, respectively. Lines
and
intersect at
, and
. What is
?
Solution
and
are vertical angles so they are congruent, as are angles
and
(both are right angles because the radius and tangent line at a point on a circle are always perpendicular). Thus,
.
By the Pythagorean Theorem, line segment
. The sides are proportional, so
. This makes
and
.
See also
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.