Difference between revisions of "2006 AMC 12A Problems/Problem 16"
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[[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 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 
and 
have radius 3 and 8, respectively. A common internal tangent intersects the circles at 
and 
, respectively. Lines 
and 
intersect at 
, and 
. What is ?
![[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]](http://latex.artofproblemsolving.com/8/9/3/8932f7dfafa96fd7b5e9bd90951c495dea0015fb.png)
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. 
