Difference between revisions of "2006 AMC 12A Problems/Problem 16"
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− | <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.