Difference between revisions of "2013 AMC 12B Problems/Problem 19"
(→Problem) |
|||
| Line 17: | Line 17: | ||
{{AMC12 box|year=2013|ab=B|num-b=18|num-a=20}} | {{AMC12 box|year=2013|ab=B|num-b=18|num-a=20}} | ||
| + | {{AMC10 box|year=2013|ab=B|num-b=22|num-a=24}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 13:17, 23 December 2014
- The following problem is from both the 2013 AMC 12B #19 and 2013 AMC 10B #23, so both problems redirect to this page.
Contents
Problem
In triangle 
, 
, 
, and 
. Distinct points 
, 
, and 
lie on segments 
, 
, and 
, respectively, such that 
, 
, and 
. The length of segment 
can be written as 
, where 
and 
are relatively prime positive integers. What is 
?
$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}}\ 27\qquad\textbf{(E)}\ 30$ (Error compiling LaTeX. Unknown error_msg)
Solution 1
Since 
, quadrilateral 
is cyclic. It follows that 
. In addition, since 
, triangles 
and 
are similar. It follows that 
. By Ptolemy, we have 
. Cancelling 
, the rest is easy. We obtain 
Solution 2
Using the similar triangles in triangle 
gives 
and 
. Quadrilateral is cyclic, implying that 
= 180°. Therefore, 
, and triangles 
and 
are similar. Solving the resulting proportion gives 
. Therefore, 
See also
| 2013 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 18 |
Followed by Problem 20 |
| 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 | |
| 2013 AMC 10B (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. 
