Difference between revisions of "2013 AMC 12B Problems/Problem 19"
m (→Solution 1) |
(→Solution 1) |
||
Line 7: | Line 7: | ||
<math> \textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 30</math> | <math> \textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 30</math> | ||
[[Category: Introductory Geometry Problems]] | [[Category: Introductory Geometry Problems]] | ||
+ | |||
+ | ==Diagram== | ||
+ | <asy> | ||
+ | size(300); | ||
+ | defaultpen(linewidth(0.4)+fontsize(10)); | ||
+ | pen s = linewidth(0.8)+fontsize(8); | ||
+ | |||
+ | pair A,B,C,D,E0,F,G; | ||
+ | A = origin; | ||
+ | C = (15,0); | ||
+ | B = IP(CR(A,13),CR(C,14)); | ||
+ | D = foot(A,C,B); | ||
+ | E0 = foot(D,A,C); | ||
+ | F = OP(CR((A+B)/2,length(B-A)/2), D--E0); | ||
+ | draw(A--C--B--A, black+0.8); | ||
+ | draw(B--F--A--D--E0); | ||
+ | dot("$A$",A,W); | ||
+ | dot("$B$",B,N); | ||
+ | dot("$C$",C,E); | ||
+ | dot("$D$",D,NE); | ||
+ | dot("$E$",E0,S); | ||
+ | dot("$F$",F,E); | ||
+ | draw(rightanglemark(B,D,A)); | ||
+ | draw(rightanglemark(B,F,A)); | ||
+ | label("$5$",D--B,NE); | ||
+ | label("$9$",D--C,NE); | ||
+ | label(Label("$13$",Rotate(B-A)), B--A); | ||
+ | </asy> | ||
==Solution 1== | ==Solution 1== |
Revision as of 12:49, 16 October 2021
- The following problem is from both the 2013 AMC 12B #19 and 2013 AMC 10B #23, so both problems redirect to this page.
Contents
[hide]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 ?
Diagram
Solution 1
Since , quadrilateral is cyclic. It follows that , so are similar. In addition, . We can easily find , , and using Pythagorean triples.
So, the ratio of the longer leg to the hypotenuse of all three similar triangles is , and the ratio of the shorter leg to the hypotenuse is . It follows that .
Let . By Ptolemy's Theorem, we have . Dividing by we get so our answer is .
~Edits by BakedPotato66
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, and our answer is .
Solution 3
If we draw a diagram as given, but then add point on such that in order to use the Pythagorean theorem, we end up with similar triangles and . Thus, and , where is the length of . Using the Pythagorean theorem, we now get and can be found out noting that is just through base times height (since , similar triangles gives ), and that is just . From there, Now, , and squaring and adding both sides and subtracting a 169 from both sides gives , so . Thus, the answer is .
Solution 4 (Power of a Point)
First, we find , , and via the Pythagorean Theorem or by using similar triangles. Next, because is an altitude of triangle , . Using that, we can use the Pythagorean Theorem and similar triangles to find and .
Points , , , and all lie on a circle whose diameter is . Let the point where the circle intersects be . Using power of a point, we can write the following equation to solve for : Using that, we can find , and using , we can find that .
We can use power of a point again to solve for : Thus, .
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.