Difference between revisions of "2013 AMC 12B Problems/Problem 19"
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<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 13:49, 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
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.