Difference between revisions of "2013 AMC 12B Problems/Problem 19"
(→See also) |
(→See also) |
||
Line 29: | Line 29: | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 14:04, 3 July 2013
- The following problem is from both the 2013 AMC 12B #19 and 2013 AMC 10B #23, so both problems redirect to this page.
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
The 13-14-15 triangle is very commonly seen in competition problems, since the altitude from the point opposite the side of length 14 (, in this case) divides the triangle into 9-12-15 and 5-12-13 right triangles. This means that
,
, and
.
We now proceed by coordinate geometry. Place the origin of the system at , let the positive x-axis be
, and the positive y-axis be
. Then consider
. It is perpendicular to
, and
has slope
. Thus
is governed by the equation
(recall that perpendicular lines' slopes are negative reciprocals of each other). This means that
must lie at a point given by
.
Now consider the vectors and
. Since
lies at
and
at
, the vectors must be
and
, respectively. If
, then
and
must be orthogonal, and their dot product must be zero. Therefore:
The first solution corresponds to , or point
. The other must be point
(since it is given that
and
are distinct). The value of
is equal to the distance from
to
, and this is clearly
. Therefore
, and it is evident that
and
, thus our answer is
. Answer choice
is correct.
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.