Difference between revisions of "2013 AMC 12B Problems/Problem 19"
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− | The first solution corresponds to <math>(0, 0)</math>, or point <math>D</math>. The other must be point <math>F</math> (since it is given <math>D</math> and <math>F</math> are distinct). The value of <math>DF</math> is equal to the distance from <math>(0, 0)</math> to <math>(4x, 3x)</math>, and this is clearly <math>5x</math>. Therefore <math>DF = 5 * \frac{16}{25} = \frac{16}{5}</math>, and it is evident that <math>m = 16</math> and <math>n = 5</math>, thus our answer is <math>m + n = 16 + 5 = 21</math>. Answer choice <math>\textbf{(B)}</math> is correct. | + | The first solution corresponds to <math>(0, 0)</math>, or point <math>D</math>. The other must be point <math>F</math> (since it is given that <math>D</math> and <math>F</math> are distinct). The value of <math>DF</math> is equal to the distance from <math>(0, 0)</math> to <math>(4x, 3x)</math>, and this is clearly <math>5x</math>. Therefore <math>DF = 5 * \frac{16}{25} = \frac{16}{5}</math>, and it is evident that <math>m = 16</math> and <math>n = 5</math>, thus our answer is <math>m + n = 16 + 5 = 21</math>. Answer choice <math>\textbf{(B)}</math> is correct. |
== See also == | == See also == | ||
{{AMC12 box|year=2013|ab=B|num-b=18|num-a=20}} | {{AMC12 box|year=2013|ab=B|num-b=18|num-a=20}} |
Revision as of 21:59, 23 February 2013
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 |