2013 AMC 12B Problems/Problem 19

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.