Question

Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is $m$ times the area of the square. The ratio of the area of the other small right triangle to the area of the square is

$\dfrac{1}{2m + 1}$

$m$

$1 - m$

$\dfrac{1}{4m}$

$\dfrac{1}{8m^2}$

Solution:

Let the square have side $1.$ One small triangle has legs $1$ and $r,$ with area $\tfrac12 r = m,$ so $r = 2m.$

The two small triangles are similar, so the other has legs $1$ and $\tfrac1r,$ with area $\tfrac12 \cdot \tfrac1r = \tfrac{1}{4m}.$

Since the square has area $1,$ the desired ratio is $\dfrac{1}{4m}.$

Thus, the correct answer is D.

2000 AMC 10 Problem 19

Solution:

Problem 19 in Other Years