2000 AMC 12 Problem 21

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. What is the ratio of the area of the other small right triangle to the area of the square?

\dfrac{1}{2m + 1}

m

1 - m

\dfrac{1}{4m}

\dfrac{1}{8m^2}

Answer: D

Concepts:similarity area ratio

Difficulty rating: 1970

Small hint:

Let the square have side

1

; one small triangle has a leg

r

along the square, so its area is

\dfrac{r}{2} = m

Big hint:

The two small triangles are similar, so the other triangle’s matching leg is

\dfrac{1}{r}

Solution:

Let the square have side

1.

The small triangle sharing one side of the square has a perpendicular leg

r,

so its area is

\tfrac12 \cdot 1 \cdot r = m,

giving

r = 2m.

The two small triangles are similar, so the other triangle’s leg along the square is

\dfrac1r,

and its area is

\frac12 \cdot 1 \cdot \frac1r = \frac{1}{2r} = \frac{1}{4m}.

Thus, the correct answer is D.