2000 AMC 10 Problem 19

Attempt Problem 19 of the 2000 AMC 10 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2000 AMC 10 solutions, or check the answer key.

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19.

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 mm times the area of the square. The ratio of the area of the other small right triangle to the area of the square is

12m+1\dfrac{1}{2m + 1}

mm

1m1 - m

14m\dfrac{1}{4m}

18m2\dfrac{1}{8m^2}

Answer: D
Concepts:similarityarea ratioright triangle
Difficulty rating: 1750
Solution:

Let the square have side 1.1. One small triangle has legs 11 and r,r, with area 12r=m,\tfrac12 r = m, so r=2m.r = 2m.

The two small triangles are similar, so the other has legs 11 and 1r,\tfrac1r, with area 121r=14m.\tfrac12 \cdot \tfrac1r = \tfrac{1}{4m}.

Since the square has area 1,1, the desired ratio is 14m.\dfrac{1}{4m}.

Thus, the correct answer is D.

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