2000 AMC 10 Problem 19

Ratios of areas do not depend on scale, so let the square have side $1$ and area $1$.

Put the right angle at the lower left. The square occupies the corner; one small triangle sits to its right, with vertical leg equal to the square's side, $1$, and some horizontal leg $a$. The other sits above the square, with horizontal leg $1$ and some vertical leg $b$.

The right-hand triangle has area $\tfrac12 \cdot 1 \cdot a = m$, so $a = 2m$.

Both small hypotenuses are pieces of the big hypotenuse, a single line, so they have the same steepness (rise over run). For the right-hand triangle that is $\frac{1}{a} = \frac{1}{2m}$; for the upper triangle it is $\frac{b}{1}$. Hence $b = \frac{1}{2m}$.

The upper triangle's area is $\tfrac12 \cdot 1 \cdot b = \frac{1}{4m}$, and the square has area $1$, so the ratio is $\frac{1}{4m}$.

The answer is $\boxed{\textbf{(D)}\ \frac{1}{4m}}$.

Lines parallel to the legs create right triangles similar to the original and to each other; sharing a hypotenuse line is the cleanest way to see that their leg ratios match. Normalizing the square to side $1$ removes a variable, and the answer's form $\frac{1}{4m}$ reflects that the two triangles are reciprocal in shape: one is stretched by $2m$ and the other compressed by $2m$.

Test a specific triangle with legs $1$ and $2$. The inscribed square has side $\frac{1 \cdot 2}{1 + 2} = \frac23$, area $\frac49$. The two leftover triangles have legs $\frac23, \frac43$ (area $\frac49$, so $m = 1$) and $\frac23, \frac13$ (area $\frac19$, ratio $\frac14$). At $m = 1$ only choice (D) gives $\frac14$.

• Guessing the second ratio is 1 - m because the pieces 'add up', or testing an isosceles right triangle, where m = 1/2 makes every choice equal 1/2.
• Setting the two small triangles' legs proportional the wrong way ($b = 2m$), which returns $m$, choice (B).