2000 AMC 10 Problem 19

1. Pick a size. The question only asks for a ratio of areas, and ratios do not change when the picture is scaled. So let the square have side $1$ and area $1$.
2. Draw it. Put the right angle at the bottom left. The square sits in that corner. One small triangle is to the right of the square: its height is the square's side, $1$, and call its base $a$. The other small triangle is on top of the square: its base is $1$, and call its height $b$.
3. Use the given area. Say the right-hand triangle is the one with area $m$ times the square. Its area is $\tfrac12 \cdot a \cdot 1 = m$, so $a = 2m$.
4. Why the two triangles are linked. The slanted sides of both small triangles are pieces of the big triangle's slanted side, which is one straight line. So they have the same steepness (rise divided by run).
- Right-hand triangle: rise $1$, run $a = 2m$, steepness $\frac{1}{2m}$.
- Top triangle: rise $b$, run $1$, steepness $b$.

Setting these equal gives $b = \frac{1}{2m}$.
5. Area of the top triangle. $\tfrac12 \cdot 1 \cdot \frac{1}{2m} = \frac{1}{4m}$. The square has area $1$, so the ratio is $\frac{1}{4m}$.
6. Check with numbers. If $m = 1$, the right triangle has base $2$, the top triangle has height $\tfrac12$ and area $\tfrac14$. Choice (D) gives $\frac{1}{4 \cdot 1} = \tfrac14$. It matches.

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

Lines drawn parallel to the legs make small right triangles with the same shape as the big one. Seeing that their slanted sides lie on one line is the easiest way to see that their sides are in the same proportion. Setting the square's side to $1$ removes one unknown. The answer $\frac{1}{4m}$ shows the two triangles balance: when one gets wider, the other gets thinner.

Test one triangle, but not an isosceles one (there $m = \tfrac12$ and all five choices equal $\tfrac12$). Take legs $1$ and $2$. The square in the corner has side $\tfrac23$ and area $\tfrac49$. The two leftover triangles have legs $\tfrac23, \tfrac43$ (area $\tfrac49$, so $m = 1$) and $\tfrac23, \tfrac13$ (area $\tfrac19$, which is $\tfrac14$ of the square). With $m = 1$, only (D) gives $\tfrac14$.