Two lines are drawn through a point on the hypotenuse of a right-angled triangle parallel to the legs of the triangle that divide the triangle into a square and two smaller triangles. The area of one of the two small triangles is $m$ times the area of the square.
Express, in terms of $m$, the ratio of the area of the other small triangle to the area of the square.
I thought that the ratio would be that of $1:1-m$, as the areas of the triangles add up to the area of the square, but this is apparently incorrect.
The two smaller triangles are themselves right and (as mentioned in the comments) similar to each other and the larger triangle. Without loss of generality, take the square's side length as 1. Let the triangle $m$ times larger than the square (i.e. with area $m$) be $A$ and the other small triangle $B$.
It is easy to see that $A$ must have side lengths 1 and $2m$ in order for it to have area $m$. Then by similar triangles, with $x$ the length of $B$'s leg that does not abut the square, $\frac{2m}1=\frac1x$ or $x=\frac1{2m}$. Then $B$'s area is $\frac1{4m}$ and its ratio with the square's area is $1:4m$.