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. The ratio of the area of the other small right triangle to the area of the square is $\textbf{(A)}\ \frac{1}{2m+1} \qquad$ $\textbf{(B)}\ m \qquad$ $\textbf{(C)}\ 1-m \qquad$ $\textbf{(D)}\ \frac{1}{4m} \qquad$ $\textbf{(E)}\ \frac{1}{8m^2}$
Solution
Let the point on the hypotenuse be $E$, let $CDEF$ be the square, and let $[BEF]:[CDEF]=m$, the goal is to find $[ADE]:[CDEF]$.
Assume $a=BC, b=AC$ and $x=CD$.
Since $EF\parallel AC$, $\triangle BEF\sim \triangle BAC$, so $BF:BC=EF:CA$, i.e. $(a-x): a=x:b$, so $x=\frac{ab}{a+b}$.
The condition says $\frac12x(a-x):x^2=m$, so $m=\frac{a-x}{2x}$. Since $a-x=a-\frac{ab}{a+b}=\frac{a^2}{a+b}$, this gives $m=\frac{a}{2b}$.
The area of one of the two small right triangles formed by drawing lines through a point on the hypotenuse of a right triangle, parallel to the legs, dividing the triangle into a square and two smaller right triangles, is $\frac13$ 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?
Let the point on the hypotenuse be $E$, let $CDEF$ be the square, and let $[BEF]:[CDEF]=m$, the goal is to find $[ADE]:[CDEF]$.