, 2013

Since $\triangle PQR$ is isosceles with $PQ=QR$ and $\angle PQR=90\degree$, then $\angle QPR=\angle QRS=45\degree$. Also in $\triangle PQR$, altitude $QS$ bisects $PR$ ($PS=SR$) forming two identical triangles, $SQP$ and $SQR$. Since these two triangles are identical, each has $\frac{1}{2}$ of the area of $\triangle PQR$. In $\triangle SQR$, $\angle QSR=90\degree$, $\angle QRS=45\degree$, and so $\angle SQR=45\degree$. Thus, $\triangle SQR$ is also isosceles with $SQ=SR$. Then similarly, altitude $ST$ bisects $QR$ ($QT=TR$) forming two identical triangles, $SQT$ and $SRT$. Since these two triangles are identical, each has $\frac{1}{2}$ of the area of $\triangle SQR$ or $\frac{1}{4}$ of the area of $\triangle PQR$. Continuing in this way, altitude $TU$ divides $\triangle STR$ into two identical triangles, $STU$ and $RTU$. Each of these two triangles has $\frac{1}{2}$ of $\frac{1}{4}$ or $\frac{1}{8}$ of the area of $\triangle PQR$. Continuing, altitude $UV$ divides $\triangle RTU$ into two identical triangles, $RUV$ and $TUV$. Each of these two triangles has $\frac{1}{2}$ of $\frac{1}{8}$ or $\frac{1}{16}$ of the area of $\triangle PQR$. Finally, altitude $VW$ divides $\triangle RUV$ into two identical triangles, $UVW$ and $RVW$. Each of these two triangles has $\frac{1}{2}$ of $\frac{1}{16}$ or $\frac{1}{32}$ of the area of $\triangle PQR$. Since the area of $\triangle STU$ is $\frac{1}{8}$ of the area of $\triangle PQR$, and the area of $\triangle UVW$ is $\frac{1}{32}$ of the area of $\triangle PQR$, then the total fraction of $\triangle PQR$ that is shaded is $\frac{1}{8}+\frac{1}{32}=\frac{4+1}{32}$ or $\frac{5}{32}$.