a) Triangle $A B C$ was folded, obviously, along the midline $D E$ (see fig.). Therefore, the "diagonals" $C E$ and $B D$ were its medians. Thus,

$B O: O D=C O: O E=2: 1$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6409e68af5a1ba5e989dg-12.jpg?height=478&width=634&top_left_y=1&top_left_x=715)

b) After the cut, three pieces are formed. Two of them are triangles $D O C$ and $B C D$, the third is composed of triangles $D O E$ and $B D E$ (after unfolding, it will also turn into a triangle). The smallest area is that of triangle $O C D$ (it is a part of triangle $B C D$, as well as a part of triangle $C D E$, which is equal in area to triangle $B D E$). Since $O D=1 / 3 B D$, then $S_{O C D}=1 / 3 S_{B C D}=1 / 6 S_{B C A}$.

## Answer

a) $2: 1$;
b) $1 / 6$.
Shapovalov A.V.

A paper rectangular triangle was folded along a straight line so that the vertex of the right angle coincided with another vertex.

a) In what ratio are the diagonals of the resulting quadrilateral divided by their point of intersection?

b) The resulting quadrilateral was cut along the diagonal coming out of the third vertex of the original triangle. Find the area of the smallest piece of paper formed.