## Task 2 - 271222

Let $A B C D$ be any plane convex quadrilateral; $k$ be any positive real number.

The points $P, Q, R, S$ are to divide the sides $A B, B C, C D, D A$ of this quadrilateral in the ratio $k: 1$ in this order.

Determine the ratio of the areas of the quadrilaterals $P Q R S$ and $A B C D$.
If $X Y \ldots Z$ is a polygon, let $I(X Y \ldots Z)$ denote its area. From the condition $A P: B P = k: 1$, it follows that

$$
A P = \frac{k}{k+1} \cdot A B \quad, \quad P B = \frac{1}{k+1} \cdot A B
$$

Since the triangles $P B Q$ and $A B Q$ have the same perpendicular height to the sides $P B$ and $A B$ respectively, it follows that

$$
I(A B Q) = \frac{k}{k+1} \cdot I(A B C) \quad \text{and thus} \quad I(P B Q) = \frac{k}{(k+1)^{2}} \cdot I(A B C)
$$

Similarly, it follows that

$$
I(R D S) = \frac{k}{(k+1)^{2}} \cdot I(C D A) \quad \text{and thus}
$$

$$
\begin{aligned}
I(P B Q) + I(R D S) & = \frac{k}{(k+1)^{2}} I(A B C D) \\
I(S A P) + I(Q C R) & = \frac{k}{(k+1)^{2}} I(A B C D)
\end{aligned}
$$

and therefore, overall,

$$
\begin{aligned}
I(P Q R S) & = I(A B C D) - I(S A P) - I(P B Q) - I(Q C R) - I(R D S) \\
& = I(A B C D) \cdot \left(1 - \frac{2 k}{(k+1)^{2}}\right) \\
\frac{I(P Q R S)}{I(A B C D)} & = \frac{k^{2} + 1}{(k+1)^{2}}
\end{aligned}
$$

## Adopted from [5]