## Task 6 - 020816

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0580.jpg?height=306&width=377&top_left_y=2234&top_left_x=314)

Given a rectangle $A B C D$, whose sides are all divided in the ratio $1: 2$ as shown in the figure. We call the division points $P, Q, R, S$ and connect them in sequence.

a) Perform this construction for the rectangle with sides $A B = 10 \mathrm{~cm}$ and $B C = 7 \mathrm{~cm}$!

b) What kind of quadrilateral is the quadrilateral $P Q R S$? (Proof!)

c) How does the area of the quadrilateral $P Q R S$ compare to that of the rectangle $A B C D$? Does this result hold for other similarly divided rectangles? (Justification!)
a) See the adjacent figure (not to scale).

b) Claim: $P Q R S$ is a parallelogram.

Proof: Since $A B = C D, B C = D A$ and the same division ratios apply to the sides, we have $A S = C Q, A P = C R, B P = D R$ and $B Q = D S$. By the SSS congruence criterion, $\triangle A P S \cong \triangle C R Q$ and $\triangle B Q P \cong \triangle D S R$. Therefore, $P Q = R S, Q R = S P$, and thus $P Q R S$ is a parallelogram.

c) With $a \equiv A B = C D$ and $b \equiv B C = D A$, we have $[A P S] = [C R Q] = \frac{1}{2} \cdot \frac{a}{3} \cdot \frac{2 b}{3} = \frac{1}{9} a b$ and $[B Q P] = [D S R] = \frac{1}{2} \cdot \frac{b}{3} \cdot \frac{2 a}{3} = \frac{1}{9} a b$. Thus,

$$
[P Q R S] = [A B C D] - ([A P S] + [C R Q] + [B Q P] + [D S R]) = a b - \frac{4}{9} a b = \frac{5}{9} a b
$$

The ratio of the areas is therefore always $\frac{5}{9}$, regardless of the specific dimensions of the rectangle.

Solutions to the 1st Round 1962 adopted from [5]

### 5.4.2 2nd Round 1962, Class 8