Problem 20: Australian Mathematics Competition 2022 – Junior Year

Let's discuss a problem from the AMC 2022 Junior Category: Problem 20 which revolves around basic algebra.

Question

Within the square \(P Q R S\), lines are drawn from each corner to the middle of the opposite sides as shown. What fraction of \(P Q R S\) is shaded?

(A) \(\frac{1}{4}\)
(B) \(\frac{1}{3}\)
(C) \(\frac{3}{8}\)
(D) \(\frac{1}{2}\)
(E) \(\frac{2}{3}\)

Solution

Let's name all the midpoints on the lines and the other joining points.

Now if we draw a line joining the points A and C then the square will get divided into two rectangles such as :

In this picture the $\triangle PES$, the area of $\triangle PES$ is $\frac {1}{4}$ of the area of rectangle $PACS$.
As the diagonals AS and PC divide the area of rectangles $PACS$ into two equal parts.

Similarly the area of $\triangle QGR$ is also $\frac {1}{4}$ of the rectangle $AQRC$.
Now if we draw a line from $D$ to $B$ we will have the same thing over there. The area of $\triangle PFQ$ is $\frac {1}{4}$ of the rectangle $PQBD$.
The area of $\triangle QGR$ is also $\frac {1}{8}$ of the square $PQRS$.
Thus the total unshaded area is $4 \times \frac {1}{8}$ of the square $PQRS$ = $\frac {1}{2}$ of the square $PQRS$.

Thus the shaded area will also be $\frac {1}{2}$ of the square $PQRS$.