Problem 17: Australian Mathematics Competition 2023 – Intermediate Year

Let's discuss a problem from the AMC 2023 Intermediate Category: Problem 17 which revolves around the area of the triangles.

Question

A \(15 \mathrm{~cm} \times 15 \mathrm{~cm}\) square of origami paper is dark blue on top and pale yellow underneath. The top-left corner is folded down so that a crease is made from the top-right corner to a point \(x \mathrm{~cm}\) above the bottom-left corner. Once folded, the visible regions of yellow and blue paper have equal areas. What is the value of \(x\)?

(A) \(5\)
(B) \(6 \frac{2}{3}\)
(C) \(3 \sqrt{3}\)
(D) \(6\)
(E) \(4 \sqrt{2}\)

Solution

Let's start this problem by joining the upper portion of the square.

Let the size of \(AE\) be \(y\)cm. The side length of \(AB\) is \(15\) cm. The area of \(\triangle AEB\), \(\triangle BEF\) and \( EDCB\) is equal to \(T\) \(cm^2\). Thus the total area of the square is being divided into three equal parts.
So the area of the sqauare is = \(15 \times 15\) \(cm^2\) = \(225\) \(cm^2\).
Thus the area of the individual 3 parts is = \(225 \div 3 = 75\) \(cm^2\).

If the area of \(\triangle AEB\) is \(75\)\(cm^2\).
Then y is : \(\frac {1}{2} \times base \times height = \frac{1}{2} \times 15 \times y = 75\)
\( y = \frac {75 \times 2}{15} = 10\)
Thus \(y = 10\) cm.

The total side length of \(AD = x + y\).

\(x = AD - y = 15 - 10 = 5\).

Thus the length of \(x\) be \(5\) cm.

Check the strategy to solve problems in Olympiads:

Can you Work Backward?