CEMC Banner

Problem of the Week
Problem C and Solution
All Equal

Problem

Using two cuts, we want to divide the \(6\) m by \(6\) m grid shown into three regions of equal area.

A square is divided into a 6 by 6 grid of identical squares.
The bottom left vertex of the square is labelled P, the top left vertex
is R, the top right vertex is Q, and the bottom right vertex is M. The
five points on top side RQ where the vertical grid lines meet RQ are
labelled S, T, U, V, and W, from left to right. The five points on right
side QM where the horizontal grid lines meet QM are labelled G, H, J, K,
and L, from top to bottom.

One way to do so is by making a horizontal cut through \(H\) and a second horizontal cut through \(K\). This method of cutting the grid works, but is not very creative.

To make things a little more interesting, we must still make two straight cuts, but each cut must start at point \(P\). Each of these two cuts will pass through a point on the outer perimeter of the grid.

Find the length of each cut. Round your answer to one decimal.

Solution

The area of the entire \(6\) m by \(6\) m square grid is \(6\times 6 = 36\mbox{ m}^2\). Since the square is divided into three regions of equal area, the area of each region must be \(\dfrac{36}{3} = 12\mbox{ m}^2\).

Consider the line through \(P\) that passes through some point on side \(QM\). Let \(A\) be the point where this line intersects \(QM\).

Since \(\angle PMQ = 90^{\circ}\), \(\triangle PMA\) is a right-angled triangle with base \(PM = 6\) m and height \(MA\).

Using the formula \(\text{area} = \dfrac{\text{base}\times\text{height}}{2}\), we have \(\mbox{area of }\triangle PMA = \dfrac{6\times MA}{2} = 3\times MA\).

We need the area of \(\triangle PMA\) to be \(12\mbox{ m}^2\). Therefore, \(3 \times MA = 12\), and so \(MA = 4\) m. Since \(H\) is the point on \(QM\) with \(MH = 4\) m, we must have \(A =H\). Therefore, one line passes through the point \(H\).

Since \(\triangle PMA\) is a right-angled triangle, using the Pythagorean Theorem we have \[\begin{aligned} PA^2 &= PM^2 + MA^2\\ &= 6^2 + 4^2\\ &= 36 + 16\\ &= 52 \end{aligned}\] Therefore, \(PA = \sqrt{52} \approx 7.2\), since \(PA >0\).

Consider the line through \(P\) that passes through some point on side \(RQ\). Let \(B\) be the point where this line intersects \(RQ\).

Since \(\angle PRQ = 90^{\circ}\), \(\triangle PRB\) is a right-angled triangle with height \(PR = 6\) m and base \(RB\).

Using the formula \(\text{area} = \dfrac{\text{base}\times\text{height}}{2}\), we have \(\mbox{area of } \triangle PRB = \dfrac{RB\times 6}{2} = 3\times RB\).

We need the area of \(\triangle PRB\) to be \(12\mbox{ m}^2\). Therefore, \(3\times RB = 12\), and so \(RB = 4\) m. Since \(V\) is the point on \(RQ\) with \(RV = 4\) m, we must have \(B =V\). Therefore, the other line passes through the point \(V\).

Therefore, one line passes through point \(H\) and the other passes through point \(V\).

Since \(\triangle PRB\) is a right-angled triangle, using the Pythagorean Theorem we have \[\begin{aligned} PB^2 &= PR^2 + RB^2\\ &= 6^2 + 4^2\\ &= 36 + 16\\ &= 52 \end{aligned}\] Therefore, \(PB = \sqrt{52}\approx 7.2\), since \(PB>0\).

Therefore, the length of each cut is approximately \(7.2\) m.

Extension:

Try dividing the grid into three regions of equal area using three cuts. (Each cut does not necessarily need to be to the outer perimeter of the grid.)