# Sliding Rectangles

The picture to the right shows a rectangle with a diagonal drawn on it.

An arbitrary point, $P$, has been placed on the diagonal.

The two shaded rectangles, $A$ and $B$, have a common vertex at $P$. **Which of these two rectangles has the greater area?**

Below is an interactive version of the picture above. You may find it helpful to click and drag point $P$, and see how the two rectangles $A$ and $B$ change:

**Can you justify your answer? **

Can you use the diagrams below to justify your results in two different ways?

You could start by drawing a $16$ cm by $24$ cm rectangle and marking off some equally spaced points on the diagonal as shown below. You could then find the areas $A$ and $B$ for the different points and compare them.

Hannah from the UK, Oli from Eltham College, Ali from Langley Grammar School, Saanvi from Newcastle under Lyme Prep School and Dylan from Brooke Weston, all in the UK, and Anil solved the problem by finding congruent rectangles and triangles.

Anil wrote:

In a rectangle, a diagonal divides a rectangle in two congruent triangles of equal area. Hence:

1. Area triangle ABC = Area Triangle ADC

2. Similarly GJIC is a rectangle, therefore Area triangle JGC = Area triangle JIC

3. Similarly AHJF is a rectangle, therefore Area triangle AFC = Area triangle AHJ

Now subtracting the sum of 2 and 3 from 1, we get Area rectangle FBGJ = Area rectangle HJID

Hannah's proof goes into more depth and includes rigourous proofs of Anil's three statements. Note that Hannah used different letters to label the shapes (click on each image to see a larger version):

Cosima from the UK used ratio:

Ali and Oli used coordinates to express the same ideas (click on the images to see a larger version). Notice that they turned the picture upside down to keep the algebra simple:

Oli's method is exactly the same, but Oli used the ideas in a different order. Click to see Oli's work.

Consider expressions for $\text A_1$ and $\text A_2,$ $$\text A_1 = u(b-v) = bu-uv \\ \text A_2 = v(a-u)=av-uv$$

Considering the line $y = mx$ at $(a,b),$ $$b = m\times a \\ m = \tfrac ba $$

Therefore, at $(u,v),$ which is on $y = mx,$ $$v=\tfrac ba u \\ av=bu$$

Substituting into expressions for $\text A_2,$ $$\text A_2 = bu-uv$$

Therefore, $$\text A_1=\text A_2$$

Dylan kept the rectangle the original way up, and worked through tougher algebra:

We can begin by drawing a graph containing the rectangle with its bottom left corner placed on the origin, and naming the width and height of the rectangle $*m$ *and $*n*.$ Drawing on the diagonal of the rectangle, we obtain the following diagram:

We can see that the diagonal contains the points $(m,0)$ and $(0,n).$ Therefore, we can find the gradient of this line, which is $-\frac nm.$ Using $y=mx+c$ with $'m'=-\frac nm$ and $'c'=n,$ we obtain the equation $$y = n - \tfrac nm x\hspace{40mm}(1)$$ We can label a general point $\text P$ on this line, with coordinates $(x,y),$ and draw out the rectangles $\text A$ and $\text B$ from this
point.

Now, we can see that the area of rectangle $\text A$ is given by $A = xy$ and the area of rectangle $\text B$ is given by $B = (m-x)(n-y)$ from the diagram. As we know that $y = n - \frac nm x$ for every point on the diagonal, we can substitute equation $(1)$ in the expressions for the areas of the rectangles. By doing this, we get $A = x\left(n-\frac nm x\right)$ and
$$\begin{split}B&=(m-x)\left(n-\left(n-\tfrac nm x\right)\right)\\&=\tfrac nm x \left(m-x\right)\\&=x\left(n-\tfrac nm x\right)=A\end{split}$$

Therefore the answer to the problem is that the two areas are equal, no matter which point P on the diagonal you choose, and no matter what the dimensions of the big rectangle are.

Quinn from the UK used algebra and ratio in a slightly different way:

1. I labelled the sides of the rectangle $l$ and $w.$

2. Because of the nature of diagonals, P is $\frac l x$ units from $w$ and $\frac w x$ units from $l.$ $x$ is an arbitrary amount but it remains the same.

3. The short side of A can therefore be labelled $\frac l x,$ and the short side of B labelled $\frac w x.$

4. The long side of A can be labelled $w - \frac w x,$ and the long side of B labelled $l - \frac l x.$

5. The area of A is therefore $\frac l x \left(w - \frac w x\right)$ and the area of B is $\frac w x \left(l -\frac l x\right).$

6. Simplifying this, we get $\frac{lw}x - \frac{lw}{x^2}$ and $\frac{lw}x - \frac{lw}{x^2}.$

7. This shows that both rectangles are equal sized, so neither is bigger than the other.

This problem has perhaps a slightly surprising result.

### Possible approach:

At the start, students could be shown the interactivity and asked to consider the areas of the rectangles $A$ and $B$ when the point $P$ is in the middle of the diagonal line.

What happens as the point $P$ slides up and down the diagonal?

Students could use dotty paper in order to investigate a particular rectangle, like the $16 \times 24$ rectangle shown in the getting started section.

There are two possible proof methods shown behind "hide and reveal" buttons at the bottom of the problem. They could be used as starting points for students to create their own proofs.

- The first method uses areas of triangles and rectangles to prove the result, and is accessible to 11-14 students.

- The second method uses coordinates and straight line graphs, or proportional reasoning, along with some algebraic manipulation, to find the areas of the two rectangles.

For the second method, students can be asked to explain why the diagram represents the same problem. Why do they think the diagonal has been drawn in the opposite direction?

### Possible support:

Students could be asked to consider a point $P$, along the diagonal of a square, before moving on to the rectangle.