Subfield: Geometry


Question: A triangle of area $770 \mathrm{~cm}^{2}$ is divided into 11 regions of equal height by 10 lines that are all parallel to the base of the triangle. Starting from the top of the triangle, every other region is shaded, as shown. What is the total area of the shaded regions?

<img_3807>

Solution:
We make two copies of the given triangle, labelling them $\triangle A B C$ and $\triangle D E F$, as shown:
<img_3909>

The combined area of these two triangles is $2 \cdot 770 \mathrm{~cm}^{2}=1540 \mathrm{~cm}^{2}$, and the shaded area in each triangle is the same.

Next, we rotate $\triangle D E F$ by $180^{\circ}$ :
<img_3429>

and join the two triangles together:

<img_3440>

We note that $B C$ and $A E$ (which was $F E$ ) are equal in length (since they were copies of each other) and parallel (since they are $180^{\circ}$ rotations of each other). The same is true for $A B$ and $E C$.

Therefore, $A B C E$ is a parallelogram.

Further, $A B C E$ is divided into 11 identical parallelograms (6 shaded and 5 unshaded) by the horizontal lines. (Since the sections of the two triangles are equal in height, the horizontal lines on both sides of $A C$ align.)

The total area of parallelogram $A B C E$ is $1540 \mathrm{~cm}^{2}$.

Thus, the shaded area of $A B C E$ is $\frac{6}{11} \cdot 1540 \mathrm{~cm}^{2}=840 \mathrm{~cm}^{2}$.

Since this shaded area is equally divided between the two halves of the parallelogram, then the combined area of the shaded regions of $\triangle A B C$ is $\frac{1}{2} \cdot 840 \mathrm{~cm}^{2}=420 \mathrm{~cm}^{2}$.
We label the points where the horizontal lines touch $A B$ and $A C$ as shown:

<img_3141>

We use the notation $|\triangle A B C|$ to represent the area of $\triangle A B C$ and use similar notation for the area of other triangles and quadrilaterals.

Let $\mathcal{A}$ be equal to the total area of the shaded regions.

Thus,

$$
\mathcal{A}=\left|\triangle A B_{1} C_{1}\right|+\left|B_{2} B_{3} C_{3} C_{2}\right|+\left|B_{4} B_{5} C_{5} C_{4}\right|+\left|B_{6} B_{7} C_{7} C_{6}\right|+\left|B_{8} B_{9} C_{9} C_{8}\right|+\left|B_{10} B C C_{10}\right|
$$

The area of each of these quadrilaterals is equal to the difference of the area of two triangles. For example,

$$
\left|B_{2} B_{3} C_{3} C_{2}\right|=\left|\triangle A B_{3} C_{3}\right|-\left|\triangle A B_{2} C_{2}\right|=-\left|\triangle A B_{2} C_{2}\right|+\left|\triangle A B_{3} C_{3}\right|
$$

Therefore,

$$
\begin{aligned}
\mathcal{A}=\mid & \triangle A B_{1} C_{1}|-| \triangle A B_{2} C_{2}|+| \triangle A B_{3} C_{3}|-| \triangle A B_{4} C_{4}|+| \triangle A B_{5} C_{5} \mid \\
& \quad-\left|\triangle A B_{6} C_{6}\right|+\left|\triangle A B_{7} C_{7}\right|-\left|\triangle A B_{8} C_{8}\right|+\left|\triangle A B_{9} C_{9}\right|-\left|\triangle A B_{10} C_{10}\right|+|\triangle A B C|
\end{aligned}
$$

Each of $\triangle A B_{1} C_{1}, \triangle A B_{2} C_{2}, \ldots, \triangle A B_{10} C_{10}$ is similar to $\triangle A B C$ because their two base angles are equal due.

Suppose that the height of $\triangle A B C$ from $A$ to $B C$ is $h$.

Since the height of each of the 11 regions is equal in height, then the height of $\triangle A B_{1} C_{1}$ is $\frac{1}{11} h$, the height of $\triangle A B_{2} C_{2}$ is $\frac{2}{11} h$, and so on.

When two triangles are similar, their heights are in the same ratio as their side lengths:

To see this, suppose that $\triangle P Q R$ is similar to $\triangle S T U$ and that altitudes are drawn from $P$ and $S$ to $V$ and $W$.
<img_3919>

Since $\angle P Q R=\angle S T U$, then $\triangle P Q V$ is similar to $\triangle S T W$ (equal angle; right angle), which means that $\frac{P Q}{S T}=\frac{P V}{S W}$. In other words, the ratio of sides is equal to the ratio of heights.

Since the height of $\triangle A B_{1} C_{1}$ is $\frac{1}{11} h$, then $B_{1} C_{1}=\frac{1}{11} B C$.

Therefore, $\left|\triangle A B_{1} C_{1}\right|=\frac{1}{2} \cdot B_{1} C_{1} \cdot \frac{1}{11} h=\frac{1}{2} \cdot \frac{1}{11} B C \cdot \frac{1}{11} h=\frac{1^{2}}{11^{2}} \cdot \frac{1}{2} \cdot B C \cdot h=\frac{1^{2}}{11^{2}}|\triangle A B C|$.

Similarly, since the height of $\triangle A B_{2} C_{2}$ is $\frac{2}{11} h$, then $B_{2} C_{2}=\frac{2}{11} B C$.



Therefore, $\left|\triangle A B_{2} C_{2}\right|=\frac{1}{2} \cdot B_{2} C_{2} \cdot \frac{2}{11} h=\frac{1}{2} \cdot \frac{2}{11} B C \cdot \frac{2}{11} h=\frac{2^{2}}{11^{2}} \cdot \frac{1}{2} \cdot B C \cdot h=\frac{2^{2}}{11^{2}}|\triangle A B C|$.

This result continues for each of the triangles.

Therefore,

$$
\begin{aligned}
\mathcal{A} & =\frac{1^{2}}{11^{2}}|\triangle A B C|-\frac{2^{2}}{11^{2}}|\triangle A B C|+\frac{3^{2}}{11^{2}}|\triangle A B C|-\frac{4^{2}}{11^{2}}|\triangle A B C|+\frac{5^{2}}{11^{2}}|\triangle A B C| \\
& \quad-\frac{6^{2}}{11^{2}}|\triangle A B C|+\frac{7^{2}}{11^{2}}|\triangle A B C|-\frac{8^{2}}{11^{2}}|\triangle A B C|+\frac{9^{2}}{11^{2}}|\triangle A B C|-\frac{10^{2}}{11^{2}}|\triangle A B C|+\frac{11^{2}}{11^{2}}|\triangle A B C| \\
& =\frac{1}{11^{2}}|\triangle A B C|\left(11^{2}-10^{2}+9^{2}-8^{2}+7^{2}-6^{2}+5^{2}-4^{2}+3^{2}-2^{2}+1\right) \\
& =\frac{1}{11^{2}}\left(770 \mathrm{~cm}^{2}\right)((11+10)(11-10)+(9+8)(9-8)+\cdots+(3+2)(3-2)+1) \\
& =\frac{1}{11^{2}}\left(770 \mathrm{~cm}^{2}\right)(11+10+9+8+7+6+5+4+3+2+1) \\
& =\frac{1}{11}\left(70 \mathrm{~cm}^{2}\right) \cdot 66 \\
& =420 \mathrm{~cm}^{2}
\end{aligned}
$$

Therefore, the combined area of the shaded regions of $\triangle A B C$ is $420 \mathrm{~cm}^{2}$.

Final answer: $420$