CEMC Euclid, 2023

Solution 1

We make two copies of the given triangle, labelling them $\triangle ABC$ and $\triangle DEF$, as shown:

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

Next, we rotate $\triangle DEF$ by
$180^\circ$:

and join the two triangles together:

We note that $BC$ and $AE$ (which was $FE$) 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 $AB$ and $EC$.

Therefore, $ABCE$ is a
parallelogram.

Further, $ABCE$ 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 $AC$ align.)

The total area of parallelogram $ABCE$ is $1540\text{ cm}^2$.

Thus, the shaded area of $ABCE$ is
$\frac{6}{11} \cdot 1540\text{ cm}^2 = 840\text{ cm}^2$.

Since this shaded area is equally divided between the two halves of the
parallelogram, then the combined area of the shaded regions of $\triangle ABC$ is $\frac{1}{2} \cdot 840\text{ cm}^2 = 420\text{ cm}^2$.

Solution 2

We label the points where the horizontal lines touch $AB$ and $AC$ as shown:

We use the notation $|\triangle ABC|$ to represent the area of $\triangle ABC$ 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, $$\mathcal{A} = |\triangle AB_1C_1| + |B_2B_3C_3C_2| + |B_4B_5C_5C_4| + |B_6B_7C_7C_6| + |B_8B_9C_9C_8| + |B_{10}BCC_{10}|$$ The area of each of these quadrilaterals is
equal to the difference of the area of two triangles. For example, $$|B_2B_3C_3C_2| = |\triangle AB_3C_3| - |\triangle AB_2C_2| = - |\triangle AB_2C_2| + |\triangle AB_3C_3|$$
Therefore, $$\begin{align*} \mathcal{A} & = |\triangle AB_1C_1| - |\triangle AB_2C_2| + |\triangle AB_3C_3| - |\triangle AB_4C_4| + |\triangle AB_5C_5| \\ & - |\triangle AB_6C_6| + |\triangle AB_7C_7|- |\triangle AB_8C_8| + |\triangle AB_9C_9| - |\triangle AB_{10}C_{10}| + |\triangle ABC|\end{align*}$$ Each of $\triangle AB_1C_1$, $\triangle AB_2C_2$, $\ldots$, $\triangle AB_{10}C_{10}$ is similar to
$\triangle ABC$ because their two
base angles are equal due.

Suppose that the height of $\triangle ABC$ from $A$ to $BC$ is $h$.

Since the height of each of the 11 regions is equal in height, then the
height of $\triangle AB_1C_1$ is
$\frac{1}{11}h$, the height of $\triangle AB_2C_2$ 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 $\triangle PQR$ is similar to $\triangle STU$ and that altitudes are drawn from $P$ and $S$ to $V$ and $W$.

Since $\angle PQR = \angle STU$,
then $\triangle PQV$ is similar to
$\triangle STW$ (equal angle; right
angle), which means that $\dfrac{PQ}{ST} = \dfrac{PV}{SW}$. In other words, the ratio of sides is equal to
the ratio of heights.

Since the height of $\triangle AB_1C_1$ is $\frac{1}{11}h$,
then $B_1C_1 = \frac{1}{11}BC$.

Therefore, $|\triangle AB_1C_1| = \frac{1}{2} \cdot B_1C_1 \cdot \frac{1}{11}h = \frac{1}{2} \cdot \frac{1}{11}BC \cdot \frac{1}{11}h = \frac{1^2}{11^2} \cdot \frac{1}{2} \cdot BC \cdot h = \frac{1^2}{11^2} |\triangle ABC|$.

Similarly, since the height of $\triangle AB_2C_2$ is $\frac{2}{11}h$,
then $B_2C_2 = \frac{2}{11}BC$.

Therefore, $|\triangle AB_2C_2| = \frac{1}{2} \cdot B_2C_2 \cdot \frac{2}{11}h = \frac{1}{2} \cdot \frac{2}{11}BC \cdot \frac{2}{11}h = \frac{2^2}{11^2} \cdot \frac{1}{2} \cdot BC \cdot h = \frac{2^2}{11^2} |\triangle ABC|$.

This result continues for each of the triangles.

Therefore, $$\begin{align*} \mathcal{A} & = \tfrac{1^2}{11^2}|\triangle ABC| - \tfrac{2^2}{11^2}|\triangle ABC| + \tfrac{3^2}{11^2}|\triangle ABC| - \tfrac{4^2}{11^2}|\triangle ABC| + \tfrac{5^2}{11^2}|\triangle ABC| \\ & - \tfrac{6^2}{11^2}|\triangle ABC| + \tfrac{7^2}{11^2}|\triangle ABC| - \tfrac{8^2}{11^2}|\triangle ABC| + \tfrac{9^2}{11^2}|\triangle ABC| - \tfrac{10^2}{11^2}|\triangle ABC| + \tfrac{11^2}{11^2}|\triangle ABC|\\ & = \tfrac{1}{11^2}|\triangle ABC| (11^2 - 10^2 + 9^2 - 8^2 + 7^2 - 6^2 + 5^2 - 4^2 + 3^2 - 2^2 + 1) \\ & = \tfrac{1}{11^2}(770\text{ cm}^2) ((11+10)(11-10) + (9+8)(9-8) + \cdots + (3+2)(3-2) + 1) \\ & = \tfrac{1}{11^2}(770\text{ cm}^2) (11+10+9+8+7+6+5+4+3+2+1) \\ & = \tfrac{1}{11}(70\text{ cm}^2) \cdot 66 \\ & = 420\text{ cm}^2\end{align*}$$ Therefore, the combined
area of the shaded regions of $\triangle ABC$ is $420\text{ cm}^2$.
Solution 1

We label five additional points in the diagram:

Since $PQ=QR=RS=1$, then $PS = 3$ and $PR = 2$.

Since $\angle PST = 90^\circ$, then
$PT = \sqrt{PS^2 + ST^2} = \sqrt{3^2 + 1^2} = \sqrt{10}$ by the Pythagorean Theorem.

We are told that $ABCD$ is a
square.

Thus, $PT$ is perpendicular to $QC$ and to $RB$.

Thus, $\triangle PDQ$ is right-angled
at $D$ and $\triangle PAR$ is right-angled at $A$.

Since $\triangle PDQ$, $\triangle PAR$ and $\triangle PST$ are all right-angled and
all share an angle at $P$, then these
three triangles are similar.

This tells us that $\dfrac{PA}{PS} = \dfrac{PR}{PT}$ and so $PA = \dfrac{3 \cdot 2}{\sqrt{10}}$. Also, $\dfrac{PD}{PS} = \dfrac{PQ}{PT}$ and so
$PD = \dfrac{1 \cdot 3}{\sqrt{10}}$.

Therefore, $$DA = PA - PD = \dfrac{6}{\sqrt{10}} - \dfrac{3}{\sqrt{10}} = \dfrac{3}{\sqrt{10}}$$ This means that the area of square $ABCD$ is equal to $DA^2 = \left(\dfrac{3}{\sqrt{10}}\right)^2 = \dfrac{9}{10}$.

Solution 2

We add coordinates to the diagram as shown:

We determine the side length of square $ABCD$ by determining the coordinates of
$D$ and $A$ and then calculating the distance
between these points.

The slope of the line through $(0,3)$ and $(3,2)$ is $\dfrac{3-2}{0-3} = -\dfrac{1}{3}$.

This equation of this line can be written as $y = -\dfrac{1}{3}x + 3$.

The slope of the line through $(0,0)$
and $(1,3)$ is 3.

The equation of this line can be written as $y = 3x$.

The slope of the line through $(1,0)$
and $(2,3)$ is also 3.

The equation of this line can be written as $y = 3(x-1) = 3x - 3$.

Point $D$ is the intersection point
of the lines with equations $y = -\dfrac{1}{3}x + 3$ and $y = 3x$.

Equating expressions for $y$, we
obtain $-\dfrac{1}{3}x + 3 = 3x$ and
so $\dfrac{10}{3}x = 3$ which gives
$x = \dfrac{9}{10}$.

Since $y = 3x$, we get $y = \dfrac{27}{10}$ and so the coordinates
of $D$ are $\left(\dfrac{9}{10}, \dfrac{27}{10}\right)$.

Point $A$ is the intersection point
of the lines with equations $y = -\dfrac{1}{3}x + 3$ and $y = 3x - 3$.

Equating expressions for $y$, we
obtain $-\dfrac{1}{3}x + 3 = 3x - 3$
and so $\dfrac{10}{3}x = 6$ which
gives $x = \dfrac{18}{10}$.

Since $y = 3x - 3$, we get $y = \dfrac{24}{10}$ and so the coordinates
of $A$ are $\left(\dfrac{18}{10}, \dfrac{24}{10}\right)$. (It is easier to not reduce these
fractions.)

Therefore, $$DA = \sqrt{\left(\dfrac{9}{10} - \dfrac{18}{10}\right)^2 + \left(\dfrac{27}{10} - \dfrac{24}{10}\right)^2} = \sqrt{\left(-\dfrac{9}{10}\right)^2 + \left(\dfrac{3}{10}\right)^2} = \sqrt{\dfrac{90}{100}} = \sqrt{\dfrac{9}{10}}$$ This means that
the area of square $ABCD$ is equal to
$DA^2 = \left(\sqrt{\dfrac{9}{10}}\right)^2 = \dfrac{9}{10}$.