Subfield: Geometry


Question: A square lattice of 16 points is constructed such that the horizontal and vertical distances between adjacent points are all exactly 1 unit. Each of four pairs of points are connected by a line segment, as shown. The intersections of these line segments are the vertices of square $A B C D$. Determine the area of square $A B C D$.

![](https://cdn.mathpix.com/cropped/2023_12_21_4165463fb3d29f83c565g-1.jpg?height=436&width=466&top_left_y=1896&top_left_x=1233)

Solution:
We label five additional points in the diagram:

![](https://cdn.mathpix.com/cropped/2023_12_21_49c4431a3b32fd4003c0g-1.jpg?height=469&width=501&top_left_y=1034&top_left_x=904)

Since $P Q=Q R=R S=1$, then $P S=3$ and $P R=2$.

Since $\angle P S T=90^{\circ}$, then $P T=\sqrt{P S^{2}+S T^{2}}=\sqrt{3^{2}+1^{2}}=\sqrt{10}$ by the Pythagorean Theorem.

We are told that $A B C D$ is a square.

Thus, $P T$ is perpendicular to $Q C$ and to $R B$.

Thus, $\triangle P D Q$ is right-angled at $D$ and $\triangle P A R$ is right-angled at $A$.

Since $\triangle P D Q, \triangle P A R$ and $\triangle P S T$ are all right-angled and all share an angle at $P$, then these three triangles are similar.

This tells us that $\frac{P A}{P S}=\frac{P R}{P T}$ and so $P A=\frac{3 \cdot 2}{\sqrt{10}}$. Also, $\frac{P D}{P S}=\frac{P Q}{P T}$ and so $P D=\frac{1 \cdot 3}{\sqrt{10}}$.

Therefore,

$$
D A=P A-P D=\frac{6}{\sqrt{10}}-\frac{3}{\sqrt{10}}=\frac{3}{\sqrt{10}}
$$

This means that the area of square $A B C D$ is equal to $D A^{2}=\left(\frac{3}{\sqrt{10}}\right)^{2}=\frac{9}{10}$.
We add coordinates to the diagram as shown:

![](https://cdn.mathpix.com/cropped/2023_12_21_61920460287a32b417ceg-1.jpg?height=529&width=528&top_left_y=294&top_left_x=907)

We determine the side length of square $A B C D$ 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 $\frac{3-2}{0-3}=-\frac{1}{3}$.

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

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

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

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

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

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

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

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

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

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

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

Therefore,

$$
D A=\sqrt{\left(\frac{9}{10}-\frac{18}{10}\right)^{2}+\left(\frac{27}{10}-\frac{24}{10}\right)^{2}}=\sqrt{\left(-\frac{9}{10}\right)^{2}+\left(\frac{3}{10}\right)^{2}}=\sqrt{\frac{90}{100}}=\sqrt{\frac{9}{10}}
$$

This means that the area of square $A B C D$ is equal to $D A^{2}=\left(\sqrt{\frac{9}{10}}\right)^{2}=\frac{9}{10}$.

Final answer: $\frac{9}{10}$