On a sheet of paper, we mark points equally spaced horizontally and vertically, so that square $A$ has an area of $1 \mathrm{~cm}^{2}$, as shown in the figure. We say that a square is legal if its vertices are four of these points; for example, squares $A$ and $B$ are legal.
![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-17.jpg?height=350&width=1210&top_left_y=1979&top_left_x=296)

a) What is the area of square $B$?

b) Draw a legal square with an area of $13 \mathrm{~cm}^{2}$.

c) Is there a legal square with an area of $41 \mathrm{~cm}^{2}$? And with an area of $43 \mathrm{~cm}^{2}$? Justify your answer.

d) Show that for each legal square, there is another legal square with twice its area.

## Solutions for Level 3

Subject Arithmetic

#
Solution

Note: To facilitate writing this solution, we will refer to the points on the grid as legal points.

a) $1^{\text {st }}$ solution: Observing the figure below, we see that square $B$ can be inscribed in a square consisting of 9 small squares.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-51.jpg?height=297&width=286&top_left_y=777&top_left_x=791)

The part outside square $B$ in the figure above can be decomposed into four equal triangles (in gray). Each triangle is half of a rectangle made of two small squares. Thus, the area of each of these triangles is $1 \mathrm{~cm}^{2}$. Therefore, the area of square $B$ is $9-4=5 \mathrm{~cm}^{2}$.

$2^{\text{nd}}$ solution: Using the figure from the previous solution again, we can also argue that square $B$ is decomposed into one small square and four triangles of area $1 \mathrm{~cm}^{2}$, so its area is $1+4=5 \mathrm{~cm}^{2}$.

$3^{\text{rd}}$ solution: We can calculate the side $P R$ of the square by observing the right triangle $P Q R$ in the figure used in the previous solutions. Its legs are $P Q$ and $Q R$, with lengths 1 and 2, respectively. By the Pythagorean theorem, we have

$$
P R^{2}=\sqrt{P Q^{2}+Q R^{2}}=\sqrt{1^{2}+2^{2}}=\sqrt{5}
$$

and it follows that the area of the square is $(\sqrt{5})^{2}=5 \mathrm{~cm}^{2}$.

b) We want to draw a legal square of area $13 \mathrm{~cm}^{2}$; its side must then measure $\sqrt{13} \mathrm{~cm}$. Observing the second solution presented in part a), we see that the side must be the hypotenuse of a right triangle with legs of lengths $a$ and $b$ that are integers and such that $a^{2}+b^{2}=13$. We can then choose $a=3$ and $b=2$ (the only solution, up to swapping the values of $a$ and $b$) and construct our square of area $13 \mathrm{~cm}^{2}$ as, for example, indicated in the figure below.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-51.jpg?height=297&width=302&top_left_y=1942&top_left_x=777)

c) If there is a legal square of area $n$, then its side is $\sqrt{n}$; to construct a segment of this length, we must, as in the previous part, find integers $a$ and $b$ such that $a^{2}+b^{2}=n$. For 41, there is no problem, since $41=4^{2}+5^{2}$; but for 43, this is impossible, as can be seen by direct listing. Indeed, since $7^{2}=49$ exceeds 43, we must test only whether 43 can be written as the sum of two squares of the numbers from 1 to 6, which does not happen because $43-1^{2}=42, 43-2^{2}=39, 43-3^{2}=34, 43-4^{2}=27, 43-5^{2}=18$, and $43-6^{2}=7$ are not perfect squares. Therefore, it is possible to construct a legal square of area $41 \mathrm{~cm}^{2}$, but it is not possible to construct one of area $43 \mathrm{~cm}^{2}$.

d) $1^{\text{st}}$ solution: The figure below shows a legal square in gray and the construction of a new square, in a thicker line, with an area equal to twice the area of the original square.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-52.jpg?height=297&width=543&top_left_y=226&top_left_x=859)

We note that, since the vertices of the original square are legal points, the vertices of the larger square are also legal points. To justify this last statement, it suffices to note that if $A$ and $B$ are legal points and $C$ is the reflection of $A$ with respect to $B$ (as in the figure above), then $C$ is also a legal point. Thus, the new square is also legal.

$2^{\text{nd}}$ solution: As we saw in part $b$, if a legal square has area $n$, then there exist integers $a$ and $b$ such that $n=a^{2}+b^{2}$. Conversely, if there exist integers $a$ and $b$ such that $n=a^{2}+b^{2}$, then there exists a legal square of area $n$. Since $(a-b)^{2}+(a+b)^{2}=2\left(a^{2}+b^{2}\right)$, we see that a right triangle with legs $a-b$ and $a+b$ will have a hypotenuse $\sqrt{2 n}$; the square constructed on this hypotenuse will have an area of $2 n$ (in other words, we have shown that if $n$ is the sum of two squares of integers, then $2 n$ is also).

Using this fact, we illustrate in the figure below a construction of a legal square of area $2 n$ (the large square in solid lines) from a legal square of area $n$ (the small square in solid lines). The dotted square serves only to indicate the horizontal and vertical directions.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-52.jpg?height=474&width=508&top_left_y=1131&top_left_x=868)

We note, as before, that since the original square is legal, all the points indicated are legal.