In the following figure, the larger square has an area of $1 \mathrm{~cm}^{2}$ and the middle square has an area of $M$. The area of the smaller square, which has a vertex on a side of the middle square, is $N$. What is the value of $N$ in terms of $M$?

![](https://cdn.mathpix.com/cropped/2024_05_01_82072e76ed187b592b1cg-07.jpg?height=428&width=426&top_left_y=584&top_left_x=887)

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Solution

Let $A Q=x \mathrm{~cm}, Q R=a \mathrm{~cm}$ and $A G=s \mathrm{~cm}$. Since the area of the larger square is $1 \mathrm{~cm}^{2}$, it follows that $A D=1 \mathrm{~cm}$. The right triangles $A R Q$ and $D Q P$ have the same angles, because

$$
\begin{aligned}
\angle A Q R & =180^{\circ}-\angle P Q R-\angle D Q P \\
& =90^{\circ}-\angle D Q P \\
& =\angle D P Q
\end{aligned}
$$

Since $Q R=Q P$, the triangles $D P Q$ and $A Q R$ are congruent. Thus, $A R=D Q=1-x$. From $Q F \| A R$, it follows that the triangles $Q G F$ and $Q A R$ are similar, and thus

$$
\begin{aligned}
\frac{s}{1-x} & =\frac{x-s}{x} \\
s x & =x-x^{2}-s+s x \\
s & =x-x^{2}
\end{aligned}
$$

By the Pythagorean Theorem applied to the triangle $Q A R$, it follows that

$$
\begin{aligned}
(1-x)^{2}+x^{2} & =a^{2} \\
1-a^{2} & =2 x-2 x^{2} \\
1-M & =2 s
\end{aligned}
$$

Therefore,

$$
N=s^{2}=\left(\frac{1-M}{2}\right)^{2}
$$