A convex quadrilateral has three sides of lengths $1 \mathrm{~cm}, 4 \mathrm{~cm}$, and $8 \mathrm{~cm}$, and its diagonals are perpendicular to each other. What could be the length of the fourth side?
Solution. Let the sides of the quadrilateral be denoted by $a, b, c$, and $d$, and the segments created by the intersection of the diagonals by $x, y, u$, and $v$, as shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_02_74fc58eb498101de36aeg-1.jpg?height=342&width=542&top_left_y=252&top_left_x=781)

We apply the Pythagorean theorem to the sides of the right triangles formed:

$$
\begin{array}{ll}
x^{2}+v^{2}=a^{2}, & y^{2}+u^{2}=c^{2} \\
x^{2}+u^{2}=b^{2}, & y^{2}+v^{2}=d^{2}
\end{array}
$$

From the first two equations:

$$
v^{2}-u^{2}=a^{2}-b^{2}
$$

From the other two equations:

$$
v^{2}-u^{2}=d^{2}-c^{2}
$$

The equality of the left sides implies the equality of the right sides, i.e.,

$$
a^{2}-b^{2}=d^{2}-c^{2}
$$

From this, we get

$$
d^{2}=a^{2}+c^{2}-b^{2}
$$

Thus, we need to choose the given sides such that $d^{2}>0$.

This condition is satisfied in two cases: if $a=8, c=4$, and $b=1$, then

$$
d=\sqrt{64+16-1}=\sqrt{79} \approx 8.89
$$

or if $a=8, c=1$, and $b=4$, then

$$
d=\sqrt{64+1-16}=7
$$

Such quadrilaterals do indeed exist, although we do not prove this here, as it would be part of the solution to the problem. We did not expect this from the contestants either.

![](https://cdn.mathpix.com/cropped/2024_05_02_74fc58eb498101de36aeg-1.jpg?height=303&width=594&top_left_y=1594&top_left_x=752)

Remarks. 1. The solutions essentially used the fact that in quadrilaterals with perpendicular diagonals, the sum of the squares of the lengths of opposite side pairs is the same. (If the sides are $a, b, c, d$ in this order, then $a^{2}+c^{2}=b^{2}+d^{2}$.) From this relationship, they calculated the fourth side.

To ensure that the diagonals are perpendicular based on the relationship, it must also be shown that the $a^{2}+c^{2}=b^{2}+d^{2}$ relationship implies the perpendicularity of the diagonals. Let $A, B, C, D$ be the vertices of the quadrilateral and let $A C$ be one of the diagonals. Let the feet of the perpendiculars from $B$ and $D$ to this diagonal be $P$ and $Q$, respectively, such that $P$ is closer to $A$. Let $|A P|=x, |P Q|=z$, and $|Q B|=y$. Let the distances from the feet to the corresponding vertices be $u$ and $v$. Then, by the Pythagorean theorem, the sums of the squares of the lengths of the opposite side pairs are $x^{2}+u^{2}$ and $y^{2}+v^{2}$, and $(x+z)^{2}+v^{2}$ and $(z+y)^{2}+u^{2}$. Thus, we get the equation

$$
\left(x^{2}+u^{2}\right)+\left(y^{2}+v^{2}\right)=\left((x+z)^{2}+v^{2}\right)+\left((z+y)^{2}+u^{2}\right)
$$

From this, $0=2 z(x+y)+2 z^{2}$ follows. Hence, $z=0$, meaning the two feet of the perpendiculars coincide, so the two diagonals are perpendicular to each other.

2. It is easily verifiable that the above reasoning is applicable even in the case of a concave quadrilateral.
3. The side $d$ opposite the sought side could theoretically be any of $a, b, c$. Due to symmetry, this could give three solutions. For example, if $a$ is opposite, then $d^{2}=b^{2}+c^{2}-a^{2}$. This must be positive because $d^{2}>0$. The expression $b^{2}+c^{2}-a^{2}$ is definitely positive unless $a>b, c$. In this case, it is also positive if $a^{2}<b^{2}+c^{2}$. This means that $a^{2}, b^{2}, c^{2}$ could be the sides of a triangle. In this case, there are three cases; otherwise, there are two.