A unit side square has one of its vertices separated from the other three by the line $e$. The products of the distances of the diagonally opposite vertex pairs from $e$ are equal. What is the distance of the square's center from $e$?
I. solution. Let the square in question be $A B C D$, the isolated vertex be $A$, and the distances of the vertices and the center $O$ from $e$ be $a, b, c, d, x$ respectively. According to the problem, $a c=b d$. We can assume that $b \leq d$, which we can achieve by swapping $B$ and $D$ if necessary. Thus, $e$ intersects the extension of the diagonal $D B$ beyond $B$, and in the case $b=d$, it is parallel to it.

![](https://cdn.mathpix.com/cropped/2024_05_02_17897d81c2da75613b0ag-1.jpg?height=609&width=1165&top_left_y=335&top_left_x=469)

1. figure
2. figure

Draw the line $e^{\prime}$ through $O$ parallel to $e$, and let the projections of the vertices on this line be $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}$. Then

thus

$$
\begin{aligned}
& a+x=A A^{\prime}=C C^{\prime}=c-x=m \\
& x-b=B B^{\prime}=D D^{\prime}=d-x=n \\
& a=m-x, c=m+x, b=x-n, d=x+n
\end{aligned}
$$

Substituting these into the assumption,

$$
m^{2}-x^{2}=x^{2}-n^{2}, \quad x^{2}=\frac{1}{2}\left(m^{2}+n^{2}\right)=\frac{1}{4}, \quad x=\frac{1}{2}
$$

If $b=d$, then $e^{\prime}$ coincides with the line $B D$, so $x=b=d, a+x=c-x=1 / \sqrt{2}$, from which $a=1 / \sqrt{2}-x$, $c=1 / \sqrt{2}+x$, and from the assumption again $x=1 / 2$.

András Hegedûs (Budapest, Apáczai Csere J. Gyak. G. III. o. t.)

II. solution. Rotate the diagram $90^{\circ}$ around the center $O$ of the square so that $A$ lands on $B$. Let the projections of $B$ and $D$ on $e$ be $B_{1}$ and $D_{1}$, and the intersection of the rotated $e$, $e^{*}$, with $e$ be $A_{2}$. Then $A_{2} B_{1}=a$, because it is equal to the distance of the rotated $A$, which is $B$, from $e^{*}$. Similarly, $A_{2} D_{1}=c$. According to the condition, $a c=b d$, so the ratio of the legs of the right triangles $A_{2} B B_{1}$ and $D A_{2} D_{1}$ is

$$
\frac{A_{2} B_{1}}{B_{1} B}=\frac{a}{b}=\frac{d}{c}=\frac{D D_{1}}{D_{1} A_{2}}
$$

Thus, the two triangles are similar. Since we can traverse the corresponding vertices of the two triangles in the same direction and the corresponding legs are perpendicular, the hypotenuses are also: $\angle D A_{2} B=90^{\circ}$. The median $O A_{2}$ of the right triangle $D A_{2} B$ is half the hypotenuse, or half the diagonal of the square.

Rotating the diagram further by $90^{\circ}-90^{\circ}$, the rotated lines of $e$ form a square centered at $O$, whose diagonal is equal to the diagonal of the square $A B C D$ according to the proven facts, so the two squares are congruent. And since their centers are the same, their inscribed circles are also, so the distance of $e$ from $O$ is half the side of the square.

III. solution. Place a coordinate system on our diagram, let the coordinates of the vertices be $A(0,0), B(1,0), C(1,1), D(0,1)$, and the equation of $e$ be $y=m x+b$, where $0<b<1$, because $e$ separates $A$ from $D$. The signed distances of the vertices from $e$ are, using the known formula,

$$
\frac{-b}{\sqrt{1+m^{2}}}, \quad \frac{-m-b}{\sqrt{1+m^{2}}}, \quad \frac{1-m-b}{\sqrt{1+m^{2}}}, \quad \frac{1-b}{\sqrt{1+m^{2}}}
$$

Our assumption concerns the absolute values of the distances, so taking the negative of the distance of $A$, according to the assumption, and immediately multiplying by the common denominator,

$$
\begin{gathered}
b(1-m-b)=(m+b)(b-1), \quad \text { from which } \\
b=\frac{1}{2}(1-m) \pm \frac{1}{2} \sqrt{m^{2}+1}
\end{gathered}
$$

Substituting this into the equation of $e$ and applying the distance formula to the point $(1 / 2,1 / 2)$, the desired distance is:

$$
x=\frac{\left|\frac{1}{2}-\frac{1}{2} m-\frac{1}{2}(1-m) \mp \frac{1}{2} \sqrt{m^{2}+1}\right|}{\sqrt{m^{2}+1}}=\frac{1}{2}
$$

Éva Rácz (Makó, József A. G. III. o. t.)

Sándor Munk (Budapest, II. Rákóczi F. G. III. o. t.)