Inside a certain square, the distances from a point $P$ to the vertices $A, B, C$ of the square are $a, b, c$. What is the area of the square? What is it in the case where $a=b=c$?
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$A$ and $C$ are opposite vertices of the square. Let the side of the square be $x$ and $\angle ABP = \varphi$. From $\triangle ABP$,

$$
\cos \varphi = \frac{b^2 + x^2 - a^2}{2bx}
$$

From $\triangle BCP$,

$$
\begin{gathered}
\cos (90^\circ - \varphi) = \sin \varphi = \frac{b^2 + x^2 - c^2}{2bx} \\
\cos^2 \varphi + \sin^2 \varphi = \\
= \left(\frac{b^2 + x^2 - a^2}{2bx}\right)^2 + \left(\frac{b^2 + x^2 - c^2}{2bx}\right)^2 = 1
\end{gathered}
$$

After removing the fractions, squaring, and rearranging:

$$
2x^4 - 2(c^2 + a^2)x^2 + (b^2 - c^2)^2 + (b^2 - a^2)^2 = 0
$$

From this,

$$
x^2 = \frac{a^2 + c^2 + \sqrt{(a^2 + c^2)^2 - 2[(b^2 - c^2)^2 + (b^2 - a^2)^2]}}{2}
$$

Assuming the roots are real, i.e.,

$$
\Delta \equiv (a^2 + c^2)^2 - 2[(b^2 - c^2)^2 + (b^2 - a^2)^2] \geq 0,
$$

both values of $x^2$ are positive because their product and sum are positive. The question is which one can satisfy the problem?

Considering the geometric relationships, $\triangle APC$ is obtuse at $P$, so it must be that

$$
AC^2 > \overline{AP}^2 + \overline{CP}^2 \text{ i.e., } 2x^2 > a^2 + c^2
$$

This implies that the area of the square is

$$
x^2 = \frac{a^2 + c^2}{2} + \frac{1}{2} \sqrt{(a^2 + c^2)^2 - 2[(b^2 - c^2)^2 + (b^2 - a^2)^2]}
$$

Let's now examine what the condition is for $x^2$ to be real. If we perform the squaring and simplification in the discriminant $\Delta$ expression, we get

$$
\begin{gathered}
\Delta \equiv 4a^2c^2 - (2b^2 - a^2 - c^2)^2 \equiv (2ac + 2b^2 - a^2 - c^2)(2ac - 2b^2 + a^2 + c^2) \equiv \\
\equiv [2b^2 - (a - c)^2][(a + c)^2 - 2b^2].
\end{gathered}
$$

$\Delta \geq 0$ if both factors are of the same sign, i.e.,

1. if

$$
(a + c)^2 \geq 2b^2 \quad \text{and} \quad 2b^2 \geq (a - c)^2
$$

2. if

$$
(a + c)^2 \leq 2b^2 \quad \text{and} \quad 2b^2 \leq (a - c)^2.
$$

However, the 2nd case, i.e.,

$$
(a + c)^2 \leq 2b^2 \leq (a - c)^2
$$

is not possible because $(a + c)^2$ cannot be less than $(a - c)^2$.

Only the 1st case is possible, where

$$
(a + c)^2 > 2b^2 \geq (a - c)^2
$$

i.e., if the sides $a, c, b\sqrt{2}$ can form a triangle.

If $a = b = c$, then $x^2 = 2a^2$, i.e., $x = a\sqrt{2}$. This can only occur if $P$ is the center of the square.[^0]

[^0]:    ${ }^{1}$ i.e., $\angle APC > \angle ABC = 90^\circ$.