In the square $A B C D$, point $P$ moves along the line segment $A B$. Between what limits does the ratio $\frac{P C}{P D}$ vary?
Solution. Let's choose the side of the square to be 2 units long and let $F$ be the midpoint of the $AB$ side, and let the distance between $P$ and $F$ be denoted by $x$. It is sufficient to examine the case when $PC \geqq PD$, i.e., when $P$ moves on the ray starting from $F$ and containing $A$, because if $P'$ is any point on the ray starting from $F$ and containing $B$, then the reflection of $P'$ over $PF$ moves on the ray containing $A$, and due to symmetry, the constraints for the ray containing $B$ are precisely the reciprocals of the constraints for the ray containing $A$.

![](https://cdn.mathpix.com/cropped/2024_05_02_493c60bb69f2640f86acg-1.jpg?height=464&width=550&top_left_y=386&top_left_x=771)

Figure 1

The triangles $PBC$ and $PAD$ are right-angled, with $BC = AD = 2$, $PB = 1 + x$, and $PA = 1 - x$ or $x - 1$, depending on whether $P$ is on the segment $FA$ or not. By the Pythagorean theorem, $PC = \sqrt{PB^2 + BC^2} = \sqrt{(1 + x)^2 + 4}$ and $PD = \sqrt{PA^2 + AD^2} = \sqrt{(1 - x)^2 + 4}$. Therefore,

$$
\frac{PC}{PD} = \frac{\sqrt{(1 + x)^2 + 4}}{\sqrt{(1 - x)^2 + 4}} = \sqrt{1 + \frac{4x}{x^2 - 2x + 5}}
$$

The value of the ratio in question thus depends only on the value of the fraction $\frac{4x}{x^2 - 2x + 5}$. Since $x \geqq 0$ and the denominator of the fraction is positive, the minimum value of the fraction is 0. If we denote the maximum value of the fraction by $m$, then $m$ is the largest number for which the equation

$$
\frac{4x}{x^2 - 2x + 5} = m
$$

has a real solution, i.e., the quadratic equation $mx^2 - (2m + 4)x + 5m = 0$ has a non-negative discriminant, $D = (2m + 4)^2 - 20m^2$. Solving the inequality $D \geqq 0$, we get

$$
\frac{1 - \sqrt{5}}{2} \leqq m \leqq \frac{1 + \sqrt{5}}{2}
$$

Thus, the maximum value of the fraction $\frac{4x}{x^2 - 2x + 5}$ is $\frac{1 + \sqrt{5}}{2}$, and therefore the maximum value of the ratio $\frac{PC}{PD}$ is

$$
\sqrt{1 + \frac{1 + \sqrt{5}}{2}} = \frac{1 + \sqrt{5}}{2}
$$

This value is indeed attained by the ratio when $PF = x = \sqrt{5}$. Therefore, if the point $P$ moves on the ray containing $A$, then

$$
1 \leqq \frac{PC}{PD} \leqq \frac{\sqrt{5} + 1}{2}
$$

If $P$ moves on the ray containing $B$, then

$$
1 \geqq \frac{PC}{PD} \geqq \frac{1}{\frac{\sqrt{5} + 1}{2}} = \frac{\sqrt{5} - 1}{2}
$$

Summarizing:

$$
\frac{\sqrt{5} - 1}{2} \leqq \frac{PC}{PD} \leqq \frac{\sqrt{5} + 1}{2}
$$

Remarks.

1) The solution to the problem becomes more comprehensible if we recognize a principle that is often useful in solving geometric extremum problems.

Suppose that a function $f(P)$ defined on the points of the plane - in this case, the ratio $\frac{PC}{PD}$ - attains its extremum values on a curve $\Gamma$.

The level sets of the function $F$ - in this case, the Apollonian circles corresponding to points $C$ and $D$ - are circles, and in the case where $\frac{PC}{PD} = 1$, they are the perpendicular bisector of $CD$.

Each of these curves divides the plane into two regions. One region is the interior of the circle, and the other is the exterior. (In the case of the perpendicular bisector, we get two half-planes.) In one of these regions, the value of $F(P)$ is less than the value on the boundary curve, and in the other, it is greater.

Thus, the quantity $F(P)$ attains its extremum values at the points on the curve $\Gamma$ where the corresponding element of the circle family - the Apollonian circles - is tangent to $\Gamma$, which in this case is the line $AB$.

To construct the point of tangency, we need to know that the perpendicular bisector of the segment $CD$ - the line $h$ - is the locus of points from which equal-length tangents can be drawn to each circle in the family. Since $C$ and $D$ are elements of the circle family - the so-called point circles - tangents of length $FC = FD$ can be drawn from $F$ to each Apollonian circle (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_493c60bb69f2640f86acg-2.jpg?height=675&width=1003&top_left_y=638&top_left_x=542)

The points of tangency of the circles tangent to the line $AB$ are symmetric to $F$ and are at a distance $FP = FC$ from $F$ on the line $AB$. Simple calculations show that for these points, the ratio $F(P) = \frac{PC}{PD}$ is $\frac{\sqrt{5} + 1}{2}$, and its reciprocal. For any other point $Q$ on the line $AB$, which is an exterior point of these circles,

$$
\frac{\sqrt{5} - 1}{2} \leqq \frac{QC}{QD} \leqq \frac{\sqrt{5} + 1}{2}.
$$

2) Many solutions only examined the case where $P$ moves on the segment $AB$. In this case, the solution is somewhat simpler, but the obtained bounds differ from the correct result, since the ratio $\frac{PC}{PD}$ attains its maximum when $x = \sqrt{5}$, i.e., when $P$ is outside the segment $AB$; therefore, these solutions deserved only 2-3 points.