A trapezoid-shaped lake has parallel shore segments of $200 \mathrm{~m}$ and $100 \mathrm{~m}$ in length, and the angles between the other two shorelines and these segments are $90^{\circ}$ and $45^{\circ}$. Two guards patrol the lake in the same direction and at the same pace, maintaining an equal distance from each other along the perimeter of the lake in both directions. What is the maximum distance between the two guards in a straight line?
In the right-angled trapezoid $ABCD$, let the projection of $B$ onto $DC$ be $B'$ (1st figure). Since $AB=DB'$, $B'C=100$, and from the isosceles right triangle $BB'C$, we have $BB'=B'C=AD=100$, and $BC=100\sqrt{2}$.

The trapezoid is thus a square with a side length of $100 \, \text{m}$ and an attached isosceles right triangle. The perimeter is: $400 + 100\sqrt{2}$.

The two guards maintain an equal distance from each other as they patrol, so the distance between them along the perimeter is half the total perimeter: $d=200+50\sqrt{2}$.

Since this distance remains constant during their movement, it is sufficient to examine the movement of one guard over half the perimeter. Let $F$ be the midpoint of segment $BC$. When one guard is at $F$, the other is at $D$, since $DA + AB + BF = 100 + 100 + 50\sqrt{2} = 200 + 50\sqrt{2} = d$.

We will examine when the two guards are farthest apart in a straight line while the first guard moves from $D$ to $F$. Instead of the distances, we will find the maximum of the squares of the distances. This is permissible since the distance is always positive.

Divide the side $CD$ such that $DP = 50\sqrt{2}$, $DQ = 50\sqrt{2} + 100$, and let $R$ be the point on segment $AD$ such that $DR = 50\sqrt{2}$ (2nd figure).

It is easy to see that (moving along the perimeter of the trapezoid in the same direction) $AQ = BP = DF = RC = 200 + 50\sqrt{2}$. We will follow the path of the first guard as it moves from $D$ to $P$, $Q$, and $C$, and then to the "opposite" $F$.

1. When the first guard is at $D$, the other is at $F$. The straight-line distance between them, from the right triangle $DFF'$ (where $F'$ is the projection of $F$ onto $DC$):

$$
DF^2 = 50^2 + 150^2 = 25000 \quad (DF \approx 158.1)
$$

2. The first guard moves along segment $DP$ to $P$. The second guard moves along segment $FB$ to $B$; when it reaches $B$, the square of the distance between the guards is:

$$
PB^2 = 100^2 + (100 - 50\sqrt{2})^2 \approx 24858
$$

At intermediate points (3rd figure), we get the square of the hypotenuse of a right triangle where the legs $a \leq 100$ and $b \leq 50\sqrt{2} + 100$, so

$$
d_1^2 \leq 100^2 + (50\sqrt{2} + 100)^2 \approx 39142
$$

3. The first guard continues to point $Q$. When the first guard is at $Q$, the second guard is at $A$, and the square of the distance between them is:

$$
QA^2 = 100^2 + (100 + 50\sqrt{2})^2 \approx 39142
$$

At intermediate points: $a = 100$, $b \leq DQ = 100 + 50\sqrt{2}$, so

$$
d_2^2 < QA^2 = 39142 \quad (4th figure)
$$

4. The first guard moves along segment $QC$, and the second guard moves along $AR$. We have seen that $C$ and $R$ are "opposite" points, so the two guards arrive at $C$ and $R$ simultaneously. We claim that the distance between the guards increases as they move along the segments $QC$ and the corresponding $AR$ (5th figure). In this case, this distance is the hypotenuse of a right triangle where the sum of the legs, $x + y$, is constant, half the perimeter.

From the identity $x^2 + y^2 = \frac{(x+y)^2 + (x-y)^2}{2}$, it follows that the hypotenuse increases with the difference between the legs. Since moving from $Q$ to $C$ and from $A$ to $R$ increases the longer leg and decreases the shorter leg,

$$
d_3 < RC^2 = RD^2 + DC^2 = 45000
$$

5. Finally, if the first guard moves along segment $CF$ and the second guard moves along $RD$, their distance is less than $RC$, since the legs of the dashed right triangle decrease compared to $RD$ and $DC$ (6th figure).

From the above, it is clear that the straight-line distance between the guards is the greatest when one is at $C$ and the other is at $R$. This distance is $CR = \sqrt{45000} \approx 222.1$ meters.

Based on the work of Béla Gajdos (Beregszász, Bethlen G. Gymnasium, 11th grade)

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