In a right-angled triangle $A B C$, from vertices $B$ and $C$, one point each starts moving simultaneously towards point $A$ along the hypotenuse $B A=c=85$ m and the leg $C A=b=75$ m, respectively. The points move at speeds of $8.5 \mathrm{~m} / \mathrm{sec}$ and $5 \mathrm{~m} / \mathrm{sec}$. At what time will the distance between the two points be $26 \mathrm{~m}$. (Without using trigonometry!)
Suppose that $x$ seconds later, the points starting from $B$ and $C$ reach positions $D$ and $E$, respectively, and $DE = 26$ meters (see the diagram).

![](https://cdn.mathpix.com/cropped/2024_05_02_726c4ef76a955fe776e8g-1.jpg?height=624&width=382&top_left_y=260&top_left_x=861)

Let the foot of the perpendicular from $D$ to the leg $AC$ be $F$. Applying the Pythagorean theorem to triangle $DFE$,

$$
D F^{2} + E F^{2} = D E^{2} = 676
$$

Here,

$$
D F^{2} = A D^{2} - A F^{2}
$$

and on one hand,

$$
A D = 85 - 8.5 x = 17\left(5 - \frac{x}{2}\right)
$$

on the other hand,

$$
\frac{A F}{A D} = \frac{A C}{A B} = \frac{75}{85} = \frac{15}{17}
$$

from which

$$
A F = \frac{15}{17} A D = 15\left(5 - \frac{x}{2}\right), \text{ and } A D^{2} - A F^{2} = \left(17^{2} - 15^{2}\right)\left(5 - \frac{x}{2}\right)^{2} = 64\left(5 - \frac{x}{2}\right)^{2}
$$

Finally,

$$
E F = A C - A F - C E = 75 - 15\left(5 - \frac{x}{2}\right) - 5 x = \frac{5 x}{2}
$$

Thus, we obtain the following equation for $x$:

$$
676 = 64\left(25 - 5 x + \frac{x^{2}}{4}\right) + \left(\frac{5 x}{2}\right)^{2} = 1600 - 320 x + \frac{89}{4} x^{2}
$$

which simplifies to

$$
89 x^{2} - 1280 x + 3696 = 0
$$

from which

$$
x_{1,2} = \frac{640 \pm 284}{89}
$$

thus

$$
x_{1} = \frac{356}{89} = 4 \quad \left[x_{2} = \frac{924}{89} > 10\right]
$$

The root greater than 10 is not valid in this case because the point starting from $B$ reaches the endpoint $A$ in $10 \text{ seconds}$.

Therefore, the distance between the two points will be $26 \text{ meters}$ after $4 \text{ seconds}$.