Given the right-angled isosceles triangle $A B C$ with unit-length legs, and points $A_{1}$ on side $A B$, $B_{1}$ on side $B C$, and $C_{1}$ on the hypotenuse $C A$. What is the minimum possible distance of $A_{1} B_{1}$, if the triangles $A B C$ and $A_{1} B_{1} C_{1}$ are similar?
Solution. The right angle of the triangle $A_{1} B_{1} C_{1}$ is either at one of the legs or on the hypotenuse.

I. Case: The right angle is on the hypotenuse. In this case, the quadrilateral $A_{1} B B_{1} C_{1}$ is a cyclic quadrilateral, since the sum of its opposite angles is $180^{\circ}$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_516578cff7e337a26665g-1.jpg?height=621&width=574&top_left_y=286&top_left_x=770)

The inscribed angles subtended by the same arc are equal, so $A_{1} B C_{1} \angle = A_{1} B_{1} C_{1} \angle = 45^{\circ}$. Therefore, the segment $B C_{1}$ bisects the angle $A B C \angle$. Since the triangle $A B C$ is isosceles, this angle bisector is also a median, and thus the point $C_{1}$ is the midpoint of the segment $A C$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_516578cff7e337a26665g-1.jpg?height=601&width=553&top_left_y=1109&top_left_x=775)

Let the point $B_{1}$ be the midpoint of the side $B C$, and the point $A_{1}$ be the midpoint of the side $A B$. Then the similarity ratio between the triangles $A_{1} B_{1} C_{1}$ and $A B C$ is 1:2. Since $B C = 1$, then $A C = \sqrt{2}$, and thus $A_{1} B_{1} = \frac{\sqrt{2}}{2}$.

Consider another isosceles right triangle with the right angle at $C_{1}$. Since $C_{1} B_{1} \perp B C$, we can only get such a triangle by intersecting the legs of the triangle $A B C$ with a circle centered at $C_{1}$ and with radius $r > C_{1} B_{1}$. The leg of the resulting triangle will be larger than $C_{1} B_{1}$, and thus its hypotenuse will also be larger than that of the triangle $A_{1} B_{1} C_{1}$: $F H > A_{1} B_{1}$.

Therefore, in this case, the minimum value of $A_{1} B_{1}$ is $\frac{\sqrt{2}}{2}$.

II. Case: The right angle is on one of the legs.

Let $A_{1} B_{1} = x$ and $B_{1} C = y$. Then $A_{1} C_{1} = x \sqrt{2}$.

Let the angle $C B_{1} C_{1}$ be $\alpha$. Then $C C_{1} B_{1} \angle = 180^{\circ} - 45^{\circ} - \alpha = 135^{\circ} - \alpha$. From this, $A C_{1} A_{1} \angle = 180^{\circ} - C C_{1} B_{1} \angle - B_{1} C_{1} A_{1} \angle = \alpha$ and thus $A A_{1} C_{1} \angle = 135^{\circ} - \alpha$. Therefore, the triangles $B_{1} C C_{1}$ and $C_{1} A A_{1}$ have equal angles, so the two triangles are similar. Thus,

$$
\frac{A C_{1}}{y} = \frac{x \sqrt{2}}{x}, \quad \text{which implies} \quad A C_{1} = y \sqrt{2} \quad (Figure 3)
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_516578cff7e337a26665g-2.jpg?height=677&width=724&top_left_y=125&top_left_x=687)

Since $B C = 1$, we have $B B_{1} = 1 - y$, and $A C = \sqrt{2}$, so $C C_{1} = \sqrt{2} - y \sqrt{2}$. We know that $A A_{1} = \sqrt{2} C C_{1}$, so $A A_{1} = \sqrt{2} (\sqrt{2} - y \sqrt{2}) = 2 - 2 y$, and thus $A_{1} B = 1 - (2 - 2 y) = 2 y - 1$.

In the triangle $A_{1} B B_{1}$, we can write the Pythagorean theorem:

$$
(2 y - 1)^{2} + (1 - y)^{2} = x^{2}, \quad \text{or} \quad 5 y^{2} - 6 y + 2 = x^{2}.
$$

It is clear that $x$ has a minimum value when the expression $5 y^{2} - 6 y + 2$ (if the value of the expression is positive at that point). The minimum of this quadratic function occurs at $y = -\frac{b}{2 a} = -\frac{-6}{2 \cdot 5} = \frac{3}{5}$, and then

$$
A_{1} B_{1} = x = \sqrt{5 \left(\frac{3}{5}\right)^{2} - 6 \cdot \frac{3}{5} + 2} = \frac{\sqrt{5}}{5} < \frac{\sqrt{2}}{2}
$$

Therefore, the distance $A_{1} B_{1}$ can be as small as $\frac{\sqrt{5}}{5}$.