3. In an isosceles right triangle \( A B C \), an isosceles right triangle \( A_{1} B_{1} C_{1} \) is inscribed. Find the minimum value of the ratio

$$
P_{A_{1} B_{1} C_{1}}: P_{A B C}
$$
Solution. Let $C$ and $C_{1}$ be the vertices of the right angles. There are two possible cases.

First case. $C_{1}$ is a point on the leg, for example, on $C B$. Let $A_{1}$ be on $A C$ and let $M$ be a point on $B C$ such that $B_{1} M$ is perpendicular to $B C$. Then the triangles $B_{1} M C_{1}$ and $C_{1} C A_{1}$ are congruent, which implies $\overline{M B_{1}}=\overline{C C_{1}}$, and therefore $\overline{B C}=2 \overline{C C_{1}}+\overline{M C}_{1}$ and ${\overline{B_{1} C_{1}}}^{2}={\overline{C C_{1}}}^{2}+{\overline{M C_{1}}}^{2}$. If we denote $\overline{C C_{1}}: \overline{B C}=x$, we get that $P_{A_{1} B_{1} \mid C_{1}}: P_{A B C}=5 x^{2}-4 x+1$. According to this, the desired minimum value in this case is the minimum value of the quadratic function $y=5 x^{2}-4 x+1$, i.e., its vertex which has coordinates $\left(\frac{4}{10}, \frac{1}{5}\right)$. For the triangle $A_{1} B_{1} C_{1}$ with $\overline{C C_{1}}=\frac{4}{10} \overline{B C}$ and $\overline{C A_{1}}=\frac{2}{10} \overline{B C}$, it holds that $P_{A_{1} B_{1} \mid C_{1}}: P_{A B C}=\frac{1}{5}$ (diagram a)).
![](https://cdn.mathpix.com/cropped/2024_06_05_777e4c6fa99d478c3e87g-4.jpg?height=348&width=984&top_left_y=804&top_left_x=364)

Second case. $C_{1}$ is a point on the hypotenuse $A B$ (diagram b)). In this case, $C_{1}$ must be the midpoint of $A B$. Then $P_{A_{1} B_{1} \mid C_{1}}$ will be the smallest if $A_{1}$ and $B_{1}$ are the midpoints of the legs. This means that in this case

$$
P_{A_{1} B_{1} \mid C_{1}}: P_{A B C}=\frac{1}{4}
$$

Finally, since $\frac{1}{5}<\frac{1}{4}$, we conclude that the desired minimum value is $\frac{1}{5}$.