# Problem 5.

B-1 From an isosceles triangle with an angle $\alpha$ at the vertex and an area of 1, a circle of maximum area is cut out, and from it, a triangle similar to the original and of maximum area is cut out. What are the greatest and least values that the area $S(\alpha)$ of the resulting triangle can take for $60^{\circ} \leqslant \alpha \leqslant 120^{\circ}$?
Answer: $\frac{1}{4} ; 7-4 \sqrt{3}$.

Solution. Let $\beta=\frac{\alpha}{2}$ and $r$ be the radius of the circle. Then the halves of the bases of the original and cut-out triangles are respectively

$$
a=\frac{r(1+\sin \beta)}{\cos \beta}, \quad b= \begin{cases}r \sin 2 \beta, & \beta \leqslant 45^{\circ} \\ r, & \beta \geqslant 45^{\circ}\end{cases}
$$

and the function

$$
S(\alpha)=S(2 \beta)=\left(\frac{b}{a}\right)^{2}, \text { where } \frac{b}{a}= \begin{cases}2 \sin \beta \cdot(1-\sin \beta), & \beta \leqslant 45^{\circ} \\ \frac{\cos \beta}{1+\sin \beta}, & \beta \geqslant 45^{\circ}\end{cases}
$$

decreases for $\alpha \in\left[60^{\circ} ; 120^{\circ}\right]$. Therefore,

$$
S_{\max }=S\left(60^{\circ}\right)=\left(2 \sin 30^{\circ} \cdot\left(1-\sin 30^{\circ}\right)\right)^{2}=\frac{1}{4}, \quad S_{\min }=S\left(120^{\circ}\right)=\left(\frac{\cos 60^{\circ}}{1+\sin 60^{\circ}}\right)^{2}=7-4 \sqrt{3}
$$

B-2 An isosceles triangle with an angle $\alpha$ at the vertex and an area of 1 is placed in the smallest circle by area, and then in the smallest triangle by area, similar to the original. What is the maximum and minimum value of the area $S(\alpha)$ of the resulting triangle for $60^{\circ} \leqslant \alpha \leqslant 120^{\circ} ?$

Answer: $7+4 \sqrt{3} ; 4$.

B-3 From an isosceles triangle with an angle $\alpha$ at the base and an area of 1, the largest circle by area is cut out, and from it, the largest triangle by area, similar to the original. What is the maximum and minimum value of the area $S(\alpha)$ of the resulting triangle for $30^{\circ} \leqslant \alpha \leqslant 60^{\circ}$ ?

Answer: $\frac{1}{4} ; 7-4 \sqrt{3}$.

B-4 An isosceles triangle with an angle $\alpha$ at the base and an area of 1 is placed in the smallest circle by area, and then in the smallest triangle by area, similar to the original. What is the maximum and minimum value of the area $S(\alpha)$ of the resulting triangle for $30^{\circ} \leqslant \alpha \leqslant 60^{\circ} ?$

Answer: $7+4 \sqrt{3} ; 4$.

B-5 On a vertical plane representing a wall, a horizontal line representing the floor is drawn. On the floor stands a freshly painted square with a side length of 1. It is cantoned (rotated by $90^{\circ}$, resting on one of its vertices). This is done four times until it stands on the same side as it did initially. During this process, the square always touched the wall, so part of it got painted (or dirtied). A similar procedure was performed on another wall, but the square had a diagonal of length 1, and it was cantoned 7 times. How much less area was painted (dirtied) in the second case?

Answer: $\frac{1}{2}+\frac{\pi}{4}$.

Solution. If a square with side $a$ is cantoned $n$ times, the entire painted figure can be represented as the union of the following intersecting only at the boundaries parts (see the figure for $n=4$).

![](https://cdn.mathpix.com/cropped/2024_05_06_eb9911c05e0d5d3c03abg-09.jpg?height=325&width=1077&top_left_y=383&top_left_x=495)

- Two isosceles right triangles with legs equal to $a$.

The total area is $a^{2}$.

- Two sectors with radius $a \sqrt{2}$ and an angle at the vertex of $45^{\circ}$.

The total area is $2 \cdot \pi \cdot 2 a^{2} / 8=\pi a^{2} / 2$.

- $(n-1)$ isosceles triangles with base $a$ and height $\sqrt{2 a^{2}-(a / 2)^{2}}=a \sqrt{7} / 2$. The total area is $a^{2}(n-1) \sqrt{7} / 4$.
- $2 n-2$ sectors with radius $a \sqrt{2}$ and an angle at the vertex of $\beta=\arcsin 1 / \sqrt{8}$.

The total area is $(2 n-2) \cdot 2 a^{2} \cdot \beta / 2=(2 n-2) a^{2} \beta$.

Thus, the difference in the areas of the first and second figures is

$\left(1+\frac{\pi}{2}+\frac{3 \sqrt{7}}{4}+6 \beta\right)-\left(\frac{1}{2}+\frac{\pi}{4}+\frac{6 \sqrt{7}}{2 \cdot 4}+\frac{12 \beta}{2}\right)=\frac{1}{2}+\frac{\pi}{4}$.

B-6 On a vertical plane representing a wall, a horizontal line representing the floor is drawn. On the floor stands a freshly painted square with a side length of 1. It is cantoned (rotated by $90^{\circ}$, resting on one of its vertices). This is done four times until it stands on the same side as it did initially. During this process, the square always touched the wall, so part of it got painted (or dirtied). A similar procedure was performed on another wall, but the square had a semiperimeter of 1, and it was cantoned 13 times. How much less area was painted (dirtied) in the second case?

Answer: $\frac{3}{4}+\frac{3 \pi}{8}$.

B-7 On a vertical plane representing a wall, a horizontal line representing the floor is drawn. On the floor stands a freshly painted square with a side length of 1. It is cantoned (rotated by $90^{\circ}$, resting on one of its vertices). This is done four times until it stands on the same side as it did initially. During this process, the square always touched the wall, so part of it got painted (or dirtied). A similar procedure was performed on another wall, but the square had a perimeter of 1, and it was cantoned 49 times. How much less area was painted (dirtied) in the second case?

Answer: $\frac{15}{16}+\frac{15 \pi}{32}$.

B-8 On a vertical plane representing a wall, a horizontal line representing the floor is drawn. On the floor stands a freshly painted square with a diagonal of length 1. It is cantoned (rotated by $90^{\circ}$, resting on one of its vertices). This is done four times until it stands on the same side as it did initially. During this process, the square always touched the wall, so part of it got painted (or dirtied). A similar procedure was performed on another wall, but the square had a perimeter of 1, and it was cantoned 25 times. How much less area was painted (dirtied) in the second case?

Answer: $\frac{7}{16}+\frac{7 \pi}{32}$.

## Lomonosov Olympiad for Schoolchildren in Mathematics

Final stage of the 2020/21 academic year for $10-11$ grades

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