In the plane, there is a line $y$ and a right-angled triangle $A B C$ with legs $A C=3 ; B C=4$. Vertex $C$ is 10 units away from the line $y$. The angle between $y$ and the direction of leg $A C$ is $\alpha$. We need to determine the angle $\alpha$ such that the surface area obtained by rotating triangle $A B C$ around the line $y$ is minimized.
Let's start by finding the surface area of the resulting solid of revolution. The lateral surface area of the cone formed by the rotation of $CD$ is $\pi \cdot CD^2 \sin \alpha$. Then the surface area formed by the rotation of segment $AC$ around the y-axis will be $\pi CD^2 \sin \alpha - \pi (CD-3)^2 \sin \alpha = \pi \sin \alpha \cdot (6 CD - 9) = \pi \sin \alpha \left(\frac{6 \cdot 10}{\sin \alpha} - 9\right) = \pi (60 - 9 \sin \alpha)$. The angle between the y-axis and the direction of the leg $BC$ is $90^\circ + \alpha$. Now, similarly, we can calculate the surface area formed by the rotation of $BC$. It is $\pi \sin (\alpha + 90^\circ) \left(\frac{8 \cdot 10}{\sin (\alpha + 90^\circ)} - 16\right) = \pi (80 - 16 \cos \alpha)$. We need to calculate the surface area formed by the rotation of segment $AB$. The angle between the y-axis and the direction of segment $AB$ is $\alpha - \beta$, where $\beta = \angle BAC$. Then $\sin (\alpha - \beta) = \sin \alpha \cos \beta - \sin \beta \cos \alpha = \frac{3}{5} \sin \alpha - \frac{4}{5} \cos \alpha$. The distance $AA_y$ is $10 - AC \sin \alpha = 10 - 3 \sin \alpha$. Now we can calculate the desired area. It is $\pi (10 \cdot (10 - 3 \sin \alpha) + 25 \cdot (\frac{3}{5} \sin \alpha - \frac{4}{5} \cos \alpha)) = \pi (100 - 15 \sin \alpha - 20 \cos \alpha)$. Therefore, the surface area of the entire figure of the triangle $ABC$ rotated around the y-axis is $\pi (60 - 9 \sin \alpha + 80 - 16 \cos \alpha + 100 - 15 \sin \alpha - 20 \cos \alpha) = \pi (240 - 12 (2 \sin \alpha + 3 \cos \alpha))$. Note that if point $A$ or $B$ were on the other side of the line passing through vertex $C$ parallel to the y-axis, the formula for the area would be different. Different signs of the terms in the formula for the surface area of the solid of revolution would result. However, all of them can be reduced to the derived formula if we allow the angle $\alpha$ to vary from $0^\circ$ to $360^\circ$. Thus, the minimum area will be achieved when the expression $2 \sin \alpha + 3 \cos \alpha = \sqrt{13} \sin (\alpha + \phi)$, where the auxiliary angle $\phi = \arctan \frac{3}{2}$, reaches its maximum. This expression reaches its maximum when $\alpha + \phi = \frac{\pi}{2}$. Hence, $\alpha = \frac{\pi}{2} - \arctan \frac{3}{2} = \arctan \frac{2}{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ac7f8c6a311c4a54edc4g-01.jpg?height=672&width=532&top_left_y=1568&top_left_x=769)

## Answer

$\alpha = \arctan \frac{2}{3}$.

## Miquel Point Inscribed and Circumscribed Circles

 Topics: [ Radical Axis [Theorem on the lengths of the tangent and secant; the product of the entire secant and its external part [ Auxiliary Circle  GSP and Inscribed AngleAuto: Riišmoouvv V...

Points $A'$, $B'$, and $C'$ are the midpoints of sides $BC$, $CA$, and $AB$ of triangle $ABC$ respectively, and $BH$ is its altitude.

## Solution

Let $A$ and $C$ be the circumcircles of triangles $AHC'$ and $CHA'$, respectively. Since points $B$ and $H$ are symmetric with respect to the midline $A'C'$ of triangle $ABC$, we have $C'H = C'B = C'A$ and $A'H = A'B = A'C$, i.e., triangles $AHC'$ and $CHA'$ are isosceles. Therefore, $A'C' (\| AC)$ is a common tangent to circles $A$ and $C$.

Let $S$ be the intersection point of lines $HM$ and $A'C'$. Then $SC'^2 = SM \cdot SH = SA'^2$, i.e., $S$ is the midpoint of $A'C'$ and

$\angle CBB' = \angle CBS$. Let $B$ be the circumcircle of triangle $BA'C'$. We will show that point $M$ lies on this circle. Indeed, if $\alpha, \beta, \gamma$ are the angles of triangle $ABC$, then $\angle C'MA' = 2\pi - \angle C'MH - \angle A'MH = 2\pi - (\pi - \alpha) - (\pi - \gamma) = \alpha + \gamma = \pi - \beta$.

Let $l$ be the perpendicular bisector of $A'C'$. Circle $B$ is symmetric with respect to $l$. Since $\angle HSC' = \angle BSC' = \angle A'SB'$, segments $SH$ and $SB'$ are symmetric with respect to $l$. Therefore, point $N$, symmetric to point $M$ with respect to $l$, lies on $B$ and on $BB'$, and arcs $C'M$ and $NA'$ are equal. Inscribed angles $\angle C'BM$ and $\angle A'BN$ subtend these arcs, i.e., they are equal. Thus, $\angle ABM = \angle C'BM = \angle A'BN = \angle CBB'$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ac7f8c6a311c4a54edc4g-02.jpg?height=686&width=849&top_left_y=843&top_left_x=41)

## Answer

## Problem