8,9,10,11 |

Avor: Teresein D.A.

Points $A_2, B_2$ and $C_2$ are the midpoints of the altitudes $A A_1, B B_1$ and $C C_1$ of an acute-angled triangle $A B C$. Find the sum of the angles $B_2 A_1 C_2, C_2 B_1 A_2$ and $A_2 C_1 B_2$.
Let $H$ be the point of intersection of the altitudes of triangle $ABC$, and $M$ be the midpoint of side $AB$. Prove that points $H, M, B_2, A_2$, and $C_1$ lie on the same circle.

## Solution

Let $H$ be the point of intersection of the altitudes of triangle $ABC$, and $M$ be the midpoint of side $AB$. Since $MB_2$ and $MA_2$ are the midlines of the right triangles $AB_1B$ and $AA_1B$, we have

$$
\angle MB_2H = \angle AB_1B = 90^\circ, \quad \angle MA_2H = \angle BA_1A = 90^\circ
$$

Therefore, from points $B_2, A_2$, and $C_1$, the segment $MH$ is seen at a right angle. This means that these points lie on a circle with diameter $MH$. The quadrilateral $A_2C_1B_2H$ is inscribed, so

$$
\angle A_2C_1B_2 = 180^\circ - \angle A_2HB_2 = 180^\circ - \angle A_1HB_1 = \angle ACB
$$

Similarly, we can prove that

$$
\angle B_2A_1C_2 = \angle BAC, \quad \angle C_2B_1A_2 = \angle ABC
$$

Therefore,

$$
\angle A_2C_1B_2 + \angle B_2A_1C_2 + \angle C_2B_1A_2 = \angle ACB + \angle BAC + \angle ABC = 180^\circ.
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_dadb5e48aa89908248cfg-48.jpg?height=509&width=577&top_left_y=2244&top_left_x=772)

## Answer

![](https://cdn.mathpix.com/cropped/2024_05_06_dadb5e48aa89908248cfg-49.jpg?height=235&width=2041&top_left_y=157&top_left_x=0)

Author: $\underline{\text{ Folklore }}$

Each pair of opposite sides of a given convex hexagon has the following property: the distance between their midpoints is $\sqrt{3} / 2$ times the sum of their lengths. Prove that all angles in the hexagon are equal.

## Solution

Proof.

Lemma.

Consider a triangle $\triangle PQR$ with angle $\angle QPR \geq \pi / 3$, and let $L$ be the midpoint of $QR$. Then $PL \leq \sqrt{3} QR / 2$, and equality is achieved only for an equilateral triangle.

Proof of the lemma. Consider an equilateral triangle $QRS$ such that points $P$ and $S$ lie on the same side of the line $QR$. Clearly, then $P$ lies inside the circumcircle of triangle $QRS$, and hence inside the circle with center $L$ and radius $\sqrt{3} QR / 2$. The lemma is proved.

Returning to the solution of the problem. Without loss of generality, we can assume that the main diagonals $AD$ and $BE$ of the hexagon $ABCDEF$ form an angle $\geq \pi / 3$. Let $P$ be their point of intersection. We will use our lemma. We have

$$
MN = \sqrt{3} / 2 (AB + DE) \geq PM + PN \geq MN
$$

($M, N$ are the midpoints of sides $AB$ and $DE$, respectively). From the lemma, it follows that triangles $ABP$ and $DEP$ are equilateral.

Diagonal $CF$ forms an angle $\geq \pi / 3$ with diagonal $AD$ or $BE$. By symmetry, without loss of generality, we can assume that $\angle AQF \geq \pi / 3$, where $Q$ is the point of intersection of diagonals $AD$ and $CF$. Similarly, using the lemma, we get that triangles $AQF$ and $CQD$ are equilateral. Consequently, $\angle BRC = \pi / 3$, where $R$ is the point of intersection of $BE$ and $CF$. Similarly, the equilateral nature of triangles $BCR$ and $EFR$ is established. The problem is solved.