10.5. In triangle $A B C$, points $M$ and $N$ are the midpoints of sides $A C$ and $B C$ respectively. It is known that the point of intersection of the medians of triangle $A M N$ is the point of intersection of the altitudes of triangle $A B C$. Find the angle $A B C$.
Answer: $45^{\circ}$.

Solution. Let $H$ be the orthocenter (the point of intersection of the altitudes) of triangle $ABC$. Then the altitude $AT$ of triangle $ABC$ contains the median of triangle $AMN$, that is, it intersects segment $MN$ at its midpoint - point $E$ (see Fig. 10.5a, b). We can reason in different ways.

First method. Since $MN \| AB$, triangles $ETN$ and $ATB$ are similar (see Fig. 10.5a), hence $\frac{TN}{TB}=\frac{TE}{TA}=\frac{EN}{AB}=\frac{1}{4}$. Let $TN=x, TE=y$, then $BN=3x, AE=3y$.

Therefore, $CT=CN-TN=2x$, and $EH=\frac{1}{3}AE=y$ (by the property of the centroid of a triangle).

Notice that in right triangles $CTH$ and $BTA$, $\frac{CT}{BT}=\frac{2x}{4x}=\frac{1}{2}$ and $\frac{HT}{AT}=\frac{2y}{4y}=\frac{1}{2}$. Thus, these triangles are similar. Therefore, $\angle TCH=\angle TBA$. But $CH$ is part of the altitude $CQ$ of triangle $ABC$, so these equal angles are the acute angles of the right triangle $CQB$, that is, each of them is $45^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_beb18cecdf66661da24eg-3.jpg?height=389&width=479&top_left_y=428&top_left_x=363)

Fig. $10.5a$

![](https://cdn.mathpix.com/cropped/2024_05_06_beb18cecdf66661da24eg-3.jpg?height=397&width=479&top_left_y=418&top_left_x=1211)

Fig. 10.5b

Second method. Mark point $K$ - the midpoint of side $AB$ (see Fig. 10.5b). Then $AMNK$ is a parallelogram, its diagonal $MK$ passes through the midpoint of $AN$, so it also passes through point $H$.

Since $MH \| BC$, triangles $EMH$ and $ENT$ are equal (by side and two adjacent angles), hence $EH=ET$. The median $CK$ of triangle $ABC$ passes through point $E$ and is bisected at it, so $CHKT$ is a parallelogram, thus $TK \| CH$.

But $CH$ is part of the altitude $CQ$ of triangle $ABC$, so $TK \perp AB$. Thus, $TK$ is both an altitude and a median of the right triangle $ATB$, meaning this triangle is isosceles, so $\angle ABC=45^{\circ}$.

There are other ways to solve this problem. In particular, it is not difficult to prove that in this triangle, the Euler line $OH$ (where $O$ is the center of the circumscribed circle of triangle $ABC$) is parallel to $AB$. Then the equality $\operatorname{tg} \angle A \cdot \operatorname{tg} \angle B = 3$ holds (see, for example, V. V. Prasolov. Problems in Plane Geometry, No. 5.110). Using this fact and some additional considerations that follow from the problem's condition, one can not only compute the angle $ABC$ but also the other angles of this triangle.

## Evaluation Criteria

+ A complete and well-reasoned solution is provided

$\pm$ A generally correct reasoning is provided, with minor gaps or inaccuracies

- Only the answer is provided
- The problem is not solved or is solved incorrectly