a) A point $M$ moves around the circumcircle of a triangle, and for each of its positions, we consider the projections of $M$ onto the sides of the triangle. We find that the points where $M$ is such that all three projections fall on the extensions of the sides (or exactly at the vertices) cover $\frac{8}{10}$ of the circumference of the circle. What is the largest angle of the triangle?

b) A point $N$ moves around the incircle of a triangle. We find that the points where $N$ is such that one of the three projections (formed in the same way as before) falls on the extension of a side cover $\frac{8}{10}$ of the circumference of the circle. What is the largest angle of the triangle? (It is sufficient to give the angle with an accuracy of $0.1^{\circ}$.)
a) The perpendicular foot from $M$ to $AB$ and $AC$ only falls on the extension of the side beyond $A$ if $\angle MAB, \angle MAC > 90^\circ$, which means that $A, C, B$ are on one side of the diameter starting from $M$ (1st figure). If they are placed in this order on the semicircle, then $\angle ACB = \gamma > 90^\circ$, and in this case, $\angle MCB > 90^\circ$, so the projection of $C$ on $CB$ is on the extension of $CB$ beyond $C$. This holds as long as $M$ is on the arc $AB'$ that does not contain $B$ and $C$, where $B'$ is the point opposite to $B$. Similarly, from the points of the arc $BA'$, the perpendiculars to $BA$ and $BC$ fall on the extension of $BA$ and $BC$ beyond $B$, and the perpendicular to $AC$ falls on the extension of $AC$ beyond $C$ ($A'$ is the point opposite to $A$ in the circle).

The arcs $AB'$ and $A'B$ are mirror images of each other with respect to the center $O$ of the circle, so each is $4/10$ of the circumference of the circle, and $\angle AOB' = 144^\circ$. This angle is also $180^\circ$ less than the central angle $2\gamma$ subtended by the arc on which the sides of $\angle ACB$ lie: $2\gamma - 180^\circ = 144^\circ$, $\gamma = 162^\circ$. This is what we were looking for.

![](https://cdn.mathpix.com/cropped/2024_05_02_37e8ae97b38949ad9798g-1.jpg?height=572&width=1220&top_left_y=560&top_left_x=431)

1st and 2nd figures

b) Perpendiculars must be drawn from the vertices to the sides such that there is a point of the inscribed circle $k$ on the opposite side of the side. This can only be a perpendicular from an obtuse angle, and each perpendicular drawn from here intersects the circle, and cuts off equal arcs, because the two perpendiculars are mirror images of each other on the angle bisector from the vertex, which is also a symmetry axis of $k$.

Let $\angle ACB = \gamma > 90^\circ$ (2nd figure), and let the perpendiculars from $C$ to $CB$ and $CA$ intersect $k$ at points $D, D'$ and $E, E'$, respectively, and let the projections of the center of $k$ on $CA$ and $CE$ be $O'$ and $O_1$, respectively. Then the shorter arcs $DD'$ and $EE'$ are equal, and the projection of $N$ on the line $CA$, taken on the arc $EE' = i$, falls on the extension of the side beyond $C$, while the projection of $N$ on $CB$ falls between $C$ and $B$, because $N$ is in the region of the angle $\angle BCE = \gamma - 90^\circ$, and the projection of $N$ on $AB$ also falls between $A$ and $B$, since $\alpha, \beta < 90^\circ$.

According to the given data, $\angle EOO_1 = 360^\circ \cdot 2 / 10 = 72^\circ$, so

$$
OO_1 = OE \cos 72^\circ, \quad CO' = OO' \cot \gamma / 2 = OE \cot \gamma / 2
$$

and from $OO_1 = CO'$,

$$
\cot \gamma / 2 = \cos 72^\circ \approx 0.3090, \quad \gamma / 2 \approx 72.83^\circ, \quad \gamma \approx 145.7^\circ
$$

The other two angles of the triangle cannot be determined from the given data in either case a) or b).