Someone measured the combined length of the two arcs on the circumference of the circle circumscribed around a triangle $ABC$, whose points, when projected onto the line $AB$, have their projection on one of the extensions of the segment $AB$. They did the same for the lines $BC$ and $CA$, and reported that the ratio of the three measurements is $1: 2: 3$. What can we say about the angles of the triangle based on this?
Draw perpendiculars to the $AB$ side at its endpoints and denote the intersection of the perpendicular raised at $A$ with the circle by $B_{1}$, and the intersection of the perpendicular raised at $B$ by $A_{1}$ (Fig. 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_7f7df74109492221cf13g-1.jpg?height=589&width=572&top_left_y=253&top_left_x=755)

Fig. 1

Thus, the points of the (smaller than a semicircle) $AB_{1}$ and $BA_{1}$ arcs - and only these points - have their projections on the line $AB$ falling on the extension of the segment $AB$ on one side or the other. According to Thales' theorem, $AA_{1}$ and $BB_{1}$ are diameters of the circle, and therefore $AB_{1} = BA_{1}$.

Performing the same for the sides $BC$ and $CA$, we get the intersection point of the diameter starting from $C$ as $C_{1}$, and again $B_{1}$ and $A_{1}$, and the points of the smaller arcs $BC_{1}$ and $CB_{1}$ on the line $BC$, and the points of the arcs $CA_{1} = AC_{1}$ on the line $CA$ give the projection on the extension. Denote the central angles corresponding to the arcs $\widehat{AB}_{1}, \widehat{BC}_{1}, \widehat{CA}_{1}$ by $\gamma_{1}$, $\alpha_{1}$, $\beta_{1}$, respectively; with these and the radius of the circle, we can express the desired arc sums as follows:

$$
\begin{gathered}
\widehat{AB}_{1} + \widehat{BA}_{1} = 2 \frac{2 \pi r 2 \gamma_{1}}{360^{\circ}} = \frac{\pi r}{45^{\circ}} \gamma_{1} \\
\widehat{BC}_{1} + \widehat{CB}_{1} = \frac{\pi r}{45^{\circ}} \alpha_{1}, \quad \widehat{CA}_{1} + \widehat{AC}_{1} = \frac{\pi r}{45^{\circ}} \beta_{1}
\end{gathered}
$$

On the other hand, these can also express the angles of the triangle $ABC$.

If the original triangle $ABC$ is acute-angled (Fig. 1), then the points $A_{1}, B_{1}, C_{1}$ lie on the smaller arcs cut by the sides $BC, CA, AB$ of the circle, and the six arcs listed in (1) fill the circumference of the circle without gaps or overlaps. Therefore, on the one hand,

$$
\frac{\pi r}{45^{\circ}}\left(\alpha_{1} + \beta_{1} + \gamma_{1}\right) = 2 \pi r, \quad \alpha_{1} + \beta_{1} + \gamma_{1} = 90^{\circ}
$$

and on the other hand, the ratio of these is given by

$$
\gamma_{1} : \alpha_{1} : \beta_{1} = 1 : 2 : 3
$$

from which $\gamma_{1} = 15^{\circ}, \alpha_{1} = 30^{\circ}, \beta_{1} = 45^{\circ}$, and from these, the angles of our triangle are:

$$
\begin{aligned}
& \angle BAC = \beta_{1} + \gamma_{1} = 60^{\circ}, \quad \angle CBA = \gamma_{1} + \alpha_{1} = 45^{\circ}, \\
& \angle ACB = \alpha_{1} + \beta_{1} = 75^{\circ}.
\end{aligned}
$$

2. Now consider the desired arc sums in an obtuse triangle $KLM$, let its angles be $\kappa$, $\lambda$, $\mu$, such that $\mu > 90^{\circ} > \lambda \geq \kappa$, and determine the size relationship of the sums $\widehat{KL}_{1} + \widehat{LK}_{1}$, $\widehat{LM}_{1} + \widehat{ML}_{1}$, and $\widehat{MK}_{1} + \widehat{KM}_{1}$, so that we can correctly match the vertices $K, L, M$ to the desired vertices $A, B, C$ (Fig. 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_7f7df74109492221cf13g-1.jpg?height=498&width=548&top_left_y=2147&top_left_x=775)

Fig. 2

The arc $\widehat{KL}$ not containing vertex $M$ is larger than a semicircle, the order of points on this arc is $K, L_{1}, M_{1}, K_{1}, L$, and the arcs between consecutive points are all smaller than a semicircle.

To find the ratio of the arcs, it is sufficient to express the inscribed angles corresponding to the arcs $\widehat{KL_{1}}$, $\widehat{L_{1}M}$, and $\widehat{KM_{1}}$:

$$
\begin{aligned}
\angle KML_{1} & = \angle KML - \angle L_{1}ML = \mu - 90^{\circ}, \\
\angle L_{1}LM & = \angle L_{1}LK + \angle KLM = \angle L_{1}MK + \lambda = (\mu - 90^{\circ}) + \lambda = 90^{\circ} - \kappa, \\
\angle KMM_{1} & = 90^{\circ} - \angle KMM_{1} = 90^{\circ} - \angle KLM = 90^{\circ} - \lambda.
\end{aligned}
$$

Since $\kappa \leq \lambda$, it follows that $\angle L_{1}LM \geq \angle KMM_{1}$, and furthermore, the angle $\angle KML_{1}$ is smaller than both, because $\widehat{KL_{1}}$ is a part of both $\widehat{L_{1}M}$ and $\widehat{KM_{1}}$. And since the terms of the $1:2:3$ ratio are different, we have

$$
\widehat{KL}_{1} : \widehat{KM}_{1} : \widehat{ML}_{1} = 1 : 2 : 3
$$

and the role of the side $KL$ should be assigned to $AB$, the role of $KM$ to $BC$, and the role of $ML$ to $CA$, that is, the roles of $K, L, M$ should be assigned to $B, A, C$, respectively. From the ratios, we have

$$
\begin{gathered}
2 \widehat{KL}_{1} = \widehat{KM}_{1} = \widehat{KL}_{1} + \widehat{L_{1}M}_{1} = \widehat{KL}_{1} + \widehat{LM} \\
\widehat{KL}_{1} = \widehat{LM}, \quad \mu - 90^{\circ} = \kappa
\end{gathered}
$$

and

$$
3 \widehat{KL}_{1} = \widehat{ML}_{1}, \quad \widehat{MK} = 2 \widehat{ML}, \quad \lambda = 2 \kappa
$$

Substituting these,

$$
\begin{gathered}
\kappa + \lambda + \mu = \kappa + 2 \kappa + (90^{\circ} + \kappa) = 180^{\circ}, \\
\kappa = \angle ABC = 22.5^{\circ}, \quad \lambda = \angle BAC = 45^{\circ}, \quad \mu = \angle ACB = 112.5^{\circ}
\end{gathered}
$$

3. According to all this, the consecutive angles of the triangle $ABC$ are either $60^{\circ}, 45^{\circ}, 75^{\circ}$ or $45^{\circ}, 22.5^{\circ}, 112.5^{\circ}$. - We can also say this: the vertices $A, B, C$ are either the first, sixth, and tenth vertices of a regular 12-sided polygon, or the first, sixth, and eighth vertices of a regular 8-sided polygon.

Remark. The triangle $ABC$ could not have been a right triangle, because then there would be no projection point on the extension of the hypotenuse, and the ratio of the corresponding arc would be 0, different from the given ratio.