Provide an estimate for the accuracy of the following procedure for trisecting acute angles. Let $k$ be any circle with radius $r$, center $O$, and diameter $D E$, and let $C$ be the point on $k$ such that $\angle C O D$ is equal to the angle to be trisected. Measure on the ray $O E$ the segments $O A_{1}=2 r, O B_{1}=\frac{5}{3} r$. Denote the intersection of $k$ and $C A_{1}$, as well as $C B_{1}$, by $A_{2}$ and $B_{2}$, respectively, and measure on the extensions of $C A_{2}$ and $C B_{2}$ the segments $A_{2} A = B_{2} B = r$. Finally, let the intersection of $A B$ and $D E$ be $T$, then $\angle C T D \approx \frac{1}{3} \angle C O D$.
Let $COD \angle = \gamma, ACO \angle = \alpha, BCO \angle = \beta, C$ be the projection of $OD$, $H$. The segments $CH, CA$, and $CB$ are denoted by $h$, $a$, and $b$ respectively. Clearly,

$$
h = r \sin \gamma, \quad a = CA_2 + r, \quad b = CB_2 + r
$$

We introduce the so-called polar coordinates: for any point $P$ in the plane, we associate its distance from point $C$ and the angle obtained by rotating the ray $CP$ in the positive direction to the ray $CO$. This assignment is one-to-one. The polar coordinates of point $P$ are denoted by $\varrho$ ( $= CP$ ) and $\Theta$ ( $= OCP$ "positive direction angle"). It is easy to see that

$$
\varrho = 2r \cos \Theta
$$

is valid only for the circle,

$$
\varrho = h / \sin (\gamma - \Theta)
$$

is valid only for the line $A_1B_1$, and

$$
\varrho = \frac{ab \sin (\alpha - \beta)}{a \sin (\alpha - \Theta) + b \sin (\Theta - \beta)}
$$

is valid only for the points on the line $AB$. (1) and (2) are trivial, for (3) the area formula for a triangle $t = \frac{1}{2} xy \sin \varphi$ must be applied. For point $T$, (2) and (3) hold, i.e.,

$$
\frac{h}{\sin (\gamma - \Theta)} = \frac{ab \sin (\alpha - \beta)}{a \sin (\alpha - \Theta) + b \sin (\Theta - \beta)}
$$

is true. From this,

$$
\frac{h}{\sin (\gamma - \Theta)} = \frac{ab \sin (\alpha - \beta)}{\cos \Theta (a \sin \alpha - b \sin \beta) + \sin \Theta (b \cos \beta - a \cos \alpha)}
$$

Our goal is to examine $\delta = 2\gamma / 3 - \Theta$, which characterizes the accuracy of the angle trisection.

We introduce the new variables $m, p, n, q$ as follows:

$$
m = \frac{ab \sin (\alpha - \beta)}{r^2}, \quad n = m \operatorname{ctg} \gamma, \quad p = \frac{b \sin \beta - a \sin \alpha}{r}, \quad q = \frac{b \cos \beta - a \cos \alpha}{r}
$$

Then,

$$
\operatorname{tg} \Theta = \frac{m + p}{n + q}
$$

The values of $a, b, \alpha, \beta$ can be expressed in terms of $\gamma$ and $r$, so (5), and then (6), can also be used to express $\Theta$.

Indeed, from (1),

$$
a = 2r \cos \alpha + r = r(1 + 2 \cos \alpha)
$$

$$
b = 2r \cos \beta + r = r(1 + 2 \cos \beta)
$$

Applying the sine rule to triangles $COA_1$ and $COB_1$, we get

$$
\begin{gathered}
\sin \alpha = 2 \sin (\gamma - \alpha) \\
\sin \beta = \frac{5}{3} \sin (\gamma - \beta)
\end{gathered}
$$

From these,

$$
\begin{gathered}
\sin \alpha = \frac{2 \sin \gamma}{P} ; \quad \cos \alpha = \frac{1 + 2 \cos \gamma}{P} \\
\sin \beta = \frac{5 \sin \gamma}{Q} ; \quad \cos \beta = \frac{3 + 5 \cos \gamma}{Q} \\
\sin (\alpha - \beta) = \frac{\sin \gamma}{P \cdot Q}
\end{gathered}
$$

where

$$
P^2 = 5 + 4 \cos \gamma \quad \text{and} \quad Q^2 = 34 + 30 \cos \gamma
$$

For a given $\gamma$, the calculation of $\Theta$ (and thus $\delta$) can proceed as follows:

1. From (7b), calculate the values of $\sin \alpha, \sin \beta, \cos \alpha, \cos \beta, \sin (\alpha - \beta)$,
2. Then, from (7a), calculate $a$ and $b$,
3. Finally, using (5), calculate $m, p, n, q$, and then from (6), express $\Theta$.

By calculating $\delta_i$ for $\gamma_i = \frac{\pi}{2} \cdot \frac{i}{100} (i=1,2, \ldots, 100)$, we get the following (to four decimal places):

$$
\begin{gathered}
\gamma_1 = 0.0157, \gamma_2, \ldots, \gamma_{52} = 1.6336 \text{ for which } \delta = 0 \\
\gamma_{53}, \gamma_{54} \ldots, \gamma_{65} \text{ for which } \delta = 0.0001 \text{ (radians) } \\
\gamma_{66}, \gamma_{67} \ldots, \gamma_{73} \text{ for which } \delta = 0.0002 \\
\gamma_{74}, \gamma_{75} \ldots, \gamma_{80} \text{ for which } \delta = 0.0003 \approx 0^\circ 1' \\
\gamma_{81}, \gamma_{82} \ldots, \gamma_{85} \text{ for which } \delta = 0.0004 \\
\gamma_{86}, \gamma_{87} \ldots, \gamma_{92} \text{ for which } \delta = 0.0005 \\
\gamma_{93}, \gamma_{94} \ldots, \gamma_{100} = \frac{\pi}{2} \text{ for which } 0.0006
\end{gathered}
$$

We observe that as $\gamma$ increases, so does $\delta$. Since $\delta = 0.0006$ when $\gamma = \frac{\pi}{2}$, we find that for acute angles, $\delta \leq 0.0006$.

G. C. Mohanty, India

Note. 1. In the final step of the solution, we did not "prove" but rather calculated by simple computational means: for a given $\gamma$, what $\delta$ corresponds to it.

2. No solution was received for the problem.