A circle $k$ has its center at $O$, and points $A, B, C, D$ are consecutive points on the circumference of $k$, such that $\angle AOB = \angle BOC = \angle COD = \alpha < 60^{\circ}$. The projection of $D$ onto the diameter $AO$ is $E$, and the point $F$ is the closer trisection point of segment $DE$ to $E$. Finally, the intersection of $AO$ and $BF$ is $G$. What does the size of angle $OGD$ approach as $\alpha \rightarrow 60^{\circ}$?
I. solution. In the construction of the described figure, the creation of the intersection point $G$ is the only uncertain part. Given the condition $\alpha0^{\circ}$, otherwise it would not be possible for $\alpha \rightarrow 60^{\circ}$. Taking $O A$ as the unit length,

$$
E F=\frac{1}{3} \sin 3 \alpha=\sin \alpha-\frac{4}{3} \sin ^{3} \alpha<\sin \alpha=B K
$$

where $K$ is the projection of point $B$ onto $O A$ (upper part of Figure 1), so the lines $A O$ and $B F$ are not parallel, and $G$ indeed forms, specifically on the part of the ray $K E$ beyond $E$.

![](https://cdn.mathpix.com/cropped/2024_05_02_52666a947bb77098e9d5g-1.jpg?height=612&width=550&top_left_y=480&top_left_x=777)

Figure 1

Let the line through $E$ parallel to $B F$ intersect $B K$ at $L$, and denote the angle $O G D$ by $\beta$. Since $\beta$ is always an acute angle according to the construction, it is sufficient to determine its tangent:

$$
\operatorname{tg} \beta=\frac{E D}{G E}=3 \frac{E F}{G E}=3 \frac{K L}{E K}
$$

According to our construction,

$$
K L=K B-E F=\sin \alpha-\frac{1}{3} \sin 3 \alpha, \quad E K=\cos \alpha-\cos 3 \alpha
$$

(In the latter statement, we are thinking of a coordinate system whose origin is $O$, and whose positive $x$-axis is the ray $O A$: in this system, the abscissas of $K$ and $E$ are $\cos \alpha$ and $\cos 3 \alpha$, respectively.) Therefore,

$$
\operatorname{tg} \beta=3 \frac{\sin \alpha-\frac{1}{3} \sin 3 \alpha}{\cos \alpha-\cos 3 \alpha}
$$

The function on the right-hand side is a continuous function of $\alpha$ (wherever it is defined, i.e., wherever $\cos \alpha \neq 0$, $\pm 1$). Thus, the limit as $\alpha \rightarrow 60^{\circ}$ is equal to its value at $\alpha=60^{\circ}$, which is $\sqrt{3}$.

The tangent of the angle $\beta$ thus approaches the tangent of the $60^{\circ}$ angle as $\alpha \rightarrow 60^{\circ}$. Since the tangent function is monotonically increasing for $0<x<90^{\circ}$, for any $\varepsilon$ there exists a $\delta$ such that if $|60^{\circ}-x|<\varepsilon$, then $|\operatorname{tg} x-\sqrt{3}|<\delta$. Choose $\alpha$ close enough to $60^{\circ}$ so that $|\operatorname{tg} \beta-\sqrt{3}|<\delta$, then $|60^{\circ}-\beta|<\varepsilon$, so if $\alpha \rightarrow 60^{\circ}$, then $\beta \rightarrow 60^{\circ}$.

II. solution. Let the projection of $B$ onto $A O$ be $K$, the reflection of $B$ across $A O$ be $B^{\prime}$, and the intersection of the lines $G D$ and $B B^{\prime}$ be $H$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_52666a947bb77098e9d5g-1.jpg?height=426&width=686&top_left_y=2156&top_left_x=709)

Figure 2

Thus, $B K \| D E$, so from $F E=D E / 3$ it follows that $B$ is the point closer to $K$ that divides the segment $K H$ into a ratio of 1:2, and $B H=2 B K=B B^{\prime}=B D$, since the isosceles triangles $B B^{\prime} O$ and $B D O$ have the angle at $O$ equal to $2 \alpha$. Therefore, the triangle $B D H$ is isosceles, and the line $B O$ bisects the exterior angle $D B B^{\prime}$ at $B$ (the angle $D B O$ and the angle $B^{\prime} B O$ are both supplementary to $\alpha$), so $B O$ is parallel to the base $D H$. Consequently, for any position of $B$, $\angle O G D = \angle A O B = \alpha$, and if $\alpha \rightarrow 60^{\circ}$, then $\angle O G D \rightarrow 60^{\circ}$. (Erzsébet Párkány (Budapest, Berzsenyi D. Gymnasium, III. grade), Zoltán Gurszky (Vác, Sztáron S. Gymnasium, III. grade))

Remarks. 1. The relationship $\alpha=\beta$ could also have been proven based on (1). This relationship can also be proven based on the statement of Exercise 1322, from which our problem originates.

2. If $\alpha \rightarrow 60^{\circ}$, then the point $D$ approaches the point $D_{0}$ on the circle $k$ that is diametrically opposite to $A$, and with $D$, the points $E, F, G$ also approach $D_{0}$: the lengths of the sides of the triangle $D E G$ approach 0, and the triangle "collapses" onto the point $D_{0}$. (Mirror image at the bottom of Figure 1.) Therefore, the angle $O G D$ we are looking for also "disappears from our sight," and we do not see its limiting position because the two points defining the line $G D$ have the same limiting position, which is the point $D_{0}$.

To prevent this, let us draw a triangle $D^{*} E^{*} G^{*}$ similar to the triangle $D E G$ for the values of $\alpha \rightarrow 60^{\circ}$, where the length of the segment $D^{*} E^{*}$ is 1, independent of $\alpha$. If $G$ approaches $D_{0}$, the line $B G$ approaches the line $B_{0} D_{0}$, where $B_{0}$ is the limiting position of the point $B$ when $\alpha \rightarrow 60^{\circ}$, i.e., $\angle A O B_{0} = 60^{\circ}$. Since $\angle A D_{0} B_{0} = \angle A O B_{0} / 2 = 30^{\circ}$, the limiting value of the angle $E G F$ is $30^{\circ}$, so with fixed $E^{*}, D^{*}$, the point $G^{*}$ approaches a point $G_{0}^{*}$ such that in the triangle $E^{*} F^{*} G_{0}^{*}$, the angle at $E^{*}$ is a right angle, and the angle at $G_{0}^{*}$ is $30^{\circ}$ (where $F^{*}$ is the point closer to $E^{*}$ that divides the segment $E^{*} D^{*}$ into a ratio of 1:2). Thus, the triangle $E^{*} G_{0}^{*} D^{*}$ is half of an equilateral triangle, and $\angle E^{*} G_{0}^{*} D^{*} = 60^{\circ}$.

We only intended to illustrate the statement of the problem with the above reasoning; we did not prove the statements appearing in the reasoning, nor did we specify what the precise meaning of these statements is. However, we note that there is no obstacle to making this sketchy "proof" exact, and the concept of a limit can be extended to geometric figures.[^0]

[^0]: ${ }^{1}$ K. M. L. 4242 (1971) 68 . page.