$\mathrm{Az} A B C D$ quadrilateral has all four of its side lines tangent to a circle, the center of which is $O$. Express the sum of the angles $A O B$ and $C O D$ in terms of the angles of the quadrilateral, as well as the sum of the angles $B O C$ and $D O A$, in the following situations:

I. Each of the four side lines touches the circle at a point on the side segment.

II. Two side lines touch the circle at a point outside the side segment.

III. All four side lines touch the circle at a point outside the side segment.
I. If the point of tangency of each of the four sides of the quadrilateral with the circle lies on the side segment of the quadrilateral, then the circle is clearly inside the quadrilateral (1st figure), and the bisectors of the internal angles of the quadrilateral pass through $O$. Let the angles at vertices $A, B, C, D$ be $\alpha, \beta, \gamma, \delta$ respectively. Thus,

$$
\begin{gathered}
A O B \varangle + C O D \varangle = 180^{\circ} - O A B \varangle - O B A \varangle + \left(180^{\circ} - O C D \varangle - O D C \varangle\right) = \\
= 360^{\circ} - \frac{\alpha + \beta + \gamma + \delta}{2} = 180^{\circ},
\end{gathered}
$$

since the sum of the angles of the quadrilateral is $360^{\circ}$.

The same value is obtained for the sum of the angles $B O C$ and $D O A$ by similar calculation, or we can also consider that the sum of the four angles in question is $360^{\circ}$, so the sum of the other two angles is again $180^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_4f448013334490fa15f2g-1.jpg?height=431&width=423&top_left_y=771&top_left_x=474)

1st figure

![](https://cdn.mathpix.com/cropped/2024_05_02_4f448013334490fa15f2g-1.jpg?height=688&width=751&top_left_y=632&top_left_x=877)

3rd figure

II. For the second type of configuration, we naturally include that the points of tangency of the other two sides are on the side segments. There are two possibilities for the mutual position of the sides that are tangent to the circle at points outside the side segments: the corresponding sides of the quadrilateral are adjacent (2nd and 3rd figures, $A B$ and $B C$), or they are opposite (4th figure, $A B$ and $C D$).

In the case of the 2nd figure, $A$ and $C$ are on the tangent segments $B E_{1}$ and $B E_{2}$ drawn from $B$, and $D$ is on the extension of the tangent segments $D E_{4}$ and $D E_{3}$ drawn from $D$, making the quadrilateral concave with the angle at $D$ greater than $180^{\circ}$. Here, the size of the angle domain containing the circle between the tangent segments drawn from $A$ is $180^{\circ} - \alpha$; similarly,

$$
E_{2} C E_{3} \varangle = 180^{\circ} - \gamma, \quad E_{3} D E_{4} \varangle = 360^{\circ} - \delta,
$$

and the rays $A O, C O, D O$ bisect these angles, respectively.

The angles in question are:

$$
\begin{gathered}
A O B \varangle = O A E_{1} \varangle - O B E_{1} \varangle = \frac{180^{\circ} - \alpha - \beta}{2}; \\
C O D \varangle = 180^{\circ} - \left(O C E_{3} \varangle + O D E_{3} \varangle\right) = 180^{\circ} - \left(\frac{180^{\circ} - \gamma + 360^{\circ} - \delta}{2}\right) = \\
= \frac{\gamma + \delta}{2} - 90^{\circ},
\end{gathered}
$$

their sum is $\gamma + \delta - 180^{\circ}$, or $180^{\circ} - \alpha - \beta$, since for a concave quadrilateral, $\alpha + \beta + \gamma + \delta = 360^{\circ}$ also holds. Furthermore, similarly,

$$
B O C \varangle + D O A \varangle = 180^{\circ} - \beta - \gamma = \alpha + \delta - 180^{\circ}
$$

The 3rd figure differs from the 2nd only in that $B$ is on the tangent segments $A E_{1}$ and $C E_{2}$, and the concave angle of the quadrilateral is at $B$. Thus,

$$
\begin{aligned}
A O B \varangle + C O D \varangle & = \left(180^{\circ} - \frac{\alpha}{2} - \frac{\beta}{2}\right) + \left(180^{\circ} - \frac{\gamma}{2} - \frac{\delta}{2}\right) = \\
& = 180^{\circ} = B O C \varangle + D O C \varangle .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_4f448013334490fa15f2g-2.jpg?height=762&width=1290&top_left_y=123&top_left_x=407)

The 4th figure results in a cyclic quadrilateral. We do not deal with this case, as the meaning of the angles of the quadrilateral becomes uncertain here.

III. In the last configuration to be examined, in the case of the 5th figure, the angle domain containing the circle at vertex $B$ is the $\beta$ angle domain, at $D$ it is the $\delta$ vertex angle domain, and at $A$ and $C$ it is the supplementary angle domain of $\alpha$ and $\gamma$, respectively. Thus,

$$
\begin{aligned}
A O B \varangle + C O D \varangle & = \left(O A E_{1} \varangle - O B E_{1} \varangle\right) + \left(O D E_{3} \varangle - O C E_{3} \varangle\right) = \\
& = 180^{\circ} - \alpha - \beta, \\
B O C \varangle + D O A \varangle & = \left(O C E_{2} \varangle - O B E_{2} \varangle\right) + \left(O D E_{4} \varangle - O A E_{4} \varangle\right) = \\
& = 180^{\circ} - \beta - \gamma .
\end{aligned}
$$

László Palla (Budapest, Piarist Secondary School) Béla Vizzvári (Budapest, Berzsenyi Dániel Secondary School)

Remark. The problem aimed to draw the readers' attention to the fact that theorems can slightly change with the change in the situation, and also that a circle tangent to each side of a quadrilateral can be outside the quadrilateral, similar to the case of a triangle.

Combining the results, we can say that if the circle is inside the quadrilateral (1st and 3rd figures), then the sums in question are equal. Furthermore, in the situations of the 2nd and 5th figures, the same expression was obtained for the two sums. In these situations, it could have been noticed that instead of the previous beautiful relationship, the differences $A O B \varangle - C O D \varangle$ and $B O C \varangle - D O A \varangle$ are equal, with their common value being $0^{\circ}$, i.e., $A O B \varangle = C O D \varangle$, and $B O C \varangle = D O A \varangle$. The proof of this is left to the reader.

We also recommend that readers examine, for the sake of completeness, whether our figures cover all possible configurations.