Consider the convex quadrilateral $A B C D$ (i.e., one where each angle is less than $180^{\circ}$). Let two opposite angles be $\alpha$ and $\gamma$. What is the acute angle formed by the angle bisectors of the angles at vertices $B$ and $D$?
I. Solution: Let us assume that among the given angles, $\alpha$ is the smaller one: $\alpha < \gamma$. On the other hand, $\varepsilon_{1} + \varepsilon_{2} = 360^{\circ}$, so $\varepsilon_{1} > 180^{\circ}$ and $\varepsilon_{2} < 180^{\circ}$. This means that $\varepsilon_{1}$ gives the sum of three of the four angles formed at point $E$, while $\varepsilon_{2}$ is the magnitude of an angle within which there are no bisectors.

We will show that $\varepsilon_{2}$ is an obtuse angle, and thus the required acute angle, $\zeta = 180^{\circ} - \varepsilon_{2}$. Indeed, from $\beta + \delta = 360^{\circ} - (\alpha + \gamma)$, we have:

\[
\varepsilon_{2} = 360^{\circ} - \gamma - \frac{\beta + \delta}{2} = 360^{\circ} - \gamma - 180^{\circ} + \frac{\alpha + \gamma}{2} = 180^{\circ} - \frac{\gamma - \alpha}{2}
\]

Due to the convexity of our quadrilateral, $\gamma$ and $\alpha$ are less than $180^{\circ}$, and the same holds for their difference. Therefore, the subtrahend in the last form of $\varepsilon_{2}$ is less than $90^{\circ}$, making $\varepsilon_{2}$ indeed greater than $90^{\circ}$. Thus, the required acute angle is: $\zeta = (\gamma - \alpha) / 2$.

Cases where the intersection point $E$ of $f_{b}$ and $f_{d}$ lies on the perimeter of the quadrilateral or outside it can be reduced to the already examined case. The point $E$ can only lie on the perimeter of the quadrilateral if it coincides with $B$ or $D$, because on the lines $B A$ and $B C$, $f_{b}$ has no other points than $B$, and on the lines $D A$ and $D C$, $f_{d}$ has no other points than $D$. Let $E$ be at $B$, or on the outward extension of $f_{b}$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_503169f760883127a48eg-1.jpg?height=561&width=645&top_left_y=1633&top_left_x=726)

Figure 2

In this case, by drawing a line through a point $B'$ on the outward extension of $f_{b}$ (beyond $E$) parallel to $B A$ and $B C$, and denoting the intersection points of these lines with $D A$ and $D C$ as $A'$ and $C'$ respectively, the angles at $B'$ and $D$ in the quadrilateral $A' B' C' D'$ are bisected by $f_{b}$ and $f_{d}$, and these bisectors intersect inside the new quadrilateral. Therefore, according to our previous result, $\zeta = (\gamma' - \alpha') / 2$, and this is equal to $(\gamma - \alpha) / 2$, because the consecutive angles of the quadrilaterals $A' B' C' D$ and $A B C D$ are corresponding and thus equal.

The required angle does not exist if $f_{b}$ and $f_{d}$ are parallel or coincident. This only happens if $\alpha = \gamma$. Let $f_{b}$ and $f_{d}$ intersect the lines $A D$ and $B C$ at points $D'$ and $B'$ respectively (Figure 3).

![](https://cdn.mathpix.com/cropped/2024_05_02_503169f760883127a48eg-2.jpg?height=385&width=422&top_left_y=133&top_left_x=838)

Figure 3

If $f_{b}$ and $f_{d}$ are identical, then $D' \equiv D$ and $B' \equiv B$, and the quadrilateral $A B C D$ is a kite, because the bisectors make $B A$ and $B C$, as well as $D A$ and $D C$, mirror images of each other with respect to $f_{b}$, and thus $\alpha = \gamma$. Otherwise, $\angle C B' D = \angle C B D' = \beta / 2$ and $\angle A D' B = \angle A D B' = \delta / 2$, because they are corresponding angles, so the triangles $B' C D$ and $B A D'$ have two equal angles, and thus their third angles are also equal, as stated. Conversely, if $\alpha = \gamma$, then $f_{b}$ and $f_{d}$ are coincident or parallel. This condition contradicts the assumption $\alpha < \gamma$ used in the above determination of $\zeta$, so the value of $\zeta$ obtained there cannot be used. However, using the above value for the case $\alpha = \gamma$ does not lead to an error, because it gives $\zeta = 0^{\circ}$, which we do not accept as the angle between two lines.

II. Solution: There is no need to distinguish cases based on the position of $E$ if we consider the angle not as a fixed position but as a rotation between directions. Translate the rays $C B$ and $C D$ to $A$ at point $C$, so their new positions are $A B' (\| C B)$ and $A D' (\| C D)$ (Figure 4).

![](https://cdn.mathpix.com/cropped/2024_05_02_503169f760883127a48eg-2.jpg?height=525&width=420&top_left_y=1147&top_left_x=836)

Figure 4

Then the (less than a straight angle) $\angle B' A D'$ is equal to $\angle B C D = \gamma$ because they are corresponding angles, and $\angle B A B'$ is equal to $\angle A B C = \beta$, and $\angle D A D'$ is equal to $\angle A D C = \delta$ because they are alternate interior angles, so these are the four angles of our quadrilateral at point $A$. Let $f_{1}$ and $f_{2}$ be the rays of the lines parallel to $f_{b}$ and $f_{d}$ through $A$ that lie in the sectors of the angles $\angle B' A B$ and $\angle D A D'$, respectively. The angles between $f_{b}$ and $f_{d}$ are equal to the angles between $f_{1}$ and $f_{2}$. The latter bisect the angles $\angle B' A B$ and $\angle D A D'$, because, for example, the angle between $f_{1}$ and $A B$ is the alternate interior angle to the angle between $f_{b}$ and $B A$, which is $\beta / 2$. Now, rotating from $f_{1}$ to $f_{2}$ in one direction, we have $\vartheta_{1} = \beta / 2 + \alpha + \delta / 2 = \alpha + (\beta + \delta) / 2$, and in the other direction, $\vartheta_{2} = \beta / 2 + \gamma + \delta / 2 = \gamma + (\beta + \delta) / 2$. Furthermore, if $\alpha = \gamma$, then $\vartheta_{1} = \vartheta_{2} = 360^{\circ} / 2 = 180^{\circ}$, so $f_{1}$ and $f_{2}$ are extensions of each other, meaning $f_{b}$ and $f_{d}$ are parallel or coincident. If $\alpha < \gamma$, then $\vartheta_{1} < \vartheta_{2}$, so the usual angle between $f_{1}$ and $f_{2}$ is $\vartheta_{1}$. This can be written as $(\alpha + \beta + \delta) / 2 + \alpha / 2 = 180^{\circ} - \gamma / 2 + \alpha / 2$, and from the above, we see that it is an obtuse angle, so its supplementary angle is required, $\zeta = 180^{\circ} - \vartheta_{1} = (\gamma - \alpha) / 2$.

Ferenc Bácskai (Kecskemét, Piarist High School, 2nd grade)

Note. We only accepted complete solutions that at least alluded to the fact that the angle bisectors can intersect outside the quadrilateral or how the size relationship between $\alpha$ and $\gamma$ affects the final result. Solutions that did not address either of these questions were considered incomplete.