In the quadrilateral $A B C D$, the diagonals $A C$ and $B D$ are perpendicular to each other, and the consecutive angles $\alpha, \beta, \gamma, \delta$ form an arithmetic sequence in the order $\alpha, \beta, \delta, \gamma$. The same applies to the consecutive angles of the triangle $A B D$. Calculate the sides, the diagonals, and the area if $A B=a=1$.
In a four-term arithmetic sequence, the sum of the two extreme and two intermediate terms is equal:

$$
\alpha+\gamma=\beta+\delta
$$

thus, our quadrilateral is a cyclic quadrilateral. Let the parts of the angles divided by the diagonals be $D A C=\alpha_{1}, C A B=\alpha_{2}, A B D=\beta_{1}, D B C=\beta_{2}, B C A=\gamma_{1}, A C D=\gamma_{2}, C D B=\delta_{1}, B D A=\delta_{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_d1b9d128945ba7a9f61fg-1.jpg?height=529&width=553&top_left_y=397&top_left_x=775)

Thus, since the sum of a three-term arithmetic sequence is equal to three times the middle term, from triangle $A B D$, $\beta_{1}=60^{\circ}$. Due to the cyclic nature of the quadrilateral and the perpendicularity of the diagonals, $\gamma_{2}=60^{\circ}, \delta_{1}=\alpha_{2}=30^{\circ}$, and $\beta_{2}=\alpha_{1}$. Therefore, the difference of the four-term sequence $\varepsilon=\beta-\alpha=\left(\beta_{1}+\beta_{2}\right)-\left(\alpha_{1}+\alpha_{2}\right)=60^{\circ}-30^{\circ}+\left(\beta_{2}-\alpha_{1}\right)=30^{\circ}$, and from (1) and the sum of the angles of the quadrilateral, $\alpha+\gamma=\alpha+(\alpha+3)=2 \alpha+90^{\circ}=180^{\circ}$, so $\alpha=45^{\circ}, \beta=75^{\circ}, \delta=105^{\circ}$, $\gamma=135^{\circ}$, and $\alpha_{1}=\beta_{2}=15^{\circ}, \gamma_{1}=\delta_{2}=75^{\circ}$. Thus, triangle $A B C$ is isosceles, $A C=A B=1$.

Now, using the Law of Sines repeatedly, the other sides and diagonals are:

from triangle $A B C$, $B C=\sin 30^{\circ}: \sin 75^{\circ} \approx 0.5176$;

from triangle $A B D$, $A D=\sin 60^{\circ}: \sin 75^{\circ} \approx 0.8966$;

from triangle $A B D$, $B D=\sin 45^{\circ}: \sin 75^{\circ} \approx 0.7321$;

from triangle $A C D$, $C D=\sin 15^{\circ}: \sin 105^{\circ} \approx 0.2679$ units,

and finally, from the product of the diagonals being half, the area of the quadrilateral $t=B D / 2=0.3660$ square units.

Máté Zsolt (Szeged, Radnóti M. g. IV. o. t.)

Remarks. 1. If we consider the values of

$$
\begin{aligned}
& \sin 75^{\circ}=\sin \left(45^{\circ}+30^{\circ}\right)=\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \cdot \frac{1}{2}=\frac{\sqrt{3}+1}{2 \sqrt{2}} \text { and } \\
& \sin 15^{\circ}=\sin \left(45^{\circ}-30^{\circ}\right)=\frac{\sqrt{3}-1}{2 \sqrt{2}}
\end{aligned}
$$

known from the addition formula of the sine function, then we can write our results in an exact, "closed" form without using decimal approximations:

$$
\begin{array}{ll}
B C=(\sqrt{6}-\sqrt{2}) / 2, & A D=(3 \sqrt{2}-\sqrt{6}) / 2, \quad B D=\sqrt{3}-1 \\
C D=2-\sqrt{3} \quad \text { and } \quad t=(\sqrt{3}-1) / 2
\end{array}
$$

Fejes László (Makó, József A. g. IV. o. t.)

2. We can also arrive at these results without trigonometric knowledge, using only repeated application of the Pythagorean theorem and some ingenuity. Denoting the intersection of the diagonals by $M$, from the right triangle $A B M$, $B M=1 / 2, A M=\sqrt{3} / 2$. Thus, $M C=A C-A M=(2-\sqrt{3}) / 2$, and from the right triangle $B C M$,

$$
B C=\sqrt{\frac{1}{4}+\frac{7-4 \sqrt{3}}{4}}=\sqrt{\frac{6-2 \sqrt{6 \cdot 2}+2}{4}}=\frac{\sqrt{6}-\sqrt{2}}{2}
$$

and from $C D M$, $C D=2 M C=2-\sqrt{3}$ and $M D=M C \sqrt{3}=(2 \sqrt{3}-3) / 2$, so $B D=B M+M D=\sqrt{3}-1$. Finally, from $D A M$,

$$
D A=\sqrt{\frac{3}{4}+\frac{21-12 \sqrt{3}}{4}}=\sqrt{\frac{18-2 \sqrt{18 \cdot 6}+6}{4}}=\frac{\sqrt{18}-\sqrt{6}}{2}
$$

Várady Gábor (Győr, Révai M. g. IV. o. t.)

3. Applying the thought process from the previous remark to any acute angle $\alpha$, we arrive at the valuable but less well-known

$$
\operatorname{tg} \frac{\alpha}{2}=\frac{1-\cos \alpha}{\sin \alpha}
$$

relation; indeed, $B M=\sin \alpha, A M=\cos \alpha, C M=1-\cos \alpha$, and $\angle C=\left(180^{\circ}-\alpha\right) / 2=90^{\circ}-\alpha / 2$ and thus $\angle C B M=\alpha / 2$.

![](https://cdn.mathpix.com/cropped/2024_05_02_d1b9d128945ba7a9f61fg-2.jpg?height=472&width=526&top_left_y=496&top_left_x=797)

4. Some solutions also worked out the cases $\alpha=60^{\circ}, \beta=80^{\circ}$, and $\alpha=67.5^{\circ}$, $\beta=82.5^{\circ}$ as the second and third solutions, respectively, by placing the $60^{\circ}$ angle of triangle $A B D$ at $A$ or $D$. However, the text does not state that the angles of triangle $A B D$ form an arithmetic sequence in any order, but rather that this applies to consecutive angles, i.e., the angles at $A, B$, and $D$.