Given a quadrilateral $A B C D$ with sides $A B$ and $A D$, the included angle $\alpha$, and $\beta=\delta=90^{\circ}$. Calculate the length of the diagonal $A C$ from the given data.
I. solution: Our quadrilateral is a cyclic quadrilateral because the sum of its opposite angles $\beta+\delta=180^{\circ}$, and the sought diagonal $AC$ is the diameter of the circumscribed circle, as $\beta$ and $\delta$ are the viewing angles of this diameter from points $B$ and $D$. For the further notations, see the diagram for all three solutions.

![](https://cdn.mathpix.com/cropped/2024_05_02_5f7d6f7f90a55f56aabag-1.jpg?height=550&width=567&top_left_y=305&top_left_x=779)

We calculate the diagonal $AC$ as the diameter of the circumcircle of the triangle $ABD$, determined by the given data, from the right triangle $A^{\prime}BD$:

$$
AC=2r=\frac{f}{\sin \alpha}=\frac{\sqrt{a^{2}+d^{2}-2ad \cos \alpha}}{\sin \alpha}
$$

(It can be shown that the used relationship is also valid for $\alpha \geqq 90^{\circ}$.)

$$
\text{Elbert Árpád (Kaposvár, public school IV. grade)}
$$

II. solution: First, let's determine the angle $\varepsilon$ that $AC$ makes with $AB$. From the right triangles $ABC$ and $ACD$,

$$
AC=\frac{a}{\cos \varepsilon}=\frac{d}{\cos (\alpha-\varepsilon)}
$$

From the second denominator, after expanding and rearranging, we get an equation where every term contains either $\sin \varepsilon$ or $\cos \varepsilon$. It is useful to express $\operatorname{tg} \varepsilon$ first:

$$
\operatorname{tg} \varepsilon=\frac{d-a \cos \alpha}{a \sin \alpha}
$$

From this, for the first expression in (2):

$$
\frac{1}{\cos \varepsilon}=\sqrt{\operatorname{tg}^{2} \varepsilon+1}=\frac{\sqrt{(d-a \cos \alpha)^{2}+a^{2} \sin ^{2} \alpha}}{a \sin \alpha}
$$

(The square root must be positive because $\varepsilon$ is certainly a positive acute angle.) Simplifying this and substituting it into (2) again leads us back to (1).

Gyôry Kálmán (Ózd, József Attila g. IV. grade)

III. solution: Ptolemy's theorem is valid for our quadrilateral:

$$
ac+bd=ef
$$

Here, the sides $c$ and $b$ are expressed with $e, a, d$ using the Pythagorean theorem, leading to this equation for $e$:

$$
a \sqrt{e^{2}-d^{2}}+d \sqrt{e^{2}-a^{2}}=ef
$$

Squaring, rearranging, and squaring again, the only positive root of the resulting equation is:

$$
e=\frac{2ad f}{\sqrt{4a^{2}d^{2}-\left(f^{2}-a^{2}-d^{2}\right)^{2}}}
$$

and this, using the already used expression for $f$, again leads to (1).

Remarks: Since we are clearly dealing with a convex quadrilateral, the triangles $ABD$ and $CBD$ can only be obtuse at $A$ or $C$. Accordingly, there are two solvability conditions:

$$
a \cos a<d \text{ and } d \cos \alpha<a \text{. }
$$

With the first condition satisfied, the intermediate result (3) of the second solution will also be positive, and the same holds for the second condition with the roles of $a$ and $d$ swapped in (3):

$$
\operatorname{tg}(\alpha-\varepsilon)=\frac{a-d \cos \alpha}{d \sin \alpha}
$$

For $\alpha \neq 90^{\circ}$, any pair of opposite sides of our quadrilateral can be extended to form a right triangle by attaching a smaller right triangle. The angle at the new vertex $E$ of this right triangle is $90^{\circ}-\alpha$ or $\alpha-90^{\circ}$, depending on whether $\alpha$ is acute or obtuse. Already at this stage, the hypotenuse $AE$ can be expressed, and by subtraction or addition, the leg $BE$ or $DE$, and finally the hypotenuse $CE$, can be calculated from the triangle $ACE$ to find $AC$. For $\alpha=90^{\circ}$, the problem simplifies to a rectangle calculation.

Elbert Árpád also submitted a coordinate geometry solution.