Let's construct the deltoid if the two unequal sides $a$ and $b$, and the radius of the inscribed circle $\varrho$ are given.
I. Solution: Imagine the problem is solved. The labeling is shown in Figure 1.

![](https://cdn.mathpix.com/cropped/2024_05_02_5115713cb34360d836ddg-1.jpg?height=767&width=705&top_left_y=248&top_left_x=688)

Figure 1

In the kite $ABCD$, draw a line through the center of the inscribed circle, $O$, parallel to the side $BC = b$. Let this parallel line intersect the side $AB = a$ at point $E$. The line $BO$ is clearly the angle bisector of $\angle ABC = \beta$. Therefore, $\angle ABO = \angle OBC = \frac{\beta}{2}$. Since $\angle OBC$, as an alternate interior angle, is equal to $\angle BOE$, the triangle $\triangle EBO$ is isosceles: $EB = EO$. Since $OE \parallel CB$, it follows that $AE : EB = AO : OC$. According to the well-known angle bisector theorem, $AO : OC = a : b$, so $AE : EB = a : b$. Therefore, the point $E$ can be constructed by simple proportional division.

The geometric locus of point $O$ is, on the one hand, a line parallel to $AB$ at a distance $\rho$ from $AB$, and on the other hand, a circle with radius $EB$ drawn around point $E$.

Since $EB : a = b : (a + b)$, it follows that $EB = \frac{ab}{a + b}$, and thus there are 2, 1, or 0 solutions depending on whether $EB = \frac{ab}{a + b} \gtreqless \rho$.

In the case of the second solution, the kite is concave (reflex). In this case, the inscribed circle touches the extensions of the sides enclosing the reflex angle $\left(C^* \angle > 180^\circ\right)$.

József Pergel (Bp. XIX., Landler Jenő g. IV. o. t.)

II. Solution: The labeling is shown in Figure 2.

![](https://cdn.mathpix.com/cropped/2024_05_02_5115713cb34360d836ddg-1.jpg?height=539&width=764&top_left_y=1747&top_left_x=667)

The area of the kite is clearly twice the area of $\triangle ABC$. The latter is, on the one hand, $a\rho + b\rho$, and on the other hand, $am$, where $m$ is the height corresponding to the side $a$. Therefore,

$$
am = (a + b)\rho,
$$

or

$$
a : \rho = (a + b) : m
$$

and thus $m$ can be constructed as the fourth proportional. With $m$ known, the construction of $\triangle ABC$ and thus the kite with the given sides $a$ and $b$ is trivial.

The number of solutions is 2, 1, or 0, depending on whether $m = \frac{(a + b)\rho}{a} \oint b$ or $\rho \oint \frac{ab}{a + b}$.

The construction process: We mark the distance $AB = a$ and extend it by the segment $BE = b$. On the perpendicular to $AB$ at point $B$, we measure the distance $BF = \rho$. The intersection of line $AF$ with the perpendicular to $AE$ at point $E$ is $C'$. $C'E : (a + b) = \rho : a$, so

$$
C'E = \frac{(a + b)}{a} \rho = m
$$

Pál Bártfai (Bp. I., Petőfi g. II. o. t.)