Problem 976

I. Solution. Let $ABCD=N$ be the quadrilateral that satisfies the conditions (part $a$ of Figure 1), where the angles $\angle BAD$ and $\angle CBA$ are equal to the prescribed angles $\alpha$ and $\beta$ at the vertices, the diagonals $AC$ and $BD$ are equal to the prescribed segments $e$ and $f$, and the angle $\angle AMB$ at the intersection point $M$ of the diagonals is the given angle $\varepsilon$.

!

Figure 1

Translate the triangle $ABC$ such that $A$ moves to $C$, and let the new position of $BD$ be $EF$. Then the quadrilateral $BEFD=P$ and the quadrilateral $BEC A$ are parallelograms. In $P$, the sides $BE=AC=e$ and $BD=f$, and the angle $\angle DBE=\varepsilon$. In the latter, the segment $BC$ is a diagonal. Thus, $\angle ECB=\angle ABC=\beta$, alternate interior angles, and $\angle ECF=\angle BAD=\alpha$, corresponding angles.

Based on this, we construct the triangle $FED$ from the data $e, f, \varepsilon$ and complete it to form $P$. We construct the circumcircle $i_1$ with angle $\alpha$ over $EF$ and the circumcircle $i_2$ with angle $\beta$ over $BE$. The (other than $E$) intersection point of these circles is the vertex $C$; finally, we translate the triangle $CEF$ such that $E$ moves to $B$, then the new position of $C$ is the vertex $A$.

The quadrilateral $ABCD$ meets the requirements because the measured data are in the prescribed positions due to the translation.

We construct $i_1$ and $i_2$ over the sides of $P$ on the side where $D$ is, unless $\alpha > 180^\circ$, in which case we construct $i_1$ on the other side of $EF$ with the supplementary angle $360^\circ - \alpha$ (if $\varepsilon > 180^\circ$, then $\varepsilon > \alpha + \beta - 180^\circ$).

The convex, concave, or crossed nature of $N$ is determined by the position of $C$ relative to $P$. A crossed solution is not acceptable because the perimeter of a crossed quadrilateral does not allow the distinction between the interior and exterior, and its angles cannot be interpreted in the usual way.

Halek Tibor (Budapest, Berzsenyi D. Gym.)

Remarks. 1. The midpoints of the consecutive sides of the quadrilateral form the consecutive vertices of a parallelogram, whose sides are half the diagonals and whose angles are equal to the angles between the diagonals. The parallelogram $A_1B_1C_1D_1$ can be constructed from our data (let $A_1$ be the midpoint of $AB$, and so on, then $\angle B_1A_1D_1 = \angle AMB = \varepsilon$, part $b$ of the figure). Thus, the vertex $A$ lies on the circumcircle $i_1$ with angle $\alpha$ over $A_1D_1$, and $B$ lies on the circumcircle $i_2$ with angle $360^\circ - \beta$ over $A_1B_1$. Since $B$ is the reflection of $A$ over $A_1$, it also lies on the reflection $i_1'$ of $i_1$ over $A_1$, so $B$ is given by the intersection of $i_2$ and $i_1'$ (other than $A_1$). Then the reflection of $B$ over $A_1$ is $A$, the reflection of $B_1$ over $C$ is $C$, and the reflection of $A$ over $D_1$ is $D$. - This solution is essentially a scaled-down version of the above solution.

Orbán Gábor (Makó, József A. Gym.)

2. If $\beta$ were replaced by the angle $\angle BCD = \gamma$, then the circumcircle $i_3$ with angle $\gamma$ over $B_1C_1$ would intersect the reflection $i_1''$ of $i_1'$ over $B_1$ to give the vertex $C$ first. It is easy to show that $i_1''$ can also be obtained by translating $i_1$ by a distance $e$ in the direction of $A_1B_1$.[^0]

!

Figure 2

II. Solution. The vertex $B$ lies on the circumcircle $i_2$ with angle $\beta$ over the segment $AC$ (we can determine in advance on which side of the line $AC$), and $A$ lies on the circumcircle $i_1$ with angle $\alpha$ over the segment $BD$. Fix the position of the segment $AC$ (Figure 2), then we need to place $i_1$ appropriately. For this, we construct the center $K$ from the fact that $BD=f$ and $\alpha$ determine the radius $r_1$ of the arc and the triangle $BDK$, and thus the angle $\angle BDK$. Therefore, $K$ must lie on the circle $k_a$ with radius $r_1$ centered at $A$; while $B$ runs through $i_2$, the direction of the segment $BD$ and thus the direction and length of $BK$ are known in every position, such as $B^*D^*$ and $K^*$. Therefore, $K$ runs along a circle $i'$ congruent to $i_2$ and obtained by translating $i_2$ in the direction and by the length of $B^*K^*$. Therefore, $K$ is the intersection point (other than $A'$) of the circle $k_a$ and the arc $i'$. Then $B$ is the intersection of the line through $K$ parallel to $K^*B^*$ with $i_2$. The discussion of the construction is left to the reader.

[^0]: ${ }^{1} \mathrm{In}$ part $b$ of the figure, the lower right $C_{1}$ should correctly be $C$.