A convex $ABCD$ quadrilateral has sides $AB=a=20$ units, $BC=b=16$, $CD=c=14$, $DA=d=8$, and the distance between the midpoints $A_{0}$, $C_{0}$ of sides $AB$, $CD$ is $A_{0}C_{0}=e=11$ units. Calculate the distance between the midpoints of sides $BC$ and $DA$, and the distance between the midpoints of segments $AA_{0}$ and $DC_{0}$.
1. Let the midpoints of segments $B C, D A, A A_{0}, D C_{0}$ be denoted by $B_{0}$, $D_{0}$, $A_{1}$, and $C_{1}$, respectively. The desired calculation is based on the procedure we would use to construct the quadrilateral from the prescribed lengths of the sides and the midline $A_{0} C_{0}$. 1

Translate the vector $D A$ to the positions $B P$ and $C Q$, and the vector $D C$ to the position $B R$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_66085508aeba9c7c3c08g-1.jpg?height=631&width=824&top_left_y=346&top_left_x=629)

Figure 1

Then $A D B P$ is a parallelogram, and since $A_{0}$ bisects its diagonal $A B$, it also bisects $D P$. Thus, $A_{0} C_{0}$ is the midline of triangle $C P D$ parallel to $C P$, so $C P=2 \cdot A_{0} C_{0}=2 e$. Therefore, the triangle $P B C$ can be constructed from its known sides. This triangle is completed to a parallelogram by the point $Q$ as the vertex opposite to $B$. Furthermore, $A D C Q$ is also a parallelogram, so $Q A=C D=c$, and point $A$ can be determined from the known distances from $B$ and $Q$. Finally, $D$ completes the parallelogram $A Q C$, thus completing the construction.

2. In the prescribed calculation, we also use the fact that $B D C R$ is a parallelogram, so its diagonal $D R$ passes through the midpoint $B_{0}$ of $B C$ and $D R=2 \cdot D B_{0}$. Therefore, $D_{0} B_{0}$, as the midline of triangle $R A D$ parallel to $R A$, is half of $R A$, and finally, $R A$ is also a diagonal of the parallelogram $A B R Q$.

By applying the Pythagorean theorem twice, it follows that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its four sides. Based on this,

$$
\begin{gathered}
B Q^{2}=2\left(B C^{2}+B P^{2}\right)-C P^{2}=2 b^{2}+2 d^{2}-4 e^{2} \\
B_{0} D_{0}^{2}=\frac{A R^{2}}{4}=\frac{1}{4}\left\{2\left(B A^{2}+B R^{2}\right)-B Q^{2}\right\}=e^{2}+\frac{a^{2}+c^{2}}{2}-\frac{b^{2}+d^{2}}{2}
\end{gathered}
$$

With our numerical data, $B_{0} D_{0}=f=\sqrt{259}=16.1$ units (the numerical data and the intermediate value of $B Q$ satisfy the required inequalities for the constructibility of triangles $B C P$ and $B Q A$).

3. Before proceeding further, it is useful to express our result (1) in a more symmetric form:

$$
f^{2}-\frac{a^{2}+c^{2}}{2}=e^{2}-\frac{b^{2}+d^{2}}{2}
$$

In the illustrative Figure 2, we also note that the midlines $e$ and $f$ bisect each other, as the endpoints of these midlines are the vertices of a parallelogram.

![](https://cdn.mathpix.com/cropped/2024_05_02_66085508aeba9c7c3c08g-1.jpg?height=540&width=525&top_left_y=2050&top_left_x=781)

Figure 2[^0]
4. According to all of the above, the second question differs from the first only in the numerical data: the consecutive sides of the quadrilateral $A D C_{0} A_{0}$ are $d, \frac{c}{2}, e, \frac{a}{2}$, and the midline between the midpoints of sides $d$ and $e$ is $D_{0} E=\frac{D_{0} B_{0}}{2}=\sqrt{64.75}$, and the other midline needs to be calculated. Applying (1) appropriately:

$$
A_{1} C_{1}^{2}=D_{0} E^{2}+\frac{A D^{2}+A_{0} C_{0}^{2}}{2}-\frac{A A_{0}^{2}+D C_{0}^{2}}{2}=82.75, \quad A_{1} C_{1}=9.1 \text { units. }
$$

Hasenfratz Anna (Budapest, Fazekas M. Gyak. Gimn., II. o. t.) Szecsői Sándor (Budapest, Berzsenyi D. Gimn., II. o. t.)


[^0]:    ${ }^{1}$ See: Horvay-Pálmay: Matematika a gimn. ... I. o. számára, 4. kiadás, Tankönyvkiadó, Budapest, 1969, 338. o. 24. d/feladat.