\section*{Problem 3B - 301043B}

In space, there are four points \(A, B, C, D\) such that the following distances occur between them: \(A B=7.2 \mathrm{~cm} ; A C=9.6 \mathrm{~cm} ; A D=15.6 \mathrm{~cm} ; B C=12.0 \mathrm{~cm} ; B D=15.6 \mathrm{~cm} ; C D=15.6 \mathrm{~cm}\).

Determine the radius \(r\) of the sphere on whose surface these four points lie!
}

Due to \(A B=3 \cdot 2.4 \mathrm{~cm} ; A C=4 \cdot 2.4 \mathrm{~cm} ; B C=5 \cdot 2.4 \mathrm{~cm}\) and \(3^{2}+4^{2}=5^{2}\), thus \(A B^{2}+A C^{2}=B C^{2}\), the triangle \(A B C\) is right-angled at \(A\) according to the converse of the Pythagorean theorem.

If, starting from the net \(A B C D_{1} D_{2} D_{3}\) of the tetrahedron \(A B C D\) (see figure), the triangles \(B C D_{1}, C A D_{2}\) and \(A B D_{3}\) are rotated around \(B C, C A\) and \(A B\) respectively, so that \(D_{1}, D_{2}\) and \(D_{3}\) meet at point \(D\), then \(D_{1}, D_{2}\) and \(D_{3}\) describe circular arcs \(k_{1}, k_{2}, k_{3}\).

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1586.jpg?height=728&width=828&top_left_y=1281&top_left_x=614)

Their images \(k_{1}^{\prime}, k_{2}^{\prime}, k_{3}^{\prime}\) under orthogonal projection onto the drawing plane are straight lines and perpendicular to \(B C, C A\) and \(A B\) respectively.

Since the triangles \(B C D_{1}, C A D_{2}, A B D_{3}\) are isosceles, \(k_{1}^{\prime}, k_{2}^{\prime}, k_{3}^{\prime}\) pass through the midpoints of \(B C, C A\) and \(A B\); i.e., they are the perpendicular bisectors of the triangle \(A B C\). Their intersection point, the image \(D^{\prime}\) of \(D\) under orthogonal projection onto the drawing plane, is thus the circumcenter of \(A B C\), and this is the midpoint of the hypotenuse \(B C\) according to the converse of Thales' theorem.

Since the center \(M\) of the sphere on whose surface \(A, B, C\) and \(D\) lie, in particular, has equal distances \(r\) to \(A, B\) and \(C\), it lies on the perpendicular erected at the circumcenter of \(A B C\) in the drawing plane; i.e., \(M\) lies on \(D^{\prime} D\) and is therefore the circumcenter of the triangle \(B C D\) (the figure shows its position \(M_{1}\) when the triangle \(B C D\) is rotated into the position \(B C D_{1}\).

Thus, and with \(B D^{\prime}=\frac{1}{2} B C=6 \mathrm{~cm}\), it follows from the Pythagorean theorem on the one hand that \(D^{\prime} D^{2}=B D^{2}-B D^{\prime 2}\), and since \(B D=13 \cdot 1.2 \mathrm{~cm} ; B D^{\prime}=5 \cdot 1.2 \mathrm{~cm}\) and \(13^{2}-5^{2}=12^{2}\), thus \(D D^{\prime}=12 \cdot 1.2 \mathrm{~cm}\); on the other hand, it follows that

\[
\left(D^{\prime} D-r\right)^{2} + B D^{\prime 2}=r^{2}, \quad\left(D^{\prime} D\right)^{2}+B D^{\prime 2}-2 r \cdot D^{\prime} D=0
\]

\[
r=\frac{B D^{2}}{2 D^{\prime} D}=\frac{(13 \cdot 1.2)^{2}}{24 \cdot 1.2} \mathrm{~cm}=8.45 \mathrm{~cm}
\]

(The formula \(r \cdot D^{\prime} D=\frac{B D}{2} \cdot B D\) also follows because \(M\) lies on the perpendicular bisector \(Q M\) of \(B D\) and thus \(\triangle M Q D \cong \triangle B D^{\prime} D\).)

\section*{Adapted from \([5]\)}