\section*{Problem 4 - 071024}

On a flat table, there are 4 wooden spheres, each with a radius of length \(r\), touching each other such that their contact points with the tabletop form the corners of a square.

A fifth wooden sphere of the same radius is placed in the resulting central gap.

Give the distance \(d\) from the highest point of this fifth sphere to the tabletop!
}

The square \(A B C D\) formed by the contact points (see plan view) has a side length of \(2 r\) and a diagonal length of \(2 r \sqrt{2}\).

A vertical section perpendicular to the table plane, containing the diagonal \(A C\), results in the elevation view.

Since the triangle \(\Delta A^{\prime} C^{\prime} E^{\prime}\) is isosceles with side lengths

\[
A^{\prime} C^{\prime}=A C=2 r \sqrt{2} ; \quad A^{\prime} E^{\prime}=A E=C^{\prime} E^{\prime}=C E=2 r
\]

it is isosceles and right-angled; the perpendicular from \(E\) to \(A C\) consequently has a length of \(r \sqrt{2}\).

The sought distance \(d\) is therefore

\[
d=r+r \sqrt{2}+r=r(2+\sqrt{2})
\]

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1250.jpg?height=445&width=1091&top_left_y=1051&top_left_x=494)

Section axis

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1250.jpg?height=597&width=585&top_left_y=1592&top_left_x=587)

Plan view

Solutions of the 2nd Round 1967 adopted from \([5]\)

\subsection*{7.9.3 3rd Round 1967, Class 10}