\section*{Problem 4 - 141024}

Given are positive lengths \(h, r, d\) with \(d < 2r\). Let \(\epsilon\) be a plane and \(k\) a circle in \(\epsilon\) with a diameter \(AB\) of length \(2r\).

On the perpendicular to \(\epsilon\) through \(A\), let \(C\) be a point with \(AC = h\). On \(k\), let \(D\) be a point with \(BD = d\).

a) Calculate the volume \(V\) of the pyramid with vertices \(C, D, A, B\)!

b) Prove that \(BD \perp CD\) holds!
}

We introduce an orthogonal Cartesian coordinate system as shown in the figure and place the center \(M\) of the circle \(k\) at the coordinate origin. The plane \(\epsilon\) is the \(x-y\) coordinate plane, so the circle \(k\) lies in it.

Then, according to the problem statement, the points \(A, B, C\) have the coordinates:

\[
A(0 ;-r ; 0) \quad ; \quad B(0 ; r ; 0) \quad ; \quad C(0 ;-r ; h)
\]

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1346.jpg?height=511&width=600&top_left_y=241&top_left_x=728)

\(D(x ; y)\) lies on the circle \(k\) with the circle equation \(x^{2}+y^{2}=r^{2} \quad\) (1). Its distance \(d\) from the point \(B\) is

\[
d=\sqrt{(0-x)^{2}+(r-y)^{2}+0^{2}}
\]

Substituting (1) rearranged for \(x^{2}\), and solving, we get the coordinate \(y\) of \(D\):

\[
d=\sqrt{r^{2}-y^{2}+(r-y)^{2}} \Rightarrow y=\frac{2 r^{2}-d^{2}}{2 r}
\]

For the \(x\)-coordinate of the point \(D\), we get

\[
x_{1 ; 2}= \pm \frac{d}{2 r} \sqrt{4 r^{2}-d^{2}}
\]

Without loss of generality, we choose the positive \(x_{1}\). For the distance \(A D=d^{\prime}\), we then get

\[
\begin{array}{r}
d^{\prime}=\sqrt{x^{2}+(-r-y)^{2}} \\
d^{\prime}=\sqrt{\left(\frac{d}{2 r} \sqrt{4 r^{2}-d^{2}}\right)^{2}+\left(-r-\frac{2 r^{2}-d^{2}}{2 r}\right)^{2}} \\
d^{\prime}=\sqrt{4 r^{2}-d^{2}}
\end{array}
\]

Without loss of generality, if we choose the positive value \(x_{1}\), the area of the base triangle of the desired pyramid \(A_{\Delta}=\frac{d \cdot d^{\prime}}{2}\), since the triangle is right-angled. The point \(D\) lies on the Thales circle of the diameter \(AB\). The volume of the pyramid is thus

\[
V=A_{\Delta} \cdot \frac{h}{3}=\frac{d \cdot h}{6} \sqrt{4 r^{2}-d^{2}}
\]

The segment \(C D\) is perpendicular to \(B D\) if the triangle \(C D B\) is right-angled, with the right angle at \(D\). Then, according to the Pythagorean theorem and its converse,

\[
C D \perp B D \quad \Leftrightarrow \quad C D^{2}+B D^{2}=C B^{2}
\]

For the squares of the three segments, we get by substituting the coordinates of \(D\) determined

\[
\begin{aligned}
C B^{2} & =4 r^{2}+h^{2} \\
D B^{2} & =d^{2} \\
C D^{2} & =\left(\frac{d}{2 r} \sqrt{4 r^{2}-d^{2}}\right)^{2}+\left(\frac{2 r^{2}-d^{2}}{2 r}+r\right)^{2}+h^{2} \\
& =-d^{2}+h^{2}+4 r^{2}
\end{aligned}
\]

From this, (2) follows immediately, and \(C D\) and \(B D\) are perpendicular to each other. w.z.b.w.