On a plane, there are 3 consecutive points on a line: $A, C, B$. $AB = a$, $AC = 2x$, and the midpoint of $AC$ is denoted as $\mathrm{O}$. Draw a semicircle around $\mathrm{O}$ with radius $OA = OC = x$. The point of tangency from $B$ to the semicircle is denoted as $D$, and the intersection of this tangent and the perpendicular line from $A$ to $AB$ is denoted as $E$. The perpendicular projection of $D$ onto $AB$ is denoted as $H$.

a) Calculate the lengths of the segments $BO, BD, AE, BE, OH$, and $AH$ as functions of $a$ and $x$.

b) Rotate the entire diagram around the $AB$ axis. Calculate the surface areas $F_{1}$ and $F_{2}$ of the surfaces of revolution formed by the segment $BE$ and the arc $AD$, respectively.

c) Determine $x$ such that $\frac{F_{2}}{F_{1}} = m$, where $m$ is a given positive number. Discuss. What is $x$ if $m = \frac{1}{2}$?
a) $B O=A B-A O=a-x$,

$$
B D=\sqrt{B A \cdot B C}=\sqrt{a(a-2 x)}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_0117dae04797d9d902b4g-1.jpg?height=295&width=418&top_left_y=259&top_left_x=859)

Since the right triangles $A E B$ and $D O B$ share the angle at vertex $B$,

$$
A E B_{\triangle} \sim D O B_{\triangle}
$$

and thus

$$
A E: A B=D O: D B
$$

which gives

$$
A E=\frac{A B \cdot D O}{D B}=\frac{a x}{\sqrt{a(a-2 x)}}
$$

Furthermore,

$$
B E: B A=B O: B D
$$

from which

$$
B E=\frac{B A \cdot B O}{B D}=\frac{a(a-x)}{\sqrt{a(a-2 x)}}
$$

In the right triangle $O D B$,

$$
O D^{2}=O H \cdot O B
$$

which gives

$$
O H=\frac{O D^{2}}{O B}=\frac{x^{2}}{a-x}
$$

Finally,

$$
A H=A O+O H=x+\frac{x^{2}}{a-x}=\frac{a x}{a-x}
$$

b) The segment $B E$, when rotated around $A B$, describes the surface of a cone, whose surface area is

$$
F_{1}=A E \cdot \pi \cdot B E=\frac{a x}{\sqrt{a(a-2 x)}} \cdot \frac{a(a-x)}{\sqrt{a(a-2 x)}} \cdot \pi=\frac{a x(a-x) \pi}{a-2 x}
$$

The surface area described by the arc $A D$ when rotated is

$$
F_{2}=2 O D \cdot \pi \cdot A H=2 x \pi \frac{a x}{a-x}=\frac{2 a x^{2} \pi}{a-x}
$$

c) $\frac{F_{2}}{F_{1}}=\frac{2 a x^{2} \pi}{a-x} \cdot \frac{a-2 x}{a x(a-x) \pi}=\frac{2 x(a-2 x)}{(a-x)^{2}}=m$.

Thus, we have obtained the following quadratic equation for $x$:

$$
(m+4) x^{2}-2 a(m+1) x+m a^{2}=0
$$

We note that, by its interpretation, $x$ can only take values between 0 and $\frac{a}{2}$.

The discriminant of our equation is

$$
D=4 a^{2}(m+1)^{2}-4 m a^{2}(m+4)=4 a^{2}(1-2 m)
$$

which is non-negative if

$$
1-2 m \geq 0
$$

or

$$
m \leq \frac{1}{2}
$$

If $0 < m \leq \frac{1}{2}$, the equation has a unique solution. If $m > \frac{1}{2}$, there is no solution.

Grätzer György (Bp., VI., Kölcsey g. III. o. t.)