Two towers stand on a horizontal plane, the ratio of their heights being $\lambda$. A straight path passes through the midpoint of the segment connecting their bases and forms a $30^{\circ}$ angle with this segment. Determine the point on the path from which the two towers are seen at equal angles. Investigate how the number of solutions depends on the value of $\lambda$.
I. solution. It is sufficient to deal with the values $\lambda>1$, because this - if necessary - can be achieved by swapping the order of the two towers, and if $\lambda=1$, then obviously the midpoint $F$ of the segment between points $A$ and $B$ satisfies the condition, and no other point does. (The mentioned swap can also be achieved by a $180^{\circ}$ rotation around $F$, and thus the path remains unchanged.)

![](https://cdn.mathpix.com/cropped/2024_05_02_8f9470f0747f230db5bbg-1.jpg?height=428&width=1025&top_left_y=339&top_left_x=520)

Denoting the tops of the towers by $A^{*}$ and $B^{*}$, and a suitable point on the path by $U$, the right-angled triangles $A^{*} U A$ and $B^{*} U B$ are similar due to the equality of the angles of view (1st figure), hence

$$
U A : U B = A A^{*} : B B^{*} = \lambda, \quad U A > U B
$$

therefore $\angle U F B = 30^{\circ}$, $\angle U F A = 150^{\circ}$. Let $A B / 2 = A F = a$ and $F U = x$, then applying the cosine theorem to triangles $U F A$ and $U F B$ we get

$$
\frac{U A^{2}}{U B^{2}} = \frac{x^{2} + a^{2} + \sqrt{3} a x}{x^{2} + a^{2} - \sqrt{3} a x} = \lambda^{2}
$$

and with the usual steps

$$
\begin{gathered}
\left(\frac{x}{a}\right)^{2} - \frac{\sqrt{3}(\lambda^{2} + 1)}{\lambda^{2} - 1} \cdot \frac{x}{a} + 1 = 0 \\
\left(\frac{x}{a}\right)_{1,2} = \frac{\sqrt{3}}{2} \cdot \frac{\lambda^{2} + 1}{\lambda^{2} - 1} \pm \sqrt{\frac{3}{4}\left(\frac{\lambda^{2} + 1}{\lambda^{2} - 1}\right)^{2} - 1}
\end{gathered}
$$

provided that the expression under the square root is non-negative. This determines the position of the points on the path with the desired property, measured in units of $F A$.

2. The number of solutions is 0, 1, or 2 depending on whether the discriminant $D$ is negative, 0, or positive. This condition needs to be transformed in such a way that we can answer the question directly from the value of $\lambda$. We note that the ratio $\frac{\lambda^{2} + 1}{\lambda^{2} - 1}$ is positive due to the assumption $\lambda > 1$, so the solution of the inequality $D \lesseqgtr 0$ for $\lambda$ can proceed without case distinction:

$$
\begin{aligned}
\frac{3}{4}\left(\frac{\lambda^{2} + 1}{\lambda^{2} - 1}\right)^{2} - 1 & \risingdotseq 0 \\
\frac{\lambda^{2} + 1}{\lambda^{2} - 1} = 1 + \frac{2}{\lambda^{2} - 1} & \risingdotseq \frac{2}{\sqrt{3}}, \\
\frac{1}{\lambda^{2} - 1} & \risingdotseq \frac{2 - \sqrt{3}}{2 \sqrt{3}} = \frac{1}{2 \sqrt{3}(2 + \sqrt{3})}, \\
\lambda^{2} - 1 & \gtreqless 4 \sqrt{3} + 6 \\
\lambda^{2} & \gtreqless 7 + 4 \sqrt{3} = (2 + \sqrt{3})^{2}
\end{aligned}
$$

Accordingly, there is no solution for $U$ if $\lambda > 2 + \sqrt{3}$, and a single solution if $\lambda = 2 + \sqrt{3}$:

$$
\frac{x}{a} = \frac{\sqrt{3}}{2} \cdot \frac{\lambda^{2} + 1}{\lambda^{2} - 1} = 1
$$

The points satisfying the condition (11) lie on the Apollonian circle $k_{\lambda}$ determined by the base points $A, B$ and the ratio $\lambda$. This circle is symmetric with respect to the line $A B$, and the points on $A B$ form a diameter.

![](https://cdn.mathpix.com/cropped/2024_05_02_8f9470f0747f230db5bbg-2.jpg?height=548&width=941&top_left_y=119&top_left_x=576)

2nd figure

For the internal division point $C_{1}$ on $A B$, $C_{1} B = \frac{2 a}{\lambda + 1}$, and for the external division point $C_{2}$, $B C_{2} = \frac{2 a}{\lambda - 1}$ (2nd figure), so the diameter $C_{1} C_{2} = C_{1} B + B C_{2} = \frac{4 \lambda a}{\lambda^{2} - 1}$, and the distance of the center $O$ from $F$ is

$$
F O = F C_{1} + \frac{C_{1} C_{2}}{2} = F B - C_{1} B + \frac{2 \lambda a}{\lambda^{2} - 1} = \frac{a(\lambda^{2} + 1)}{\lambda^{2} - 1}
$$

and the distance from the path is $F O \sin 30^{\circ} = F O / 2$. The number of common points of the circle and the path is 2, 1, or 0 depending on whether this distance is less than, equal to, or greater than the radius of the circle:

$$
\frac{a(\lambda^{2} + 1)}{2(\lambda^{2} - 1)} \lesseqgtr \frac{2 a \lambda}{\lambda^{2} - 1}
$$

which is equivalent to

$$
\lambda^{2} + 1 \lesseqgtr 4 \lambda
$$

From this, we again obtain the above result.