On the Gellért Hill, an observer stands. Their eye is in the plane defined by the southern edges of the two arches of the Erzsébet Bridge. They see the vertical edge of one arch's southern side (from the road to the highest point) at an angle of $\alpha$, and the other arch at an angle of $\beta$. How high above the horizontal plane passing through the bases of the arches is the observer, if the distance between the two arches is $d$ meters and the height of the arches considered is $m$ meters? $(d=290 \text{~m}, m=40 \text{~m}$, $\left.\alpha=11.4^{\circ}, \beta=4.7^{\circ}.\right)$
I. solution. Let the sought height be $h$, the orthogonal projection of the observer's eye onto the plane mentioned in the problem be $u$ distance from the Buda tower - let the angle of view of this be $\alpha$ - and the line of sight leading to the lower end of this point form an angle $\varphi$ with the vertical direction. Then

$$
\begin{aligned}
\operatorname{tg} \varphi & =\frac{u}{h} \\
\operatorname{tg}(\varphi+\alpha) & =\frac{u}{h-m}=\frac{\operatorname{tg} \varphi+\operatorname{tg} \alpha}{1-\operatorname{tg} \varphi \operatorname{tg} \alpha}=\frac{u / h+\operatorname{tg} \alpha}{1-(u \operatorname{tg} \alpha) / h}=\frac{u+h \operatorname{tg} \alpha}{h-u \operatorname{tg} \alpha}
\end{aligned}
$$

From this, the following equation is obtained for $u$ and $h$:

$$
u h-u^{2} \operatorname{tg} \alpha=u h+h^{2} \operatorname{tg} \alpha-u m-h m \operatorname{tg} \alpha
$$

or rearranged and multiplied by $\operatorname{cotg} \alpha$:

$$
u^{2}+h^{2}-u m \operatorname{cotg} \alpha-h m=0
$$

For the Pest tower, the same equation is obtained, with $\alpha$ and $u$ replaced by $\beta$ and $u+d$:

$$
(u+d)^{2}+h^{2}-(u+d) m \operatorname{cotg} \beta-h m=0
$$

By taking the difference of the two equations, $u$ can be determined, and knowing this, the sought height $h$ can be found as the positive root of the first equation:

$$
\begin{gathered}
(u+d)^{2}-u^{2}-[u(\operatorname{cotg} \beta-\operatorname{cotg} \alpha)+d \operatorname{cotg} \beta] m= \\
=u[2 d-m(\operatorname{cotg} \beta-\operatorname{cotg} \alpha)]-d(m \operatorname{cotg} \beta-d)=0 \\
u=\frac{d(m \operatorname{cotg} \beta-d)}{2 d-m(\operatorname{cotg} \beta-\operatorname{cotg} \alpha)} \\
h=\frac{m}{2}+\sqrt{\frac{m^{2}}{4}+u m \operatorname{cotg} \alpha-u^{2}}
\end{gathered}
$$

Among our data, only the larger angle can be the angle of view of the Buda tower and the smaller one the angle of view of the Pest tower, because by swapping the angles, $m \cdot \operatorname{cotg} 11.4^{\circ}<40 \cdot 5=200$ would result in a negative value for $u$. Performing the calculations with a 4-digit function table:

$$
\begin{gathered}
m \operatorname{cotg} \beta=196.4, \quad m(\operatorname{cotg} \beta-\operatorname{cotg} \alpha)=288 \\
2 d-m(\operatorname{cotg} \beta-\operatorname{cotg} \alpha)=292 \\
\lg u=2.2902, \quad u m \operatorname{cotg} \alpha=38700 \\
(m / 2)^{2}=400, \quad u^{2}=38050 \\
h=20+\sqrt{1050}=52.4 \approx 52 \mathrm{~m}
\end{gathered}
$$

(Since some of the data is only known to 2 significant figures, the result can also have at most 2 significant figures.)

![](https://cdn.mathpix.com/cropped/2024_05_02_27e9493fb91da1f910dcg-1.jpg?height=415&width=984&top_left_y=2243&top_left_x=560)[^0]

II. solution. The observer's point of view $P$ is on the one hand on the circle of view from which the closer tower is seen at an angle $\alpha$, and on the other hand on the circle of view from which the farther tower is seen at an angle $\beta$. Let the centers of the circles be $O_{1}$, $O_{2}$, their distances from the respective towers be $t_{1}$, and $t_{2}$, their radii be $r_{1}$, and $r_{2}$, and the distance of $P$ from the line $O_{1} O_{2}$ be $l$; then the sought height is $m / 2+l$. We determine $l$ as the height of the triangle $O_{1} O_{2} P$. The sides can be calculated: $O_{1} P=r_{1}$, $O_{2} P=r_{2}$, and $O_{1} O_{2}=t_{1}+d-t_{2}$; let us denote the latter by $t$. From the relationship between the central and peripheral angles, it follows that $r_{1}, t_{1}$, and $m / 2$ are the sides of a right triangle with an acute angle $\alpha$, and $r_{2}, t_{2}$, and $m / 2$ are the sides of a right triangle with an acute angle $\beta$, so

$$
\begin{gathered}
r_{1}=\frac{m}{2 \sin \alpha}, \quad r_{2}=\frac{m}{2 \sin \beta} \\
t_{1}=\frac{m}{2} \operatorname{cotg} \alpha, \quad t_{2}=\frac{m}{2} \operatorname{cotg} \beta, \quad t=d+t_{1}-t_{2}
\end{gathered}
$$

Now, calculating the area of the triangle using the height $l$ and Heron's formula, and expressing $l$ - if we denote $(r_{1}+r_{2}+t) / 2$ by $s$ -

$$
l=\frac{2 \sqrt{s\left(s-r_{1}\right)\left(s-r_{2}\right)(s-t)}}{t}
$$

Using a 4-digit function table, $r_{1}=101.2, r_{2}=244.1, t=146, l=31.5$ is obtained, so $h=51.5 \approx 52 \mathrm{~m}$.

Remark. With our data, $s-r_{2}=1.5$ is obtained, so the last two digits of the calculated 4-digit values play a significant role. This explains the inaccuracy of the result, the difference in the values obtained in the two solutions.


[^0]:    ${ }^{1}$ The notation of $h$ on the diagram should be corrected.