How would the result of the 1st problem of the 2nd round of the 1965 National High School Mathematical Competition look if we consider the angle data rounded to $0.1^{\circ}$, and, as is customary, we consider the rounded digit to be even if the neglected part is exactly equal to half the place value of the last retained digit, that is, if

$$
11.35^{\circ} \leq \alpha \leq 11.45^{\circ}, \quad \text { and } \quad 4.65^{\circ}<\beta<4.75^{\circ}
$$

1[^0]


[^0]:    ${ }^{1}$ For the text of the competition problem, see K. M. L. 31 (Nov. 1965) p. 104.
Let's delve into the second solution of the competition problem mentioned in the quoted section, focusing on the $11.4^{\circ}$ and $4.7^{\circ}$ visual angles of the observer's position relative to the two arches of the bridge. Due to the allowed variation in visual angles, each value that satisfies (1) corresponds to a circular arc. This slightly differs from the used arc because increasing the angle brings the center of the arc closer to the arch, and the radius decreases. It is sufficient to take the arcs corresponding to the maximum and minimum visual angles, which define a crescent-shaped, circular-curved, two-vertex figure (Figure 1). The vertices represent the endpoints of the segment representing the arch, and every intermediate angle's visual arc passes through the interior of the figure and ends at the two vertices.

![](https://cdn.mathpix.com/cropped/2024_05_02_1b05c84c91c071a819d3g-1.jpg?height=470&width=596&top_left_y=513&top_left_x=751)

# 1. $a b r a$ 

In the original solution, the common point $P$ of the two arcs is replaced by the common points of the two crescent figures, because every common point is intersected by an $\alpha$-arc and a $\beta$-arc. The common points are expected to fill a figure, whose points will be at different heights above the $O_{1} O_{2}$ plane, which makes it reasonable to modify the competition problem's question as follows: what is the maximum and minimum possible height of the observer above the horizontal plane through the bases of the arches?

![](https://cdn.mathpix.com/cropped/2024_05_02_1b05c84c91c071a819d3g-1.jpg?height=1016&width=681&top_left_y=1324&top_left_x=711)

For clarity, let's create a scaled diagram of the $P O_{1} O_{2}$ triangle from the second solution, using the partial results $t_{1}=99.2 \mathrm{~m}, r_{1}=101.2 \mathrm{~m}, t_{2}=243.2 \mathrm{~m}, r_{2}=244.1 \mathrm{~m}$, and the data $d=290 \mathrm{~m}$ (Figure 2), then calculate the position of the centers and the radii of the arcs defining the crescent figures, and draw these arcs as well.

$$
\begin{array}{lcc}
\alpha_{\min }=11.35^{\circ} \text { from } & t_{1 \max }=99.7 \mathrm{~m}, & r_{1 \max }=101.7 \mathrm{~m} \\
\alpha_{\max }=11.45^{\circ} \text { from } & t_{1 \min }=98.7 \mathrm{~m}, & r_{1 \min }=100.7 \mathrm{~m} \\
\beta_{\min }=4.65^{\circ} \text { from } & t_{2 \max }=245.9 \mathrm{~m}, & r_{2 \max }=246.8 \mathrm{~m} \\
\beta_{\max }=4.75^{\circ} \text { from } & t_{2 \min }=240.6 \mathrm{~m}, & r_{2 \min }=241.5 \mathrm{~m}
\end{array}
$$

(It is not significant that $\beta$ cannot take these two values, as it can still be arbitrarily close to them; otherwise, the maximum and minimum lengths were rounded up and down, respectively.)

According to the extended diagram, for the highest point $P^{*}$ of the common part of the crescent figures, $h_{\max } \approx 73$ meters, and (considering the symmetry) there are common points at every height between $h=-73 \mathrm{~m}$ and $h=73 \mathrm{~m}$. Indeed, the distance of the points on the arcs from the line $O_{1} O_{2}$, which are farthest from the closer arch, is $u=t_{1}+r_{1}, u=t_{2}+r_{2}-d$, which from the above data are 201.4 m and 199.4 m, or 202.7 m and 192.1 m, respectively, so the crescent figure corresponding to $\alpha$ does not intersect the outer boundary of the figure corresponding to $\beta$. For the point $P^{*}$, repeating the calculations of the competition problem with the visual angle pair $\alpha_{\min }, \beta_{\max }$ according to both solutions, $h \approx 73 \mathrm{~m}$ is obtained.

Remarks. 1. Our result shows that the allowed uncertainty in determining $\beta$ would render the entire procedure unusable. (However, $0.05^{\circ}=3^{\prime}$ is much larger than the $10-20^{\prime \prime}$, or even $1^{\prime \prime}$ accuracy achievable with typical theodolites!)

From our diagram, we also see that if the $\beta=4.7^{\circ}$ data were assumed to be rounded down to $0.1^{\circ}$, then only the height values $45<h<73$ would be possible for the observer's position, since in this case the outer boundary of the crescent figure corresponding to $\beta$ would intersect the inner boundary of the figure corresponding to $\alpha$ at $P^{* *}$.

The uncertainty decreases rapidly with the decrease in the visual angle; the width of the $\alpha$-figure is only about 2 meters near $P$, while that of the $\beta$-figure is already about 10 meters, and even in the case of modification, it is about 5 meters. The maximum size of the common part is still about 30 meters even in the latter case.

2. Most of the solutions in the competition tried to infer the variation of $u$ and, indirectly, $h$ from the formulas for the coordinates of the observer's position in the first solution of the competition problem, based on the variation of $\alpha$ and $\beta$. The investigations - although mostly correct in essence - are quite complex due to the two variables and use knowledge beyond the high school curriculum. Almost every solution ignored or relegated to the background the geometric content of the problem.
3. Some solutions incorrectly identified (1) as incorrect because there is no solution for $\beta=4.65^{\circ}$. We do not claim that $\beta$ runs through the interval (1) at the observation point, as the value of $\beta$ is a single definite value. We only claim that the data $4.7^{\circ}$ may be inaccurate due to the tools and methods used, but the error does not exceed $0.5^{\circ}$ in either direction.

[^0]:    ${ }^{1}$ See the solution of the competition problem in K. M. L. 31 (1965. Nov.), p. 104. Reading the solution there is necessary for understanding the new question.

    ${ }^{2} \mathrm{~A}$ The scale of the presented diagram is 1:600. The complete $P O_{1} O_{2}$ triangle, scaled down by 1:1000, requires a sheet of $50 \times 10 \mathrm{~cm}$. A regular compass is too small for drawing the larger arc; this arc can be drawn by making two holes at a distance of $r_{2} / 1000$ on a 25 cm ruler, one for the pin and the other for the pencil.