USAAAO 2022, II кръг — Задача 9. The Sundial I

• (a) Top curve: winter solstice; middle curve: equinoxes; bottom curve: summer solstice.
• (b) $\delta = 0$
• (e) The wall is angled Southwest to Northeast.

(a) The Sun is highest on the summer solstice, and lowest on the winter solstice. Therefore shadows for sundials on the ground are shorter in summer than in winter. However, this sundial is on a vertical wall, so this trend is actually reversed: the top curve (shortest shadows) corresponds to the winter solstice and the bottom curve (longest shadows) corresponds to the summer solstice.

Also notice the visible shadow close to the top curve; as the picture was taken in December 2020, this shadow corresponds to the winter solstice and again the top curve.

The equinoxes therefore correspond to the curve in the middle.

(b) On the equinoxes, the declination of the Sun is $\delta = 0$.

(c) There are many possible solutions to this part; here we present one.

First note that, as the Sun’s path is symmetric across the meridian, by symmetry, the line must run perfectly East to West.

Consider the Sun’s position at solar noon on any day; if the Sun does not rise, no line is possible. But if the Sun does rise, the altitude of the Sun as it crosses the meridian is nonzero. For a finite stick height, the shadow length is finite. Therefore, the line must be a finite distance from the stick.

Now consider the azimuth of the Sun at sunset. If $\alpha \neq 270$, then there is some component of the shadow that runs North-South. But as the Sun’s altitude goes to 0, the shadow length goes to $\infty$. As the line runs East-West, the line must therefore be an infinite distance from the stick, which is a contradiction.

Therefore, this is only possible for $\alpha = 270$ at sunset, implying $\delta = 0$.

(d) We present a variety of solutions. First we present the method described in the problem statement:

i) From the previous part, we know the line must run perfectly East to West, at a finite distance from the stick.

ii) If the Sun has altitude $h$, then if the stick has length $h_0$, then the length of the shadow is given by $l = \frac{h_0}{\tan h}$.

iii) Let $a = \mathrm{azimuth} - 90$, such that the Sun rises at $a = 0$ and sets at $a = 180$. Let the line be a distance $x_0$ from the stick; then we need $l \sin a = x_0$.

iv) Therefore, $\frac{h_0 \sin a}{\tan h} = x_0$, which we need to hold for all positions $(a, h)$ of the Sun over the entire day. We have therefore derived the constant quantity $\frac{\sin a}{\tan h}$.
To finish off the proof, we need to show this is a constant quantity:

Using the spherical law of cosines:

Using the spherical right triangle equality $\cos d = \cos h \cos a$:

Noticing the first term has the same form as the original:

Using $1 - \cos^2 a = \sin^2 a$ and cancelling terms:

Using the spherical law of sines $\frac{\sin h}{\sin(90 - \phi)} = \sin d$:

Therefore, this is a constant quantity, which means the tip of the shadow is always on this straight line.

If the previous part was skipped, it is also necessary to prove the line is perfectly East to West, and that it is a finite distance from the stick, as those assumptions were used in the derivation above.

The inverse would also need to be correctly justified. For example, one way to do this would be to simplify $\frac{\sin a}{\tan h}$ without the $\delta = 0$ assumption.

Using the spherical law of cosines to derive the coordinate transformation:

Applying this transformation:

The latitude $\phi$ is a constant, and the declination $\delta$ is approximately unchanged over the course of a day. However, $\frac{1}{\sin h}$ varies over the course of a day as the Sun rises and sets. Therefore, for this to be a conserved quantity, we need $\sin\delta = 0$, implying $\delta = 0$.

However, this problem can be solved much faster geometrically. A simple solution using nothing but properties of great circles and planes goes as follows:

Place the tip of the stick at the center of the celestial sphere. On the equinoxes, the declination of the Sun is 0, so the path of the Sun is a great circle; therefore the path of the Sun through the day and the tip of the stick lie in the same plane. Therefore, the shadow of the tip of the stick must be in the same plane. Intersecting this plane with the plane of the ground gives a straight line for the shadow’s path. $\square$

Notice that this solution quickly generalizes to any other intersecting plane, be it a vertical wall or any other orientation the sundial is mounted in.

(e) Consider two positions of the Sun on the equinox, equidistant from noon. By symmetry, the altitudes are the same, and the azimuths are equally offset from due South.

Consider the shadow cast by the Sun at each of these times, which is the intersection of the line connecting the Sun’s position and the tip of the sundial with the wall. As the Sun is up in the South, these lines slope down as they head North.

Looking at the line in the picture, the line traveling West drops less compared to the line traveling East. This implies the wall juts out towards the observers perspective on the left, meaning the wall is angled Southwest to Northeast.

Looking at the actual building, Fisher Hall, on Google Maps, it indeed is oriented that way!

Problem adapted and extended from Roberto Bozcko and Lucas Carrit Delgado Pinheiro.