Solution

(a) The first important observation to make is, because the stick is vertical and perpendicular to the ground, the line of the shadow points in a direction directly opposite the azimuth of the Sun at that moment. Therefore, we can compare angles between lines by comparing changes in azimuth of the Sun between two times.

As March 20th is the spring equinox (which can also be deduced as the problem mentions the Sun rising at 6am and setting at 6pm), the Sun travels along a great circle in the sky at a constant rate of $15^\circ$ per hour. Let $C_{12},C_1,C_5,C_6$ be points along the great circle where the Sun is at 12pm, 1pm, 5pm, 6pm respectively. As the Sun rises at 6am and sets at 6pm, we know solar time corresponds with local time (which is also confirmed as $75^\circ\mathrm W$ is perfectly in the center of the UTC-5 time zone). Therefore, the Sun is on the meridian at 12pm and on the horizon due west at 6pm.

Extend the arcs from $Z$ through $C_{12},C_1,C_5,C_6$ to meet the horizon at $H_{12}=S,H_1,H_5,H_6=C_6=W$. Then changes in azimuth correspond to arcs along the horizon. The arc $H_{12}H_1$ is the angle between 12pm and 1pm and has length $\theta_1$, and the arc $H_5H_6$ is the angle between 5pm and 6pm and has length $\theta_2$. The arc lengths $\theta_1,\theta_2$ also are the same as the corresponding angles from $Z$.

As one possible solution, consider the triangle $ZC_{12}C_1$. As $\angle ZC_{12}C_1=90^\circ$, then by the spherical law of sines:

But as $\sin90^\circ=1$ and $ZC_1<90^\circ$, we have that the right-hand side is greater than 1 and therefore $\theta_1>15^\circ$.

Many other solutions involving spherical triangles are possible (arguments from other parts of the solution can also be adapted to this part).

(b) As one possible solution, consider the triangle $C_5H_5W$. As $\angle C_5H_5W=90^\circ$, then by the spherical law of cosines:

But as $\cos C_5H_5<1$, we then have $\cos\theta_2>\cos15^\circ$ and therefore $\theta_2<15^\circ$.

Again, other choices of spherical triangles leading to a correct solution are possible.

(c) From this point onwards, there are two similar but slightly different ways to solve the problem, depending on choice of origin. Points on subparts differ between the methods, but the overall amount of work needed is about the same. We will be calling these methods ”Tip of stick” and ”Base of stick”.

Tip of stick: Drop a perpendicular from the tip of the stick to the ground, and set the point directly below the tip of the stick to be the origin.

While this definition initially seems more contrived, note that the position of the tip of the shadow depends only on the position of the tip of the stick, and the Sun. Therefore, we can directly write down positions in our coordinate system of tips of shadows.

Base of stick: Set the base of the stick to be the origin.

This may be the more immediately apparent choice, and has the advantage that we immediately know what directions shadows point in. However, this method is more difficult to follow through with; an astute observation is needed later to make progress.

(d) Tip of stick:

In order to determine lengths of shadows of the ground, we first need to know the height of the stick tip. Therefore, introduce a variable $h_0$ for the height of the tip of the stick above the ground.

We will carry through $h_0$ for the rest of the solution. However, note that the problem is invariant under rescaling (no specific lengths are given, and $h_0$ divides out at the end). Therefore, it is valid to set $h_0$ to an arbitrary positive value, as long as it is divided out at the end.

Start by computing the altitude and azimuth of the Sun at a given time. For some time $T$ hours after noon, where $T\in(-6,6)$: Let $C$ be the position of the Sun (along the celestial equator) at the time. Then, as the Sun passes $15^\circ$ of angle per hour, we have:

As angle $\angle ZEC$ subtends arc $ZC_{12}=\phi$ where $EC_{12}=ZE=90^\circ$, we must have $\angle ZEC=\phi$. Then, we can apply the spherical law of cosines on triangle $ZEC$:

To get the angle of arc $ZC$. Then, the altitude is just $90^\circ$ minus this angle:

To find the altitude, use spherical triangle $ECH$, where we extend $ZC$ to meet the horizon at $H$. As $\angle CHE$ is right, the $\sin a\sin b\cos C$ term in the spherical law of cosines disappears, so we just have:

Then:

And as $EH$ is the arc along the horizon starting from an azimuth of $90^\circ$, we have $a=90^\circ+EH$.

As the data given in the problem is given for every hour of the day, any integer choice of $T$ where the Sun is above the horizon $(T\in\{-5,-4,\ldots,5\})$ is valid. As one example, evaluating these equations through for $T=1$ hour after noon gives $(h,a)=(47.72648^\circ,202.62906^\circ)$.

If the stick were perfectly vertical, the base of the stick would be at the origin, and the length of the shadow on the ground would be given by:

where $h_0$ is the height of the stick tip, and $h$ is the altitude of the Sun.

For the specific coordinate axes given, where the $x$-axis points due North and the $y$-axis points due East, azimuths correspond perfectly with angles from the $x$-axis, first rotating towards the $y$-axis. Therefore, to convert between azimuths and (cartesian) coordinates, no angle offset is necessary.

Then, as the shadow is directed opposite the Sun, the coordinates of the shadow tip would be:

But this holds regardless of where the base of the stick is! Then we have, for the coordinates of the shadow tip:

Which for $T=1$ hour after noon gives:

Base of stick:

As the shadow is a projection of a line (the stick) onto the plane of the ground, it must be a straight line. In addition, the line of the shadow must contain the base of the stick (which is on the ground) and the tip of the shadow.

Again, by a rescaling argument, we introduce a length $l_0$ of a shadow on the ground. The shadow on the ground has an angle given by the data: as the 6am line points due west, the azimuth of the shadow is given by the angle from 6am in the last column of the table, less 90 degrees. Then for an angle from 6am measurement $\theta$, the shadow coordinates are:

Then for $T=1$ hour after noon gives:

(e) Tip of stick:

Again, shadows are straight lines on the ground having an angle given by the data. Just as azimuths can be converted directly to polar coordinates, they can also be converted directly to slopes:

Then, to find the intercept $b$, we just need to substitute in our known coordinate of the shadow tip $(x_0,y_0)$:

Values for $T=1$ hour after noon:

Base of stick:

To work towards finding the actual tilt, consider an ideal stick, with the same tip, but standing perfectly vertical. Then the ideal shadows cast would have the same tip as our stick, but would be at a different angle on the ground. These shadows would intersect at a different point, the base of the ideal stick. As the base of the ideal stick is directly below the tip of the ideal (and our) stick, this gives us the means to find the tip of the stick in our coordinates.

We therefore want the angle of the ideal shadows. However, this is just an azimuth calculation (described in the previous part)! Therefore:

Values for $T=1$ hour after noon:

(f) Tip of stick:

In order to solve for the coordinates $(x,y)$, we need a second condition. At this point, we could just choose another time during the day, compute altitude and azimuth using the same process, and find the coordinates of another shadow tip. However, there are some clever choices and ideas that we can use to avoid fully running through another calculation.

Consider the position of the Sun at noon, or $T=0$. Then the Sun is on the meridian, implying that it is due South. As it is also still on the celestial equator, we can immediately write down horizontal coordinates for the Sun:

Again converting this to shadow tip coordinates:

The $y$-coordinate is 0 as expected, as the shadow should have no deviation in the East-West direction when the Sun faces due South. However, note that the $x$-coordinate is the same as before! It turns out (and is noted at the start of the problem text) this holds true for any choice of $T$: the tip of shadows trace out a straight line running West to East on the equinoxes! (for more details as well as a proof, see last year’s second round problem with a very similar name, where this was the core concept tested!)

Then finding the slope and line for $T=0$:

Now that we have a pair of linear equations, we can solve for $(x,y)$. As they are already in slope-intercept form, it is easiest to eliminate; doing so and substituting in gives:

Base of stick:

In order to solve for the coordinates $(x,y)$, we need a second condition. Again, we can use the data to get an angle for the shadow on the ground. However, in order to find the coordinates of the tip, we need the length of the second shadow, which we somehow need to relate to the length of the first.

But this is where the ”shadow tips draw out a straight line” property proves useful! Regardless of coordinate origin, for coordinate axes aligned the same way, the $x$-coordinate of the shadow tip remains constant. As we know shadows are drawn from the origin at an angle given by the data, this gives us enough information to find the tip of another shadow.

Taking $x_0=0.94992l_0$ from earlier and taking the Sun’s position at noon $(T=0)$:

Then the azimuth of the Sun at noon is simply $a=180^\circ$, and the ideal shadow runs perfectly North-South. Then, the base of the ideal stick must have the same $y$-coordinate (no East-West deviation). This gives:

At this point, we can substitute our known value of $y$ into the previous equation, giving the solution:

(g) Tip of stick:

We now have the coordinates of the stick base, when the stick tip is at the origin. To get a base to tip vector, we must first invert this vector:

Then, the direction of tilt is just the azimuth of this vector, or $\boxed{240^\circ}$.

To compute the angle o f tilt from the vertical, first compute the length of the deviation on the ground:

Then, as $h_0$ is the vertical height of the tip, we get the tilt from the vertical:

Base of stick:

We now have the coordinates of the stick tip, where the stick base is at the origin. Again, we can just take the azimuth of the base to tip vector to get the tilt direction, which gives the same answer of $\boxed{240^\circ}$.

The stick tilt is a bit more difficult, as we don’t yet have $h_0$, the height of the stick; we have $l_0$, the length of a real shadow. To compute the height of stick, we first need the length of an ideal shadow at some time. This computation is easiest for noon, or $T=0$:

Then stick height:

Finally, we compute the length of the deviation on the ground:

And the tilt from the vertical, which gives the same answer: