This week’s Riddler classic is a neat geometry problem.

Two people climb two of the tallest towers on an planet, which happen to be in neighboring cities. You both travel 100 meters up each tower on a clear day. Due to the curvature of the planet, they can barely make each other out. The first person returns to the ground floor of their tower. How high up their tower must the second person be you can barely make each other out again?

My solution:

[Show Solution]

Suppose the planet has radius $r$. Assuming the planet is a perfect sphere, when both people are 100 meters above ground, the line connecting them must be tangent to the sphere. This produces an isosceles triangle. When the first person goes to the bottom of their tower, we obtain a right-angle triangle. Here is a picture of what this might look like, for a planet with radius $r\approx 1000\,\text{m}$, here is what the picture looks like. We can see the height we seek (shown in red) is about 530 meters.

But this isn’t the whole story. The answer depends on the planet’s radius! Here is what we get when we use a larger radius or a smaller radius:

It appears that the larger the radius of the planet, the smaller the red height (HG). Let’s figure out the exact answer in terms of the radius. Looking at triangle $OEC$, we see that

\[

\cos(\theta) = \frac{OE}{OC} = \frac{r}{r+100}

\]Now looking at triangle $OBG$, we see that

\[

\cos(2\theta) = \frac{OB}{OG} = \frac{r}{r+x}

\]where $x$ is the unknown height we seek. Substituting these expressions in the the double-angle cosine identity $\cos(2\theta)=2\cos^2(\theta)-1$, we obtain:

\[

\frac{r}{r+x} = 2\left(\frac{r}{r+100}\right)^2-1

\]Solving for $x$, we obtain:

$\displaystyle

x=HG = \frac{400r^2+20000r}{r^2-200r-10000}

$

Here is what this function of $r$ looks like when plotted:

The vertical asymptote occurs when $r=100(1+\sqrt{2}) \approx 241.42$. This is the point at which the two towers are at 90 degrees to each other. In this case, when the first person goes the the bottom of their tower, their line of sight is parallel to the other tower, so the other tower will never be visible, no matter how tall it is!

The horizontal asymptote occurs in the limit $r\to\infty$, which leads to a height of 400 meters. So in the limit of a large planet, this is what we can expect to see. We can also develop an asymptotic approximation by expanding as a series about $r=\infty$. This leads to:

\[

x = 400+\frac{10^5}{r} + O(r^{-2})

\]

For some context, the Earth has a radius of $6.371\times 10^6$ meters. Using this value of $r$, we find height of $x\approx 400.01569671$. Using the approximation with the extra $r^{-1}$ term, we obtain $x\approx 400.01569612$, which is correct to nine significant figures!

I believe that at the vertical asymptote, the towers are 90 degrees apart, not on opposite sides of the planet. The line of sight of the person at the bottom of the tower is parallel to the other tower.

Thanks — you’re absolutely right! I corrected my post.