Tag: geometry

Suppose the base has radius $r_0$ and we start a distance $\rho_0$ from the center (point $A$). Suppose the slant height is $r_1$ and our goal is to reach a point $\rho_1$ from the top (point $B$). We can break the journey into two parts. No matter how we get from $A$ to $B$, we must reach the edge of the base (point $X$) at some point on our journey. Suppose $X$ is located at an angle $\theta$ from the center of the base.

In the diagram above, we show the 3D cone and the corresponding unfolded 2D version to the right. The corresponding points have matching labels (click the diagram to enlarge) The first part of the journey, from $A$ to $X$, should be a straight line, since this is the shortest distance connecting two points on a flat surface. By the law of cosines, the distance $AX$ is
\[
|AX| = \sqrt{ r_0^2 + \rho_0^2-2r_0\rho_0\cos(\theta) }
\]The second part of the journey, from $X$ to $B$, will be a curved path along the surface of the cone. However, since we want the shortest possible path, this will be a straight line if we unfold the cone. This unfolded cone is a circular sector with radius $r_1$. Based on the diagram below, we can apply the law of cosines again to find the distance $XB$:
\[
|XB| = \sqrt{ r_1^2 + \rho_1^2-2r_1\rho_1\cos(\phi) }
\]But what is $\phi$? The arclengths along the edge of the base for both parts of the path must sum to 90 degrees. So the sum of both arclengths must equal one quarter of the circumference of the base. In other words, the points $X’$ and $R’$ must coincide with $X$ and $R$, respectively, when we fold the cone back together. Therefore,
\[
r_0 \theta + r_1 \phi = \tfrac{\pi}{2}r_0
\]Putting these three ingredients together, our goal is to solve the optimization problem:
\begin{align}
\underset{\theta,\phi}{\text{minimize}} \quad & \sqrt{ r_0^2 + \rho_0^2-2r_0\rho_0\cos(\theta) } + \sqrt{ r_1^2 + \rho_1^2-2r_1\rho_1\cos(\phi) } \\
\text{subject to} \quad & r_0 \theta + r_1 \phi = \tfrac{\pi}{2}r_0 \\
& 0 \leq \theta \leq \tfrac{\pi}{2}
\end{align}

Eliminating $\phi$ from the objective, we can write a single expression for the total distance $f(\theta)$ traveled in terms of the location $\theta$ of the crossover point:
\[
f(\theta) = \sqrt{ r_0^2 + \rho_0^2-2r_0\rho_0\cos(\theta) }
+ \sqrt{ r_1^2 + \rho_1^2-2r_1\rho_1\cos\Bigl[\tfrac{r_0}{r_1}\bigl(\tfrac{\pi}{2}-\theta\bigr)\Bigr] }
\]Here is a plot of what this function looks like when we use the parameters given in the problem statement ($r_0=2$, $\rho_0=1$, $r_1=4$, $\rho_1=2$)

We can see the minimum occurs at around 30 degrees. Next, we will dig a bit deeper and show that the minimum occurs at exactly 30 degrees, and we will derive a solution in the more general case.

General solution

For general values of $(r_0,\rho_0,r_1,\rho_1)$, we can look for a value of $\theta$ that minimizes $f(\theta)$ by taking the derivative and setting it equal to zero. However, the algebra is a bit simpler if we do not eliminate $\phi$ and write down optimality conditions using the method of Lagrange multipliers. Doing this and simplifying, we obtain two conditions:
\begin{gather}
\frac{\rho_0\sin\theta}{\sqrt{ r_0^2+\rho_0^2-2 r_0 \rho_0\cos(\theta)}}
= \frac{\rho_0\sin\phi}{\sqrt{ r_1^2+\rho_1^2-2 r_1 \rho_1\cos(\phi)}}
\\
\theta + \frac{r_1}{r_0} \phi = \frac{\pi}{2}
\end{gather}By the law of sines, the first equation says that $\alpha = \angle OXA = \angle PX’B$. Intuitively, this has the following interpretation: If we imagine rolling one circle around the other until points $X$ and $X’$ coincide, the points $B,X’,X,O$ will all lie on a straight line (the two red segments form a straight line). By writing out the law of sines on the third side, we can obtain slightly simpler expressions now involving $(\theta,\phi,\alpha)$:
\begin{align}
\rho_0 \sin(\theta+\alpha) &= r_0 \sin(\alpha) \\
\rho_1 \sin(\phi+\alpha) &= r_1 \sin(\alpha) \\
\theta + \tfrac{r_1}{r_0}\phi &= \tfrac{\pi}{2}
\end{align}Expanding the first two equations using angle sum identities and eliminating $\alpha$, we obtain an even simpler set of equations:
\[
\frac{\tfrac{r_0}{\rho_0}-\cos\theta}{\sin\theta} = \frac{\tfrac{r_1}{\rho_1}-\cos\phi}{\sin\phi}
\quad\text{and}\quad
\theta + \tfrac{r_1}{r_0}\phi = \tfrac{\pi}{2}
\]This is pretty difficult to solve in general, but certain special cases make things easier. For example, if $\tfrac{r_0}{\rho_0}=\tfrac{r_1}{\rho_1}$, then we must have $\theta=\phi$. This is precisely the case in the problem statement, because we have $\tfrac{r_0}{\rho_0}=\tfrac{r_1}{\rho_1}=2$. Furthermore, we have $\tfrac{r_1}{r_0}=2$, so the second condition becomes $\theta + 2\theta = \tfrac{\pi}{2}$, from which we conclude that $\theta=\tfrac{\pi}{6}=30^\circ$. Substituting back into the formula for total distance, we find the length of the shortest path:

$\displaystyle
\text{Minimum distance} = f\bigl(\tfrac{\pi}{6}\bigr) = 3 \sqrt{5-2 \sqrt{3}} \approx 3.71794
$

So we can find an analytic solution to this problem whenever the start and end points are the same fraction of the way from the centers of their respective circles!

Let’s solve a slightly more general version of the problem. Suppose the triangle has angle $\theta$ at point $A$, and suppose Amare’s next three waypoints are at $D$, $E$, and $F$, as shown below.

Reflect the triangle about the line $AC$, so $B \mapsto B’$ and $E \mapsto E’$, as shown:

Now reflect the new triangle about the line $AB’$, so $C\mapsto C’$ and $F \mapsto F’$:

The key insight is that since the reflections preserve lengths, the path $BD+DE+EF$ followed by Amare has the same length as the path $BD+DE’+E’F’$ shown in red below:

If we move the points $E$ and $F$, then the points $E’$ and $F’$ move accordingly, and we obtain another possible path:

Rather than picking $D,E,F$, we can instead pick $D,E’,F’$. Since the goal is to minimize the total distance, it’s clear that we should place $F’$ such that $AF’ \perp BF’$, and $D$ and $E’$ should be placed so that all three points lie on a line. This produces the figure:

So the shortest distance Amare can travel can be found by examining the right triangle $ABF’$. Since $AB$ has length $1$, we conclude that that

$\displaystyle
\text{Minimum distance } = \sin(3\theta)
$

In the case where $\theta = 15^\circ$, we get a length of $\sin(45^\circ) = \frac{1}{\sqrt{2}} \approx 0.7071$.

Note that this solution only works if $3\theta \lt 90^\circ$, i.e. $0 \lt \theta \lt 30^\circ$. In the case that $\theta \geq 30^\circ$, we obtain the degenerate solution $D = E = F = A$, so Amare should head directly to point $A$ and the total distance traveled is $1$.

We can also use calculus to solve the problem. Let’s start with the same picture as before:

Suppose $|AD|=x$, $|AE|=y$, and $|AF|=z$. From the law of cosines:
\begin{align}
|BD| &= \sqrt{1+x^2-2x\cos(\theta)} \\
|DE| &= \sqrt{x^2+y^2-2xy\cos(\theta)} \\
|EF| &= \sqrt{y^2+z^2-2yz\cos(\theta)}
\end{align}Let $f(x,y,z)$ be the sum of these three distances, which is the total distance traveled by Amare. We want to find $x,y,z$ such that $f(x,y,z)$ is minimized. A necessary condition for minimality is that the partial derivatives of $f$ with respect to $x,y,z$ should be zero. Let’s start with $z$:
\[
\frac{\partial}{\partial z}f(x,y,z) = \frac{z-y \cos (\theta )}{\sqrt{y^2+z^2-2 y z \cos (\theta )}}
\]Setting this equal to zero, we conclude that $z=y\cos(\theta)$. We can now substitute this value of $z$ into the definition for $f$ and we obtain a simpler expression in only two variables:
\begin{align}
g(x,y) &= f(x,y,y\cos(\theta)) \\
&= \sqrt{1+x^2-2x\cos(\theta)} + \sqrt{x^2+y^2-2xy\cos(\theta)} + y\sin(\theta)
\end{align}To make sure there is no funny business going on, let’s plot this function to see what it looks like for $\theta=15^\circ$. Here is a contour plot:

We can clearly see that there is a unique minimum that occurs in the region near $(x,y) \approx (0.8,0.7)$. Ok, so let’s proceed. Consider now the derivative with respect to $y$:
\[
\frac{\partial}{\partial y} g(x,y) = \frac{y-x \cos (\theta )}{\sqrt{x^2+y^2-2 x y \cos (\theta )}} + \sin (\theta )
\]Setting this equal to zero (I’ll spare you the algebra), we obtain $y = x \cos(2\theta)\sec(\theta)$. Substituting this into $g(x,y)$, we obtain a function of just $x$:
\begin{align}
h(x) &= g(x,x \cos(2\theta)\sec(\theta)) \\
&= \sqrt{1+x^2-2 x \cos (\theta )}+x \sin (2 \theta )
\end{align}Finally, we can take the derivative with respect to $x$:
\[
\frac{\partial}{\partial x}h(x) = \frac{x-\cos (\theta )}{\sqrt{1+x^2-2 x \cos (\theta )}} + \sin (2 \theta )
\]Solving for $x$, (sparing you the algebra again!) we obtain $x = \cos (3 \theta ) \sec (2 \theta )$. Putting everything together, the $(x,y,z)$ that minimize the total distance traveled by Amare is
\[
x = \cos (3 \theta ) \sec (2 \theta ),\quad
y = \cos(3\theta)\sec(\theta),\quad
z = \cos(3\theta)
\]Substituting into either $h(x)$, $g(x,y)$, or $f(x,y,z)$, we obtain the minimum distance traveled by Amare (after simplification):

$\displaystyle
\text{Minimum distance } = \sin(3\theta)
$

If you read the first solution, then it should come as no surprise that the total distance is also equal to $\sqrt{1-z^2}$, since (based on the last figure of the first solution), we have $z=|AF|=|AF’|$ and $|BF’|^2 + |AF’|^2 = 1$ by the Pythagorean theorem.