This article is translated (while omitting some tedious calculations) from a Chinese article on my Zhihu account. The original article was posted at 2019-05-15 20:28 +0800.

Suppose $P$ is a point on the circle $\odot C$ with radius $r$. Now we study the feature of the position of the point $P$ when the sum of the distances from $P$ to the two edges of $\angle O$ is extremal.

Set up Cartesian plane coordinates with the origin at $O$ and the $x$-axis pointing from $O$ to $C$. Suppose the coordinates of $C$ are $\left(x_C,0\right)$, the coordinates of $P$ is $\left(x_C+r\cos\theta,r\sin\theta\right)$, and the slope of the two sides of $\angle O$ are $k_1$ and $k_2$ respectively. Then, we have

\[\begin{align*} l_1&:k_1x-y=0,\\ l_2&:k_2x-y=0. \end{align*}\]

Suppose

\[\begin{align*} d_1&:=\frac{k_1\left(x_C+r\cos\theta\right)-r\sin\theta}{\sqrt{k_1^2+1}},\\ d_2&:=\frac{k_2\left(x_C+r\cos\theta\right)-r\sin\theta}{\sqrt{k_2^2+1}}, \end{align*}\]

and then $\left|d_1\right|$ and $\left|d_2\right|$ are the distances from $P$ to $l_1$ and $l_2$ respectively. The sum of the distances

\[\left|d\right|=\left|d_1\right|+\left|d_2\right|\]

(the definition of $d$ here are discussed case by case below).

Now, we discuss case by case.

Case 1: $d_1$ and $d_2$ have the same sign

In this case, $P$ is on the same “side” of $l_1$ and $l_2$, i.e. $P$ is in the interior of the adjacent supplementary angle of $\angle O$.

Suppose

\[d:=d_1+d_2,\]

and then $\left|d\right|$ is the sum of distances from $P$ to $l_1$ and $l_2$, so we just need to discuss the case when $d$ is extremal.

Let

\[d_C:=x_C\left(\frac{k_1}{\sqrt{k_1^2+1}}+\frac{k_2}{\sqrt{k_2^2+1}}\right),\] \[A:=r\sqrt{\left(\frac{k_1}{\sqrt{k_1^2+1}}+\frac{k_2}{\sqrt{k_2^2+1}}\right)^2 +\left(\frac1{\sqrt{k_1^2+1}}+\frac1{\sqrt{k_2^2+1}}\right)^2},\] \[\phi:=\arctan\frac{\frac1{\sqrt{k_1^2+1}}+\frac1{\sqrt{k_2^2+1}}}{\frac{k_1}{\sqrt{k_1^2+1}}+\frac{k_2}{\sqrt{k_2^2+1}}}.\]

Then, we have

\[\begin{align*} d&=d_C+A\cos\phi\cos\theta-A\sin\phi\sin\theta\\ &=d_C+A\cos\left(\phi+\theta\right). \end{align*}\]

Therefore, we find that, $d$ is extremal iff $\theta=n\pi-\phi$ ($n\in\mathbb Z$). Then we study what are the features of $\theta$ when $d$ is extremal.

Let

\[\begin{align*} \theta_1&:=\arctan k_1,\\ \theta_2&:=\arctan k_2. \end{align*}\]

Then, we have after some calculations

\[\tan\theta=-\frac{\frac1{\sqrt{k_1^2+1}}+\frac1{\sqrt{k_2^2+1}}}{\frac{k_1}{\sqrt{k_1^2+1}}+\frac{k_2}{\sqrt{k_2^2+1}}} =\tan\frac{\theta_1+\theta_2+\pi}2.\]

Therefore,

\[\theta=n\pi+\frac{\theta_1+\theta_2+\pi}2.\]

This means that $\tan\theta$ is the slope of the bisector of the adjacent supplementary angle of $\angle O$.

Therefore, we get such a method of construction of $P$ for extremal $\left|d\right|$: draw the bisector $l$ of the adjacent supplementary angle of $\angle O$; draw a line passing $C$ parallel to $l$, whose intersection with $\odot C$ is $P$ (there are two such intersection points, corresponding to $n$ being even and odd respectively, and $d$ takes maximal and minimal values respectively).

Case 2: $d_1$ and $d_2$ have different signs

Now, $P$ are on different “sides” of $l_1$ and $l_2$, i.e. $P$ is in the interior of $\angle O$ or its opposite angle.

Similarly, let

\[d:=d_1-d_2,\]

and then $\left|d\right|$ is the sum of distances from $P$ to $l_1$ and $l_2$.

Let

\[d_C:=x_C\left(\frac{k_1}{\sqrt{k_1^2+1}}-\frac{k_2}{\sqrt{k_2^2+1}}\right),\] \[A:=r\sqrt{\left(\frac{k_1}{\sqrt{k_1^2+1}}-\frac{k_2}{\sqrt{k_2^2+1}}\right)^2 +\left(\frac1{\sqrt{k_1^2+1}}-\frac1{\sqrt{k_2^2+1}}\right)^2},\] \[\phi:=\arctan\frac{\frac1{\sqrt{k_1^2+1}}-\frac1{\sqrt{k_2^2+1}}}{\frac{k_1}{\sqrt{k_1^2+1}}-\frac{k_2}{\sqrt{k_2^2+1}}}.\]

Then, we have

\[d=d_C+A\cos\!\left(\theta+\phi\right).\]

Therefore, we find that $d$ is extremal iff $\theta=n\pi-\phi$ ($n\in\mathbb Z$). Then we study what are the features of $\theta$ when $d$ is extremal.

Similarly, let

\[\begin{align*} \theta_1&:=\arctan k_1,\\ \theta_2&:=\arctan k_2. \end{align*}\]

Then, we have after some calculations

\[\theta=n\pi+\frac{\theta_1+\theta_2}2.\]

In other words, $\tan\theta$ is the slope of the bisector of $\angle O$.

Therefore, we get such a method of construction of $P$ for extremal $\left|d\right|$: draw the bisector $l$ of $\angle O$; draw a line passing $C$ parallel to $l$, whose intersection with $\odot C$ is $P$ (there are two such intersection points, corresponding to $n$ being even and odd respectively, and $d$ takes maximal and minimal values respectively).

Case 3: $d_1=0$ or $d_2=0$

In this case, $P$ is on either $l_1$ or $l_2$.

Without loss of generality, we assume $d_2=0$. Then, $P$ is the intersection of $\odot C$ and $l_2$, the number of cases is reduced to finite. To avoid confusion, we denote $\theta$ now $\theta_0$. Now, $d_1$ and $d_2$ are functions of $\theta$, while $\theta_0$ is the $\theta$ at which $d_2=0$.

Subcase 1: $d_1=0$

Obviously, in this case, when $\theta=\theta_0$, the sum of distances from $P$ to $l_1$ and $l_2$ takes minimal. This case occurs only when $l_1,l_2,\odot C$ intersect at the same point.

Subcase 2: $d_1\neq 0$

Then, according to the property of continuous functions, in some neighborhood of $\theta_0$, $d_1\ne0$.

Subsubcase 1: $\odot C$ intersects but is not tangent to $l_2$

Then, in some neighborhood of $\theta_0$, for the two cases $\theta<\theta_0$ and $\theta>\theta_0$, the sign of $d_2$ is different. We can define in this neighborhood

\[d:=\begin{cases} d_1+d_2,&\text{if $d_1d_2>0$,}\\ d_1,&\text{if $\theta=\theta_0$,}\\ d_1-d_2,&\text{if $d_1d_2<0$.} \end{cases}\]

It is easy to see that the left and right derivative of the continuous function $d$ both exist at $\theta_0$. It can be proved (how?) that, if the two derivatives have different signs, then $d$ is extremal at $\theta=\theta_0$.

To examine whether the two derivatives have different signs, we can write the product of them and see whether the result is positive or negative (the case of it being $0$ will be discussed later).

Find the derivatives of $d_1$ and $d_2$ respectively.

\[\begin{align*} \theta_1&:=\arctan k_1,\\ \theta_2&:=\arctan k_2. \end{align*}\]

Then, we have

\[\begin{align*} \frac{\mathrm dd_1}{\mathrm d\theta} &=\frac{-k_1r\sin\theta-r\cos\theta}{\sqrt{k^2+1}}\\ &=-r\cos\!\left(\theta-\theta_1\right), \end{align*}\] \[\frac{\mathrm dd_2}{\mathrm d\theta}=-r\cos\!\left(\theta-\theta_2\right).\]

Therefore, the left and right derivatives of $d$ at $\theta_0$ are respectively

\[\begin{align*} \left.\frac{\mathrm dd}{\mathrm d\theta}\right|_{\theta=\theta_0^\pm} &=\left.\frac{\mathrm dd_1}{\mathrm d\theta}\right|_{\theta=\theta_0} +\left.\frac{\mathrm dd_2}{\mathrm d\theta}\right|_{\theta=\theta_0}\\ &=-r\cos\!\left(\theta_0-\theta_1\right)-r\cos\!\left(\theta_0-\theta_2\right)\\ &=-2r\cos\!\left(\theta_0-\frac{\theta_1+\theta_2}2\right)\cos\frac{\theta_1-\theta_2}2, \end{align*}\] \[\left.\frac{\mathrm dd}{\mathrm d\theta}\right|_{\theta=\theta_0^\mp} =-2r\sin\!\left(\theta_0-\frac{\theta_1+\theta_2}2\right)\sin\frac{\theta_1-\theta_2}2.\]

Then, the product of the two derivatives is

\[\nu_0:= \left.\frac{\mathrm dd}{\mathrm d\theta}\right|_{\theta=\theta_0^\pm} \cdot\left.\frac{\mathrm dd}{\mathrm d\theta}\right|_{\theta=\theta_0^\mp} =r^2\sin\!\left(\theta_1-\theta_2\right)\sin\left(2\left(\theta_0-\frac{\theta_1+\theta_2}2\right)\right).\]

According to this equation, we can determine the sign of $\nu_0$ by merely knowing which quadrant the angle $\theta_0-\frac{\theta_1+\theta_2}2$ is in, and can therefore determine whether $d$ is extremal.

(When $\nu_0=0$, $P$ is the intersection of the three object: the bisector of $\angle O$ or its adjacent supplementary angle, $\odot C$, and $l_2$. In this case, $d$ may take extremal or not. How do we discuss this case now?)

Subsubcase 2: $\odot C$ is tangent to $l_2$

This case is easy. You only need to see whether $P$ is in the interior of $\angle O$ or its opposite angle when $P$ is moving near the tangent point. If it is in the interior, then the case is identical to Case 2; if it is in the exterior, then the case is identical to Case 1.

Summary

We finally got the method of determining whether the point $P$ on $\odot C$ has the extremal sum of distances to $l_1$ and $l_2$:

  • If $P$ is in the interior of $\angle O$ or its opposite angle (or, $P$ is on the tangent point of $\odot C$ and one of the sides of $\angle O$ while $P$ is in the interior of $\angle O$ or its opposite angle when it moves near the tangent point), then we can see whether it is the intersection of $\odot C$ and the bisector of $\angle O$. If it is, then the sum of distances is extremal; if it is not, then the sum of distances is not extremal.
  • If $P$ is in the interior of an adjacent supplementary angle of $\angle O$ (or, $P$ is on the tangent point of $\odot C$ and one of the sides of $\angle O$ while $P$ is in the interior of an adjacent supplementary angle of $\angle O$ when it moves near the tangent point), then we can see whether it is the intersection of $\odot C$ and the bisector of the adjacent supplementary angle $\angle O$. If it is, then the sum of distances is extremal; if it is not, then the sum of distances is not extremal.
  • If $P$ is on the intersection of $\odot C$ and one of the edge $l_2$ of $\angle O$, then we can divide the plane into four parts by drawing the bisector of $\angle O$ and that of the adjacent supplementary angle of $\angle O$, call the union of the two divided parts with $l_2$ passing through as $D_2$. Then, translate $D_2$ to make the intersection of its boundary lines overlap with $C$, and see whether $P$ belongs to the translated $D_2$. If it is an interior point of the region, then the sum of distances is extremal; if it is an exterior point of the region, then the sum of distances is not extremal; if it is a boundary point of the region, then the sum of distances may be extremal or not.