Kvant Math Problem 1533
Let $\ell$ be a line through $C$.
Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 28m24s
Source on kvant.digital
Problem
Three points $A$, $B$, $C$ are given on the plane. Draw a line through $C$ such that the product of the distances from $A$ and $B$ to this line is maximized. Is such a line always unique?
N. B. Vasilyev
Exploration
Let $\ell$ be a line through $C$. Denote by $d_A$ and $d_B$ the distances from $A$ and $B$ to $\ell$.
Choose an oriented angle parameter $\alpha$ as follows. Let $\alpha$ be the oriented angle from $CA$ to $\ell$, and let $\phi=\angle ACB$, with $0\le\phi\le\pi$. Then the angle from $CB$ to $\ell$ equals $\alpha-\phi$.
The distance from a point to a line through $C$ is the length of the corresponding segment times the sine of the angle with the line. Hence
$$d_A=CA,|\sin\alpha|, \qquad d_B=CB,|\sin(\alpha-\phi)|.$$
Thus
$$d_A d_B = CA\cdot CB\cdot |\sin\alpha,\sin(\alpha-\phi)|.$$
Since $CA$ and $CB$ are fixed, the problem reduces to maximizing
$$|\sin\alpha,\sin(\alpha-\phi)|.$$
Problem Understanding
The task is to find geometrically all lines through $C$ for which
$$|\sin\alpha,\sin(\alpha-\phi)|$$
is maximal, and then determine whether the maximizing line is unique.
Because a line is unchanged when rotated by $\pi$, it suffices to regard $\alpha$ modulo $\pi$.
Proof Architecture
The key step is a trigonometric reduction:
$$\sin\alpha,\sin(\alpha-\phi) = \frac{\cos\phi-\cos(2\alpha-\phi)}2.$$
Since $\cos(2\alpha-\phi)$ can take any value in $[-1,1]$, the maximum possible absolute value of the right-hand side is obtained when $\cos(2\alpha-\phi)$ is an endpoint of that interval.
This yields both the maximal value and the corresponding directions of $\ell$.
Solution
Using
$$\sin\alpha,\sin(\alpha-\phi) = \frac{\cos\phi-\cos(2\alpha-\phi)}2,$$
put
$$x=\cos(2\alpha-\phi).$$
Then
$$|\sin\alpha,\sin(\alpha-\phi)| = \frac{|\cos\phi-x|}{2}, \qquad -1\le x\le 1.$$
For a fixed number $c=\cos\phi\in[-1,1]$, the quantity $|c-x|$ is maximal when $x$ is the endpoint of $[-1,1]$ farthest from $c$.
If $c\ge0$, that farthest endpoint is $-1$, and
$$\max |c-x| = c+1.$$
If $c\le0$, that farthest endpoint is $1$, and
$$\max |c-x| = 1-c.$$
In both cases,
$$\max |\sin\alpha,\sin(\alpha-\phi)| = \frac{1+|\cos\phi|}{2}.$$
Consequently
$$\boxed{\max(d_A d_B) = \frac{CA\cdot CB}{2},(1+|\cos\phi|).}$$
It remains to identify the maximizing lines.
Assume first that $\cos\phi>0$, equivalently $\phi<\frac{\pi}{2}$.
Then equality is attained precisely when
$$\cos(2\alpha-\phi)=-1,$$
hence
$$2\alpha-\phi\equiv\pi\pmod{2\pi},$$
or
$$\alpha\equiv\frac{\phi+\pi}{2}\pmod{\pi}.$$
Thus there is exactly one line through $C$ realizing the maximum. This line bisects the supplementary angle between the rays $CA$ and $CB$. Indeed, its angle with $CA$ is
$$\frac{\pi+\phi}{2},$$
while its angle with $CB$ is
$$\frac{\pi-\phi}{2},$$
so the two angles on the same side add to $\pi-\phi$ and are equal.
Now assume that $\cos\phi<0$, equivalently $\phi>\frac{\pi}{2}$.
Then equality is attained precisely when
$$\cos(2\alpha-\phi)=1,$$
hence
$$2\alpha-\phi\equiv0\pmod{2\pi},$$
or
$$\alpha\equiv\frac{\phi}{2}\pmod{\pi}.$$
Again there is exactly one maximizing line. This line is the internal bisector of the angle $\angle ACB$.
Finally, consider the case $\phi=\frac{\pi}{2}$. Then
$$|\sin\alpha,\sin(\alpha-\phi)| = \frac{|,\cos(2\alpha-\phi),|}{2}.$$
The maximum value is $\frac12$, attained when
$$\cos(2\alpha-\phi)=\pm1.$$
Hence
$$2\alpha-\phi\equiv0\pmod{\pi},$$
which gives two distinct lines:
$$\alpha\equiv\frac{\phi}{2}\pmod{\pi}, \qquad \alpha\equiv\frac{\phi+\pi}{2}\pmod{\pi}.$$
For a right angle at $C$, these are exactly the two angle bisectors.
Thus:
If $\angle ACB\ne90^\circ$, there is a unique maximizing line. It is the bisector of the smaller angle between $CA$ and $CB$ when $\angle ACB>90^\circ$, and the bisector of the larger angle between them when $\angle ACB<90^\circ$.
If $\angle ACB=90^\circ$, there are two maximizing lines, namely the two angle bisectors.
Verification of Key Steps
The maximizing value is
$$\frac{CA\cdot CB}{2}(1+|\cos\phi|).$$
Using the law of cosines,
$$|\cos\phi| = \frac{|CA^2+CB^2-AB^2|}{2,CA,CB},$$
so the maximum is always nonnegative, as required.
For $\phi=\frac{\pi}{2}$,
$$\max(d_A d_B) = \frac{CA\cdot CB}{2},$$
which is attained by either angle bisector, giving
$$d_A=\frac{CA}{\sqrt2}, \qquad d_B=\frac{CB}{\sqrt2}.$$
This agrees with the formula.
Alternative Approaches
A coordinate solution is possible by placing $C$ at the origin and writing the line through $C$ with direction vector $(\cos\alpha,\sin\alpha)$. The distances become linear combinations of sines and cosines of $\alpha$, and the same trigonometric expression arises. The angle formulation yields the geometric construction immediately: choose the appropriate angle bisector according to whether $\angle ACB$ is obtuse or acute, and in the right-angle case both bisectors work.