An Equation Involving Four Collinear Points

Here's a problem by Leo Giugiuc that was offered at the 2015 Romanian Regional Olympiad for grade 10 (Problem 3.)

Solve the following equation in complex numbers:

(*)

$|z|+|z-5i|=|z-2i|+|z-3i|.$

Leo supplied the first three solutions. I have added one more, and Grégoire Nicollier added the fifth, one line, solution. Later Leo sent a solution by Seclaman Dan Radu.

Solution 1

Note that

(1)

$|z-2i|=|\frac{2}{5}(z-5i)+\frac{3}{5}z|\le \frac{2}{5}|z-5i|+\frac{3}{5}|z|.$

Similarly,

(2)

$\displaystyle |z-3i|\le \frac{3}{5}|z-5i|+\frac{2}{5}|z|.$

The combination of (1) and (2) tells us that

(3)

$|z|+|z-5i|\le |z-2i|+|z-3i|.$

Now, we have an equality in (3) only if both (1) and (2) are equalities which requires

(4)

$\overline{z}(z-5i)\ge 0.$

Let $z=a+bi,$ $a,b\in\mathbb{R}.$ Then (4) becomes $(a-bi)(a+bi-5i)\ge 0,$ i.e., $a^{2}+b(b-5)-5ai\ge 0.$ This is only possible when $a=0,$ implying $b(b-5)\ge 0.$ This is equivalent to $b\lt 0$ or $b\gt 5.$

Thus the solution is the set $\{bi:\; b\lt 0\;\mbox{or}\;b\gt 5\}.$

Solution 2

This solution makes use of the well-known Hlawka's inequaility:

(5)

$|x+y|+|y+z|+|z+x|\le |x|+|y|+|z|+|x+y+z|$

which holds for any there complex numbers $x,y,z$ (and has an analog in vector spaces with scalar product.) To use the inequality rewrite (*) as

$|z|+|-2i|+|-3i|+|z-5i|=|z-2i|+|-2i-3i|+|z-3i|.$

By Hlawka's inequaility,

$|z|+|-2i|+|-3i|+|z-5i|\ge |z-2i|+|-2i-3i|+|z-3i|$

which is (3).

For there to be equality, it is necessary that $\overline{(ab)}c(a + b + c)\ge 0.$ So, setting $a = -2i,$ $b = -3i$ and $c = z$ we have to require $\overline{(2i\cdot 3i)}z(z - 5i)\ge 0,$ or $z(z - 5i)\ge 0.$ Let $z = a + bi,$ where $a,b\in\mathbb{R}.$ Then $(a + bi)[a + (b - 5)i]\ge 0,$ or equivalently, $a=0$ and $b(b - 5)\ge 0$ which gives $b\in (-\infty , 0] \cup [5, \infty).$

Solution 3

Here's a more general formulation:

Let points $A, B, C, D$ lie on line $d$ (in this order), with $AB=CD;$ $M$ a point in the plane. Then

(**)

$MA+MD=MB+MC$

if and only if $M\in d\setminus \{AD\}.$

Using vectors, we have $\overrightarrow{MB} = k\overrightarrow{MA} + (1 - k)\overrightarrow{MD},$ where $k \in (0 , 1).$ $\overrightarrow{MC} = k\overrightarrow{MD} + (1 - k)\overrightarrow{MA}.$ It follows that

(6)

$MB = |\overrightarrow{MB}| = |k\overrightarrow{MA} + (1 - k)\overrightarrow{MD}| = kMA + (1 - k)MD$

and, similarly

(7)

$MC = kMD + (1 - k)MA.$

Combining (6) and (7) gives

(8)

$MB + MC \le MA + MD.$

(8) becomes an equality only if (6) and (7) are equalities, i.e., when vectors $\overrightarrow{MA}$ and $\overrightarrow{MD}$ are collinear and have the same orientation. This is exactly equivalent to $M\in d\setminus \{AD\}.$

Solution 4

As in the Solution #3, the four points $A,B,C,D$ need not lie on the $y$-axis, but may be collinear on any lie. For convenience I choose the $x$-axis, with points $A,B,C,D$ having coordinates $a,b,c,d,$ respectfully (so that $a\lt b\lt c\lt d).$ The situation is reminiscent of an optimization problem of building a house on a straight road. I start with giving a general formulation:

Let point $M$ in the plane have coordinates $(x,y).$ Assume $AB=CD.$ Then (**) holds if and only if $y=0$ and $x\in (-\infty, a]\cup [d,\infty).$

First we establish that for points outside the $x$-axis the equality in (**) cannot hold. Let $M'$ be the reflection of $M$ in the midpoint of $AD,$ or $BC,$ which is the same because $AB=CD:$

an equation with four points on a line

By the construction, the parallelogram $MBCM'$ is located entirely within the parallelogram $MADM'$ such that its perimeter is less than that of the latter:

$2(BM+CM)\lt 2(AM+DM).$

Thus, we have only inspect points $M$ on the $x$-axis. For $b\le x\le c,$ $BM+CM=c-b$ while $AM+DM=d-a.$ The latter still holds for, say $a\lt x\lt b.$ However,

$\begin{align} MB+MC &= (b-x)+(c-x)\\ &\lt (b-a)+[(c-b)+(b-x)]\\ &\lt [(b-a)+(c-b)]+(b-a)\\ &\lt (c-a)+(d-c)\\ &=d-a\\ &=AM+DM. \end{align}$

By analogy, the equality in (**) can't hold for $c\lt x\lt d$ either. Let now $x\le a.$ In this case,

$\begin{align} MB+MC &= (b-x)+(c-x)\\ &= (b-a)+ (c-a)+2(a-x)\\ &= (d-c)+ (c-a)+2(a-x)\\ &= (d-a)+2(a-x)\\ &= (d-x)+(a-x)\\ &=AM+DM, \end{align}$

and, similarly, for $x\ge d.$

Solution 5

The solutions are the points common to two ellipses with the same major axis and are thus the possible end points of the major axis.

an equation with four points on a line with two ellipse

Solution 6

Square the two sides of (*) to obtain first

$z\cdot\overline{z} + (z - 5i)(z + 5i) + 2|z^{2}-5iz| = (z - 2i)(z + 2i) + (z - 3i)(z + 3i) + 2|z^{2} - 5iz - 6|.$

which is then simplified to

$|z^{2}-5iz| + 6 = |z^{2} - 5iz - 6|.$

Denoting $Z=z^{2}-5iz$ we obtain $|Z|+6=|Z-6|.$ Due to the triangle inequality, this implies that $Z$ is real and not positive: $Z\in\mathbb{R}$ and $Z\le 0.$ Thus, for $z$ we have a quadratic equation $z^{2} - 5iz=a\le 0,$ with a real $a.$ Solving this equation we obtain

$\displaystyle z=\frac{5i\pm\sqrt{-25+4a}}{2}=\frac{i}{2}(5\pm\sqrt{25-4a})=Ai,$

where $A=\displaystyle\frac{1}{2}(5\pm\sqrt{25-4a}),$ $a\le 0.$ Choosing sign "+" gives $A=\displaystyle\frac{1}{2}(5+\sqrt{25-4a})\ge 5.$ Choosing sign "-" gives $A=\displaystyle\frac{1}{2}(5-\sqrt{25-4a})\le 0.$

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