Two points $A$ and $B$ are given in the plane. A point $X$ is called their preposterous midpoint if there is a Cartesian coordinate system in the plane such that the coordinates of $A$ and $B$ in this system are non-negative, the abscissa of $X$ is the geometric mean of the abscissae of $A$ and $B$, and the ordinate of $X$ is the geometric mean of the ordinates of $A$ and $B$. Find the locus of all the preposterous midpoints of $A$ and $B$. (K. Tyschu)
Problem
Source: Tuymaada 2017 Seniors Q8 (day 2 P4)
Tags: Locus, coordinate geometry, analytic geometry, geometry
don2001
03.11.2017 18:00
Any solution.
Mindstormer
06.12.2017 23:44
I got that required locus is disc with diameter $AB$ without its center. Indeed, it’s not hard to verify that $MX \leq {AB \over 2}$, and that by taking $A=(1,0)$ with arbitrary $B$ we can get all described points.
This is motivated by the extreme case where $A$ and $B$ are located on different axes, so that $X=(0,0)$ and $\angle BXA=90^\circ$.
ACGNmath
24.07.2020 15:18
We claim that the answer is the disk with diameter $AB$, without the center. Let $M$ denote the midpoint of $AB$, and $X$ a preposterous midpoint. Claim: $MX\leq \frac{AB}{2}$. Proof: Let $A=(a_1,a_2)$, $B=(b_1,b_2)$, $X=(\sqrt{a_1b_1},\sqrt{a_2b_2})$. We wish to show that $MX^2\leq \frac{AB^2}{4}$. \begin{align*} \Leftrightarrow \left(\frac{a_1+b_1}{2}-\sqrt{a_1b_1}\right)^2+\left(\frac{a_2+b_2}{2}-\sqrt{a_2b_2}\right)^2 &\leq \frac{(a_1-b_1)^2+(a_2-b_2)^2}{4} \\ \Leftrightarrow (\sqrt{a_1}-\sqrt{b_1})^4+(\sqrt{a_2}-\sqrt{b_2})^4 &\leq (a_1-b_1)^2+(a_2-b_2)^2 \\ \Leftrightarrow a_1^2+a_2^2+b_1^2+b_2^2-4a_1^{3/2}b_1^{1/2}-4a_1^{1/2}b_1^{3/2}-4a_2^{3/2}b_2^{1/2}-4a_2^{1/2}b_2^{3/2}+6a_1b_1+6a_2b_2&\leq a_1^2+b_1^2+a_2^2+b_2^2-2a_1b_1-2a_2b_2 \\ \Leftrightarrow 8a_1b_1+8a_2b_2 &\leq 4a_1^{3/2}b_1^{1/2}+4a_1^{1/2}b_1^{3/2}+4a_2^{3/2}b_2^{1/2}+4a_2^{1/2}b_2^{3/2} \end{align*}which is true by AM-GM. Hence, all preposterous midpoints in the disk with diameter $AB$. To construct the coordinates, suppose $C$ lies in the disk. If $ACB$ is anti-clockwise oriented, set $A=(1,0)$, $C=(\sqrt{x},0)$ and $B=(x,y)$. Clearly all points can be achieved in the semicircle. Similarly all points can be achieved in the other semicircle. Now consider the diameter. For $A$, $C$, $B$ collinear, $A=(1,0)$, $C=(\sqrt{x},0)$ and $B=(x,0)$ works, except for the midpoint $M$. If $M$ were to be a preposterous midpoint, then $\sqrt{a_1b_1}=\frac{a_1+b_1}{2}$, which means $a_1=b_1$. Similarly $a_2=b_2$, hence $A=B$. Contradiction. Hence we have found the locus of all preposterous midpoints.