locus with coordinates from distances about 2 points

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

locus with coordinates from distances about 2 points

G H J

Reveal topic tags

coordinate geometry

analytic geometry

L

G H BBookmark kLocked kLocked NReply

Source: Tuymaada 2017 Seniors Q8 (day 2 P4)

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

30629 posts

#1 h Sep 26, 2017, 11:25 PM • 2 Y

Y by Adventure10, Mango247

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)

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

170 posts

#2 h Nov 3, 2017, 3:00 PM • 2 Y

Y by Adventure10, Mango247

Any solution.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

102 posts

#3 h Dec 6, 2017, 8:44 PM • 2 Y

Y by mohammad14101381, Adventure10

answer

This post has been edited 1 time. Last edited by Mindstormer, Dec 6, 2017, 9:16 PM
Reason: spelling

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

327 posts

#4 h Jul 24, 2020, 12:18 PM • 1 Y

Y by jatrophalouvre

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.

This post has been edited 1 time. Last edited by ACGNmath, Jul 24, 2020, 12:20 PM

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a