Prove that there is a unique point satisfying conditions

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

Prove that there is a unique point satisfying conditions

G H J

Reveal topic tags

analytic geometry

geometry unsolved

L

G H BBookmark kLocked kLocked NReply

Source: Vietnam TST 1997, Problem 1

The post below has been deleted. Click to close.

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

1270 posts

April

#1 h Jul 28, 2008, 2:01 AM • 2 Y

Y by Adventure10, Mango247

Let $ABCD$ be a given tetrahedron, with $BC = a$ , $CA = b$ , $AB = c$ , $DA = a_1$ , $DB = b_1$ , $DC = c_1$ . Prove that there is a unique point $P$ satisfying
$PA^2 + a_1^2 + b^2 + c^2 = PB^2 + b_1^2 + c^2 + a^2 = PC^2 + c_1^2 + a^2 + b^2 = PD^2 + a_1^2 + b_1^2 + c_1^2$
and for this point $P$ we have $PA^2 + PB^2 + PC^2 + PD^2 \ge 4R^2$ , where $R$ is the circumradius of the tetrahedron $ABCD$ . Find the necessary and sufficient condition so that this inequality is an equality.

Z K Y

The post below has been deleted. Click to close.

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

21 posts

#2 h Apr 22, 2009, 12:51 AM • 1 Y

Y by Adventure10

this problem was have nice solution with algbe ! See at vietnam tst of nguyen van mau.

Z K Y

The post below has been deleted. Click to close.

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

666 posts

#3 h Apr 22, 2009, 5:05 PM • 2 Y

Y by Adventure10, Mango247

$\mbox{In tetrahedron } A_1A_2A_3A_4 \mbox{ denote, with: } a_{ij} = |A_iA_j|; \\ \mbox{for } i;j\in\{1;2;3;4\}; i\neq j. \mbox{Prove that there is a unique point P satisfying: } \\ |PA_1|^2 + a_{12}^2 + a_{13}^2 + a_{14}^2 = |PA_2|^2 + a_{12}^2 + a_{23}^2 + a_{24}^2 = \\ = |PA_3|^2 + a_{13}^2 + a_{23}^2 + a_{34}^2 = |PA_4|^2 + a_{14}^2 + a_{24}^2 + a_{34}^2.$ (1)
1). $\mbox{Using barycentric absolute coordonate, if: } P(\alpha; \beta;\gamma; \delta); \alpha + \beta + \gamma + \delta = 1,$
$\mbox{ then have: }$
$ |PA_1|^2 = - (\alpha - 1)\beta .a_{12}^2 - (\alpha - 1)\gamma .a_{13}^2 - (\alpha - 1)\delta .a_{14}^2 - \beta \gamma .a_{23}^2 - \beta\delta .a_{24}^2 - \gamma \delta .a_{34}^2; \\ |PA_2|^2 = - \alpha(\beta - 1).a_{12}^2 - \alpha \gamma .a_{13}^2 - \alpha \delta .a_{14} - (\beta - 1)\gamma .a_{23}^2 - (\beta - 1)\delta .a_{24}^2 - \gamma \delta .a_{34}^2; \dots \\ \mbox{and (1) becomes: } \\ - \alpha \beta .a_{12}^2 - \alpha \gamma .a_{13}^2 - \alpha \delta .a_{14}^2 - \beta \gamma .a_{23}^2 - \beta\delta .a_{24}^2 - \gamma \delta .a_{34}^2 + \\ + (\beta + 1)a_{12}^2 + (\gamma + 1)a_{13}^2 + (\delta + 1)a_{14}^2 = \\ = - \alpha \beta .a_{12}^2 - \alpha \gamma .a_{13}^2 - \alpha \delta .a_{14}^2 - \beta \gamma .a_{23}^2 - \beta\delta .a_{24}^2 - \gamma \delta .a_{34}^2 + \\ + (\alpha + 1).a_{12}^2 + (\gamma + 1).a_{23}^2 + (\delta + 1).a_{24}^2 = \dots \Leftrightarrow \\ \Leftrightarrow (\beta + 1)a_{12}^2 + (\gamma + 1)a_{13}^2 + (\delta + 1)a_{14}^2 = (\alpha + 1).a_{12}^2 + (\gamma + 1).a_{23}^2 + (\delta + 1).a_{24}^2 = \\ = (\alpha + 1).a_{13}^2 + (\beta + 1).a_{23}^2 + (\delta + 1).a_{34}^2 = (\alpha + 1).a_{14}^2 + (\beta + 1).a_{24}^2 + (\gamma + 1).a_{34}^2. \\ \Rightarrow \mbox{The barycentric coordonates of point P is solutions of system: } \\ \left\{\begin{array}{c}( - (\alpha + 1).a_{14}^2 + ( \beta + 1)(a_{12}^2 - a_{24}^2) + (\gamma + 1)(a_{13}^2 - a_{34}^2) + (\delta + 1)a_{14}^2 = 0 \ (\alpha + 1)(a_{12}^2 - a_{14}^2) - (\beta + 1).a_{24}^2 + (\gamma + 1)(a_{23}^2 - a_{34}^2) + (\delta + 1).a_{24}^2 = 0 \ (\alpha + 1)(a_{13}^2 - a_{14}^2) + (\beta + 1)(a_{23}^2 - a_{24}^2) - (\gamma + 1).a_{34}^2 + (\delta + 1).a_{34}^2 = 0 \ (\alpha + 1) + (\beta + 1) + (\gamma + 1) + (\delta + 1) = 5 \end{array} \right ;$
$ \mbox{but } \\ \left| \begin{array}{cccc}{cccc} { - a_{14}^2} & {a_{12}^2 - a_{24}^2} & {a_{13}^2 - a_{34}^2} & {a_{14}^2} \ {a_{12}^2 - a_{14}^2} & { - a_{24}^2} & {a_{23}^2 - a_{34}^2} & {a_{24}^2} \ {a_{13}^2 - a_{14}^2} & {a_{23}^2 - a_{24}^2} & { - a_{34}^2} & {a_{34}^2} \ 1 & 1 & 1 & 1 \end{array}\right| = \left| \begin{array} {ccccc} a_{14}^2 & a_{24}^2 & a_{34}^2 & 0 & 1 \ { - a_{14}^2} & {a_{12}^2 - a_{24}^2} & {a_{13}^2 - a_{34}^2} & {a_{14}^2} & 0 \ {a_{12}^2 - a_{14}^2} & { - a_{24}^2} & {a_{23}^2 - a_{34}^2} & {a_{24}^2} & 0 \ {a_{13}^2 - a_{14}^2} & {a_{23}^2 - a_{24}^2} & { - a_{34}^2} & {a_{34}^2} & 0 \ 1 & 1 & 1 & 1 & 0 \end{array}\right| = \ = \left| \begin{array} {ccccc} a_{14}^2 & a_{24}^2 & a_{34}^2 & 0 & 1 \ 0 & {a_{12}^2} & {a_{13}^2} & {a_{14}^2} &{ 1 \ {a_{12}^2}} & 0 & {a_{23}^2} & {a_{24}^2} & 1 \ {a_{13}^2} &{ {a_{23}^2}} & 0 & {a_{34}^2} & 1 \ 1 & 1 & 1 & 1 & 0 \end{array}\right| = - \left| \begin{array} {ccccc} 0 & {a_{12}^2} & {a_{13}^2} & {a_{14}^2} &{ 1 \ {a_{12}^2}} & 0 & {a_{23}^2} & {a_{24}^2} & 1 \ {a_{13}^2} &{ {a_{23}^2}} & 0 & {a_{34}^2} & 1 \ a_{14}^2 & a_{24}^2 & a_{34}^2 & 0 & 1 \ 1 & 1 & 1 & 1 & 0 \end{array}\right| = \\ = - 288.V^2\neq 0 \Rightarrow \\ \mbox{ Solution: } (\alpha + 1; \beta + 1; \gamma + 1; \delta + 1) \mbox{ exist and unique } \Rightarrow \mbox{ the point P unique.}$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a