Polish tetrahedron

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

Polish tetrahedron

G H J

Reveal topic tags

geometry solved

L

G H BBookmark kLocked kLocked NReply

Source: Poland 2001

The post below has been deleted. Click to close.

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

278 posts

#1 h Sep 22, 2004, 1:49 PM • 2 Y

Y by Adventure10, Mango247

Given a regular tetrahedron $ABCD$ with edge length $1$ and a point $P$ inside it.
What is the maximum value of $\left|PA\right|+\left|PB\right|+\left|PC\right|+\left|PD\right|$ .

Z K Y

The post below has been deleted. Click to close.

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

7849 posts

#2 h Sep 28, 2004, 12:43 AM • 2 Y

Y by Adventure10, Mango247

I assume $P$ can also be on the border of $ABCD$ , because otherwise the maximum can't be reached (in case the maximum is really what we want; are you sure the problem didn't ask for the minimum?).

I'll use the fact that the distance between two points in (or on) the regular tetrahedron with side $\ell$ is at most $\ell$ . Through $P$ draw a plane parallel to $(BCD)$ , cutting $AB,AC,AD$ in $B',C',D'$ respectively. In the plane $(B'C'D')$ draw a parallel through $P$ to $C'D'$ , cutting $B'C',B'D'$ in $C'',D''$ respectively.

We now have $PA\le AB'\ (1)$ , $PB\le PB'+BB'\le B'C''+BB'\ (2)$ , $PC+PD\le DD'+D'D''+D''C''+C''C'+C'C\ (3)$ . If you draw the picture properly you'll see that the sum of the three RHS's is exactly $3\ell$ ( $3$ , in our case, since $\ell=1$ ). The equality is reached only when $P$ is one of the vertices of the tetrahedron.

Z K Y

The post below has been deleted. Click to close.

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

278 posts

#3 h Oct 2, 2004, 3:33 PM • 2 Y

Y by Adventure10, Mango247

Really P can be on the boundary.
Nice solution.

Z K Y

The post below has been deleted. Click to close.

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

6555 posts

#4 h Nov 2, 2004, 5:29 PM • 2 Y

Y by Adventure10, Mango247

hardsoul wrote:

Given a regular tetrahedron ABCD with edge length 1 and a point P inside it.
What is the maximum value of |PA|+|PB|+|PC|+|PD|.

I am just seeing that this is a corollary of the well-known fact that the maximum of a convex function inside a convex polyhedron is reached in a vertex of the polyhedron. This time, the convex function is |PA| + |PB| + |PC| + |PD| (the distance function is convex!). Actually, Grobber's solution was just an elementary formulation of this.

EDIT: See http://www.mathlinks.ro/Forum/viewtopic.php?t=19485 for details.

Darij

This post has been edited 1 time. Last edited by darij grinberg, Jun 13, 2006, 1:58 PM

Z K Y

The post below has been deleted. Click to close.

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

26 posts

armanf

#5 h Jun 13, 2005, 2:21 PM • 2 Y

Y by Adventure10, Mango247

Excuse me .
i wanted to know why the function |PA| + |PB| + |PC| + |PD| is convex .
thanks for your help.

Z K Y

The post below has been deleted. Click to close.

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

33 posts

#6 h Oct 14, 2019, 4:04 PM • 2 Y

Y by Adventure10, Mango247

hardsoul wrote:

Given a regular tetrahedron $ABCD$ with edge length $1$ and a point $P$ inside it.
What is the maximum value of $\left|PA\right|+\left|PB\right|+\left|PC\right|+\left|PD\right|$ .

We claim that the maximum value is $3$ . Equality occurs at any one of the vertices. Let $f(P) = AP + BP + CP + DP$ . Suppose $P$ is the point for which $f$ attains it maximum value.

Claim 1: $P$ lies on one of the face of the tetrahedron

Proof. Suppose not then consider the ellipsoid $\mathcal{C}_1, \mathcal{C}_2$ with foci $(A,B) ; (C,D)$ . with sum of distance from foci $AP + BP$ and $CP +DP$ respectively. Note that there exists another ellipsoid $\mathcal{C}_3$ with foci $(C,D)$ and sum of focal distance slightly larger than that of $\mathcal{C}_2$ such that $\mathcal{C}_3$ meets $\mathcal{C}_1$ inside the tetrahedron. This leads to a contradiction and we get the desired result. $\square$

Claim 2: $P$ lies on the sides of equilateral triangle $\triangle ABC$ if $P$ maximizes $g(P) = AP + BP + CP$

Proof. Similar to proof of Claim 1. $\square$

By our Claim 1 and Claim 2, we have $P$ lies on the edges of the tetrahedron let’s say $AB$ . Note that $AP + BP + CP + DP = 1 + CP + DP \leq 3$ . $\blacksquare$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a