many points on a sphere :|

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#1 h Apr 23, 2007, 6:43 PM • 2 Y

Y by Adventure10, Mango247

Let $ABCD$ be a tetrahedron and let $E,F,G,H,K,L$ be points lying on the edges $AB,BC,CD$ $,DA,DB,DC$ respectively, in such a way that
$AE \cdot BE = BF \cdot CF = CG \cdot AG= DH \cdot AH=DK \cdot BK=DL \cdot CL.$
Prove that the points $E,F,G,H,K,L$ all lie on a sphere.

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#2 h Apr 24, 2007, 8:37 PM • 1 Y

Y by Adventure10

Thank you pohoatza for a nice problem.

I think that the points $E,$ $F,$ $G,$ $H,$ $K,$ $L$ and also the points $E',$ $F',$ $G',$ $H',$ $K',$ $L',$ as their isotomic points with respect to side segments $AB,$ $BC,$ $AC,$ $AD,$ $BD,$ $BC$ respectively, lie on a sphere taken the same center as the circumsphere of the given tetrahedron.

I will search my notes and if I am not mistaken, I will post here later the proof I have in mind.

Kostas vittas.

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gemath

#3 h Apr 25, 2007, 10:00 AM • 4 Y

Y by dizzy, Adventure10, Mango247, and 1 other user

I think solution is simple as following:
Let $(O)$ be circumsphere of $ABCD$ we have
$AE \cdot BE =-P_{E/(O)}=R^{2}-OE^{2}$ similarly for $F,G,H,K,L$ ( $P_{X/(O)}$ is power of $X$ w.r.t $(O)$ )thus
$R^{2}-OE^{2}=R^{2}-OF^{2}=....=R^{2}-OL^{2}\Rightarrow OE=OF=...=OL$ thus they lie on a sphere center $O$
@Vittasko:what's isotomic points with respect to side segments ?

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#4 h Apr 25, 2007, 2:55 PM • 2 Y

Y by Adventure10, Mango247

Let the triangle $\bigtriangleup ABC$ be, as the face of the given tetrahedron $ABCD.$

Let also $E,$ $G$ be, the points on side segments $AB,$ $AC$ respectively, as the problem states, such that $(EA)\cdot (EB) = (GA)\cdot (GC) = \kappa^{2}$ $,(1)$

We denote as $P,$ $P',$ the intersection points of the circumcircle $(O_{1})$ of $\bigtriangleup ABC,$ from the segment line $EG.$

So, we have $(EP)\cdot (EP') = (EA)\cdot (EB) = (GA)\cdot (GC) = (GP')\cdot (GP)$

$\Longrightarrow$ $(EP)\cdot (EP') = (GP')\cdot (GP)$ $\Longrightarrow$ $\frac{EP}{GP}= \frac{GP'}{EP'}$

$\Longrightarrow$ $\frac{EP}{GP-EP}= \frac{GP'}{EP'-GP'}$ $\Longrightarrow$ $\frac{EP}{EG}= \frac{GP'}{EG}$

$\Longrightarrow$ $EP = GP'$ $\Longrightarrow$ $O_{1}E = O_{1}G$ $,(2)$

Because of $(EP)\cdot (EP') = \kappa^{2}= R_{1}^{2}-(O_{1}E)^{2}$ where $R_{1},$ is the circumradius of $\bigtriangleup ABC,$ we conclude that $O_{1}E = O_{1}G = \sqrt{R_{1}^{2}-\kappa^{2}}$ $,(3)$

That is the points $E,$ $G,$ lie on a circle $(O_{1}),$ centered at $O_{1},$ with radius given by $(3).$

By the same way we can also prove that the point $F$ lies on $BC$ and such that $(FB)\cdot (FC) = \kappa^{2},$ belongs to $(O_{1}).$

Also the points $E',$ $F',$ $G',$ as the isotomic points of $E,$ $F,$ $G,$ with respect to $AB,$ $BC,$ $AC$ respectively, lie on the circle $(O_{1}).$

$\bullet$ We consider now, the triangle $\bigtriangleup BCD,$ as the other face of the given tetrahedron $ABCD,$ whose the side segment $BC,$ is the intersection line of the planes $[ABC]$ and $[BCD].$

We denote as $O_{2}$ the circumcenter of the triangle $\bigtriangleup BCD$ and it is easy to show that the center $O,$ of the circumsphere of $ABCD,$ is the intersection point of the lines through the points $O_{1},$ $O_{2}$ and perpendicular to the planes $[ABC],$ $[BCD],$ respectively.

Let $K,$ $L$ be, the points on $BD,$ $CD$ respectively, such that $(KB)\cdot (KD) = (LC)\cdot (LD) = \kappa^{2}$ and also let $K',$ $L'$ be, their isotomic points with respect to $BD,$ $CD,$ respectively.

Because of $O_{1}E = O_{1}E' = O_{1}F = O_{1}F' = O_{1}G = O_{1}G'$ and $OO_{1}\perp [ABC],$ we conclude that $OE = OE' = OF = OF' = OG = OG'$ $,(4)$

By the same way we have that $OK = OK' = OF = OF' = OL = OL'$ $,(5)$

From $(4),$ $(5)$ we conclude that the points $E,$ $F,$ $G,$ $K,$ $L$ and also the points $E',$ $F',$ $G',$ $K',$ $L',$ as their isotomic point with respect to $AB,$ $BC,$ $AC,$ $BD,$ $BC$ respectively, lie on a sphere $(S'),$ taken the same center $(O),$ as the circumsphere $(S)$ of the given tetrahedron $ABCD.$

By the same way ( consider the triangles $\bigtriangleup ABD,$ $\bigtriangleup ACD,$ as the other two faces of $ABCD$ ), we can prove that the points $H,$ $H',$ isotomic conjugates with respect to $AD$ and such that $(HA)\cdot (HD) = (H'A)\cdot (H'D) = \kappa^{2},$ lie on the sphere $(S')$ and the proof is completed.

Kostas Vittas.

PS. We say two points as isotomic conjugates with respect to the corrspondent sidesegment of a given triangle, if they are symmetrical each other wrt its midpoint.

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gemath

#5 h Apr 25, 2007, 3:12 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

I have inspired a similar problem as this problem
Given regular tetrahedron $ABCD$ inscribed $(O)$ and $A',B',C',D'$ are arbitrary points on (inside) the face $BCD,CDA,DAB,ABC$ inface $(BCD)$ let $[A'BC]=S_{ad},[A'CD]=S_{ab},[A'DB]=S_{ac}$ similarly in others face, prove that $A',B',C',D'$ lie on a sphere center $O$ iff $S_{ad}^{2}+S_{ab}^{2}+S_{ac}^{2}=S_{bc}^{2}+S_{bd}^{2}+S_{ba}^{2}=S_{cd}^{2}+S_{cb}^{2}+S_{ca}^{2}=S_{da}^{2}+S_{db}^{2}+S_{dc}^{2}$

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