THEOREMS IN TETRAHEDRON
LEONARD GIUGIUC
Lemma 0.1. Let ABCD be a tetrahedron. Then G = O if and only if ABCD is
equifacial.
Lemma 0.1’s proof (Leonard Giugiuc):
⇐ Let AB = CD = a, AC = BD = b and AD = BC = c. We have:
−−→ −→ −−→
−−→ −→ −−→ !2
−→ AB + AC + AD
AB + AC + AD
2
⇒
⇒ AG =
AG =
4
4
−−→ −→
−→ −−→
−−→ −−→
AB 2 + AC 2 + AD2 + 2AB · AC + 2AC · AD + 2AB · AD
=
16
a2 + b2 + c2
a2 + b2 + c2 + a2 + b2 − c2 − a2 + b2 + c2 + a2 − b2 + c2
=
⇒
=
16
8
r
a2 + b2 + c2
⇒ AG =
8
r
a2 + b2 + c2
Due to the symmetry of the expression, get BG = CG = DG =
8
⇒ G = O, hence
−−→ −→ −−→
−−→ −→ −−→ !2
−→ AB + AC + AD
AB + AC + AD
2
AO =
⇒R =
⇒
4
4
AG2 =
−−→ −→
−→ −−→
−−→ −−→
AB 2 + AC 2 + AD2 + 2AB · AC + 2AC · AD + 2AB · AD
=
16
3(AB 2 + AC 2 + AD2 ) − (BC 2 + BD2 + CD2 )
=
; similarly,
16
3(AB 2 + BC 2 + BD2 ) − (AC 2 + AD2 + CD2 )
R2 =
=
16
3(AC 2 + BC 2 + CD2 ) − (AB 2 + AD2 + BD2 )
=
=
16
3(AD2 + BD2 + CD2 ) − (AB 2 + AC 2 + BC 2 )
=
.
16
All of these gives us AB = CD, AC = BD and AD = BC.
R2 =
√
Theorem 0.2. In any tetrahedron ABCD, AG + BG + CG + DG ≤ 4 R2 − d2 ,
d = GO and the notations are as usual. Equality holds if f ABCD is equifacial.
1
2
LEONARD GIUGIUC
Theorem 0.2’s proof (Leonard Giugiuc):
−→ −−→ −−→ −−→ →
−→ −−→ −−→ −−→
−
We have: AG + BG + CG + DG = 0 ⇒ (AG + BG + CG + DG)2 = 0 ⇒
X
X −→ −−→
AG·BG = 0 ⇒ 4(AG2 +BG2 +CG2 +DG2 ) =
AB 2 .
AG2 +BG2 +CG2 +DG2 +2
all
all
−→ −−→ −−→ −−→
−−→
−→ −−→ −−→ −−→
On the other hand, AO + BO + CO + DO = 4GO ⇒ (AO + BO + CO + DO)2 =
= 16GO2 ⇒
X
16(R2 − d2 ) =
AB 2 ⇒ 4(AG2 + BG2 + CG2 + DG2 ) = 16(R2 − d2 ). But
all
p
4(AG2 + BG2 + CG2 + DG2 ) = 4 R2 − d2 . We are done.
p
But AG + BG + CG + DG = 4(AG2 + BG2 + CG2 + DG2 )
if f AG = BG = CG = DG and by the lemma AG = BG = CG = DG
if f ABCD is equifacial.
AG+BG+CG+DG ≤
p
Theorem 0.3. In any tetrahedron ABCD, 8R2 ≥ AB · CD + AC · BD + AD · BC.
Equality holds if f ABCD is equifacial.
Theorem 0.3’s proof (Leonard Giugiuc):
We have: 2AB · CD ≤ AB 2 + CD2 , 2AC · BD ≤ AC 2 + BD2 and
2AD · BC ≤ AD2 + BC 2 ⇒
X
2(AB · CD + AC · BD + AD · BC) ≤
AB 2 = 16(R2 − d2 ) ≤ 16R2 .
all
Equality holds iff AB = CD, AC = BD and AD = BC.
Theorem 0.4. In any tetrahedron ABCD,
P
√
all
AB ≤ 4R 6.
Theorem 0.4’s proof (Leonard Giugiuc):
!2
X
We have:
AB ≤ 3(AB + CD)2 + 3(AC + BD)2 + 3(AD + BC)2 =
all
!
=3
X
AB
2
+6(AB · CD + AC · BD + AD · BC) ≤ 48R2 + 48R2 ⇒
all
⇒
X
√
AB ≤ 4R 6
all
From above, clearly equality holds if f ABCD is equifacial.
Mathematics Department, ”Theodor Costescu” National Economic College, Drobeta
Turnu - Severin, MEHEDINTI.
E-mail address: [email protected]