Sum of face angles of a tetrahedron - ISL 1986

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

w

15 users

TOPICS

No tags match your search

M

No tags match your search

M

Sum of face angles of a tetrahedron - ISL 1986

G H J

Reveal topic tags

congruent triangles

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

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

5452 posts

#1 h Aug 31, 2010, 10:33 AM • 1 Y

Y by Adventure10

Prove that the sum of the face angles at each vertex of a tetrahedron is a straight angle if and only if the faces are congruent triangles.

Z K Y

The post below has been deleted. Click to close.

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

8 posts

JGU

#2 h Jun 13, 2013, 3:14 PM • 2 Y

Y by Adventure10, Mango247

If the faces of tetrahedron $ABCD$ are congruent triangles,
then $AB = CD$ , $AC = BD$ , $AD = BC$

$\triangle BCD \cong \triangle ADC \cong \triangle DAB \cong \triangle CBA$

sum of the face angles at $A$ : $\angle BAC + \angle CAD + \angle DAB = \angle CDB + \angle DBC + \angle BCD = \pi$
similarly $\angle ABC + \angle CBD + \angle DBA = \pi$ , $\angle ACB + \angle BCD + \angle DCA = \pi$ , $\angle ADB + \angle BDC + \angle CDA = \pi$

Conversely, if the sum of the face angles at each vertex of tetrahedron $ABCD$ is $\pi$
$\angle BAC + \angle CAD + \angle DAB = \pi$ ,
$\angle ABC + \angle CBD + \angle DBA = \pi$ ,
$\angle ACB + \angle BCD + \angle DCA = \pi$ ,
$\angle ADB + \angle BDC + \angle CDA = \pi$

1. by trihedral angle theorem, $\angle CAD + \angle DAB > \angle BAC$ , then we have $\angle BAC < \pi/2$
this holds for all 12 angles above, that is, they are acute angles. Note if $\theta$ and $\psi$ are acute angles, then $\theta \ge \psi$ if and only if sin $\theta \ge$ sin $\psi$

2. denote the circumradius of $\triangle BCD$ as $R_A$ , the circumradius of $\triangle ADC$ as $R_B$ , the circumradius of $\triangle DAB$ as $R_C$ , the circumradius of $\triangle CBA$ as $R_D$

without loss of generality, assume $R_A$ is the largest

by law of sines $2R_A$ sin $\angle BDC = BC = 2R_D$ sin $\angle BAC$ , and since $R_A \ge R_D$ , then

$\angle BAC \ge \angle BDC$ and similarly we can deduce
$\angle CAD \ge \angle CBD$
$\angle DAB \ge \angle DCB$
adding above we get $\pi \ge \pi$ so all inequality are equality, $R_A = R_B = R_C = R_D$ and

$\angle BAC = \angle BDC$ ,
$\angle CAD = \angle CBD$ ,
$\angle DAB = \angle DCB$ ,

by law of sines again $2R_B$ sin $\angle DCA = DA = 2R_C$ sin $\angle DBA$ , and since $R_B = R_C$ , then

$\angle DCA = \angle DBA$ , and similarly
$\angle ACB = \angle ADB$ ,
$\angle ABC = \angle ADC$

3. now we know the 12 angles are in 6 pairs, then since all 12 angles add to $4\pi$ , picking one angle in each pair:

$\angle BAC + \angle CAD + \angle DAB + \angle DBA + \angle ADB + \angle ABC = 2\pi$

substract $\angle BAC + \angle CAD + \angle DAB = \pi$ , we have $\angle DBA + \angle ADB + \angle ABC = \pi$

compare to angles of $\triangle ABD$ results $\angle ABC = \angle BAD$

then $AC = 2R_D$ sin $\angle ABC = 2R_C$ sin $\angle BAD = BD$

similarly we can deduce $AB=CD$ , $AD=BC$

so $\triangle BCD \cong \triangle ADC \cong \triangle DAB \cong \triangle CBA$ (SSS)

Z K Y

The post below has been deleted. Click to close.

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

76 posts

duanby

#3 h Jun 14, 2013, 4:49 AM • 2 Y

Y by Adventure10, Mango247

The if part is obvious
the only if part:
make the faces on a plane,that is on plane
ABC let D1,D2,D3 be the point that BCD1 congruent to BCA
D2,D3 the same way
Then D2D3A are collinear so ABC is the midpoint triangle of
D1D2D3 then we get the result

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a