equifacial tetrahedron and circumcircles of faces

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

equifacial tetrahedron and circumcircles of faces

G H J

Reveal topic tags

trig identities

L

G H BBookmark kLocked kLocked NReply

Source: RMO District 2005, 8th Grade, Problem 3

The post below has been deleted. Click to close.

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

Valentin Vornicu

7301 posts

Valentin Vornicu

#1 h Mar 5, 2005, 10:24 PM • 2 Y

Y by Adventure10, Mango247

Prove that if the circumcircles of the faces of a tetrahedron $ABCD$ have equal radii, then $AB=CD$ , $AC=BD$ and $AD=BC$ .

Z K Y

The post below has been deleted. Click to close.

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

1072 posts

#2 h Mar 6, 2005, 7:43 AM • 2 Y

Y by Adventure10, Mango247

(A picture might help follow)

Let <CAB=a, <ABC=b, <BCA=c, so that a+b+c=180. From Law of Sines, sin(BAC)=BC/(2R)=sin(BDC), so either <BDC=a or <BDC=180-a. Let's dispose of the second case.

Assume <BDC=180-a. We must also have <ADB=c or 180-c and <ADC=b or 180-b. We can't have both 180-b and 180-c, cause then at vertex D the three angles would add up to 360, i.e. vertex D would be flat. Furthermore, we cant have both b and c because then at vertex D, b+c=180-a, i.e. vertex D is flat again. So, WLOG, assume it is 180-b and c.

We must have (180-a)+c>(180-b), which (remembering that c=180-a-b) is equivalent to a<90. Likewise, b is acute. Also, we must have (180-a)+(180-b)+c<360, which is equivalent to c<90. Since <BDC=180-a and <ADC=180-b are obtuse, <DBC and <DAC are both acute, and since sin(DBC)=sin(DAC), the two angles are the same. Call their measure d. Also let <DCB=e and <DCA=f, which are both acute. Since (180-a)+d+e=180, e=a-d, and likewise, f=b-d. Let's look at triangle DAB.

If <DAB=e and <DBA=f, then c+(a-d)+(b-d)=180 implies d=0, which isn't possible. Also, if <DAB=180-e and <DBA=180-f, then triangle DAB has two obtuse angles, not possible. If <DAB=180-e and <DBA=f, then c+(180-e)+f=180 is equivalent to a=90, contradicting the fact that a is acute. So we have reached our contradiction, showing that <BDC=a and not 180-a.

In the same way, <ADC=b, <ADB=c, <BAD=e, and <ABD=f. We have 2(d+e+f)=(a+d+e)+(b+d+f)+(c+e+f)-(a+b+c)=180+180+180-180=360, implying d+e+f=180. Thus, f=180-d-e=a, and likewise c=d and b=e. So triangles BAD and DCB have the same angles and the same corresponding side BD, so they are congruent, so AB=CD. Likewise, AC=BD and AD=BC.

Z K Y

The post below has been deleted. Click to close.

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

392 posts

kueh

#3 h Mar 10, 2005, 9:37 AM • 2 Y

Y by Adventure10, Mango247

augh, i can't visualise these things

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a