six angles subtended at an interior point of a tetrahedron

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six angles subtended at an interior point of a tetrahedron

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Source: IMO 1980 Austria-Poland, problem 3

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#1 h May 6, 2004, 4:10 PM • 2 Y

Y by Adventure10, Mango247

Prove that the sum of the six angles subtended at an interior point of a tetrahedron by its six edges is greater than 540°.

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#2 h Jun 9, 2004, 10:03 PM • 2 Y

Y by Adventure10, Mango247

I will give you a solution due to Klamkin to that problem :

Lemma.
A point O lies in the base ABC of a tetrahedron PABC.
Then, the sum of the angles formed by the line OP and the edges PA, PB, PC is less than the sum of the faces angles at vertex P.

Proof.
We will use the duality between trihedral angles and spherical triangles. First note that it suffices to assume that PO is an interior ray of a trihedral angle whose edges are the rays PA, PB, PC. With $\alpha , \beta , \gamma$ denoting the angles BPC, CPA, APB respectively, and $x,y,z$ the angles OPA, OPB, OPC respectively, we want to prove that $x+y+z < \alpha + \beta + \gamma$ .

Draw an unit sphere centered at P, intersecting the ray PO at O'. The points A',B',C' are defined in a similar way.
Clearly the angle OPB is equal to the angle O'PB', and we have similar equalities for the other angles.
Now, the intersection of the sphere with the trihedral angle is a spherical triangle A'B'C' which can be triangulated by great circle arcs meeting at O'. If all the faces angles of the trihedral angles are mesured in radians, then (the radius of the sphere being unity) the lengths of all connecting arcs are :
arc B'C' = $\alpha$ , arc C'A' = $\beta$ and arc A'B' = $\gamma$ ,
arc O'A' = $x$ , arc O'B' = $y$ and arc O'C' = $z$ .
Let arc O'B' and arc A'C' meet at B''.
Since all the spherical triangles we are concerned with here are convex, we have :
arc O'B'' + arc B'C' > $z$ and $\gamma$ + arc A'B'' > $y$ + arc OB''
Adding theses two inequalities, we obtain $\beta + \gamma > y + z$ .
We deduce similar inequalities for the other angles. Summing leads to the desired result.

Now, we return to the original problem :
Let ABCD be the tetrahedron and P the interior point. let A' be the intersection of line AP with face BCD.
Let S be the sum of the angles APB, APC and APD.
Let S' be the sum of the angles BPC, CPD and DPB.
Let S'' be the sum of the angles A'PB, A'PC and A'PD.
Then :
Clearly S + S'' = 3 $\pi$ and, using the lemma, we have S' > S''.
Thus S + S' > 3 $\pi$ , and we are done.

Pierre.

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#3 h Jun 10, 2004, 7:45 AM • 2 Y

Y by Adventure10, Mango247

Pierre, thanks but there is a point I don't quite understand:

pbornsztein wrote:

If all the faces angles of the trihedral angles are mesured in radians, then (the radius of the sphere being unity) the lengths of all connecting arcs are :
arc B'C' = $\alpha$ , arc C'A' = $\beta$ and arc A'B' = $\gamma$ ,
arc O'A' = $x$ , arc O'B' = $y$ and arc O'C' = $z$ .
Let arc O'B' and arc A'C' meet at B''.
Since all the spherical triangles we are concerned with here are convex, we have :
arc O'B'' + arc B'C' > $z$ and $\gamma$ + arc A'B'' > $y$ + arc OB''
Adding theses two inequalities, we obtain $\beta + \gamma > y + z$ .

How? Maybe, there are some typos, or I am just blind?

Also, some very fundamental questions (sorry, I am a beginner in stereometry):

- When is a spherical triangle called convex?
- How to prove the triangle inequality for spherical triangles?

Thanks!
Darij

This post has been edited 1 time. Last edited by darij grinberg, Feb 20, 2005, 1:18 AM

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#4 h Jun 10, 2004, 8:25 AM • 2 Y

Y by Adventure10, Mango247

1) I think that since we start from a tetrahedron and an interior point, the resulting spherical triangle is contained in an half-sphere. Thus, it is convex.

2) The great circles arcs are paths of minimal length between two points on the surface of a sphere. Thus :
arc O'C' < arc O'B'' + arc B''C', since O'C' is a great circle arc.
and
arc O'B'' + $y$ = arc O'B'' + arc O'B' = arc B'B'' < arc B'A' + arc A'B'' = $\gamma$ + arc A'B'', since B'B'' is a great circle arc, which pass through O'.

Adding these two inequalities, we obtain :
arc O'C' + arc O'B'' + $y$ < arc O'B'' + arc B''C' + arc A'B'' + arc A'B'.
But arc A'B'' + arc B''C' = arc A'C' = $\beta$ , arc O'C' = $z$ and arc A'B' = $\gamma$
Thus, eliminating arc O'B'', it leads to $z + y < \beta + \gamma$ .

Pierre.

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#5 h Jun 10, 2004, 9:36 AM • 2 Y

Y by Adventure10, Mango247

Thanks a lot, I have now understood your proof. And a polygon on the sphere is called convex if for any two points P and P' in its interior, the smaller of the two great circle arcs PP' completely lies inside the polygon, isn't it?

Darij

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