platonic love

[radosr]

A pyramid and a tetrahedron walk into a bar. The pyramid has a square base and four equilateral triangles as its four sides. The tetrahedron has four equilateral triangles which are of the same size as the triangles of the pyramid. After having several drinks, they decide they like each other and one side of the tetrahedron gets glued onto one triangular side of the pyramid, forming another solid. How many faces does this new solid have?

 

[kuberkhan]

7?

 

[hariadya]

4?

 

[JinWJ]

5?

 

[Tsotne]

5 is the answer...

1 base (merged)
2 sides tied together
2 independent sides left which are in front of the 'beloved' sides

 

[pruse]

Don't we need to mention dihedral angles here? The reason it's 5 and not 7 is because the dihedral angle between an equilateral face of the tetrahedron and an equilateral face of the pyramid is actually 180 degrees, which can be seen by some trig calculations. This is why you lose two more faces in addition to the two glued together.

 

[pratikpoddar]

agree with @peterruse
Can you please elaborate on this calculation?

 

[Tsotne]

Yes it is true. I also assumed the same implicitly.

 

[Devdeep]

agree with @peterruse
Can you please elaborate on this calculation?

@pratikpoddar When two triangular sides (one of the pyramid and one of the tetrahedron) are joined together, you lose 2 sides... so we are already at 9-2=7. Now, following the arguments of @peterruse and @Tsotne, there are 2 occasions where one triangular side of the pyramid and one triangular side of the tetrahedron are on the same plane and side-by-side... and hence they look like just 1 surface. Thus these 4 sides actually make 2 sides. Hence the final number of sides would be 9-2-2=5.

 

[pruse]

@pratikpoddar When two triangular sides (one of the pyramid and one of the tetrahedron) are joined together, you lose 2 sides... so we are already at 9-2=7. Now, following the arguments of @peterruse and @Tsotne, there are 2 occasions where one triangular side of the pyramid and one triangular side of the tetrahedron are on the same plane and side-by-side... and hence they look like just 1 surface. Thus these 4 sides actually make 2 sides. Hence the final number of sides would be 9-2-2=5.

that's not the calculation he was asking for; that's the trivial part. he was asking about the trig.

by drawing perpendiculars to the common edge, you can see that the dihedral angle (which you want to show is a straight angle), is made up of the vertex angles of two isosceles triangles, one with sides (2,\sqrt{3},\sqrt{3}) and another with sides (2\sqrt{2},\sqrt{3},\sqrt{3}) (assuming all edges have length 2). Find the tangents of these vertex angles (using the double angle formula for tan, after dropping heights from the vertex angles) and check that they are negatives of each other; this shows that the two angles are in fact supplementary.

 

[MiloRambaldi]

This can also be shown without doing trigonometric calculations:

Let P be the plane containing the base of the pyramid, let T be the top vertex of the pyramid and let U be the vertex of the tetrahedron opposite the pyramid.

(1) Observe that TU is parallel to P: Let M1 be the midpoint of the edge of the pyramid base shared with the tetrahedron, and let M2 be the midpoint of the opposite base edge. Then T,U,M1 and M2 are coplanar, TU and M1M2 both have length 2 and UM1 and TM2 both have length sqrt(3). Thus UTM2M1 is a parallelogram.

(2) Let Q and R be the two planes coincident with the two sides of the pyramid adjacent to the tetrahedron. Their intersection is a line L that is parallel to P and whose projection onto P passes through M1M2. Since there is only one such line, TU must be a line segment of L by (1). This proves that U is in both Q and R, and this shows that the tetrahedron has faces incident with both Q and R. Q.E.D.

 

[pruse]

This can also be shown without doing trigonometric calculations:

Let P be the plane containing the base of the pyramid, let T be the top vertex of the pyramid and let U be the vertex of the tetrahedron opposite the pyramid.

(1) Observe that TU is parallel to P: Let M1 be the midpoint of the edge of the pyramid base shared with the tetrahedron, and let M2 be the midpoint of the opposite base edge. Then T,U,M1 and M2 are coplanar, TU and M1M2 both have length 2 and UM1 and TM2 both have length sqrt(3). Thus UTM2M1 is a parallelogram.

(2) Let Q and R be the two planes coincident with the two sides of the pyramid adjacent to the tetrahedron. Their intersection is a line L that is parallel to P and whose projection onto P passes through M1M2. Since there is only one such line, TU must be a line segment of L by (1). This proves that U is in both Q and R, and this shows that the tetrahedron has faces incident with both Q and R. Q.E.D.

or notice that the two triangles I mention above, with sides (2,\sqrt{3},\sqrt{3}) and (2\sqrt{2},\sqrt{3},\sqrt{3}) both divide into (1\text{-}\sqrt{2}\text{-}\sqrt{3}) right triangles by drawing heights from the vertex angles; this proves it immediately. no trig ;]

now we've beaten this poor little problem to death.

 

[radosr]

now we've beaten this poor little problem to death.

Indeed, there is no love left here.

 

[Barny]

I would say 5, but that was just from intuition/visualisastion of the fact that two of the sides would merge into one on each side. Pretty impressive stuff from peterruse.