3 replies to this topic

[BMAD]

A quadrilateral is given in space, such that its edges are tangent to a sphere.  Prove that all the points of tangency lie in one plane.

[bonanova]

Spoiler for nice problem

I saw this problem a while back.
It's most easily solved using physics.
You imagine weights are placed on the vertices of sizes whose pair wise centers of mass lie at the tangent points.
Suppose vertices A and B are adjacent.
The weights placed on those vertices have a center of mass at the tangent point.
Similarly for points C and D, and the other two adjacent vertex pairs.

The center of mass of all four weights lies at a point that is the intersection of two lines connecting the pair wise centers of mass.
Thus the points of tangency describe two intersecting lines, which in turn describe a plane

[k-man]

Spoiler for nice problem

I saw this problem a while back.
It's most easily solved using physics.
You imagine weights are placed on the vertices of sizes whose pair wise centers of mass lie at the tangent points.
Suppose vertices A and B are adjacent.
The weights placed on those vertices have a center of mass at the tangent point.
Similarly for points C and D, and the other two adjacent vertex pairs.

The center of mass of all four weights lies at a point that is the intersection of two lines connecting the pair wise centers of mass.
Thus the points of tangency describe two intersecting lines, which in tern describe a plane

 

That's an interesting approach to solving a problem, but I guess what remains to be proven is

Spoiler for

that the described distribution of weights can actually be done. You can certainly pick the 3 weights for vertices A, B and C that will place the center of mass on the tangent points of the AB and BC edges, but what guarantees that there exists the 4th weight for D that will place the center of mass on the tangent point of the remaining 2 sides AD and CD? 

[bonanova]

Spoiler for nice problem

I saw this problem a while back.
It's most easily solved using physics.
You imagine weights are placed on the vertices of sizes whose pair wise centers of mass lie at the tangent points.
Suppose vertices A and B are adjacent.
The weights placed on those vertices have a center of mass at the tangent point.
Similarly for points C and D, and the other two adjacent vertex pairs.

The center of mass of all four weights lies at a point that is the intersection of two lines connecting the pair wise centers of mass.
Thus the points of tangency describe two intersecting lines, which in tern describe a plane

 

That's an interesting approach to solving a problem, but I guess what remains to be proven is

Spoiler for

that the described distribution of weights can actually be done. You can certainly pick the 3 weights for vertices A, B and C that will place the center of mass on the tangent points of the AB and BC edges, but what guarantees that there exists the 4th weight for D that will place the center of mass on the tangent point of the remaining 2 sides AD and CD? 

 

Spoiler for Here is the process

First, note that from an external point every line tangent to a sphere has the same length.

So for each vertex of the quadrilateral, the distances to its two associated points of tangency are equal.

Second, note that the distance from either of two point masses to their center of mass is inversely proportional to its mass.

So by placing a mass on each vertex that is equal to the reciprocal of the distance to its associated points of tangency,

the center of mass of the masses at the ends of each edge will occur at the points of tangency.

 

These four equivalent masses can be taken in either of two pairwise groups to find the cm of all four masses.

Since the points of tangency thus describe two lines that intersect, they are coplanar.

 

It's easy enough to make a sketch of this.