1962 IMO Problems/Problem 7
Problem
The tetrahedron has the following property: there exist five spheres, each tangent to the edges , or to their extensions.
(a) Prove that the tetrahedron is regular.
(b) Prove conversely that for every regular tetrahedron five such spheres exist.
Solution
Part (a)
Let points be the points where the smallest sphere is tangent to the edges respectively.
For each face of the tetrahedron there is a circular cross section of the smallest sphere. Since that circle also needs to be tangent to the edges, then that circle is the incircle of each triangular face.
From the properties of an incircle we know that:
, , and in
There is a larger circle that is tangent to and the extensions of and .
This larger circle is a cross section of the sphere that's also tangent to the extension of away from and the edges of
Therefore, this larger circle and the incircle of are part of that same sphere.
In order for these two circles to be part of the same sphere and also tangent to line , then the point of the tangent of this larger circle needs to be the same as point
The only way these to circles can share the same tangent point on edge is if
Using the same argument with the larder circle of edge then
and with the larger circle of edge then
This results in:
in
which means that
in , thus is an equilteral triangle.
Likewise,
in , thus is an equilteral triangle.
in , thus is an equilteral triangle.
in , thus is an equilteral triangle.
Since all four faces are equilateral triangles, then tetrahedron is a regular tetrahedron.
Part (b)
Let's consider a regular tetrahedron in the cartesian space with center at with as the circumradius.
We can write the coordinates for the four vertices as:
And the midpoints of all edges as:
Now we calculate the following six dot products:
Since all these dot producs equal to that means that the lines from the center to each of the midpoints are all perpendicular.
Now we calculate the distances from to all the midpoints:
Since all distances are all the same and all dot products are , then we have our first sphere at with a radius of
Now we will look at the other 4 spheres.
Let be the point of tangent of the larger sphere on the extension of line in the direction of to and beyond for ; ; and
Since , then , thus
Using this formula we calculate the following:
We will start with the sphere below the base of the tetrahedron opposite of vertex below
The center of this larger sphere is at and it is tangent at points
We calculate the following dot products:
Since all these dot producs equal to that means that the lines from the center to each of the point are all perpendicular.
Now we calculate the distances from to points
Since all distances are all the same and all dot products are , then we have one of the larger spheres at with a radius of
Then, the other three larger spheres which are the same size as the sphere with center at are congruent and tangent to their respective sides near the other faces of the tetrahedron.
and this proves that a tetrahedron with any circumradius will have these 5 spheres, one with radius , at the center of the tetrahedron and the other 4 with radius at centers that are at a distance of away from any of the vertices of the tetrahedron in the direction from that vertex to the center of its opposite face.
Thus these five spheres exist for any regular tetrahedron.
~Tomas Diaz. orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1962 IMO (Problems) • Resources | ||
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