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1. Introduction

There are many problems of plane geometry for which we can find corresponding problems in the stereometry. In this paper, I will consider two of such problems. I'm going to find spheres with the properties similar to the properties of the nine-point circle and the eight-point circle. After that, I will show a nice proof of the existence of one of these using the existence of another.

2. The nine-point circle

Given a triangle ABC, let K, L and M denote midpoints of its sides and let P, Q, R be feet of altitudes from the vertices A, B and C respectively. Furthermore, let H be triangle's orthocenter and O - the circumcenter. We denote also midpoints of segments from the orthocenter to the triangle's vertices by X, Y and Z (fig. 1).

Fig. 1

Definition/Theorem 1. The nine point circle is a circle passing through the points K, L, M, P, Q, R, X, Y and Z. The center of this circle I lies in the middle of the segment HO.

It's remarkable that nine particular triangle points lie on one and the same circle because we know that three noncollinear points determine a circle and if there are four or more points given then in general they don't lie on the same circle.

Another interesting object is the Euler line. That's a line passing through the orthocenter H and circumcenter O. As we already know it contains also the center of the nine point circle. Moreover the center of gravity G lies on it and

.

(2.1)

Proofs of above properties including all details can be found in [1].

3. The eight-point circle

Another, less known circle is the eight-point circle. This was given in 1944 by Louis Brand of Cincinnati. Let ABCD be a quadrilateral with perpendicular diagonals and let K, L, M and N be the midpoints of its sides (fig. 2). Since KL and MN are parallel to AC and LM and NK are parallel to BD, KLMN is a parallelogram. Furthermore since AC and BD are perpendicular, so do any two adjacent sides of KLMN and it's a rectangle. Since KM and LN are the diameters of its circumcircle, the circle passes through the feet of the perpendiculars from K, L, M and N upon the opposite sides - through the points , , and .

Fig. 2

We proved that eight particular quadrilateral points belong to one circle. The center of this circle O is the midpoint of KM. Since K is the center of gravity of AB and M is the center of gravity of CD, O is the center of gravity of the quadrilateral.

4. Simple proof of the nine-point circle theorem

Using the eight-point circle definition, we can easily prove the existence of the nine-point circle. Let ABC be a triangle and let all points be defined like in figure 3.

Fig. 3

Notice that ABCH is a quadrilateral with perpendicular diagonals. This is because the altitude BH is perpendicular to the side CA. Using definition of the eight-point circle we get that the midpoints of the quadrilateral sides (these are K, M, X and Z) and the feet of the perpendiculars from these midpoints upon the opposite sides (these are P and R) are concyclic. The center of the circle - point I - is the center of gravity of the points A, B, C and H. Since the center of gravity of the first three points is G, I lies on the line GH and

.

(4.1)

Now we can consider one of the two other quadrilaterals, for example BCAH, and get another circle with center in I passing through K, L, P, Q, X and Y. Since the circles have three points in common, they're equal and they're the nine point circle of the triangle.

In this simple way, we've proved that the nine-point circle exists.

5. The nine point circle and space - hypotheses

We're going to find a sphere with properties similar to the properties of the nine-point circle. As we know, the nine-point circle is associated with a triangle. We can expect, that a body corresponding to the nine point circle (we want it to be a sphere) will be associated with a tetrahedron which corresponds to a triangle in three dimensions. To determine this sphere we have to find in a tetrahedron points corresponding to particular triangle points from the definition 2.1.

Centers of the triangle sides correspond to centers of gravity of tetrahedron's faces. We could also understand a center of a face as its circumcenter but this would not lead to any interesting results. Points P, Q, R correspond supposedly to feet of tetrahedron's altitudes. Finally points X, Y and Z correspond to midpoints of segments from the tetrahedron's orthocenter to its vertices, but not every tetrahedron has an orthocenter! It has one if every two opposite edges are perpendicular (which means that there exists a plane containing one of them and perpendicular to the another) or if foot of every tetrahedron's altitude coincides with the orthocenter of the face on which it lies.

Definition 2. Tetrahedra, which have an orthocenter, are called orthocentric tetrahedra.

In the next part of this paper, I'll consider only orthocentric tetrahedra.

We need to know two more things. Firstly, medians in a tetrahedron intersect in the ratio 1:3. The proof of it is easy, hence not presented here. Secondly, which also follows from this proof, the tetrahedron given by the centers of gravity of faces of another tetrahedron is similar to the original tetrahedron and the ratio is .

6. The twelve-point sphere

Let A₁A₂A₃A₄ be an orthocentric tetrahedron. We introduce the following notation (fig. 4):

i	- tetrahedron's face opposite to the vertex Ai or its plane
Hi	- orthocenter of the face i and the foot of the altitude from Ai
Gi	- center of gravity of the face i
Oi	- circumcenter of the face i
H	- tetrahedron's orthocenter
G	- tetrahedron's center of gravity
O	- center of tetrahedron's circumsphere
pri (X)	- image of X in perpendicular projection onto plane i

Fig. 2

Points A₄, H₄ and O₄ are non-collinear (H₄ and O₄ lie on the plane ₄ and A₄ doesn't) so they determine a plane. It's perpendicular to ₄ because A₄H₄ is an altitude. This plane contains also H (HA₄H₄ because H is the common point of all altitudes), O (OO₄ - radius of the circumsphere is perpendicular to the circle A₁A₂A₃ because it goes through its center), G₄ (from the properties of the Euler line we know that G₄H₄O₄) and G (GA₄G₄ because G is a medians' intersection). Figure 5 shows this plane section of the tetrahedron.

Fig. 5

Using properties of the Euler line, we know that

.

(6.1)

We know also that G divides the median A₄G₄ in the ratio 3:1 which means that

.

(6.2)

Fig. 6

From this equation using similarity we get

(6.3)

where is an image of G in the perpendicular projection onto line H₄O₄ (). From this and (4.1) we get

(6.4)

which also means that

(6.5)

Since in an analogous way we can define three other planes A_iH_iO_i and each one contains H, G and O it follows that these planes have to intersect in a line containing H, G and O and hence these points are collinear. On the other hand we know that and H₄, , O₄ are images of H, G, O respectively in a perpendicular projection on the line H₄O₄ (, , ) so using similarity we get

(6.6)

Let S be a sphere circumscribed on the tetrahedron G₁G₂G₃G₄ (these points are always non-coplanar, fig. 6). I'm going to check if the points H_i belong to S. We already know that tetrahedra G₁G₂G₃G₄ and A₁A₂A₃A₄ are similar and the ratio is 1:3. Let denote the homothety with center in G and ratio . Points A_i transform under into points G_i (medians A_iG_i intersect in G and divide themselves in ratio 1:3). Since homothety doesn't change ratios of distances, point equidistant from A₁, ..., A₄ - center of circumsphere, point O - transforms into a point equidistant from G₁, ..., G₄ - center of sphere S, point (, fig. 7). Points G and O lie on the above-considered plane so also has to lie on it.

Fig. 7

Moreover, we know that . From this and (4.6) we get

(6.7)

which means that

(6.8)

(6.9)

which according to (6.5) implies that

.

(6.10)

Since is perpendicular to H₄G₄ it follows that

(6.11)

so H₄ S (G_i S by definition) and analogously H_i S.

Now let M₄ be the intersection point of the altitude A₄H₄ and the sphere S. so M₄G₄ is a sphere's diameter which implies that it passes through so that

(6.12)

(6.13)

but

and

(6.14)

so

(6.15)

This ratio doesn't depend on the plane A_iH_iO_i we took, so we can say that for every M_i such that holds, M belongs to S.

Finally we got a sphere consisting of 12 particular tetrahedron's points (faces' centers of gravity G_i, altitudes' feet H_i and points M_i dividing segments HA_i in ratio 1:2).

7. The sixteen-point sphere

This time, I'm going to find a sphere corresponding to the eight-point circle. It appears that the body associated with the sphere could be octahedron.

Let ABCDEF be an octahedron, whose diagonals are perpendicular (they don't have to intersect each other, fig. 8).

Let G, H, I and J be the midpoints of the segments AB, BC, CD and DA respectively (fig. 9). Then GH and IJ are parallel to AC. Similarly HI and JG are parallel to BD and GHIJ is a parallelogram. Furthermore, since AC and BD are perpendicular, so do every two adjacent sides of GHIJ, which means that GHIJ is a rectangle.

Now, denote the centroids of the faces EAB, EBC, ECD and EDA by K, L, M and N respectively. The quadrilateral KLMN is then the image of the rectangle GHIJ in the homothety with center in E and ratio (a centroid of a triangle lies on the medians and divides them in the ratio 1:2), and so it is a rectangle. In analogous way, we find out that OPQR is a rectangle, and after that, also that KLPO, LMQP, MNRQ and NKOR are rectangles (fig. 10). Such hexahedron KLMNOPQR consisting of six rectangular faces has to be a cuboid, and therefore it can be circumscribed by a sphere.

Fig. 8

Since all the vertices of the cuboid are the centroids of the octahedron faces, cuboid circumcenter - center of the cuboid's diagonals - is the octahedron's center of gravity. Furthermore since cuboid's diagonals are the sphere's diameters, the feet of the perpendiculars from these centers of gravity upon opposite faces (points , , , , , and ) also belong to that sphere.

Fig. 9

We've got a sphere passing through sixteen particular octahedron points. These are centers of gravity of octahedron faces and the feet of the perpendiculars from these centers of gravity upon opposite faces.

Fig. 10

8. Twelve-point sphere for the second time

It appears that there is a much simpler proof of the existence of the twelve-point sphere than the one above. It uses the definition of the sixteen-point sphere.

Fig. 11

Let A₁A₂A₃A₄ be an orthocentric tetrahedron and denote the points like before. Moreover let be a point on the segment A₄H₄ such that

.

(8.1)

Note, that the diagonals of the concave octahedron are perpendicular to each other (fig. 11). This is because face altitude A₂H₄ is perpendicular to the side A₁A₃ of this face and altitude A₄H₄ is perpendicular to the face A₁A₂A₃. Using the properties of the sixteen-point sphere we get, that the centers of gravity of the octahedron faces and the feet of the perpendiculars from these centers of gravity upon the opposite faces lie on one sphere.

The centers of gravity of the faces A₁A₂A₄ and A₂A₃A₄ are the points G₃ and G₁. Since (8.1) holds the centers of gravity of the faces and are the points dividing the segments A₁H and A₃H in the ratio 2:1 (A₁H and A₃H are the medians of the faces, and the centers of gravity divide them in the ratio 2:1) - points M₁ and M₃.

Since the center of gravity of the face lies on the segment A₁H (that's its median) and H is the orthocenter, the foot of the perpendicular from that center of gravity upon the opposite face A₂A₃A₄ is the foot of the altitude A₁H i.e. point H₁. Similarly, the foot of the perpendicular from the center of gravity of the face upon the opposite face A₁A₂A₄ is the foot H₃ of the altitude A₃H.

The sixteen point sphere for the octahedron passes through the points G₁, G₃, H₁, H₃, M₁ and M₃. The center of the sphere is the center of gravity of the octahedron. Since the center of gravity of the points A₁, A₂, A₃, A₄ is G and the center of gravity of the points H₄ and is H (), the center O lies on GH and

.

(8.2)

Choosing in the same way other octahedra in the tetrahedron (there are possibilities) we get more spheres. Since all of them have the center in O and for every one there exists exactly ten others, which have at least three points in common, all these spheres are equal and they're the twelve point sphere of the tetrahedron (they pass through all the points G_i, H_i and M_i).

9. Conclusion

The aim of this project was to find spheres corresponding to the nine-point circle and the eight-point circle. The twelve-point sphere has the similar properties as the first of the circles. It passes through twelve particular points of every orthocentric tetrahedron (these are centers of gravity, faces' orthocenters and points dividing segments from the tetrahedron orthocenter to vertices in ratio 1:2). The sixteen-point sphere corresponds to the eight-point circle and is defined for every octahedron whose diagonals are perpendicular. It passes through centers of gravity of its faces and feet of the perpendiculars from these centers of gravity upon the opposite faces.

We found also a nice proof of the existence of the twelve-point sphere using properties of the sixteen-point sphere. This parallels the proof, which holds for the nine-point circle using the eight-point circle in the plane.

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