# Elementary proof of Pavillet tetrahedron properties.

1 Introduction

The orthocentric tetrahedron of triangle ,  is a simple construction which gives students a great opportunity for in-depth study of perpendicularity in space and testing of various methods of proof. In this paper, we give elementary proofs of the main properties of a Pavillet tetrahedron. We use simple and sometimes coarse methods of proofs. Students may create their own short and elegant methods based on inversion or other transformations of the space.

2 Description of a Pavillet tetrahedron and its main property

Let's consider a triangle ABC, named the base triangle, be given. Let its incircle be centered in I and [A.sub.1][B.sub.1][C.sub.1] be the contact triangle of ABC. Let segments AA, BB', CC' be perpendicular to the plane ABC, where AA' = [AB.sub.1], BB' = [BC.sub.1],CC' = [CA.sub.1]. Let us name the solid IA'B'C' as the Pavillet tetrahedron of the triangle ABC [9, [section] 2], . Let us name triangle A'B'C' as upper triangle. This tetrahedron is shown on figure 1.

3 The tetrahedron is orthocentric

Theorem 1 The segment B'C' is perpendicular to the plane IA'[A.sub.1].

Proof.

Let [A'.sub.1] be the foot of the perpendicular dropped from [A.sub.1] on B'C' (fig. 2) and denote x = A[B.sub.1] = [AC.sub.1] = AA', y = [BC.sub.1] = [BA.sub.1] = BB' and z = [CA.sub.1] = C[B.sub.1] = CC'. Hence

[A.sub.1][B'.sup.2] = 2 [y.sup.2], [A.sub.1][C'.sup.2] = 2 [z.sup.2] [??] A'[B'.sup.2] = (x + y)[sup.2] + (x - y)[sup.2] = 2 ([x.sup.2] + [y.sup.2]).

We use [A'.sub.1][A.sub.1] [perpendicular to] B'C' and we get

[A'.sub.1][B'.sup.2] - [A'.sub.1][C'.sup.2] = [A.sub.1][B'.sup.2] - [A.sub.1][C'.sup.2] = 2 ([y.sup.2] - [z.sup.2])

[FIGURE 1 OMITTED]

We calculate A'[B'.sup.2] - A'[C'.sup.2] = 2 ([y.sup.2] - [z.sup.2]), it implies similarly that A'[A'.sub.1] is an altitude of the triangle A'B'C'. Now we also have B'[I.sup.2] - C'[I.sup.2] = 2 ([y.sup.2] - [z.sup.2]). Hence I[A'.sub.1] [perpendicular to] B'C', combined with [A'.sub.1][A.sub.1] [perpendicular to] B'C' it yields that the points I, A', [A.sub.1]) lie on the same plane perpendicular to B'C'. This plane contains the point [A'.sub.1] on B'C'.

We proved that the plane IA'[A'.sub.1] is perpendicular to the boundary plane BB'C'.

[FIGURE 2 OMITTED]

Theorem 2 The Pavillet tetrahedron is orthocentric.

Proof.

From Theorem 1, the plane IA'[A'.sub.1] is perpendicular to the boundary plane BB'C'. Therefore IA' [perpendicular to] B'C', similarly we get IB' [perpendicular to] C'A' and IC'LA'B'. From [1, Definition 208, p. 62] this tetrahedron orthocentric.

4 Altitudes of the tetrahedron

Theorem 3 The straight line joining a vertex of the upper triangle to the corresponding vertex of the contact triangle of the base triangle (e.g. A'[A.sub.1]) is an altitude of the tetrahedron.

Proof.

* It is clear, that A'[A.sub.1] [perpendicular to] LB'C' as a line belonging to the plane I A' [A'.sub.1].

* We consider the quadrilateral IA'[A'.sub.1][A.sub.1] and compute the quantity A'[I.sup.2] - [A.sub.1][I.sup.2]. We get

A'[I.sup.2] - [A.sub.1][I.sup.2] = A'[B.sup.2.sub.1] + [B.sub.1][I.sup.2] - [A.sub.1][I.sup.2] = A'[B.sup.2.sub.1] = 2 [x.sup.2]. (1)

[FIGURE 3 OMITTED]

* In the same quadrilateral, we compute the quantity A'[A'.sup.2.sub.1] - [A.sub.1][A'.sup.2.sub.1]. Let E and F lie on BC such that [A'.sub.l]E [perpendicular to] BC, AF [perpendicular to] BC (fig. 3). Then

A'[A'.sup.2.sub.1] - [A.sub.l][A'.sup.2.sub.1] = (A'A - [A'.sub.1]E)[sup.2] + A[F.sup.2] + E[F.sup.2] - [A.sub.l][A'.sup.2.sub.1].

We use the right triangle [A.sub.l]B'C' with legs [A.sub.l]B' = y[square root of (2)], [A.sub.l]C' = z[square root of (2)]. We get the square of the height

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

We use the trapezium (British definition) BB'C'C and get

[A'.sub.1]E = yz(y + z)/[y.sup.2] + [z.sup.2].

From the right triangle [A.sub.l][A'.sub.1]E, we get the leg [A.sub.l]E = yz [absolute value of y - z]/[y.sup.2] + [z.sup.2]. We use the formulas for the area of triangle ABC and get the square of the altitude A[F.sup.2] = 4 x y z(x + y + z)./(y + z)[sup.2]. We use the formula for distance between base of height and tangent point for incircle and get [A.sub.l]F = x[absolute value of y - z]. Finally, we use formula for square of incircle radius [A.sub.l][I.sup.2] = [C.sub.l][I.sup.2] = x y z/x + y + z.

After substitutions and simplifications we get

A'[A'.sup.2.sub.1] - [A.sub.1][A'.sup.2.sub.1] = 2[x.sup.2].

* We recall, to finish the proof, that "for any orthodiagonal quadrilateral, the sum of the squares of two opposite sides equals that of the other two opposite sides and conversely".

Now, because we have found that 2 [x.sup.2] = A'[I.sup.2] - [A.sub.1][I.sup.2] = A'[A'.sup.2.sub.1] - [A.sub.1][A'.sup.2.sub.1], the sides of the quadrilateral IA'[A'.sub.1][A.sub.1] comply with the equation

A'[I.sup.2] + [A.sub.1][A'.sup.2.sub.1] = A'[A'.sup.2.sub.1] + [A.sub.1][I.sup.2]

so that I A'[A'.sub.1][A.sub.1] is orthodiagonal and A'[A.sub.l] [perpendicular to] I[A'.sub.1].

We have proved that A'[A.sub.1] being perpendicular to I[A'.sub.1] and B'C', hence to the plane IB'C', is an altitude of I A'B'C'.

4.1 Orthogonal Projection of the Orthocenter

[FIGURE 4 OMITTED]

Theorem 4 The orthogonal projection of the orthocenter of the Pavillet tetrahedron on the plane of its base triangle is the Gergonne point of this base triangle.

Proof.

Theorem 3 shows that the lines A'[A.sub.1], B'[B.sub.1] and C'[C.sub.1] are three altitudes of the orthocentric tetrahedron. They intersect at the point [H.sub.o] orthocenter of the tetrahedron (fig. 4). Each of these altitudes lies in a plane perpendicular to the plane A'B'C', e.g. [A.sub.1]A' in the plane [AA.sub.1]A' which contain AA' [perpendicular to] ABC. Let G be the point of intersection of the lines [AA.sub.l] and [BB.sub.l], orthogonal projections of the lines A'[A.sub.l] and B'[B.sub.l] on the base plane ABC. Then [H.sub.0]G is perpendicular to the base plane. Now, because C'[C.sub.l] also goes through G, the line [CC.sub.l] which is the orthogonal projection of C'[C.sub.l] is concurrent with A[A.sub.1] and B[B.sub.1].

As the reader sees, we have proved the existence of the Gergonne point. We also have proved that the orthogonal projection of the orthocenter of the Pavillet tetrahedron on the plane of its base triangle is the Gergonne point of its base triangle.

5 Axis of perspective

Theorem 5 The corresponding sides of the base triangle, the upper triangle and the contact triangle are concurrent.

[FIGURE 5 OMITTED]

Proof.

We use Menelaus' theorem for the sides of the triangle ABC cut by the line [A.sub.1][B.sub.1] (fig. 5). We get the distance BK from the point of intersection K of the lines AB and [A.sub.l][B.sub.l] as BK = AB x [A.sub.1]B/[absolute value of AC - BC]. We use proportion for the trapezium ABB'A', we get the distance B[K.sub.l] from the point [K.sub.1], intersection of the lines AB and A'B' as B[K.sub.1] = AB x [A.sub.1]B/[absolute value of AC - BC]. Hence BK = B[K.sub.1] and the points K and [K.sub.1] are coincident.

Theorem 6 The plane of the base triangle and the plane of the upper triangle intersect along the Gergonne line of the base triangle.

Proof.

The line AB is a projection of the line A'B' on the base plane. Hence the point of intersection of the lines AB and A'B' lies on the line of intersection of the planes ABC and A'B'C' (fig. 6). Similarly, K', point of intersection of the lines AC and A'C', and K", point of intersection of the lines BC and B'C' lie on the line of intersection of the planes ABC and A'B'C'. So points K, K' and K" are collinear. These points are also the crossing points of the lines AB, [A.sub.1][B.sub.1], and similar. According to the definition, these points lie on the polar of the Gergonne point of ABC, the point of intersection of the Cevians A[A.sub.1], B[B.sub.1] and C[C.sub.1].

Hence we proved that these points are collinear and lie on the Gergonne line of ABC.

[FIGURE 6 OMITTED]

6 Soddy line an circles.

Theorem 7 The Soddy line is perpendicular to the Gergonne line. This property is found in [6, p. 324 [section] 6].

[FIGURE 7 OMITTED]

Proof.

By definition, the Soddy line contains the Gergonne point G and the incenter I [6, p. 319]. The plane IGH0 is perpendicular to the base plane ABC because G[H.sub.0] [perpendicular to]ABC (fig. 7). The plane IG[H.sub.0] is also perpendicular to the upper plane A'B'C' because I[H.sub.0] is an altitude of the orthocentric Pavillet tetrahedron. Hence the Gergonne line KK' which, from Theorem 6, belongs to both planes is perpendicular to the plane IG[H.sub.0] and therefore to the line IG.

Theorem 8 The Soddy inner and outer centers lie on the Soddy line.

Proof.

Let S be the inner Soddy center (as defined in [2, [section] 2-3] or ). Let [S.sub.0] belongs to the same half-space define by the base plane as the Pavillet tetrahedron, such that S[S.sub.0] [perpendicular to] ABC and S[S.sub.0] = [rho] where [rho] is the radius of the inner Soddy circle. We call [S.sub.a] the point of contact of the inner Soddy circle with the circle centered at A with radius x. The points A, [S.sub.a], S are collinear. The triangle [S.sub.0][S.sub.a]A' is right, hence [S.sub.a]

[S.sub.0][A'.sup.2] = [S.sub.0][S.sup.2.sub.a] + [S.sub.a][A.sup.'2] = 2 [[rho].sup.2] + 2[x.sup.2]

and we find similarly [S.sub.0]B' and [S.sub.0]C'. Using (1), we find that

A'[I.sup.2] - B'[I.sup.2] = 2 [x.sup.2] - 2[y.sup.2] = A'[S.sup.2.sub.0] - B'[S.sup.2.sub.0]

and

A'[I.sup.2] - C'[I.sup.2] = 2[x.sup.2] - 2[z.sup.2] = A'[S.sup.2.sub.0] - C'[S.sup.2.sub.0].

[ILLUSTRATION OMITTED]

So we proved that points I and [S.sub.0] lie on the same perpendicular to the plane A'B'C' (fig. 8). The point S being the projection of [S.sub.0] on the plane ABC lies on the Soddy line, projection of I[H.sub.o] on the base plane. The reader can make a similar proof for the outer Soddy center.

[FIGURE 8 OMITTED]

7 The circumcircle sphere

Definition 9 We define the circumcircle sphere as the sphere having the circumcircle of the base triangle as diametral circle.

Theorem 10 The trace of the circumcircle sphere of the base triangle on the plane of the upper triangle is the Euler circle of this triangle.

Proof.

Let [C.sub.2] be the midpoint of the side AB of the base triangle and [C'.sub.2] be the midpoint of the side A'B' of the upper triangle (fig. 9). Let O be the circumcenter of the base triangle so that [OC.sub.2] [perpendicular to] AB. From the trapezium AA"B'B, we understand that

[C.sub.2][C'.sub.2]= [C.sub.2]A = x + y/2.

Hence, OA = O[C'.sub.2] = R where R is the radius of the circumcircle of the base triangle ABC. Similarly O[A'.sub.2] = O[B'.sub.2] = R. Therefore all midpoints of the sides of the upper triangle lie on the circumcircle sphere of the base triangle (fig. 10).

Corollary 11 The centroid of the Pavillet tetrahedron lies on the line joining the circumcenter of the base triangle to the center of the Euler circle of the upper triangle.

Proof.

The center of the first twelve point sphere of an orthocentric tetrahedron is its centroid [1, [section] 797-798]. Now, because the trace of the first twelve point sphere on the face of the upper triangle is also the Euler circle of this triangle, its center lies on a perpendicular to the upper face going through the center of this circle, but the center of the circumcircle sphere also lies on this perpendicular.

[FIGURE 9 OMITTED]

[FIGURE 10 OMITTED]

8 Conclusions

We have given elementary geometric or algebraic proofs of the fundamental properties of the Pavillet tetrahedron. Moreover we have found elementary proofs for a number of properties of the Soddy line of a triangle. Lots of geometric properties remained to be discovered about this new object and from the use of inversion or other transformations we probably could also derive interesting results.

Though it is not new to use solid geometry to solve plane geometry problems, it is rather unusual in triangle geometry (may be with the exception of [4, [section] 486 p. 290]). Hence the discovery of the Pavillet tetrahedron gives a new dimension to triangle geometry. There is not only a correspondence between the base triangle and the tetrahedron, we have also found a correspondence between the base triangle and the upper triangle. The Pavillet tetrahedron could be a new type of triangle transformation which would need to be formalized. In the meantime, it allows finding new short ways to prove some well known results about triangle properties usually found with quite labor-intensive methods. Students which study geometry, get one more tool to address known problems in an unusual way and to demonstrate the unity of the world of mathematics.

References

 N. Altshiller-Court. Modern pure solid geometry. The Macmillan company, 1935.

 D. Eppstein. Tangent spheres and triangle centers. The American Mathematical Monthly, 108(1):63-66, 2001.

 Richard Guy. Five-point circles, the 76-point sphere, and the Pavillet tetrahedron. (preprint), October 2011.

 Roger A. Johnson. Advanced Euclidean Geometry. Dover Publications Inc., Mineola, N.Y., reprint from modern geometry - 1929 edition, 2007. QA474J6 2007.

 Dergiades Nikolaos. The soddy circles. FORUM GEOMETRICUM, 2007.

 Adrian Oldknow. The Euler-Gergonne-Soddy triangle of a triangle. The American Mathematical Monthly, 103:319-329, April 1996.

 Axel Pavillet. The orthocentric tetrahedron of a triangle, new properties and inverse problem. In Proceedings of the 15 th International Conference on Geometry and Graphics (ICGG 2012),

pages 569-578, 08 2012.

 Axel Pavillet. The orthocentric tetrahedron of a triangle. Forum Geometricorum, forthcoming.

 Gunter Weiss. Is advanced elementary geometry on the way to regain scientific terrain? In Proceedings of the 15 th International Conference on Geometry and Graphics (ICGG 2012), pages 793-804, 08 2012.

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Axel Pavillet

axel.pavillet@polytechnique.org

vvsss@rambler.ru

Murmansk State University

Russia
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