62 Line Tangents to Four Thin Triangles

H. Brönnimann, O. Devillers, S. Lazard, and F. Sottile
Consider the four triangles in R3 whose vertices are as follows:
t1   (-10.5, 1, -10.5),
  (.5628568345479573470378601, 1, .5628568345479573470378601),
  (.56285683454726874605620706, .99999999999822994290647247, .56285683454726874605620706)
t2   (-10.5, -1, 10.5),
  (1.394218989475, -1, -1.394218989475),
  (1.3942406911811439954597161, -1.0000237884694881275439271, -1.3942406911811439954597161)
t3   (-9.5, -9.5, .25),
  (.685825, .685825, .25),
  (.69121730616063647303519136, .69121730616063647303519136, .26069756890079842876805653)
t4   (9.5, 0, 0),
  (-.51, 0, 0),
  (-1.0873912730501133759642956, 0, -.51645811088049333541289247)

    These triangles are quite `thin', their smallest angles are (in degrees), respectively
t1 : .64823e-11     t2 : .81028e-4     t3 : .42527e-1     t4 : 2.7927.

Theorem. There are exactly 62 lines in R3 that are tangent to the four triangles t1, t2, t3, and t4.

    The vertices above are decimal approximations to rational numbers obtained in our construction (outlined below). They are given to 26 decimal places. This precision is necessary: If truncated to 25 decimal places, then there will be only 60 common tangents. Here is the last coordinate of the third point of t1:
70357104318408593257025882167/125000000000000000000000000000
    While this Theorem may be verified directly by simply computing the lines (that is the origin of the remark about decimal precision), it is more illuminating to contemplate our construction of these four triangles. This construction also explains the exponential shrinkage of the smallest angle, as we progress from t4 down to t1.
    Our construction of four triangles having 62 common tangent is a pertubation of four lines having two common transversals. The construction begins with the four lines, in blue, red, green, and magenta. The blue, red, and magenta lines lie in one ruling of the yellow hyperbolic paraboloid. There are two black lines that meet all four lines (and also lie on this hyperboloid).
    More precisely, the four lines are parametrized as follows:
l1 : (t,1,t),     l2 : (t,-1,-t),     l3 : (t,t,1/4),    and    l4 : (t,0,0).

    Next, we perturb each line into a triangle, two of whose sides are close to the line with the third (shorter) side chosen judiciously. We do this by first considering a plane through each line. These planes (P1 --- P4) are
P1 : x=z,     P2 : x=-z,     P3 : x=y,    and    P4 : y=0.
In each plane Pi consider the conic Ci that is its intersection with the doubly-ruled hyperboloid on which the other lines lie. The plane Pi was chosen so that the two intersections of this conic with the line li are in the same connected component (branch) of Ci . The picture in each plane is shown below. The coordinates (t,u) in each plane are chosen so that the line li is u=0, and t is the parameter given above when describing the lines.
     
P1   P2   P3   P4
    These conics meet the lines (l1 --- l4) at the same points as the two transversals. For each i=1,2,3,4, write li now for some segment of the line which meets the conic Ci in two points. If we rotate each segment li slightly in the plane Pi , keeping one vertex fixed, we obtain a nearby segment ki . These two lines li and ki  form a triangle ti  whose third side is mi . If the rotation is sufficiently small and the common point far enough away from Ci , then each of the 16 four-tuples of segments (choosing either li or ki from each triangle) will give 2 lines tangent to the four triangles, a total of 32 in all.

    We can get more common tangents by carefully chosing the third sides (mi) of the triangles. We first introduce some notation. The choices of li or ki give different conics in each of the planes, which we suggestively write as follows. Set Cllli := Ci , and for instance, set Cklk2 to be the conic in the plane P2 which is the intersection of the lines meeting each of k1, l3, and k4 with the plane P2 . Similarly, Cllk3 is the conic in the plane P3 where the lines meeting each of l1, l2, and k4 meet the plane P3 .

    These new conics are `near' to the original conics. For example, on the left below, we display the picture in the fourth plane P4 for the pertubation giving us 62 common tangents. There, the horizontal coordinate is the usual x and the vertical coordinate is the usual z coordinate. In this picture, k4 is red, Clll4 is brown, and Ckkk4 is green. The other conics are not displayed, as they cannot be distinguished from one of these two. Since the two conics are so close, the configuration is better shown in different coordinates. On the right below, we display the part of the conics between the two lines l4 and k4 on the left of the first picture, but in different coordinates. In these coordinates, l4 remains horizontal, but the vertical direction is taken to be the blue line, which was secant to the conics Clll4 and Ckkk4 . This, and the exaggeration of the horizontal scale were chosen to accentuate the curvature of the conics.

    The other conics are not shown for they lie very close to one of these two and would not appear even in the second picture. There are four near Clll4, all come from l3 (and have the form Cxxl4), and there are four near Ckkk4, and all come from the perturbed segment k3 (and have the form Cxxk4),. Note that all 8 cut the secant line in two points each. If we let t4 be the triangle formed by l4, k4, and the secant line m4, then there are now 48 common tangents to the four triangles.
    One might naively expect that this construction may be repeated in each of the other three planes. However, that it not the case. In order for the conics in P4 to be close enough so that the segment m4 between l4 and k4 can meet all 8, the perturbed lines k1, k2, and k3 had to be rather close to the original lines l1, l2, and l3. In fact, they have to be significantly closer to the lines l1, l2, and l3 than k4 was to l4. (Which was not very close, as there needed to be enough curvature in the conics Cxxx4 for it to be possible to have m4 cut all of them.) A result of this is that the line segent m3 can meet at most four of the conics in P3.
    In fact it turns out that the four conics Cxxl3 are `inside' of the four conics Cxxk3, and these are the ones that m3 can meet.
    In order for those conics to be close enough together, l2 and k2 must be significantly closer together than were the previous two pairs. Then, m2 can be chosen to meet the 2 conics Cxkk2, which are the outermost of the 8 conics in P2.
    Finally, in the same way l1 and k1 must be microscopically close enough together to enable m2 can be chosen to meet the 2 conics Cxkk2. Then it is only possible for m1 to meet the conic Clkk1, which turns out to be the outermost.
    This gives 32 + 16 + 8 + 4 + 2 = 62 common tangent to the four triangles t1, t2, t3, and t4, proving the Theorem.
Here are some files to accomplish the construction, check that there are indeed 62 comon tangents, and draw some pictures.
construction.maple  This is the skeleton file which actually contains the construction.
procedures These are procedures which are necessary for computing the common tangents to 4 triangles.
plotH1.maple This draws the hyperboloid on this page.
plane1.maple Draws the picture in the first plane, P1
plane2.maple Draws the picture in the second plane, P2
plane3.maple Draws the picture in the third plane, P3
plane4.maple Draws the picture in the fourth plane, P4
plotConics.maple This draws some of the pictures on this page.
plane4.www.maple This is a version of plane4.maple for drawing some more of the pictures.
Based upon work supported by the National Science Foundation under CAREER Grant DMS-0134860.
Modified since: 4 January 2005 by Frank Sottile