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Is there anything known about the existence of Heronian triangles ABC (i.e. with rational side lengths and rational area) that can be decomposed into three Heronian triangles ABD, BCD, CAD? Equivalently, is there a degenerate Heronian tetrahedron such that one vertex is in the interior of the triangle formed by the three other ones?

(Edit: wrong alternative formulation removed)

Note that it is easy to find solutions for the problem of dividing ABC into three triangles whose common vertex is not an inner point, but one of A, B, C. For example, take (36,30,30) and divide 36=11+14+11, so the two "chords" have length 25 and all triangles are Heronian.

Edit: sorry, I have posted this too quickly. In fact, all so-called Bis triangles in Buchholz' Thesis, i.e. Heronian triangles with 3 rational bisectors, satisfy the condition. Those triangles can be equivalently characterized by the condition that the sides and $AO,BO,CO$ are rational where $O$ is the incenter. (Then their area is necessarily rational, too). Equivalently, in the well-known Carmichael parametrization $a=n(m^{2}+k^{2}) ,b=m(n^{2}+k^{2}),c=(m+n)(mn-k^{2})$, both $m^{2}+k^{2}$ and $n^{2}+k^{2}$ must be squares.

Also, if $H$ denotes the orthocenter of a Heronian triangle, it is easy to see that $AH,BH,CH$ are always rational.

So the following question is more interesting:

Do there exist Heronian triangles ABC that can be decomposed into three Heronian triangles ABD, BCD, CAD where D is an inner point other than the incenter or the orthocenter of ABC?

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Including the definition of Heronian triangle would be very helpful. –  Alexandre Eremenko Dec 29 '12 at 23:28
    
A heronian triangle is one with rational side lengths and rational area, see mathworld.wolfram.com/HeronianTriangle.html This definition should be a part of the question. –  Igor Rivin Dec 30 '12 at 0:46
    
Also, the OP's alternative formulation seems to be wrong, since it is not clear how this specifies that the area is rational. –  Igor Rivin Dec 30 '12 at 0:48
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2 Answers

up vote 5 down vote accepted

Yes, for example the 13-14-15 triangle can be scaled by 11 to find a point $D$ at distance $80$, $91$, $102$ from the vertex opposite the side of length $11 \cdot 13$, $11 \cdot 14$, $11 \cdot 15$ respectively:

This was found by fixing $A,B,C$ and searching through the points $D$ of low height, using the parametrization of Pythagorean triples $x^2+y^2=d_1^2$ to search through only those $D$ that are already at rational distance from $(0,0)$, and then testing whether the distances to $B$ and $C$ are rational as well.

In fact, for any choice of triangle $ABC$, such points $D$ should be dense in the Euclidean plane, and thus in the interior of the triangle, because they're parametrized by a K3 surface with enough structure that standard tricks apply.

Let $A,B,C$ be the vertices of any rational Heronian triangle. We may choose Euclidean coordinates so that $A,B,C = (x_i,y_i)$ ($i=1,2,3$) with all $x_i$ and $y_i$ rational (for example, put $A$ at the origin and $B$ at $(x_2,0)$, etc.). Then the points $D=(x,y)$ at rational distances $d_i$ from $x_i$ correspond to solutions $(x,y,d_1,d_2,d_3)$ of the three Diophantine equations $(x-x_i)^2 + (y-y_i)^2 = d_i^2$ with each $d_i$ positive. Thus we seek rational points on the intersection $S$ of three quadrics in 5-space. Here the singularities of this intersection are mild enough that $S$ is birationally a K3 surface, as it would be if the intersection were smooth. The geometry yields several elliptic fibrations on $S$; e.g. for any line through (say) $A$ whose slope comes from a Pythagorean triangle, the points $D$ on that line that that are also at rational distance from $B$ and $C$ are parametrized by a genus-$1$ curve with rational points at infinity. Starting from those rational points (or those for which $D$ is the orthocenter, or indeed one of the vertices $A,B,C$), it should be straightforward to bounce around a few elliptic fibrations to find a dense set of rational points.

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There should be many examples where the interior vertex lies on the perpendicular bisector of one of the sides. It's best to redefine a Heronian triangle to be one with rational sides and rational area. I'll call such a T "standard" if its vertices are at (-2,0), (2,0) and (r,s) where r and s are rational. Every Heronian triangle is similar to a standard one.

Now let T be standard and P be (0,(x)-(1/x)) where x is rational. Then P has rational distance from 2 vertices of T. The condition that it have rational distance from the third is that there exists a rational y such that (xx-sx-1)^2 +(rx)^2 =y^2. If the elliptic curve one gets in this way has positive rank then there will be a dense set of points (0,(x)-(1/x)), all lying on the perpendicular bisector of the base of T, each giving the desired decomposition of T.

I worked out the case r=-2, s=3. Unfortunately the curve one gets is one of conductor 15 with 8 rational torsion points and rank 0. But there must be lots of choices of r and s where the rank is positive.

EDIT: Here's another construction which should give many examples where the interior point lies on an altitude. Consider a Heronian triangle with the base extending from (0,0) to (a+b.0), and the foot of the altitude to the base at (a,0). Let P be (a,x) where x is rational. Then P is at a rational distance from one vertex, and is at a rational distance from the other two when there are rational u and v with xx+aa=uu and xx+bb=vv. These equations again define an elliptic curve and one will get a dense set of points (a,x) on the altitude, each giving a desired decomposition, when the curve has positive rank.

The interesting question then seems to be the existence of a point, that lies neither on an altitude nor on the perpendicular bisector of a side, and that yields the desired decomoposition.

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