19
$\begingroup$

Let's call a point $P$ which satisfies the following condition 'a rational point'.

Condition: Each distance $PA, PB, PC$ from a point $P$ to three vertices $A, B, C$ of an equilateral triangle $ABC$ which has edge-length $1$ is rational number.

I'm interested in this point because I found the following three:

1. There exist an infinite number of rational points on an edge of the equilateral triangle $ABC$.

2. There exist an infinite number of rational points on the circumference of the circumscribed circle of the equilateral triangle $ABC$.

3. There exists a rational point in the equilateral triangle $ABC$.

Supposing that $A(0,\frac{\sqrt3}{2}), B(-\frac12,0), C(\frac12,0)$, one example for the above 3 is the following: $$P\left(\frac{61}{2058},\frac{220}{1029}\sqrt3\right), \left(PA, PB, PC\right)=\left(\frac{73}{147}, \frac{95}{147}, \frac{88}{147}\right).$$

I found this example by using computer. I found the other examples, but I don't know there exist an infinite number of rational points in the equilateral triangle $ABC$. Then, here are my questions.

Question 1: Do there exist an infinite number of rational points in the equilateral triangle $ABC$ ?

Question 2: Can we find all rational points in the equilateral triangle $ABC$ ?

$\endgroup$
5
  • 3
    $\begingroup$ It can be shown that rational points correspond to integer solutions of the equation $a^4 + b^4 + c^4 + d^4 = a^2b^2 + a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2 + c^2d^2$. For example, $(73,88,95,147)$ is such a quadruplet. $\endgroup$
    – mercio
    Sep 9, 2013 at 15:53
  • $\begingroup$ @mericio: Nice info. Does every integer solution correspond to every rational point? $\endgroup$
    – mathlove
    Sep 9, 2013 at 16:04
  • $\begingroup$ Yes, since this equation tells you if it's possible to have a point $P$ with $AP = a, BP = b, CP = c$ when $ABC$ is equilateral with $AB = AC = BC = d$. $\endgroup$
    – mercio
    Sep 9, 2013 at 16:06
  • $\begingroup$ @mercio: Thanks, but it seems hard to find every solution. $\endgroup$
    – mathlove
    Sep 9, 2013 at 16:51
  • $\begingroup$ For proofs of mercio's formula, see math.stackexchange.com/questions/467952 $\endgroup$ Sep 12, 2013 at 21:23

2 Answers 2

5
+50
$\begingroup$

For the record, I will solve the much easier problem of finding infinitely many rational points on the circumcircle. Let's consider the case of a point $P$ on the arc from $A$ to $B$. By Ptolemy's theorem, $|PA| + |PB| = |PC|$, so if $|PA|$ and $|PB|$ are rational then so is $|PC|$. Also, $P$ is on the arc from $A$ to $B$ if and only if $P$ is on the right side of $\overline{AB}$ and $\angle APB = 120^{\circ}$. By the Law of Cosines, $\angle APB = 120^{\circ}$ is equivalent to $$|PA|^2 + |PA| |PB| +|PB|^2=1.$$

This conic can be paramterized in the usual way: Put $|PA|=1+t$, $|PB|=kt$. Solve for $t$ in terms of $k$; the result is $t=-(k+2)/(k^2+k+1)$. So $$|PA| = \frac{k^2-1}{k^2+k+1} \quad |PB| = \frac{-k^2-2k}{k^2+k+1} \quad |PC|=\frac{-2k-1}{k^2+k+1}$$ or, in other words, $$a=k^2-1 \quad b= -k^2-2k \quad c=-2k-1 \quad d=k^2+k+1.$$ We want to have $-2 < k < -1$ to get the right signs.

I don't want the bounty for this though; I want mathlove to explain how the heck he or she found his or her solution.


I have figured out a way to get mathlove's answer. I'll write $PA=a/d$, $PB=b/d$ and $PC=c/d$. As described here, these obey the relation $$a^4+b^4+c^4+d^4 = a^2 b^2 + a^2 c^2 + a^2 d^2 + b^2 c^2 + b^2 d^2 + c^2 d^2.$$ Let $\Sigma$ be the surface in $\mathbb{P}^3$ cut out by this degree $4$ equation. Notice that $\Sigma$ has $16$ singular points: The $4$ points $(\pm 1 : \pm 1 : \pm 1 : 0)$ and the other $12$ which come from putting the zero in the other possible positions. The three points $(1:1:0:1)$, $(1:0:1:1)$ and $(0:1:1:1)$ are the vertices of the triangle; the other $13$ singularities involve negative or infinite values for $(PA, PB, PC)$.

The technical term for this is a Kummer surface and, in fact, $\Sigma$ is the special kind of Kummer surface called a tetrahedroid. But we don't need to know this to follow the rest of the argument.

Take a plane through any three of the singularities. The resulting planar slice of $\Sigma$ will be a degree $4$ plane curve with $\geq 3$ nodes. If there are exactly $3$ nodes, the resulting curve is genus $0$, and thus has a rational parametrization over $\mathbb{C}$. That parmetrization doesn't have to have rational coefficients but, in some lucky cases, it does. We also aren't promised that the resulting values of $(PA, PB,PC)$ will be positive, let alone inside the triangle but, again, sometimes we get luck.

Mathlove uses the plane $b+d=2a$, passing through the points $(0,1,1,-1)$, $(0,-1,1,1)$ and $(1,1,0,1)$. The complete list of planes, up to permuting $(a,b,c,d)$ and switching signs, is $a=0$, $a+b=0$, $a+b+c=0$, $a+b+2c=0$ and $a+b+2c+3d=0$. These symmetries of $\Sigma$ do not respect the condition that the points actually correspond to physical points inside the triangle, so you have to keep track of more possibilities if you want that to hold.

Several of these planes correspond to interesting geometric configurations:

The equations $a+d=b$, $a+b=d$ and $b+d=a$ are $PA+1=PB$, $PA+PB=1$ and $PB+1=PA$ respectively, which say that $P$ lies on the line $AB$ (either in the two unbounded rays or in the line segment $AB$.)

The equation $a+b=c$ means $PA+PB=PC$, so (by Ptolemy's theorem) $P$ is on the arc of the circumcircle from $A$ to $B$; the other arcs of the circumcircle are described similarly.

The equation $a=b$ means that $P$ is on the perpendicular bisector of $AB$. This doesn't actually contribute any points; the intersection of $\Sigma$ with $\{ a=b \}$ the product of two conics, neither of which has rational coefficients.

Other planes, such as mathlove's choice $PB+1=2PA$ have no clear geometric meaning, but we can still rationally parametrize them and, at least in some cases, it seems we win.

As far as I can tell from skimming papers, there is an enormous literature on rational points on Kummer surfaces, but there isn't one simple answer. I had hoped to use this question as an opportunity to teach myself about Kummer surfaces (and, to some extent, I have) but it looks it's a big field, so I'll stop here.

$\endgroup$
5
$\begingroup$

I've just got the following:

The answer for Question 1 : There exist an infinite number of rational points in the equilateral triangle ABC.

$\left(PA, PB, PC\right)=$ $$\left(\frac{n^4+10n^2+9}{n^4+4n^3+10n^2-12n+9}, \frac{n^4-4n^3+10n^2+12n+9}{n^4+4n^3+10n^2-12n+9}, \frac{8n^3+24n}{n^4+4n^3+10n^2-12n+9}\right)$$ where $n=6k\pm4$ for $k\in\mathbb N$.

Proof : Supposing that $PA=a, PB=b, PC=c$ for a point $P$ and an equilateral triangle $ABC$ which has edge-length $m$, let's prove that $P$ exists in the $ABC$ where

$$a=n^4+10n^2+9, \ \ \ b=n^4-4n^3+10n^2+12n+9,$$$$c=8n^3+24n, \ \ \ m=n^4+4n^3+10n^2-12n+9.$$

Letting $\gamma=e^{i\pi/3}=\frac{1+\sqrt3i}{2}$, note that $\gamma+\bar\gamma=\gamma\bar\gamma=1, {\gamma}^2=\gamma-1, |\gamma|=1$.

Letting $z=\{(n^2-3)+4n\gamma\}^2$, then $$z=(n^4-22n^2+9)+8n(n^2+2n-3)\gamma.$$ Since $$|z|^2=z\bar z=\{(n^2-3)+4n\gamma\}^2\{(n^2-3)+4n\bar\gamma\}^2,$$then $$|z|=\{(n^2-3)+4n\gamma\}\{(n^2-3)+4n\bar\gamma\}=(n^2-3)^2+4n(n^2-3)+16n^2=m.$$

Since $$z-a=-32n^2+8n(n^2+2n-3)\gamma=8n\{-4n+(n^2+2n-3)\gamma\},$$ then $$|-4n+(n^2+2n-3)\gamma|^2=16n^2-4n(n^2+2n-3)+(n^2+2n-3)^2=(n^2+3)^2.$$ Hence, $$|z-a|=8n(n^2+3)=c.$$

Noting that $$(z{\gamma}^{-1}-a){\gamma}^2=\{(n^4-22n^2+9)+8n(n^2+2n-3)\}\gamma-a{\gamma}^2$$$$=(n^4-22n^2+9)\gamma+\{8n(n^2+2n-3)-a\}(\gamma-1)$$$$=\{a-8n(n^2+2n-3)\}+\{n^4-22n^2+9-a+8n(n^2+2n-3)\}\gamma$$$$=(n^4-8n^3-6n^2+24n+9)+8n(n^2-2n-3)\gamma$$ and that $$\{(n^2-4n-3)+4n\gamma\}^2=(n^2-4n-3)^2+8n(n^2-4n-3)\gamma+16n^2{\gamma}^2$$$$=(n^4-8n^3-6n^2+24n+9)+8n(n^2-2n-3)\gamma,$$ then $$|z{\gamma}^{-1}-a|=|(z{\gamma}^{-1}-a){\gamma}^2|=|(n^2-4n-3)+4n\gamma|^2$$$$=(n^2-4n-3)^2+4n(n^2-4n-3)\gamma+16n^2=n^4-4n^3+10n^2+12n+9=b.$$

Note that both $ABC$ and $APQ$ are equilateral triangles where $A(0), B(z{\gamma}^{-1}), C(z), P(a), Q(a\gamma)$.

Since $$QC=PB=b,$$ then $$QC+CA=b+m=2a=QP+PA.$$

Hence, since the point $C$ is outside of the circumscribed circle of $APQ$, $\angle ACP\lt\frac{\pi}{3}.$ Hence, we know that the point $P$ is in the equilateral triangle $ABC$.

Here, let's prove that the GCD of $a,b,c,m$ is $1$ for $n=6k\pm4 (k\in\mathbb N)$.

Suppose that $a,b,c,m$ has a common prime factor $p\ge5$. In mod $p$, since $a\equiv b\equiv c\equiv 0, $ $$4n^3\equiv12n\rightarrow c\equiv 48n\equiv0\rightarrow n\equiv0\rightarrow a\equiv9\equiv0$$ leads a contradiction. If $n\equiv0$ (mod $2$), then $a\equiv b\equiv m\equiv 1, c\equiv 0$. If $n\equiv1$ (mod $3$), then $a\equiv c\equiv 2, b\equiv 1, m\equiv 0$. If $n\equiv2$ (mod $3$), then $a\equiv 2, b\equiv 0, c\equiv m\equiv 1$.

Hence, we know that the GCD of $a,b,c,m$ is $1$ for $n=6k\pm4 (k\in\mathbb N)$. Now the proof is completed.

PS: $(PA, PB, PC)=\left(\frac{65}{73}, \frac{57}{73}, \frac{112}{73}\right)$ is the $n=2$ case. I expect that this would be the minimum solution about $c$.

Question 2 still remains unsolved.

$\endgroup$
6
  • 1
    $\begingroup$ How the heck did you find this? In retrospect, I can explain what you did as follows: Let $\Sigma$ be the surface in $\mathbb{P}^3$ given by mercio's degree $4$ equation. You have discovered that $a=b+c/2$ is a tritangent plane to $\Sigma$, so $\Sigma \cap \{ a=b+c/2 \}$ is degree $4$ plane curve with $3$ nodes, hence rational, and you found its parametrization. But finding tritangent planes by brute force requires solving an equation of degree $3200$, with no expectation of rational solutions. So how did you do it? $\endgroup$ Sep 12, 2013 at 15:32
  • 1
    $\begingroup$ @DavidSpeyer this is magic $\endgroup$
    – Norbert
    Sep 12, 2013 at 21:38
  • $\begingroup$ Sorry, got the linear relation wrong above. Should be $b+d=2a$ $\endgroup$ Sep 12, 2013 at 22:19
  • $\begingroup$ I would like to contribute with a simple observation: through the diagonalization of the suitable quadratic form, the integer solutions of $3(a^2+b^2+c^2+d^2)=(a+b+c+d)^2$ are the same of $6(a-b)^2+2(a+d-2c)^2+(3b-a-c-d)^2=(a+b+c+d)^2$, so for every solution of $6r^2+2s^2+t^2=u^2$, we may simply set $a=\frac{6r+2s-7t+3u}{12},b=\frac{u+t}{4},c=-\frac{4s+7t+3u}{12},d=\frac{-6r+2s-7t+3u}{12}.$ $\endgroup$ Sep 14, 2013 at 2:51
  • $\begingroup$ @JackD'Aurizio But we want solutions to $3 (a^4+b^4+c^4+d^4) = (a^2+b^2+c^2+d^2)^3$. We want the distances to be rational, not just their squares. $\endgroup$ Sep 17, 2013 at 0:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.