# Is there a triangle like this?

I want to find the numbers $a$, $b$, $c$, $d$ of the function $y = \dfrac{a x + b}{c x + d}$ so that the triangle $ABC$ with three points $A$, $B$, $C$ have integer coordinates and lies on the graph of the given function, then the centroid of the triangle $ABC$ (have also integer coordinates) is also lies on the graph. I tried

Clear[f, m, n, p];
f[x_] := (a x + b)/(c x + d);
m = {h, k};
n = {e, f};
p = {(m[[1]] + n[[1]] + x)/3, (f[m[[1]]] + f[n[[1]]] + f[x])/3}
Reduce[{p[[2]] == f[p[[1]]], f[h] == k, f[e] == f,
Sequence @@ Thread[0 < {a, b, c, d, e, f, h, k, x} < 10],
a < b < c < d}, {a, b, c, d, e, f, h, k, x}, Integers]


With my code, I get the answer is False. Is there a triangle like this?

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Are there any examples if you relax the integer constraint? – bill s Sep 14 '13 at 16:18
This seems like it would fit well in the Mathematics site. – David Carraher Sep 14 '13 at 16:54

Edit

A new solution (no luck is needed now :P ):

g[x_] := (a x + b)/(c x + 1); (*d is a scale factor or zero*)
cent[x1_, x2_, x3_] := 1/3  ({x1, g[x1]} + {x2, g[x2]} + {x3, g[x3]});
fi = First@FindInstance[{
g@First@cent[x1, x2, x3] == Last@cent[x1, x2, x3],
a/b != c != 0, x1 != x2 != x3,
Element[{x1, x2, x3, g[x1], g[x2], g[x3],Sequence @@ cent[x1, x2, x3]}, Integers]},
{x1, x2, x3, a, b, c}, Reals];
Grid@{
{"p1", {x1, g[x1]}},
{"p2", {x2, g[x2]}},
{"p3", {x3, g[x3]}},
{"Centroid", cent[x1, x2, x3]}} /. fi


\begin{array}{cc} \text{p1} & \{6,-6585\} \\ \text{p2} & \{0,-85\} \\ \text{p3} & \{9,1540\} \\ \text{Centroid} & \{5,-1710\} \\ \end{array}

x = {{x1, g[x1]}, {x2, g[x2]}, {x3, g[x3]}};
IntegerQ /@ (Flatten@Join[x, {cent[x1, x2, x3]}] /. fi)


{True, True, True, True, True, True, True, True}

Grid@{{"a", a}, {"b", b}, {"c", c}} /. fi // TeXForm


\begin{array}{cc} \text{a} & -59 \\ \text{b} & -85 \\ \text{c} & -\frac{7}{45} \\ \end{array}

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Why don't you have it print out the x,y coordinates of the vertices. – David Carraher Sep 14 '13 at 17:27
@DavidCarraher Done – belisarius has settled Sep 14 '13 at 17:28
Amazing, weird result. I'm impressed. – m_goldberg Sep 14 '13 at 23:04
@m_goldberg I really started trying to find a stronger evidence for the nonexistence, but ... – belisarius has settled Sep 15 '13 at 17:13
Very nice, indeed. – Eckhard Sep 15 '13 at 17:30

My student found a function $y = \dfrac{93 x + 6}{x+2}$. Then, $A(-6,138)$, $B(8,75)$, $C(-62,96)$ and centroid $G(20,103)$. And he found a general formulas in order to find the coefficients $a$, $b$, $d$ if choice $c= 1$. Another functions from the following code

{y == (a x + b)/(c x + d)} /. Solve[{a d - b == 180, 90 <= a <= 180, c == 1, 1 <= b <= 10, 1 <= d <= 10}, {a, b, d}, Integers]


Try this code

Clear[f, a, b, c, g];
f[x_] := (19 x + 10)/(x + 10);
a = {x1, f[x1]};
b = {x2, f[x2]};
c = {x3, f[x3]};
g = 1/3 {x1 + x2 + x3, f[x1] + f[x2] + f[x3]}
{a, b, c, g} /.
Solve[{g[[2]] == f[g[[1]]],
Sequence @@ Thread[-200 < {x1, x2, x3} < 200], x1 < x2 < x3}, {x1, x2, x3}, Integers]


I want to the numbers a, b, d and coordinates of point A, B, C, G are smaller. Another function is f[x_] := (11 x + 1)/(7 x + 17);

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