geometry
I want to find the length of the sides of a triangle (not right triangle) with center $C(1; 2)$ and inradius $R=5$. How can I do?
I tried.
We also know that $a=k(u^2-v^2)$, $b=2kuv$, $c=k(u^2+v^2)$ for some integers $k,u,v$, so $$ 2k^2(u^2-v^2)uv=2009k\cdot(u^2-v^2+2uv+u^2+v^2)$$ i.e. $$ k(u-v)v = 2009.$$ Now match the factors on the left with factors of $2009$.
Hint: A sketch of the situation should be helpful:
What can we say about the position of the center of a circle with those two tangents?
What is its radius?
The equation of a circle with radious $r$ and center $(x_0,y_0)$ is: $$ (x - x_0)^2 + (y - y_0)^2 = r^2 $$
Solution:
a) Center line: The center points lie on a line in the middle of the lines, thus on \begin{align} y_c(x) &= \frac{1}{2}(y_1(x) + y_2(x)) \\ &= \frac{1}{2}((x + \sqrt{2}) + (x - 2 \sqrt{2})) \\ &= x - \frac{1}{\sqrt{2}} \end{align}
b) Radius: The radius of the circle is two times the distance of the tangent lines.
An orthogonal line to both tangents is $y = - x$ it intersects $y_1$ if $$ x + \sqrt{2} = - x \iff 2x = -\sqrt{2} \iff x = - 1/\sqrt{2} $$ thus at $P = (-1/\sqrt{2}, 1/\sqrt{2})$. The distance to the origin is $\lVert OP \rVert = \sqrt{1/2 + 1/2} = 1$
It intersects $y_2$ if $$ x - 2 \sqrt{2} = -x \iff 2x = 2 \sqrt{2} \iff x = \sqrt{2} $$ thus at $Q = (\sqrt{2}, -\sqrt{2})$. The distance to the origin is $\lVert OQ \rVert = \sqrt{2 + 2} = 2$. So both tangents are at a distance $3$ and the radius is $3/2$.
c) Equation of any circle with those tangents:
So all circles with those two tangents have the equation $$ (x - x_0)^2 + \left(y - x_0 + \frac{1}{\sqrt{2}}\right)^2 = 9/4 $$
d) The circle with those tangents that contains $(0, \sqrt{2})$:
We want the one that contains $(0, \sqrt{2})$ thus $$ 9/4 = x_0^2 + (\sqrt{2} - x_0 + 1/\sqrt{2})^2 = x_0^2 + (3/\sqrt{2} - x_0)^2 = x_0^2 + 9/2 + x_0^2 - 3\sqrt{2} x_0 \Rightarrow \\ -9/4 = 2 x_0^2 - 3\sqrt{2} x_0 \Rightarrow \\ -9/8 = x_0^2 - (3/\sqrt{2}) x_0 = (x_0-3/(2\sqrt{2}))^2 - 9/8 \Rightarrow \\ x_0 = \frac{3}{2\sqrt{2}} $$ This gives the center $C=(3/(2\sqrt{2}), 3/(2\sqrt{2}) - 1/\sqrt{2}) = (3/(2\sqrt{2}), 1/(2\sqrt{2}))$. And we have the equation $$ \left(x - \frac{3}{2\sqrt{2}} \right)^2 + \left(y - \frac{1}{2\sqrt{2}} \right)^2 = 9/4 $$
Best Answer
To make this question easier, I'll put the center of the incircle in the origin. Any integral solution can be easily transferred to the location you indicated. Let $M,N,P$ be the corners of the triangle, as used in your question. Let the coordinates of $M$ be $(M_x,M_y)\in\mathbb Z^2$. One important quantity in your scenario is
$$d:=M_x^2+M_y^2-25$$
As you can see, as long as $M$ is outside the circle, that quantity will be positive. Its square root can be used to describe the tangents from $M$ to the circle:
\begin{align*} t_1 &= \left\{(x,y)\;\middle\vert\; (5M_x+\sqrt dM_y)x + (5M_y-\sqrt dM_x)y = 5\left(M_x^2+M_y^2\right) \right\} \\ t_2 &= \left\{(x,y)\;\middle\vert\; (5M_x-\sqrt dM_y)x + (5M_y+\sqrt dM_x)y = 5\left(M_x^2+M_y^2\right) \right\} \end{align*}
The way I obtained these tangents is by performing the dual of a conic-line intersection, as described in section 11.3 of Perspectives on Projective Geometry by Richter-Gebert.
If these tangents are to cross any points with integer coordinates, their slopes have to be rational (or $\infty$ for vertical lines). These slopes are
$$-\frac{5M_x\pm\sqrt dM_y}{5M_y\mp\sqrt dM_x}$$
There are two conceivable ways these slopes might be rational: either because $\sqrt d$ is rational, or because numerator and denominator are rational multiples of the same irrational number. If one treats $\mathbb Q[\sqrt d]$ as a two-dimensional $\mathbb Q$-vectorspace, then this translates into the requirement that the two vectors have to be linearily dependent, which here means
$$\det\begin{pmatrix} 5M_x & \pm M_y \\ 5M_y & \mp M_x \end{pmatrix}=\mp5\left(M_x^2+M_y^2\right)=0$$
So the only other way would mean $M$ is the origin, which is of course impossible. So we are left with the requirement that $\sqrt d$ be rational, so $d$ must be a square number.
Now one possible way to tackle this problem is iterating over all integral coordinates in a given range, and checking whether the resulting $d$ is square. One can build up a collection of possible corner points, and using the coordinates of the tangents, one can see which corners could belong to the same triangle. The $t_1$ of one point must match the $t_2$ of the other edge, but scalar multiples of the coefficients describe the same line. I've implemented such an enumeration. Many results contain $5$ as one of the coordinates of one of the points. But not all do, so one of the more interesting results is
\begin{align*} M&=(-11, -2) & N&=( 1, 7) & P&=( 13, -9) \end{align*}
After moving $C$ back to where you defined it, this becomes
\begin{align*} M&=(-10, 0) & N&=( 2, 9) & P&=( 14, -7) \end{align*}
However, the solution above has aright angle at $N$, so as your question asks for a slution without a right angle, this one is not acceptable. Thanks to @coffeemath for making me aware of this condition. It seems that at least if the absolute value of the coordinates is restricted to no more than $1000$, then all solution without a right angle neccessarily have either a horizontal or a vertical tangent. So here is another go, again first in the untranslated situation:
\begin{align*} M&=(-11, -10) & N&=( -3, 5) & P&=( 25, 5) \end{align*}
After moving $C$ back to where you defined it, this becomes
\begin{align*} M&=(-10, -8) & N&=( -2, 7) & P&=( 26, 7) \end{align*}
I also adapted my program to filter out the right-angled triangles automatically. There are still many possible solutions remaining, but none with all coordinates less than $25$ in absolute value, in the coordinate system where the circle center is the origin.