The nine-point-circle

Frankly speaking, I did not know much about nine-point circles. But, due to this question I Googled it and found out that if $R$ is the circumradius of a triangle and $R_{0}$ is the radius of the nine point cirle.

Then they are related by the following formula,

$R=2R_{0}$

Now simply we can find out the third side of the triangle by the Pythagorean theorem, and it comes out to be $14$ units.

Figuring out the are of the triangle( $\bigtriangleup$ ),

$\bigtriangleup= 0.5(14)(12)$

$\bigtriangleup= 84$ square units.

Now we also know the relation,

$\frac{abc}{4\bigtriangleup}=R$

Where a,b and c are the sides of the triangle.

Placing values,

$\frac{(13)(14)(15)}{4(84)}=R$

$R=\frac{65}{8}$ units.

And,

$R_{0}=\frac{65}{16}$ units.

$R_{0}=4+\frac{1}{16}$ units.

Thanks for sharing this question, I learned a new thing out of it.

If the altitude of length $h$ lies between two sides of length $a$ and $b$ , and if the altitude divides the opposite side of the triangle into signed lengths $x$ and $y$ respectively, so that the third side of the triangle is $c = x+y$ and $a^2 = h^2 + x^2$ and $b^2 = h^2 + y^2$ , then the sine of the angle of the triangle between the two sides $a$ and $b$ is (splitting the angle as the sum of the two angles formed by the altitude): $\sin C \; = \; \frac{x}{a}\times\frac{h}{b} + \frac{h}{a} \times \frac{y}{b} \; = \; \frac{h(x+y)}{ab} \; = \; \frac{hc}{ab} \;.$ Using the extended Sine Rule, the radius of the outcircle of the triangle is $R \; =\; \frac{c}{2\sin C} \; = \; \frac{ab}{2h}$ and so the radius of the nine-point circle is $r = \tfrac12R = \frac{ab}{4h}$ .

Even without the rider that the altitude in question lies between the sides of length $13$ and $15$ , this question has a unique solution, since we are told that the radius of the nine-point circle is rational. While there are six possible triangles that have sides of $13$ and $15$ and with one altitude equal to $12$ (two each for each of the three possibilities of where the altitude is in relation to the two known sides; in each case the altitude can meet the opposite side within the side itself, or else on the side extended), there are only two for which the nine-point circle has rational radius. These are the ones where (in my notation) $a=13$ , $b=15$ and $h=12$ , in which case $r = \tfrac{65}{16} = 4\tfrac{1}{16}$ .

I used co-ordinate geometry & a fact that:

"one and only one unique circle can pass through 3 given non-collinear points"

Let given triangle be ABC. AE is altitude from A meeting BC at E. AB is 13, AC is 15 and AE is 12. Hence BE is 5, EC is 9. Let D be midpoint of BC and F be midpoint of AC

Let BC become X-Axis and AE become Y-Axis, with origin at E(0,0). Hence A(0,12) ; B(-5,0) ; C(9,0) ; D(2,0) ; F(4.5,6) are the co-ordinates of various points.

Let equation of given circle be:

x²+y²+2gx+2fy+c=0

This passes through E(0,0), D(2,0) and F(4.5,6). Substituting the above co-ordinates we get:

c=0, g=-1, f=-63/16

The required radius being, r=√(g²+f²-c)=65/16

r=4+(1/16). Hence a+m+n=4+1+16=21

Note: we can verify that the above circle also passes through 6 other points by finding co-ordinate of all those points. I have used the easiest available co-ordinates for calculation, presuming that the answer is unique & question is correctly framed in a way that such a circle is actually possible with the given dimensions.