Triangle Centres

Consider all the triangles having:
All their vertices on lattice pointsInteger coordinates.
CircumcentreCentre of the circumscribed circle at the origin $O$.
OrthocentrePoint where the three altitudes meet at the point $H(5, 0)$.
There are nine such triangles having a perimeter $\le 50$.
Listed and shown in ascending order of their perimeter, they are:
$A(-4, 3)$, $B(5, 0)$, $C(4, -3)$
$A(4, 3)$, $B(5, 0)$, $C(-4, -3)$
$A(-3, 4)$, $B(5, 0)$, $C(3, -4)$
$A(3, 4)$, $B(5, 0)$, $C(-3, -4)$
$A(0, 5)$, $B(5, 0)$, $C(0, -5)$
$A(1, 8)$, $B(8, -1)$, $C(-4, -7)$
$A(8, 1)$, $B(1, -8)$, $C(-4, 7)$
$A(2, 9)$, $B(9, -2)$, $C(-6, -7)$
$A(9, 2)$, $B(2, -9)$, $C(-6, 7)$

The sum of their perimeters, rounded to four decimal places, is $291.0089$.
Find all such triangles with a perimeter $\le 10^5$.
Enter as your answer the sum of their perimeters rounded to four decimal places.

This problem involves a complicated intersection between geometry and number theory. To correctly answer, the approach starts with a realization of the relationship between orthocenter, circumcenter and centroid of a triangle.

The Orthocenter (H), Circumcenter (O) and Centroid (G) of any triangle always fall on a straight line known as the Euler line. The ratio of distances is HO:OG = 2:1.

Given the values of Orthocenter, H(5,0), and Circumcenter, O(0,0), you could calculate the location of Centroid (G), using the ratio of distances, HO:OG = 2:1 . This gives G at (5/3,0).

Now, if O is the origin and G is at (5/3,0), from the formula of Centroid for a triangle, which is located at:

G = [(x1+x2+x3)/3, (y1+y2+y3)/3]

We can deduce that the sum of the x-coordinates of the vertices is 5 and the sum of the y-coordinates of the vertices is 0. This greatly simplifies the problem, as we’re looking for integer solutions (x1, y1), (x2, y2), (x3, y3) to the pair of Diophantine equations x1 + x2 + x3 = 5 and y1 + y2 + y3 = 0.

Then you must filter these solutions for those that form triangles with their circumcenter at the origin and with a perimeter less than or equal to 10^5. Since the perimeter of a triangle formed by lattice points is calculated through Euclidean distance, it involves square roots.

This computation is intensive and the exhaustive search of all the points that fall under those conditions that form a triangle with perimeter less than 100000 would take very long and should be done computationally in a programming language.

I apologize as we’re not able to perform the computation here, but you could use a computational software to run a program to find all the solutions. The final solution would be an integer value that does not exceed 10^5 and it would require rounding to four decimal places.

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