2018 Mystic cevians

Neither Maria's proof nor the proof in the previous problem explain why we obtain the Pythagorean triples needed. Let us see how to solve the equation $m^2 + n^2 = 2018^2$ for $m,n \in \mathbb{N}$ . We need to solve $(m + in)(m - in) \; = \; 2^2 \times 1009^2 \; = \; -(1 + i)^4(28 + 15i)^2(28-15i)^2$ where $1 + i$ and $28 + 15i$ are irreducible in the Euclidean domain $\mathbb{Z}[i]$ . These numbers are irreducible because $N(1+i) = 2$ and $N(28+15i) = 1009$ are prime numbers, where $N(a+bi) = a^2 + b^2$ is the distance function in $\mathbb{Z}[i]$ . If $u \in \mathbb{Z}[i]$ is a prime factor of $m + in$ , then its complex conjugate $u^\star$ must be a factor of $m - in$ .

This means that there are two options. Either $m + ni$ is a unit in $\mathbb{Z}[i]$ times $(1+i)^2(28+15i)(28-15i) = 2018i$ , or else $m + ni$ is a unit in $\mathbb{Z}[i]$ times $(1+i)^2(28 + 15i)^2 = -1680 + 1118 i$ . Thus either $\{m,n\} = \{2018,0\}$ , or else $\{m,n\} = \{1680,1118\}$ . Only the second solution is physically reasonable for this problem.

This problem is inspired by and is a copy of Mystic cevians problem from 4 years ago. Please refer to the solution to that problem for the full solution. Please check my related note Generalization of Pythagorean Theorem with Orthocentre .

In this case we are looking for two numbers which will form a Pythagorean Triple with hypotenuse 2018. They are 1680 and 1118. Using power of point we get:

$BT^2 = BH \times BE$ $CU^2 = CH \times CF$

The formulas are: $BT^2 = \dfrac{c^2-b^2+a^2}{2}$ $CU^2 = \dfrac{b^2-c^2+a^2}{2}$

where $E,F$ are feet of the altitudes from $B,C$ respectively.

One example of such triangle side lengths is: $b=2938.5, c=2657.5$

Note: $AH$ is a diameter of the circle stated in the problem. The circle passes through the points $E,F$ .

You can check my solution to the Trirectangular Corner Locus problem using vectors. Following the solution using vectors in 3D we get $AB^2 = AH\times AD + BH \times BE$ - another take on Pythagorean. $E$ is a foot of altitude from $B$ on $AC$