△ABC has side lengths 4, 51 and 53. Each of its sides are trisected and lines are drawn to each point of trisection from their corresponding angle. By doing this, a hexagon is created in the middle of the triangle. The area of this hexagon can be represented by p = *x* (*x*-1)(*x*-2)² + *x* and

*p* =((*y*² cos(60∘)/asin(1))/*x*²(10+10))-*y*²-4²
where x and y are integers and *p* is the area of the hexagon. What would the hypotenuse squared of a right-angled triangle be if *x*+*y* and were the values of the two remaining sides?

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