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This is a continuation of finding a vector as it relates to the Product of 2 Prime numbers. If you have a radius x which is the smallest of 2 Prime products you can rotate it around the x and y axis at an angle of y/x radians; where y is the larger of the Prime products. You have 2 sides of the triangle; both of which equal to x. The only question is how many times around the coordinate plane a radius of x while revolve around a 2 Pi circular revolution. Find the area of this triangle (and include all the times it revolves around the circle; 3 and 2347 for example) and you have N, the product of 2 Prime numbers. The only downfall is that you must isolate x and it has trigonometric functions such as Sine, Cosine, and Tangent. So it is not solved, but it should be easier to find equations and geometry to tinker with a solution.

the area of a triangle = 1/2 (base * height)

1/2 *[[cos (N/x^2) * x + x] * [sin (N/x^2) * x] = N

But this equation will not work until you know what quadrant y/x resides in.

test x = 5 ; N = 85

1/2 *[[cos (85/5^2) * 5 + 5] * [ Pi + sin ((85/x^2)) * 5] = 85

1/2 *[[cos (85/5^2) * 5 + 5] * [ Pi + sin ((85/x^2)) * 5] approximately = 83.5184

*It is important to note that this above equation worked because the angle of radians y/x falls in the second quadrant and the cosine is positive in the equation.

If we were in the 1st quadrant the equitation 1/2 *[[cos (N/x^2) * x] * [sin (N/x^2) * x] = N

As above in the 2nd quadrant: 1/2 *[[cos (N/x^2) * x + x] * [sin (N/x^2) * x] = N

In the 3rd quadrant: 1/2 *[[cos (N/x^2) * x + x] * [sin (N/x^2) * x + x] = N

In the 4th quadrant: 1/2 *[[cos (N/x^2) * x + 2x] * [sin (N/x^2) * x + x] = N

where sine and cosine are in radians and always positive.

We will use the definition of sine: Opposite / Hypotenuse

and

A trigonometric Pythagorean Identity (sin^2 angle ) + (cos^2 angle ) = 1.

There is a remainder problem here. We can divide the angle into sectors of Pi but mathematically finding the remainder of a number not between -1 < 0 < 1. Restated the problem is that we don't know x and y so we do not know in what quadrant y/x radians will fall in. We would have to test each of the four quadrant equations above. Also if you have a larger Prime number for y and a very small Prime number in x, the number of quadrants or revolutions of 2 Pi will be very large and there would be no way to calculate area. Let's see if we can still savage this equation.

(y/x) / PI = angle of last quadrant in radians

[(y/x) / Pi] is the angle in radians. It is the number of quadrants that y encircles a circle of radius x. The area of this circle covered is x * y or N. (Yes Prime numbers again. But the relation is by multiplication and not a pattern of Prime numbers.)

sin(angle) = Opposite / Hypothenus = ( ( (y/x) / Pi) / x)

Given the area of a triangle = 1/2 (base * height)

1/2 *[[cos (N/x^2) * x + x] * [sin (N/x^2) * x] = N

(sin^2 (angle)) + (cos^2 (angle)) = 1 That is the Pythagorean identity.

transforms the equation to

1/2 * [ [ (1- (( (y/x) / Pi) / x)^2) * x + x] * [ ( ( (y/x) / Pi) / x) * x] ] = N

knowing y = N/x or y/x = N / x^2

1/2 * [ [( 1- (( ( (N / x^2) /x) / Pi) / x)^2) * x + x] * [ ( ( ( (N / x^2) /x) / Pi) / x) * x] ] = N

where ((N / x^2) / x) = N / (x^3)

1/2 * [ [( 1- (( ( N / (x^3) / Pi) / x)^2) * x + x] * [ ( ( ( N / (x^3) /
Pi) / x) * x] ] = N

Remember the entire revolution of y/x radians must be added to the sine and
cosine. For example 23 and 3 would is in the 1st quadrant after making one
revolution. So that would be the area of one entire circle of radius 3 plus
the area of the triangle in the 1st quadrant.

Ok so we are stuck in knowing how many quadrants the angle of y/x radians will pass through. It seems like a challenging roadblock. But what if we were to investigate in determining not how many quadrants the angle passes through, but the percentage of N that is left in the last quadrant?

Again knowing

(N / (Pi/2))

and

(N / (y/x) = (N / x^2)

So

[(N / (Pi / 2)) - ((N / x^2) / (Pi / 2)) ]

So the angle is in one quadrant is:

1/2 *[[cos (N/x^2) * x] * [sin (N/x^2) * x] = N

in any quadrant:

1/2 * [ [( 1- (( ( N / (x^3) / Pi) / x)^2) * x] * [ ( ( ( N / (x^3) / Pi) / x) * x] ] = [(N / (Pi / 2)) - ((N / x^2) / (Pi / 2)) ]

Of course it needs to be tested but it is a start. The only problem is the quadrant problem which we are working towards to fix. We are assuming that Because the numbers are Prime the angle will be unique. So to find x we don't need the complete area N to find x.