A seemingly impossible problem from my teacher

[Kbecker]

No its not homework, but my teacher (pre-cal) couldn't figure it out and neither can I, so see if you can!

I can't draw it so i'll explain it, pretty simple.

You have a square (ABCD) with a random point closest to the bottom-left corner of the square (it doesn't really matter which corner). The point is 3 units from A (the closest corner), 7 units from B (the point directly above A), and 5 units from D (the point directly right of A)

Keep in mind you don't no any angles except of course the 90 degrees of each corner.

Solve for - any side length!

 

[Tide]

The point lies at the intersection of 3 circles centered on the given vertices with their respective radii. You have 3 equations and 3 unknowns (x, y, a) where a is the length of each side of the circle. You're only interested in a so you want to eliminate x and y.

After a quick run through I get $a = 4$ or $a = 3 \sqrt {11}$.

 

[AKG]

Let X be your point. Let x be the angle BXA, y is the angle DXA, and z is the angle DXB. Let s be the side length of your square. Using the cosine law, you know:

1) 2a² = 25 + 49 - 2(5)(7)cos(z)
2) a² = 9 + 25 - 2(3)(5)cos(y)
3) a² = 9 + 49 - 2(3)(7)cos(x)

Treat a², cos(x), cos(y), and cos(z) as your unknowns, then you have three linear equations in terms of 4 unknowns. You can then solve for all of them in terms of one of them, say cos(x).

But you also know that x + y + z = 2π, and that π/2 < x < π. With a bunch of messy algebra (it will boil down to solving a cubic polynomial with integer coefficients), or using a computer program, you can find a solution for cos(x), and hence a², and hence a.

 

[Robokapp]

well, one triangle has sides 7, 3, a
another has sides 3, 5, a
a third has sides 7,5,a* sqrt {2}

you can use a lot of law of cosines, a big ugly heron formula for the whole square or some plain geometry. If i were to gwt this problem i'd go for the Cosines.