EPTI thought I would devote this entry to some interesting mathematics rather than a problem.
All of us know how to solve quadratic equations, and if you know Galois theory you know it is impossible to solve a quintic. Recently I saw a very cool way to solve the general quartic equation, relying on a bit of geometry. It does assume that we can already solve cubic equations.
Suppose P(x)=x^4+ax^3+bx^2+cx+d is an arbitrary quartic polynomial.
Let Q(x,y)=x-y^2 and let R(x,y)=y^2+axy+by+cx+d.
To find the zeros of P, it will suffice to find the common zeros of Q and R, two conics.
Let S(x,y,z)=xz-y^2 and let T(x,y,z)=y^2+axy+byz+cxz+dz^2
If we find the common zeros of S and T, we can set z=1 and find the common zeros of Q and R.
If C is any complex number, then CS+(1-C)T shares the same common zeros as S and T.
Our strategy will be to find a C which makes CS+(1-C)T particularly simple.
We can view CS+(1-C)T as a quadratic form on 3-space. The symmetric matrix representing this form will have entries that are linear polynomials in C.
By Sylvester's theorem, the form is degenerate if and only if the matrix has determinant 0. But the determinant will be a degree 3 polynomial in C, and since we can solve degree 3 polynomials we can find a C which makes CS+(1-C)T degenerate.
Geometrically, this means that the zeros of CS+(1-C)T are two planes. Finding the planes is a linear problem of completing the square, so we can assume the planes are known.
We want to find the common zeros of S, CS+(1-C)T, and z-1.
But the common zeros of the second two polynomials listed above are just two lines, and it is easy to find the intersections of a line with a quadratic like S!
. It could've been higher (a guys got to defend himself, right) but I lingered in this rather long problem and didn't have time for the others. If only I would've started with the trigonomical functions first, I would've gotten a higher grade. It also takes into account that I was kind of slacking off. To tell you that I forgot how to rationalize the freaking denominator, jeez! Oh well. 
