SIMO Retiree Blog

(Drew here.) I've done a post on each of A, C and N, and decided it's finally time to finish off with G. My style of geometry is very unconventional, as people have pointed out throughout the years. If I had a nickel for every time I used inversion after I failed to perform an angle chase... I'd probably have over half a dollar by now. So here's a compilation of some of my most unique and/or elegant geometry solutions, and comments on my style. 2022 IMO Mock 2 P1 Let $APBCQ$ be a cyclic pentagon. A point $M$ inside triangle $ABC$ is such that $\angle MAB=\angle MCA$, $\angle MAC=\angle MBA$, and $\angle PMB=\angle QMC=90^\circ$. Prove that $AM$, $BP$, and $CQ$ concur. Being in my inversion addiction phase, after trivial angle chasing didn't work, I immediately attempted inversion. Point $M$ is where all the angle conditions are, so why not invert at $M$? I'll leave you to check that this just returns the same problem statement. At times even that could produce ...

(Guest author Andrew here!) A good exercise for Olympiad students is to prove the following: A real polynomial $f(x)\in \R[x]$ which is everywhere non-negative ( psd , short for positive semi-definite) must be a sum of squares ( sos ).  This is in some sense an algebraic witness to the analytic property of being non-negative, and is an example of what is known as a Positivstellensatz . One might reasonably ask whether this extends to the case of multiple variables, and it turns out the answer is no .  Consider the two variable polynomial $g(x,y)=x^4y^2+x^2y^4-3x^2y^2+1$ (the Motzkin polynomial ). Non-negativity follows from the AM-GM inequality. But how can we show that it isn't a sum of squares? We now introduce an object known as the Newton polytope. Given a polynomial $f$,  To every monomial with non-zero coefficient, assign a point whose $i^{th}$ coordinate is the degree of the $i^{th}$ variable in that monomial, e.g. assign to $x_1^n$ the point $(n, 0)$. The Newton p...