Chords $ AB$ and $ CD$ of a circle intersect at a point $ E$ inside the circle. Let $ M$ be an interior point of the segment $ EB$. The tangent line at $ E$ to the circle through $ D$, $ E$, and $ M$ intersects the lines $ BC$ and $ AC$ at $ F$ and $ G$, respectively. If \[ \frac {AM}{AB} = t, \] find $\frac {EG}{EF}$ in terms of $ t$.

Let $ n \geq 3$ and consider a set $ E$ of $ 2n - 1$ distinct points on a circle. Suppose that exactly $ k$ of these points are to be colored black. Such a coloring is good if there is at least one pair of black points such that the interior of one of the arcs between them contains exactly $ n$ points from $ E$. Find the smallest value of $ k$ so that every such coloring of $ k$ points of $ E$ is good.

Determine all integers $ n > 1$ such that \[ \frac {2^n + 1}{n^2} \]is an integer.

Let $ {\mathbb Q}^ +$ be the set of positive rational numbers. Construct a function $ f : {\mathbb Q}^ + \rightarrow {\mathbb Q}^ +$ such that \[ f(xf(y)) = \frac {f(x)}{y} \] for all $ x$, $ y$ in $ {\mathbb Q}^ +$.

Click for solution It suffices to construct such a function satisfying $f(ab)=f(a)f(b),\ \forall a,b\in\mathbb Q^+\ (*)$ (this implies $f(1)=1$) and $f(f(x))=\frac 1x,\ \forall x\in\mathbb Q^+\ (**)$. All we need to do is define $f(p_i)$ s.t. $(*)$ whenever $x=p_i$ for some $i\ge 1$, where $(p_n)_{n\ge 1}$ is the sequence of primes, and then extend it to the rest of $\mathbb Q^+$ so that $(**)$ holds. Then it's clear that $(*)$ will automatically hold.

Given an initial integer $ n_0 > 1$, two players, $ {\mathcal A}$ and $ {\mathcal B}$, choose integers $ n_1$, $ n_2$, $ n_3$, $ \ldots$ alternately according to the following rules : I.) Knowing $ n_{2k}$, $ {\mathcal A}$ chooses any integer $ n_{2k + 1}$ such that \[ n_{2k} \leq n_{2k + 1} \leq n_{2k}^2. \] II.) Knowing $ n_{2k + 1}$, $ {\mathcal B}$ chooses any integer $ n_{2k + 2}$ such that \[ \frac {n_{2k + 1}}{n_{2k + 2}} \] is a prime raised to a positive integer power. Player $ {\mathcal A}$ wins the game by choosing the number 1990; player $ {\mathcal B}$ wins by choosing the number 1. For which $ n_0$ does : a.) $ {\mathcal A}$ have a winning strategy? b.) $ {\mathcal B}$ have a winning strategy? c.) Neither player have a winning strategy?

Prove that there exists a convex 1990-gon with the following two properties : a.) All angles are equal. b.) The lengths of the 1990 sides are the numbers $ 1^2$, $ 2^2$, $ 3^2$, $ \cdots$, $ 1990^2$ in some order.