$E$ and $F$ are interior points of convex quadrilateral $ABCD$ such that $AE = BE$, $CE = DE$, $\angle AEB = \angle CED$, $AF = DF$, $BF = CF$, $\angle AFD = \angle BFC$. Prove that $\angle AFD + \angle AEB = \pi$.

$a$ and $b$ are natural numbers such that $b > a > 1$, and $a$ does not divide $b$. The sequence of natural numbers $\{b_n\}_{n=1}^\infty$ satisfies $b_{n + 1} \geq 2b_n \forall n \in \mathbb{N}$. Does there exist a sequence $\{a_n\}_{n=1}^\infty$ of natural numbers such that for all $n \in \mathbb{N}$, $a_{n + 1} - a_n \in \{a, b\}$, and for all $m, l \in \mathbb{N}$ ($m$ may be equal to $l$), $a_m + a_l \not\in \{b_n\}_{n=1}^\infty$?

For a given natural number $k > 1$, find all functions $f:\mathbb{R} \to \mathbb{R}$ such that for all $x, y \in \mathbb{R}$, $f[x^k + f(y)] = y +[f(x)]^k$.

Click for solution The additivity, along with $f(x^k)=f^k(x)$ for some $k > 1$, implies at once that $f(x)=x$ for all $x \in \mathbb{R}$. Indeed, for $n \in \mathbb{Z}$, we have $f(x+n)=f(x)+n$, so \[ \sum_j \binom{k}{j}f^j(x)n^{k-j} = (f(x)+n)^k = f^k(x+n) = f\big( (x+n)^k \big) = \sum_j \binom{k}{j}f(x^j)n^{k-j}. \] Fixing $x$, we may view this as a polynomial identity with respect to $n$. Comparing the coefficients of $n^{k-1}$, we see that $f(x^2) = f(x)^2 \geq 0$, so $f(x) \geq 0$ for $x \geq 0$ which, along with the additivity, implies that $f(x)$ increases monotonously. Hence, $f(x) \equiv x$.

For a given natural number $n > 3$, the real numbers $x_1, x_2, \ldots, x_n, x_{n + 1}, x_{n + 2}$ satisfy the conditions $0 < x_1 < x_2 < \cdots < x_n < x_{n + 1} < x_{n + 2}$. Find the minimum possible value of \[\frac{(\sum _{i=1}^n \frac{x_{i + 1}}{x_i})(\sum _{j=1}^n \frac{x_{j + 2}}{x_{j + 1}})}{(\sum _{k=1}^n \frac{x_{k + 1} x_{k + 2}}{x_{k + 1}^2 + x_k x_{k + 2}})(\sum _{l=1}^n \frac{x_{l + 1}^2 + x_l x_{l + 2}}{x_l x_{l + 1}})}\] and find all $(n + 2)$-tuplets of real numbers $(x_1, x_2, \ldots, x_n, x_{n + 1}, x_{n + 2})$ which gives this value.

In the equilateral $\bigtriangleup ABC$, $D$ is a point on side $BC$. $O_1$ and $I_1$ are the circumcenter and incenter of $\bigtriangleup ABD$ respectively, and $O_2$ and $I_2$ are the circumcenter and incenter of $\bigtriangleup ADC$ respectively. $O_1I_1$ intersects $O_2I_2$ at $P$. Find the locus of point $P$ as $D$ moves along $BC$.

Click for solution It should be equilateral $\triangle ABC$. Note that now the answer is still an algebraic curve of degree 2. (the locus is the curve $y^2-\frac{x^2}{3}=1$, where BC=2 is the x-axis with the midpoint as the origin)

Let $F = \max_{1 \leq x \leq 3} |x^3 - ax^2 - bx - c|$. When $a$, $b$, $c$ run over all the real numbers, find the smallest possible value of $F$.

Click for solution Okay, I am back home from school.. problem doesn't look that hard, which means I am doing something wrong.. someone please check my soln? EDIT: The last part was bad, but I think its fixed now. Use above trick, substituting y = x+ 2. Now the coefficents of y are independent, so we can consider the equivalent problem of minimizing G, the absolute value of a general monic cubic $y^3 + ay^2 + by + c$ on the domain $|y| \leq 1$, over all reals a,b,c. Write it in the form: $ay^2 + c + (y^3 + by)$. It is easy to check that $(a,b,c) = (0,\frac{-3}{4},0)$ gives us a maximum of $1/4$, because $y^3 - (3/4)y = 1/4 \Leftrightarrow (y-1)(y+ 1/2)^2$. Suppose [for a contradiction] that (a,b,c) is a tuple that gives a lower maximum. By differentiation, it is clear that the extremal values of $y^3 + by$ occur at $y = \pm \sqrt{-b/3}$ (if b<0) and $y = \pm 1$, (in the domain $|y| \leq 1$). Now, $ay^2 + c \leq 0$ whereever maximums of $y^3 + by$ occur, since if it was $> 0$, we would have the maximum larger. Also, $ay^2 + c \geq 0$ whereever minimums of $y^3 + by$ occur. By symmetry, both in the places where extremal values of $y^3 + by$ occur, as well as the graph of $ay^2 + c$, we have to conclude $ay^2 + c = 0$ at the maximums of $y^3 + by$ (**). -- If b < -3/4, then y = 1 gives us a maximum > 1/4, and y = -1 gives us a minimum < 1/4. Since $ay^2 + c$ is an even function, if it is nonzero, then for some value of y in $\{-1, 1\}$, the signs of $ay^2 + c$ and $y^3 + by$ must be the same, and the magnitude of the entire cubic exceeds 1/4, if $ay^2 + c$ doesn't have roots at $y = \pm 1$. Now suppose $ay^2 + c$ has roots at $y = \pm 1$. Clearly, $y = \sqrt{-b/3}$ is a local maximum of $y^3 + by$, and by the supposition b < -3/4, we have infact that it is a real number in [0, 1/2]. Now consider $\sqrt{-b/3}^3 + -ab/3 + b\sqrt{-b/3} + c$. By "$ay^2 + c$ has roots at $y = \pm 1$", we have $-a = c$; by considering the cubic at $y=0$, we have $|a| < 1/4$. So we have $\sqrt{-b/3}(4b/3) + c(b/3 + 1) < (1/2)(-1) + c(3/4) = 3c/4 - 1/2$. Now $|3c/4| < 3/16$, whereby $3c/4 - 1/2 < -5/16$. So the general cubic has a magnitude above $1/4$ at that point, and it is inadmissible. Similarily, consider b > -3/4 at the point $y = \sqrt{-b/3}$. Similarily, we must have $ay^2 + c$ with roots at $y = \pm \sqrt{-b/3}$. This implies ab = 3c. Consider $y = 1$ in the general cubic. We have $1 + a + b + c > 1/4 + a + c$. Now $3c = ab > -3a/4$, whereby $c > -a/4$. So $1/4 + a + c > 1/4 + 3a/4$. Obviously, due to the symmetry (oddness) of $y^3 + by$, it doesn't matter whether a was positive or negative. So the magnitude of the entire cubic is clearly $> 1/4$, and so it is inadmissible. So $b = -3/4$. By (**), $ay^2 + c$ is zero at 4 places, thus it must be the zero function. So we have $(a,b,c) = (0, -3/4, 0)$, with a maximum of 1/4. This is the same maximum as in the original problem, so we are done.