The triangle $ABC$ has $CA = CB$. $P$ is a point on the circumcircle between $A$ and $B$ (and on the opposite side of the line $AB$ to $C$). $D$ is the foot of the perpendicular from $C$ to $PB$. Show that $PA + PB = 2 \cdot PD$.

Click for solution I note the middlepoint $M$ of the side $AB$. $\widehat {CAB}\equiv \widehat {CPB}\equiv \widehat {CPD}\Longrightarrow \triangle CAM\sim\triangle CPD\Longrightarrow$ $\frac{CA}{AM}=\frac{CP}{PD}\Longrightarrow PC=\frac{2a}{c}\cdot PD$. From the Ptolemeu's theorem results the relation $PA\cdot BC+PB\cdot AC=PC\cdot AB\Longrightarrow a\cdot (PA+PB)=c\cdot PC\Longrightarrow$ $PA+PB=\frac ca\cdot PC=\frac ca\cdot \frac{2a}{c}\cdot PD\Longrightarrow PA+PB=2PD$.

Two circles with centers $O_{1}$ and $O_{2}$ meet at two points $A$ and $B$ such that the centers of the circles are on opposite sides of the line $AB$. The lines $BO_{1}$ and $BO_{2}$ meet their respective circles again at $B_{1}$ and $B_{2}$. Let $M$ be the midpoint of $B_{1}B_{2}$. Let $M_{1}$, $M_{2}$ be points on the circles of centers $O_{1}$ and $O_{2}$ respectively, such that $\angle AO_{1}M_{1}= \angle AO_{2}M_{2}$, and $B_{1}$ lies on the minor arc $AM_{1}$ while $B$ lies on the minor arc $AM_{2}$. Show that $\angle MM_{1}B = \angle MM_{2}B$. Ciprus

Find all positive integers which have exactly 16 positive divisors $1 = d_1 < d_2 < \ldots < d_{16} =n$ such that the divisor $d_k$, where $k = d_5$, equals $(d_2 + d_4) d_6$.

Prove that for all positive real numbers $a,b,c$ the following inequality takes place \[ \frac{1}{b(a+b)}+ \frac{1}{c(b+c)}+ \frac{1}{a(c+a)} \geq \frac{27}{2(a+b+c)^2} . \] Laurentiu Panaitopol, Romania

Click for solution Another solution is : From AM-GM ineq we have $\frac{2}{b(a+b)}+\frac{2}{c(b+c)}+\frac{2}{a(c+a)}\geq\frac{6}{XY}$ where $X=\sqrt[3]{abc}$ and $Y=\sqrt[3]{(a+b)(b+c)(c+a)}$ By AM-GM again we have that $X\leq\frac{a+b+c}{3}$ and $Y\leq\frac{2a+2b+2c}{3}$ The result follows