# Weitzenbock by Sanchez

### Solution 1

Let $A=(0,2),\,$ $B=(-2u,0),\,$ $C=(2v,0),\,$ where $u=\cot B,\,$ $v=\cot C.\,$ Obviously, $u+v\gt 0.\,$ We have $[ABC]=2(u+v),\,$ $[NAB]=u^2+1,\,$ $[PAC]=v^2+1,\,$ $[MBC]=(u+v)^2,\,$ implying

$[MBNAPC]=[2+u^2+v^2+(u=v)^2]+2(u+v).$

Hence, we need to prove that $2\sqrt{3}(u+v)\le 2+u^2+v^2+(u+v)^2.\,$ Denote $u+v=2s\,$ and $uv=p.\,$ Our inequality becomes $2\sqrt{3}s\le 4s^2-p+1,\,$ or, equivalently, $p\le 4s^2-2\sqrt{3}s+1.,$ But $p\le s^2,\,$ hence, suffice it to show that $s^2\le 4s^2-2\sqrt{3}s+1,\,$ i.e., $(\sqrt{3}s-1)^2\ge 0,\,$ which is true.

Equality iff $\displaystyle u=v=\frac{1}{\sqrt{3}},\,$ i.e., when $\Delta ABC\,$ is equilateral.

### Solution 2

Denote $BC=a,\,$ $AC=b,\,$ $AB=c,\,$ $\displaystyle p=\frac{a+b+c}{2},\,$ $S=[ABC].\,$ Using Heron's formula, $S=\sqrt{p(p-a)(p-b)(p-c)},\,$

(1)

### Acknowledgment

Leo Giugiuc has kind posted a problem from the Peru Geometrico facebook group (originally by Miguel Ochoa Sanchez) at the CutTheKnotMath facebook page, along with his solution (Solution 1). Solution 2 is by Marian Cucoanes; Solution 3 is by Rachid Moussaoui.