Weitzenbock by Sanchez

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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)

$16S^2=2(a^2b^2+b^2c^2_c^2a^2)-(a^4+b^4+c^4)

By the Pythagorean Theorem, $\displaystyle BM=CM=\frac{a}{\sqrt{2}},\,$ $\displaystyle AP=CP=\frac{b}{\sqrt{2}},\,$ $\displaystyle NA=NB=\frac{c}{\sqrt{2}}\,$ so it follows that

(2)

$\displaystyle [MBC]=\frac{a^2}{4},\,[PAC]=\frac{b^2}{4},\,[NAB]=\frac{c^2}{4}.$

Then the required inequality is equivalent to

$\displaystyle\begin{align} &S+S\sqrt{3}\le S+[MBC]+[PAC]+[NAB]\,\Longleftrightarrow\\ &S\sqrt{3}\le\frac{a^2+b^2+c^2}{4}\,\Longleftrightarrow\\ &3\cdot 16\cdot S^2\le (a^2+b^2+c^2)^2\,\Longleftrightarrow\\ &(a^2+b^2+c^2)^2\ge 3[2(a^2b^2+b^2c^2+c^2a^2)-(a^4+b4+c^4)], \end{align}$

which is true because

$\begin{align}&(a^2+b^2+c^2)^2-3[2(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4)]\\ &\qquad\qquad =2[(a-b)^2+(b-c)^2+(c-a)^2]\ge 0. \end{align}$

Solution 3

Denote $BC=a,\,$ $AC=b,\,$ $AB=c.\,$

$\displaystyle\begin{align} [MBNAPC]&=[ABC]+\frac{1}{4}(a^2+b^2+c^2)\\ &\ge [ABC]+\frac{1}{4\cdot 3}(a+b+c)^2, \end{align}$

by the AM-GM inequality. Using the well inequality $\displaystyle S\le\frac{\sqrt{3}}{36}p^2,\,$ where $S\,$ is the area, $p\,$ the perimeter, we deduce

$\displaystyle\begin{align} [MBNAPC] &\ge [ABC]+\frac{1}{4\times 3}\cdot\frac{36}{\sqrt{3}}[ABC]\\ &=[ABC](1+\sqrt{3}). \end{align}$

Solution 4

$\displaystyle\begin{align} [MBNAPC]&=[ABC]+\frac{1}{4}(a^2+b^2+c^2)\\ &\ge [ABC]+\frac{1}{4}\cdot 4\sqrt{3}[ABC]\\ &= [ABC](1+\sqrt{3}) \end{align}$

by Weitzenböck's Inequality.

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.

 

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