An Area Inequality in Right Triangle


An Area Inequality in Right Triangle, source


An Area Inequality in Right Triangle, problem

Solution 1

Since $\angle DAE=45^{\circ}$ and angles $ADE$ and $AED$ are acute, then, clearly, both are greater than $45^{\circ}.$ Let's choose $A=(0,1),$ $D=(-b,0),$ $E=(c,0),$ where $b=\cot\angle ADE$ and $c=\cot\angle AED,$ so that $0\lt b,c\lt 1.$

We also get $\displaystyle B=\left(-\frac{2b}{1-b^2},0\right)$ and $\displaystyle C=\left(\frac{2c}{1-c^2},0\right).$ We need to prove that $[\Delta ABC]\ge (\sqrt{2}+1)[\Delta ADE]$ which is equivalent to showing that

$\displaystyle \frac{2b}{1-b^2}+\frac{2c}{1-c^2}\ge(\sqrt{2}+1)(b+c),$


$\displaystyle \frac{2(1-bc)}{(1-b^2)(1-c^2)}\ge \sqrt{2}+1.$

But $\displaystyle 1=\cot\angle ADE=\frac{1-bc}{b+c}.$ Thus, for $b+c=2s$ and $bc=p^2,$ we have $s\ge p$ and $\displaystyle \frac{1-p^2}{2}=s,$ so that $\displaystyle \frac{1-p^2}{2}\ge p,$ or, $p\le\sqrt{2}-1.$

This leads to a sequence of equivalent inequalities:

$\displaystyle \begin{align} &\frac{2(1-bc)}{(1-b^2)(1-c^2)}\ge\sqrt{2}+1\;&\Leftrightarrow\\\\ &\frac{2(1-p^2)}{(1+p^2)^2-4s^2}\ge\sqrt{2}+1\;&\Leftrightarrow\\\\ &\frac{2(1-p^2)}{(1+p^2)^2-4(1-p^2)^2}\ge\sqrt{2}+1\;&\Leftrightarrow\\\\ &\frac{1-p^2}{2p^2}\ge\sqrt{2}+1\;&\Leftrightarrow\\\\ &p^2(2\sqrt{2}+3)\le 1\;&\Leftrightarrow\\\\ &p^2\le (\sqrt{2}-1)^2\;&\Leftrightarrow\\\\ &\sqrt{2}-1\le p \le \sqrt{2}-1. \end{align}$

But, by the definition, $p\gt 0,$ so that last inequality is equivalent to the plain $p \le \sqrt{2}-1$ which is true as was found earlier.

Solution 2

Simplest proof. Suppose the altitude equals to $1$ and set $\theta=t.$ It suffices to estimate $\displaystyle \frac{a_1+a_2}{a_3}$ for the lengths $a_1, a_2, a_3$ on $BC.$ One finds with the Law of Sines

$\displaystyle \begin{align} \frac{a_1+a_2}{a_3} &= \frac{\displaystyle \frac{\tan t}{\cos 2t}+\frac{\tan\left(\frac{\pi}{4}-t\right)}{\sin 2t}}{\tan t + \tan\left(\frac{\pi}{4}-t\right)}\\ &= -1 +\frac{1}{\sqrt2\sin(2t+\frac{\pi}{4})-1} \end{align}$

after standard simplifications. The minimum is clearly $\displaystyle -1 + \frac{1}{\sqrt2-1} = \sqrt2$ for $t = \displaystyle \frac{\pi}{8}.$

Solution 3

Since $\displaystyle \angle ADC=\frac{\pi}{2}-\theta=\ DAC,$ we get $CD=AC=b.$

An Area Inequality in Right Triangle, proof 3

Similarly, $AB-BE=c.$ From which $DE=b+c-a,$ $BD=a-b$ and $CE=a-c.$ We have

$\displaystyle\begin{align} &\frac{A_1+A_2}{A_3}\ge \sqrt{2}&\Leftrightarrow\\ &\frac{AH(2a-b-c)}{AH(b+c-a)}\ge \sqrt{2}&\Leftrightarrow\\ &\sqrt{2}(\sqrt{2}+1)a\ge(\sqrt{2}+1)(b+c)&\Leftrightarrow\\ &a\sqrt{2}\ge b+c&\Leftrightarrow\\ &\sqrt{2(b^2+c^2)}\ge b+c, \end{align}$

which is jut the AM-QM inequality.


The problem that was originally posted at the Peru Geometrico facebook group by Miguel Ochoa Sanchez was kindly brought to my attention by Leo Guiguic. Solution 1 is by Leo; Solution 2 is by Grégoire Nicollier. Additional solutions can be found at the above link. Solution 3 is by Eleodor Popescu.


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