How Do Angle Trisectors Divide the Area?

Problem

how angle trisectors divide the area? - problem

Solution 1

Let $\displaystyle m=\tan\frac{A}{3}.\,$ Then $0\lt m\lt\sqrt{3}.\,$ Choose $A=(0,0),\,$ $B=(b,-mb),\,$ $E=(c,mc),\,$ where $b,c,\gt 0,\,$ such that $\displaystyle D=\left(\frac{2bc}{b+c},0\right).\,$ We also get that $BC\,$ is defined by $mx(b+c)+y(b-c)=2mbc\,$ and $AC\,$ by $2mx=(1-m^2)y\,$ which meet at $C=\displaystyle \left(\frac{2(1-m^2)bc}{(3-m^2)b-(1+m^2)c},\frac{4mbc}{(3-m^2)b-(1+m^2)c)}\right).\,$ From here,

$\displaystyle\begin{align}2[\Delta ABC]&=\left|\begin{array}{ccc} 0 & 0 & 1\\b & -mb & 1\\ \displaystyle\frac{2(1-m^2)bc}{(3-m^2)b-(1+m^2)c}&\displaystyle\frac{4mbc}{(3-m^2)b-(1+m^2)c)}&1\end{array}\right|\\ &=\frac{2mb^2c(3-m^2)}{(3-m^2)b-(1+m^2)c} \end{align}$

and

$\displaystyle\begin{align}2[\Delta ADE]&=\left|\begin{array}{ccc} 0 & 0 & 1\\\displaystyle \frac{2bc}{b+c} & 0 & 1\\ c & mc & 1\end{array}\right|\\ &=\frac{2mbc^2}{b+c}. \end{align}$

Hence, we need to prove

$\displaystyle \frac{mb^2c(3-m^2)}{(3-m^2)b-(1+m^2)c}\ge\frac{3mbc^2}{b+c},$

or, equivalently,

$\displaystyle \frac{b(3-m^2)}{(3-m^2)b-(1+m^2)c}\ge\frac{3c}{b+c}.$

If we denote $\displaystyle \frac{b}{c}=\alpha\,$ and $\displaystyle \frac{1+m^2}{3-m^2}=u,\,$ then $\displaystyle \alpha\gt u\gt \frac{1}{3}\,$ and the latest inequality becomes $\displaystyle \frac{\alpha}{\alpha-u}\ge\frac{3}{\alpha+1}\,$ which is equivalent to $\alpha^2-2\alpha+3\ge 0,\,$ which is clearly true since the discriminant $4(1-3u)\lt 0.$

Solution 2

Using the Law of Sines in $\Delta ABD,\,$ $\displaystyle AD=\frac{AB\cdot\sin B}{\displaystyle \sin \left(B+\frac{A}{3}\right)}.\,$ From $\Delta ACE,\,$ $\displaystyle AE=\frac{AC\cdot\sin C}{\displaystyle \sin \left(C+\frac{A}{3}\right)}.\,$ Thus

$\displaystyle\begin{align} [\Delta ADE] &= \frac{1}{2}AD\cdot AE\\ &=\frac{1}{2}\cdot\frac{AB\cdot\sin B}{\displaystyle \sin \left(B+\frac{A}{3}\right)}\cdot\frac{AC\cdot\sin C}{\displaystyle \sin \left(C+\frac{A}{3}\right)}\cdot\sin\frac{A}{3} \end{align}$

and $\displaystyle [\Delta ABC]=\frac{1}{2}AB\cdot AC\cdot\sin A.\,$ The required inequality is then equivalent to

$\displaystyle \begin{align} &\sin A\cdot\sin\left(B+\frac{A}{3}\right)\cdot\sin\left(C+\frac{A}{3}\right)\ge 3\sin\frac{A}{3}\cdot\sin B\cdot\sin C\,\Leftrightarrow\\ &\frac{1}{2}\sin A\cdot\left(\cos (B-C)-\cos\left(B+C+\frac{2A}{3}\right)\right)\\ &\qquad\qquad\qquad\qquad\ge\frac{3}{2}\sin\frac{A}{3}[\cos (B-C)-\cos (B+C)]\,\Leftrightarrow\\ &\sin A\cdot\left(\cos (B-C)+\cos\frac{A}{3}\right)\\ &\qquad\qquad\qquad\qquad\ge 3\sin\frac{A}{3}[\cos (B-C)+\cos A]\,\Leftrightarrow\\ &\left(3\sin\frac{A}{3}-4\sin^3\frac{A}{3}\right)\left(\cos (B-C)+\cos\frac{A}{3}\right)\\ &\qquad\qquad\qquad\qquad\ge 3\sin\frac{A}{3}(\cos (B-C)+\cos A)\,\Leftrightarrow\\ &\left(3-4\sin^2\frac{A}{3}\right)\left(\cos (B-C)+\cos\frac{A}{3}\right)\\ &\qquad\qquad\qquad\qquad\ge 3\cos (B-C)+3\cos A\,\Leftrightarrow\\ &\left(3-4\left(1-\cos^2\frac{A}{3}\right)\right)\left(\cos (B-C)+\cos\frac{A}{3}\right)\\ &\qquad\qquad\qquad\qquad\ge 3\cos (B-C)+3\left(4\cos^3\frac{A}{3}-3\cos\frac{A}{3}\right)\,\Leftrightarrow\\ &\left(4\cos^2\frac{A}{3}-1\right)\left(\cos (B-C)+\cos\frac{A}{3}\right)\\ &\qquad\qquad\qquad\qquad\ge 3\cos (B-C)+12\cos^3\frac{A}{3}-9\cos\frac{A}{3}\,\Leftrightarrow\\ &4\cos^2\frac{A}{3}\cos (B-C)-\cos (B-C)+4\cos^3\frac{A}{3}-\cos\frac{A}{3}\\ &\qquad\qquad\qquad\qquad\ge 3\cos (B-C)+12\cos^3\frac{A}{3}-9\cos\frac{A}{3}\,\Leftrightarrow\\ &8\cos^3\frac{A}{3}-4\cos^2\frac{A}{3}\cdot\cos (B-C)-8\cos\frac{A}{3}+4\cos (B-C)\le 0\,\Leftrightarrow\\ &4\left(\cos^2\frac{A}{3}-1\right)\left(2\cdot\cos\frac{A}{3}-\cos (B-C)\right)\le 0\,\Leftrightarrow\\ &4\sin^2\frac{A}{3}\left(2\cdot\cos\frac{A}{3}-\cos (B-C)\right)\ge 0\,\Leftrightarrow\\ &2\cdot\cos\frac{A}{3}-\cos (B-C)\ge 0\;\;(1). &\end{align}$

Now, since $0\lt A\lt \pi,$ $\displaystyle 0\lt\frac{A}{3}\lt\frac{\pi}{3},\,$ and, since the function $cos\,$ is strictly decreasing on $[0,\pi],\,$ we have $\displaystyle \cos\frac{A}{3}\gt\cos\frac{\pi}{3}=\frac{1}{2},\,$ .i.e.,

(2)

$\displaystyle 2\cos\frac{A}{3}-1\gt 0.$

Obviously, also $1-\cos (B-C)\ge 0\,\,(3).\,$ Combining (2) and (3) gives

$\displaystyle 2\cos\frac{A}{3}-\cos (B-C)\gt 0,$

thus proving (1). Note that the inequality is strict: $\displaystyle [\Delta ADE]\lt\frac{1}{3}[\Delta ABC].\,$ However, it can't be improved because in the limit, when the distance from $A\,$ to $BC\,$ increases, the ratio of the two areas tends to $\displaystyle \frac{1}{3}.$

Solution 3

Let $\displaystyle x=2[\Delta ABD]=AB\cdot AD\cdot\sin\frac{A}{3},\,$ $\displaystyle y=2[\Delta ADE]=AD\cdot AE\cdot\sin\frac{A}{3},\,$ $\displaystyle z=2[\Delta AEC]=AE\cdot AC\cdot\sin\frac{A}{3}.\,$ We have

$x+y+z=2[\Delta ABC]=AB\cdot AC\cdot\sin A$

and

$\displaystyle xz=\left(AB\cdot AC\cdot\sin\frac{A}{3}\right)\left(AD\cdot AE\cdot\sin\frac{A}{3}\right)=y(x+y+z)\cdot\frac{\displaystyle \sin\frac{A}{3}}{\sin A}.$

Note that $\displaystyle \sin A\le 3\sin\frac{A}{3}.\,$ Indeed, consider function $\displaystyle f(A)=\sin A-3\sin\frac{A}{3}.\,$ $\displaystyle f'(A)=\cos A-\cos\frac{A}{3}\lt 0,\,$ for $cos\,$ is strictly decreasing for $0\lt A\lt \pi.\,$ It follows that $f(A)\le f(0)=0.\,$ To continue,

$\displaystyle xz=\frac{\displaystyle \sin\frac{A}{3}}{\sin A}\ge\frac{y(x+y+z)}{3},$

or, equivalently, $\displaystyle xyz\ge y^2\left(\frac{x+y+z}{3}\right),\,$ implying

$\displaystyle \left(\frac{x+y+z}{3}\right)^3\ge xyz\ge y^2\left(\frac{x+y+z}{3}\right),$

from which $\displaystyle \frac{x+y+z}{3}\ge y,\,$ which is the required inequality.

Solution 4

Let $\theta=\displaystyle \frac{A}{3},\,$ $x=AD\,$ $y=AE.\,$ From $[\Delta ABC]=[\Delta ABD]+[\Delta ADC],$

$\displaystyle x=\frac{bc\sin 3\theta}{b\sin 2\theta+c\sin\theta}.$

From $[\Delta ABC]=[\Delta ABE]+[\Delta AEC],$

$\displaystyle y=\frac{bc\sin 3\theta}{b\sin \theta+c\sin 2\theta}.$

From a property of angle bisectors, $BD:DE=c:y\,$ and $DE:EC=x:b,\,$ combining which $BD:DE:EC=cx:xy:by,\,$ implying

$\displaystyle\begin{align}\frac{[\Delta ADE]}{[\Delta ABC]}&=\frac{xy}{cx+xy+by}\\ &=\frac{1}{\displaystyle \frac{b}{x}+\frac{c}{y}+1}. \end{align}$

Now,

$\displaystyle\begin{align} \frac{b}{x}+\frac{c}{y}+1 &= \left(\frac{c}{b}+\frac{b}{c}\right)\cdot\frac{\sin 2\theta}{\sin 3\theta}+2\frac{\sin\theta}{\sin 3\theta}+1\\ &\gt 2\cdot\frac{2}{3}+2\cdot\frac{1}{3}+1=3. \end{align}$

The inequality is strict; tends to equality when $\theta\rightarrow 0^{+}.$

Solution 5

how angle trisectors divide the area?, proof 5

For the areas, $\displaystyle \frac{[\Delta AED]}{[\Delta ACB]}\le\frac{1}{3}\,$ is the same as $\displaystyle \frac{[\Delta ACB]}{[\Delta AED]}\ge 3$ which is equivalent to

$\displaystyle \begin{align} &\frac{[\Delta ADB]}{[\Delta AED]}+1+\frac{[\Delta ACE]}{[\Delta AED]}\ge 3\,\Longleftrightarrow\\ &\frac{[\Delta ADB]}{[\Delta AED]}+\frac{[\Delta ACE]}{[\Delta AED]}\ge 2\,\Longleftrightarrow\\ &\frac{AB}{AE}+\frac{AC}{AD}\ge 2\,\Longleftrightarrow\,\frac{CG}{EC}+\frac{KG}{CG}\ge 2. \end{align}$

$\displaystyle \begin{align} \frac{CG}{EC}+\frac{\lambda\cdot EC}{CG}&=\frac{GC}{EC}+\frac{[1+(\lambda-1)]EC}{CG}\\ &=\frac{GC}{EC}+\frac{EC}{CG}+\frac{(\lambda -1)EC}{CG}\ge\frac{GC}{EC}+\frac{EC}{CG}\ge 2. \end{align}$

Acknowledgment

Solution 1 is by Leo Giugiuc; Solution 2 is by Marian Cucoaneş; Solution 3 is by Marian Dincă; Solution 4 is by Kunihiko Chikaya; Solution 5 is by Hlias Aggelakos.

 

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