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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A Cyclic Inequality in Triangle for Integer Powers $\left(\displaystyle \sum_{cycl}\frac{a^{n+1}}{b+c-a}\ge\sum_{cycl}a^n\right)$
R and r When G on Incircle $(8R\ge 25r)$
An Inequality for Cevians And The Ratio of Circumradius and Inradius $\left(\displaystyle \frac{XB}{XY}\cdot\frac{YC}{YZ}\cdot\frac{ZA}{ZX}\le\frac{R}{2r}\right)$
Centroid on the Incircle in Right Triangle $\left(648Rr\ge 25 (a^2+b^2+c^2)\right)$
An Inequality with Cotangents And the Circumradius $\left(\displaystyle \sum_{cycl}a^2\cot B\cot C\le 4R^2\right)$
An Inequality with Inradius and Excenters $\left(\displaystyle \sum_{cycl}\frac{1}{II_a^2}+\sum_{cycl}\frac{1}{I_aI_b^2}\le\frac{1}{4r^2}\right)$
An Inequality with Inradius and Side Lengths $\left(\displaystyle \sum_{cycl}(b+c-a)^2\cdot\sum_{cycl}(b+c-a)^3\ge 2592\sqrt{3}r^5\right)$
An Inequality with Exradii and an Altitude $\left(\displaystyle \sqrt{\frac{1}{r_b^2}+\frac{1}{r_c}+1}+\sqrt{\frac{1}{r_c^2}+\frac{1}{r_b}+1}\ge 2\sqrt{\frac{1}{h_a^2}+\frac{1}{h_a}+1}\right)$
Leuenberger's Inequality for Medians, Inradius and Circumradius $\left(\displaystyle m_a+m_b+m_c\le 4R+r\right)$
Adil Abdulayev's Inequality With Angles, Medians, Inradius and Circumradius $\left(\displaystyle \frac{A}{m_a}+\frac{B}{m_b}+\frac{C}{m_c}\le \frac{3\pi}{4R+r}\right)$
An Inequality with Sides, Cosines, and Semiperimeter $\left(\displaystyle \sum_{cycl}a^2(b\cos B+c\cos C)\le \frac{8s^3}{9}\right)$
Seyran Ibrahimov's Inequality $\left(\displaystyle \sqrt{3}s\cdot\sum_{cycl}m_a\le 20R^2+r^2\right)$
An Inequality in Triangle, with Sides and Medians III $\left(\displaystyle \sum_{cycl}\frac{(m_b+m_c-m_a)^3}{m_a}\ge\frac{3}{4}(a^2+b^2+c^2)\right)$
Leo Giugiuc's Inequality in Triangle, Solely with Cotangents $\left(\cot A +\cot B+\cot C\ge 2\sqrt{2}-1\right)$
An Inequality in Triangle with Side Lengths and Circumradius $\left(\displaystyle\displaystyle\frac{ab}{\sqrt{a^2+b^2}}+\frac{bc}{\sqrt{b^2+c^2}}+\frac{ca}{\sqrt{c^2+a^2}}\le\frac{3\sqrt{6}R}{2}\right)$
All Trigonometric Inequality in Triangle $\left(\displaystyle 3\sum_{cycl}\cos A\ge 2\sum_{cycl}\sin A\sin B\right)$
An Inequality with Two Sets of Cevians $\left(\displaystyle \frac{27[A'B'C']}{[A''B''C'']}\leq \Bigr(\frac{BA'}{BA''}+\frac{CB'}{CB''}+\frac{AC'}{AC''}\Bigr)^3\right)$
An Inequality with the Most Important Cevians $\left(\displaystyle \frac{m_am_bm_c}{h_ah_bh_c}\ge\frac{\ell_a^2+\ell_b^2+\ell_c^2}{\ell_a\ell_b+\ell_b\ell_c+\ell_c\ell_a}\right)$
An Inequality in Triangle with a Constraint $(a\sqrt{2}\ge b+c)$
Problem 11984 From the American Mathematical Monthly $\left(\displaystyle a^6+b^6+c^6\ge 5184\cdot r^6\right)$
A Long Cyclic Inequality of Degree 4 $\left(\displaystyle 4\cdot\sum_{cycl}ab\cdot\sum_{cycl}a-\left(\sum_{cycl}a\right)^3\ge\frac{\displaystyle 3\sum_{cycl}ab\left[4\sum_{cycl}ab-\left(\sum_{cycl}a\right)^2\right]}{\displaystyle \sum_{cycl}a}\right)$
An Inequality of Degree 3 with Inradius $\left(\displaystyle \sum_{cycl}(a+b-c)^3+24abc\ge 648\sqrt{3}r^3\right)$
A Cyclic Inequality from the 6th IMO, 1964 $\left(\displaystyle \sum_{cycl}a^2(b+c-a)\le 3abc\right)$
An Area Inequality in Right Triangle $\left(\displaystyle \frac{[\Delta ABD]+[\Delta ACE]}{[\Delta ADE]}\ge\sqrt{2}\right)$
An Angle Inequality in Triangle with Perpendicular Medians $\left(\displaystyle \cos A\ge \frac{4}{5}\right)$
An Inequality in Triangle form the 1996 APMO $\left(\sqrt{a+b-c}+\sqrt{b+c-a}+\sqrt{c+a-b}\le\sqrt{a}+\sqrt{b}+\sqrt{c}\right)$
An Inequality in a Nonobtuse Triangle $\left(R\sqrt{2}\le h_a\right)$
An Inequality in Triangle with Radicals, Semiperimeter and Inradius $\left(\displaystyle \sqrt{\frac{a+b}{s-b}}+ \sqrt{\frac{b+c}{s-c}} +\sqrt{\frac{c+a}{s-a}}\le \frac{\sqrt{a^2+b^2+c^2}}{r}\right)$
Dan Sitaru's Inequality with Radicals and Cosines $\left(\displaystyle (a^2+b^2+c^2)^3\ge 6^3(abc)^2\cos A\cos B\cos C\right)$
Lorian Saceanu's Cyclic Inequalities with Three Variables $\left(\displaystyle 2+\sum_{cycl}\frac{a}{b}\ge \sum_{cycl}\frac{a}{c}+2\frac{ab+bc+ca}{a^2+b^2+c^2}\right)$
An Inequality for the Tangent to the Incircle $\left(\displaystyle DE\le\frac{1}{8}(AB+BC+CA)\right)$
Lorian Saceanu's Sides And Angles Inequality $\left(\displaystyle \frac{\pi}{3}\le \frac{a\alpha + b\beta + c\gamma}{a+b+c}\le \arccos\left(\frac{r}{R}\right)\right)$
An Inequality in Triangle with Radicals, Semiperimeter, Incenter and Inradius $\left(\displaystyle \frac{AI+BI+CI}{r}+3\ge\left(\sum_{cycl}\sqrt{s-a}\right)\left(\sum_{cycl}\frac{1}{\sqrt{s-a}}\right)\right)$
An Inequality in Triangle for Side Lengths, Cycled in Two Ways $\left(\displaystyle 3\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-1\right)\ge 2\left(\frac{b}{a}+\frac{a}{c}+\frac{c}{b}\right)\right)$
Lorian Saceanu's Inequality for All Triangles $\left(\displaystyle \sin 2A+\sin 2B+\sin 2C\ge 4\sin 2A\cdot\sin 2B\cdot\sin 2C\right)$

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