Geometric proof for $\tan{(3x)}=\tan{(x)}\tan{\left(\frac{\pi}{3}-x\right)}\tan{\left(\frac{\pi}{3}+x\right)}$

Question

Are there geometric proofs for the identitity $$\tan{(3x)}=\tan{(x)}\tan{\left(\frac{\pi}{3}-x\right)}\tan{\left(\frac{\pi}{3}+x\right)}$$

My try:

Thank in advances.

Answer 1

I have another simple proof.

Assume $\angle AFD = \frac{\pi}{3}$, $\angle FAD = 4x$, and $\angle FDA = 4y$. Obviously, $y = \frac{\pi}{6}-x$, $\textit{i.e.}$, $x+y = \frac{\pi}{6}$.

$AQ$ bisects $\angle FAD$, and $AC$ bisects $\angle FAQ$. Also, $DP$ bisects $\angle FDA$, and $DB$ bisects $\angle FDP$. And $R = AQ \cap DP$, $H = AQ \cap DB$, $G= AC \cap DP$, $E = AC \cap DB$.

Then, $\angle AED = \angle AFD + \angle FAE + \angle FDE = \frac{\pi}{3} + x + y = \frac{\pi}{2}$. Similarly, $\angle ARD = \frac{2\pi}{3}$.

$\text{area of } \triangle ADP =\frac{1}{2} AR \cdot DP \sin \frac{2\pi}{3} = \frac{1}{2} AP \cdot DF \sin \frac{\pi}{3}$. Thus, $AR\cdot DP = AP \cdot DF$. It means that $AR:AP = DF:DP$.

We know that $AR:AP=RG:GP$, and $DF:DP=FB:BP$. Thus $RG:GP = FB:BP$. Therefore $RF \parallel GB$.

Because $R$ is the incenter of $\triangle AFD$, $\angle RFP = \frac{\pi}{6}$. Hence, $\angle GBP = \frac{\pi}{6}$.

Thus, $\angle BGE = \angle BAG + \angle ABG = \frac{\pi}{6} + x$, and $\angle EBC = \angle EBG = \frac{\pi}{2} - \angle BGE = \frac{\pi}{3} - x$. Additionally, $\angle ECD = \frac{\pi}{2} - y = \frac{\pi}{2} - \left(\frac{\pi}{6} -x \right) = \frac{\pi}{3} + x$. Therefore, $\square ABCD$ satisfies the condition of problem, and we showed that $\angle EAD = 3x$.

Answer 2

I do not know if there is a simpler geometrical incarnation, however this triplication formula (or, better, the analogous triplication formula for the sinus) is at the heart of the trigonometric proof of the celebrated Morley's trisector theorem.

Answer 3

This is not a geometric proof (the one I mentioned in the comments still eludes me) but I hope you find it interesting nonetheless. By the Fourier series for $\log\sin$ and $\log\cos$

$$ \log\tan x = -2\sum_{k\geq 0}\frac{\cos((4k+2)x)}{2k+1} = -2\,\text{Re}\sum_{k\geq 0}\frac{e^{(2k+1)2ix}}{2k+1} \tag{1}$$ holds for any $x\in\left(0,\frac{\pi}{2}\right)$, so the given identity is a simple consequence of the discrete Fourier transform, $\mathbb{1}_{\equiv 0\!\!\pmod{3}}(n)=\frac{1+\omega^n+\omega^{2n}}{3}$.

Answer 4

$A,B,C,D,E$ are the same points in the problem.

$F= AB \cap CD$.

$G$ and $C$ are symmetric with respect to $E$. $H$ and $B$ are symmetric with respect to $E$.

$P = GD \cap AF$, $Q = AH \cap DF$. $R = GD \cap AH$.

(In my figure, I forgot to write $R$. Sorry.)

We want to show that $R$ is the incenter of $\triangle ADF$.

because $\angle AFD = \angle AED - \angle FAC - \angle FDB$,

$\angle AFD = \frac{\pi}{3}$.

$\angle EBC = \angle EBG = \angle EHC = \frac{\pi}{3} - x$.

and $\angle ECB = \angle EGB = \angle ECH = \frac{\pi}{6} + x$.

Because $\angle ABG = \angle EGB - \angle GAB$,

$\angle ABG = \frac{\pi}{6}$.

Also, because $\angle HDC = \frac{\pi}{6} -x$, and $\angle DCH = \angle EHC - \angle HDC$,

$\angle DCH = \frac{\pi}{6}$.

$\triangle DCH \equiv \triangle DGH$, and $\triangle ABG \equiv \triangle AHG$.

Thus, $\angle AHG = \angle DGH = \frac{\pi}{6}$. Thus, $RG=RH$.

Additionally, $\angle PRQ = \pi - \angle RGH - \angle RHG = \frac{2\pi}{3}$, and $\angle PRQ + \angle PFQ = \pi$. Hence, four points $R, P, F, Q$ are on the same circle.

Because $\square BCHG$ is a rhombus, let $BG=CH=l$.

$\angle BPG = \pi - \angle AFD - \angle FDP = \frac{\pi}{3}+2x$. In $\triangle BPG$,

$$\frac{GP}{\sin\frac{\pi}{6}} = \frac{l}{\sin(\frac{\pi}{3}+2x)}$$

Similarly, $\angle CQH = \pi - \angle AFD - \angle FAQ = \frac{2\pi}{3}-2x$. In $\triangle CQH$,

$$\frac{HQ}{\sin\frac{\pi}{6}} = \frac{l}{\sin(\frac{2\pi}{3}-2x)}$$

$\sin(\frac{\pi}{3}+2x) = \sin(\frac{2\pi}{3}-2x)$, thus $GP=HQ$.

Hence, $RP = RG + GP = RH + HQ = RQ$.

Because $RP=RQ$ and $R,P,F,Q$ are on the same circle, $\angle RFP = \angle RFQ$.

Thus, we know $\angle RAF = 2x$, $\angle RFA = \angle RFD = \frac{\pi}{6}$, $\angle RDF = \frac{\pi}{3}-2x$, and $\angle RAD + \angle RDA = \frac{\pi}{3}$.

Finally, we use the trigonometric form of Ceva's theorem.

$$\frac{\sin{\angle RAD}}{\sin{\angle RAF}} \times \frac{\sin{\angle RDF}}{\sin{\angle RDA}} \times \frac{\sin{\angle RFA}}{\sin{\angle RFD}} = 1$$

It is equal to

$$ \frac{\sin{\angle RAD}}{\sin 2x} \times \frac{\sin{\frac{\pi}{3}-2x}} {\sin{\frac{\pi}{3}- \angle RAD}} = 1 $$

$\angle RAD$ should be $2x$. Because if $\angle RAD < 2x$, then the left-hand side of the above equation is less than $1$, and if $\angle RAD > 2x$, then the left-hand side of the above equation is greater than $1$.

Thus $R$ is the incenter of $\triangle FAD$, and $\angle EAD = 3x$.