Solving $\tan x= \tan 2x\tan3x\tan4x$ and a related geometric problem

Question

My friend asked me a geometric problem.

In $\triangle ABC$, $\angle B=\angle C=70^\circ$. $D$ is an interior point of the triangle such that $\angle BCD=40^\circ$ and $\angle CBD=20^\circ$. Find $\angle BAD$.

If $\angle BAD=\theta$, it is not difficult to see that $\dfrac{\tan(40^\circ-\theta)}{\tan40^\circ}=\dfrac{\tan30^\circ}{\tan70^\circ}$ and I managed to show that $\tan(40^\circ-\theta)=\tan 10^\circ$ using trigonometric identities.

I actually proved that $\tan10^\circ\tan70^\circ=\tan30^\circ\tan40^\circ$, or equivalently, $\tan10^\circ=\tan20^\circ\tan30^\circ\tan40^\circ$. This result is so beautiful and make me interested in the equation $\tan x= \tan2x\tan3x\tan4x$, but I have difficulty in solving it. By plotting the graph, I can see that the solution is $180n^\circ$ or $60n^\circ\pm10^\circ$.

My questions are

1. How to solve the original geometric problem without using trigonometry?

2. How to solve the equation $\tan x= \tan2x\tan3x\tan4x$?

Remark Just find a solution to the second question a few minutes after posting it. But I still want to see if there are other ways to solve it.

Answer 1

Geometric solution

Construct the point $E$ as a reflection of the point $D$ w.r.t. the vertical line through $A$, $\angle DAE=\theta-(40^\circ-\theta)=2\theta-40^\circ$.

Point $H=BD\cap CE$.

$\triangle BDE$, $\triangle CDE$ and $\triangle EHD$ are isosceles, $\angle EBD=\angle BDE=\angle DEC=\angle ECD=20^\circ$, $|BE|=|CD|=|DE|$.

Point $D_1:DD_1\perp AC,\ |DD_1|=|DE|$, point $F=AC\cap DD_1$. From $\triangle CDF$, $|DF|=|D_1F|=\tfrac12\,|CD|=\tfrac12\,|DE|$.

Similarly, point $E_1:DD_1\perp AB,\ |EE_1|=|DE|$, point $G=AB\cap EE_1$. From $\triangle BEG$, $|EG|=|E_1G|=\tfrac12\,|BE|=\tfrac12\,|DE|$.

Noe we have $\angle D_1AD=\angle EAE_1=\angle DAE$, $\angle FAD=\angle EAG=\tfrac12\,\angle DAE$,

so

\begin{align} 2\angle DAE&=\angle CAB ,\\ 2(2\theta-40^\circ)&=40^\circ ,\\ \theta&=30^\circ . \end{align}

Answer 2

2) If $\tan3x=0$, so we can check it easily.

Let $\tan3x\neq0.$

Thus, we need to solve: $$\tan{x}\cot3x+1=\tan2x\tan4x+1$$ or $$\frac{\sin4x}{\cos{x}\sin3x}=\frac{\cos2x}{\cos2x\cos4x}$$ or $$\sin4x\cos4x=\sin3x\cos{x}$$ or $$\sin8x=\sin4x+\sin2x$$ or $$\sin8x-\sin4x=\sin2x$$ or $$\sin2x(2\cos6x-1)=0.$$ Can you end it now?

The first problem.

Take $\Delta BFG$ such that $BG=GF$ and $\measuredangle G=20^{\circ}.$

Let $K\in GF$, $M\in GK$ and $E\in GB$ such that $BF=BK=EK=EM.$

Thus, $\measuredangle KBF=20^{\circ}$ and $$\measuredangle EBK=80^{\circ}-20^{\circ}=60^{\circ},$$ which gives $$BE=BK=EK=EM.$$ Also, $$\measuredangle EMK=\measuredangle EKM=180^{\circ}-\measuredangle EKB-\measuredangle BKF=180^{\circ}-60^{\circ}-80^{\circ}=40^{\circ}$$ and since $\measuredangle G=20^{\circ},$ we obtain: $$\measuredangle GEM=40^{\circ}-20^{\circ}=20^{\circ},$$ which gives $$GM=ME=EK=BK=EB=BF.$$ Thus, $$\measuredangle EBM=\measuredangle EMB=\frac{1}{2}\measuredangle GEM=10^{\circ},$$ which gives $$\measuredangle MBF=80^{\circ}-10^{\circ}=70^{\circ},$$ $$\measuredangle MBK=50^{\circ},$$ $$\measuredangle KBF=20^{\circ}.$$ Also, we have: $$\measuredangle BKM=60^{\circ}+40^{\circ}=100^{\circ}$$ and $$\measuredangle BMK=40^{\circ}-10^{\circ}=30^{\circ}.$$ Now, let $EK\cap BF=\{C\}$ and $N$ be placed on the line $BC$ such that $B$ is a mid-point of $NF$.

But $BN=BE$ and $$\measuredangle NBE=180^{\circ}-80^{\circ}=100^{\circ}=\measuredangle MEK,$$ which gives $$\Delta NBE\cong\Delta MEK,$$ which says $$NE=MK.$$ In another hand, $$\measuredangle ECN=180^{\circ}-\measuredangle N-\measuredangle NEC=180^{\circ}-40^{\circ}-100^{\circ}=40^{\circ},$$ which gives $$EC=NK=MK.$$ Thus, $$\Delta MEC\cong\Delta BKM,$$ which gives $$MC=BM$$ and $$\measuredangle MCB=\measuredangle MBC=70^{\circ}.$$ Id est, $$\Delta MBC\cong\Delta ABC,$$ which gives $M\equiv A$ and since $\measuredangle KCB=40^{\circ},$ we obtain: $K\equiv D,$ which says $$\measuredangle BAD=\measuredangle BMK=30^{\circ}.$$

Answer 3

Rewrite $\tan x= \tan2x\tan3x\tan4x$ as

$$\sin x \cos 2x\cos3x\cos 4x = \cos x \sin 2x\sin 3x\sin 4x$$

and factorize,

$$\sin x\cos 2x (\cos3x\cos 4x -4\cos^2 x\sin 3x\sin 2x)=0$$

Further factorize with $\cos 3x = \cos x(2\cos 2x -1)$,

$$\sin x\cos 2x \cos x [(2\cos 2x -1)\cos 4x -4\cos x\sin 3x\sin 2x)]=0\tag 1$$

Recognize $\cos x \ne 0$, $\cos 2x \ne 0$ and

$$(2\cos 2x -1)\cos 4x =\cos2x+\cos6x-\cos4x$$ $$4\cos x\sin 3x\sin 2x=2(\sin4x+\sin2x)\sin2x= \cos2x-\cos6x+1-\cos4x$$

to reduce the equation (1) to,

$$\sin x(2\cos 6x -1)=0 $$

which leads to $\sin x =0$ and $\cos6x=\frac12$. Thus, the solutions are

$$x=n\pi,\>\>\>\>\> x = \frac{n\pi}3\pm\frac\pi{18}$$

Answer 4

There are two nice solutions to my second question. Here I would like to share mine. I am not intended to answer my own question and I have yet to solve the first problem.

When $x,2x,3x,4x\notin\{(n+\frac12) \pi:n\in\mathbb{Z}\}$, we have

\begin{align*} \sin x\cos 4x \cos 2x\cos 3x&=\cos x\sin 4x \sin2x\sin3x\\ \frac12(\sin5x-\sin3x)\cdot\frac12(\cos x+\cos 5x)&=\frac12(\sin5x+\sin3x)\cdot\frac12(\cos x-\cos5x)\\ \sin5x\cos5x-\sin3x\cos x&=0\\ \sin10x-\sin4x-\sin2x&=0\\ 2\cos6x\sin4x-\sin4x&=0\\ \sin4x(2\cos6x-1)&=0 \end{align*}

So, $\displaystyle x=\frac{n\pi}{4}$ or $\displaystyle \frac{n\pi}3\pm\frac{\pi}{18}$.

As $x,2x,3x,4x\notin\{(n+\frac12) \pi:n\in\mathbb{Z}\}$, we have $\displaystyle x=n\pi$ or $\displaystyle \frac{n\pi}3\pm\frac{\pi}{18}$.