MHB Challenge question on equilateral triangle: Prove ∠DBA=42°

[anemone]

In an equilateral triangle $ABC$, let $D$ be a point inside the triangle such that $\angle BAD=54^\circ$ and $\angle BCD=48^\circ$. Prove that $\angle DBA=42^\circ$.

 

[Olinguito]

I managed to get the answer using some complicated calculation – but given that the solution is a nice-looking answer, it’s likely much of the complicated calculation is unnecessary. I’m sure there is a less complicated solution than mine. (Wondering)

Spoiler

Since the problem involves only angles, the actual size of the equilateral triangle is immaterial; for convenience, let us take it to have side length $2$.

Let A, B, C, D have co-ordinates $(0,0)$, $(2,0)$, $(1,\sqrt3)$, $(a,b)$ respectively. Then we immediately have
$$b\ =\ a\tan54^\circ.$$
Let M be the midpoint of AB and N be the foot of the perpendicular from D to CM. Then we have $|\mathrm{DN}|=1-a$, $|\mathrm{CN}|=\sqrt3-b$, and $\angle\mathrm{DCN}=\angle\mathrm{DCB}-\angle\mathrm{NCB}=18^\circ$ and so
$$\tan18^\circ\ =\ \frac{1-a}{\sqrt3-b}\ =\ \frac{1-a}{\sqrt3-a\tan54^\circ}$$
$\displaystyle\implies\ a\ =\ \frac{1-\sqrt3\tan18^\circ}{1-\tan18^\circ\tan54^\circ}.$

Finally, the positive value of the slope of the line segment DB, which is $\tan\angle\mathrm{DBA}$, is
$$\frac b{2-a}\ =\ \frac{a\tan54^\circ}{2-a}$$
and substituting for $a$ should give this value as the tangent of 42° (calculator possibly needed). (Thinking)

 

[Olinguito]

I think I’ve got it!

Spoiler

I showed that in my post above that
$$a\ =\ \frac{1-\sqrt3\tan18^\circ}{1-\tan18^\circ\tan54^\circ}.$$
But
$$\tan54^\circ\ =\ \frac1{\tan36^\circ}\ =\ \frac{1-\tan^218^\circ}{2\tan18^\circ}$$
$\implies\ \tan18^\circ\tan54^\circ\ =\ \dfrac{1-\tan^218^\circ}2$

$\implies\ 1-\tan18^\circ\tan54^\circ\ =\ \dfrac{1+\tan^218^\circ}2$

$\implies\ a\ =\ \dfrac{2(1-\sqrt3\tan18^\circ)}{1+\tan^218^\circ}$

$\implies\ 2-a\ =\ 2-\dfrac{2(1-\sqrt3\tan18^\circ)}{1+\tan^218^\circ}\ =\ \dfrac{2\tan18^\circ(\sqrt3+\tan18^\circ)}{1+\tan^218^\circ}.$

$\implies\ \dfrac a{2-a}\ =\ \dfrac{(1-\sqrt3\tan18^\circ)}{\sqrt3+\tan18^\circ}\cdot\dfrac1{\tan18^\circ}$.

Also
$$\tan{78^\circ}\ =\ \tan(60+18)^\circ\ =\ \frac{\sqrt3+\tan18^\circ}{1-\sqrt3\tan18^\circ}$$
and the quantity $\dfrac b{2-a}$ in my post above is supposed to be $\tan42^\circ$. So my entire solution reduces to proving the following:
$$\boxed{\tan42^\circ\ =\ \frac{\tan54^\circ}{\tan78^\circ\cdot\tan18^\circ}}.$$

It’s quite simple, really: just show that
$$\tan x^\circ\tan(60+x)^\circ\tan(60-x)^\circ\ =\ \tan(3x)^\circ.$$
Proof:
$$\begin{array}{rcl}\tan x^\circ\tan(60+x)^\circ\tan(60-x)^\circ &=& \tan x^\circ\cdot\dfrac{\sqrt3+\tan x^\circ}{1-\sqrt3\tan x^\circ}\cdot\dfrac{\sqrt3-\tan x^\circ}{1+\sqrt3\tan x^\circ} \\\\ {} &=& \dfrac{3\tan x^\circ-\tan^3x^\circ}{1-3\tan^2x^\circ} \\\\ {} &=& \tan(3x)^\circ.\end{array}$$
Now put $x=12$:
$$\tan12^\circ\tan72^\circ\tan48^\circ\ =\ \tan36^\circ.$$
Now use the fact that $\tan(90-\theta)^\circ=\dfrac1{\tan\theta^\circ}$:
$$\frac1{\tan78^\circ\tan18^\circ\tan42^\circ} =\ \frac1{\tan54^\circ}$$
$\implies\ \tan42^\circ\ =\ \dfrac{\tan54^\circ}{\tan78^\circ\cdot\tan18^\circ}$

– QED! (Clapping)

 

[WMDhamnekar]

I think I’ve got it!

Spoiler

I showed that in my post above that
$$a\ =\ \frac{1-\sqrt3\tan18^\circ}{1-\tan18^\circ\tan54^\circ}.$$
But
$$\tan54^\circ\ =\ \frac1{\tan36^\circ}\ =\ \frac{1-\tan^218^\circ}{2\tan18^\circ}$$
$\implies\ \tan18^\circ\tan54^\circ\ =\ \dfrac{1-\tan^218^\circ}2$

$\implies\ 1-\tan18^\circ\tan54^\circ\ =\ \dfrac{1+\tan^218^\circ}2$

$\implies\ a\ =\ \dfrac{2(1-\sqrt3\tan18^\circ)}{1+\tan^218^\circ}$

$\implies\ 2-a\ =\ 2-\dfrac{2(1-\sqrt3\tan18^\circ)}{1+\tan^218^\circ}\ =\ \dfrac{2\tan18^\circ(\sqrt3+\tan18^\circ)}{1+\tan^218^\circ}.$

$\implies\ \dfrac a{2-a}\ =\ \dfrac{(1-\sqrt3\tan18^\circ)}{\sqrt3+\tan18^\circ}\cdot\dfrac1{\tan18^\circ}$.

Also
$$\tan{78^\circ}\ =\ \tan(60+18)^\circ\ =\ \frac{\sqrt3+\tan18^\circ}{1-\sqrt3\tan18^\circ}$$
and the quantity $\dfrac b{2-a}$ in my post above is supposed to be $\tan42^\circ$. So my entire solution reduces to proving the following:
$$\boxed{\tan42^\circ\ =\ \frac{\tan54^\circ}{\tan78^\circ\cdot\tan18^\circ}}.$$

It’s quite simple, really: just show that
$$\tan x^\circ\tan(60+x)^\circ\tan(60-x)^\circ\ =\ \tan(3x)^\circ.$$
Proof:
$$\begin{array}{rcl}\tan x^\circ\tan(60+x)^\circ\tan(60-x)^\circ &=& \tan x^\circ\cdot\dfrac{\sqrt3+\tan x^\circ}{1-\sqrt3\tan x^\circ}\cdot\dfrac{\sqrt3-\tan x^\circ}{1+\sqrt3\tan x^\circ} \\\\ {} &=& \dfrac{3\tan x^\circ-\tan^3x^\circ}{1-3\tan^2x^\circ} \\\\ {} &=& \tan(3x)^\circ.\end{array}$$
Now put $x=12$:
$$\tan12^\circ\tan72^\circ\tan48^\circ\ =\ \tan36^\circ.$$
Now use the fact that $\tan(90-\theta)^\circ=\dfrac1{\tan\theta^\circ}$:
$$\frac1{\tan78^\circ\tan18^\circ\tan42^\circ} =\ \frac1{\tan54^\circ}$$
$\implies\ \tan42^\circ\ =\ \dfrac{\tan54^\circ}{\tan78^\circ\cdot\tan18^\circ}$

– QED! (Clapping)

Hello,

It can be proved using sine and cosine theorem in a very short period of time and without unnecessary work.

 

[anemone]

Hello,

It can be proved using sine and cosine theorem in a very short period of time and without unnecessary work.

Hi Dhamnekar Winod, can you enlighten us with your quick and neat solution?(Wondering)

 

[Wilmer]

I've been trying this with NO trigonometry.

                       F             A

               E

           
          D     C                 B

CD is extended to F (E is on AB) creating right triangle CFB.
Since angleBCF=48, then angleBFC=42.

I then extended BD to G (not shown in my poor diagram!),
such that angleGAB = 90.

Now if it could be shown that triangleABG is similar to triangleCFB,
then angleABD = 42.

But can't wrap this up...any ideas you guys?