MHB Challenge Problem #6: Prove tan 18°=√(1-2/√5)

[Olinguito]

Prove that
$$\tan18^\circ\ =\ \sqrt{1-\dfrac2{\sqrt5}}.$$
No calculator, computer program, Excel, Google, or any other kind of cheating tool allowed. (Smirk)

Have fun!

 

[topsquark]

It's true. Proof: It's intuitively obvious.

-Dan

 

[Opalg]

Prove that
$$\tan18^\circ\ =\ \sqrt{1-\dfrac2{\sqrt5}}.$$
No calculator, computer program, Excel, Google, or any other kind of cheating tool allowed. (Smirk)

[sp]This comes from the geometry of the pentagon.
[TIKZ][scale=0.75]
\coordinate [label=above: $C$] (C) at (90:5cm) ;
\coordinate [label=above right: $D$] (D) at (18:5cm) ;
\coordinate [label=above left: $B$] (B) at (162:5cm) ;
\coordinate [label=below: $A$] (A) at (234:5cm) ;
\coordinate [label=below: $E$] (E) at (306:5cm) ;
\coordinate [label=below: $P$] (P) at (-4.755,-4.045) ;
\coordinate [label=below: $R$] (R) at (4.755,-4.045) ;
\coordinate [label=above right: $Q$] (Q) at (0.45,0.6) ;
\draw (A) -- node

{$1$} (B) -- node[above left] {$1$} (C) -- node[above right] {$1$} (D) -- node

{$1$} (E) -- node[below] {$1$} (A) -- (P) -- (B) -- (D) -- (R) -- (E) ;
\draw (C) -- (E) -- node[above right] {$d$}(B) ;
\draw (-3.4,-3.75) node {$72^\circ$} ;[/TIKZ]
If $ABCDE$ is a regular pentagon with side $1$ then it is a standard result that the diagonals have length $d = \frac12(\sqrt5+1)$, the golden ratio. A quick way to see that is to notice that in the above diagram the triangles $CQD$ and $BQE$ are similar (with angles $36^\circ$ and $108^\circ$), from which $QD = \frac1d$. Then $BD = BQ+QD$, so that $d = 1 + \frac1d$, a quadratic equation whose positive root is the golden ratio.

If $BP$ and $DR$ are the perpendiculars from $B$ and $D$ to $AE$ then $d = PR = PA + AE + ER = 1+2PA$, from which $PA = \frac12(d-1) = \frac14(\sqrt5-1)$.

By Pythagoras in the triangle $BPA$, $BP^2 = 1 - \frac1{16}(\sqrt5-1)^2 = \frac18(5+\sqrt5)$. Therefore $$\frac{PA^2}{BP^2} = \frac{\frac18(3-\sqrt5)}{\frac18(5+\sqrt5)} = \frac{(3-\sqrt5)(5-\sqrt5)}{(5+\sqrt5)(5-\sqrt5)} = \frac{20-8\sqrt5}{20} = 1 - \frac2{\sqrt5}.$$ But $\frac{PA}{BP} = \tan 18^\circ$, and therefore $\tan 18^\circ = \sqrt{1 - \frac2{\sqrt5}}.$

[/sp]​

 

[Olinguito]

Thanks, Opalg! It’s a neat geometric solution. (Clapping)

My own solution is entirely analytic. (Nerd)

Spoiler

Consider $\tan54^\circ$. On the one hand:
$$\tan54^\circ\ =\ \frac1{\tan36^\circ}\ =\ \frac{1-\tan^218^\circ}{2\tan18^\circ}.$$
On the other hand:
$$\tan54^\circ\ =\ \tan3(18)^\circ\ =\ \frac{3\tan18^\circ-\tan^318^\circ}{1-3\tan^218^\circ}$$
using the triple-angle formula for tangent. Hence:
$$\frac{1-\tan^218^\circ}{2\tan18^\circ}\ =\ \frac{3\tan18^\circ-\tan^318^\circ}{1-3\tan^218^\circ}$$
$\implies\ 1-4\tan^218^\circ+3\tan^418^\circ\ =\ 6\tan^218^\circ-2\tan^418^\circ$

$\implies\ 5\tan^418^\circ-10\tan^218^\circ+1\ =\ 0$

$\implies$ $\tan^218^\circ$ is a root of the quadratic equation $f(x)=5x^2-10x+1=0$

$\implies\ \tan^218^\circ=\dfrac{10\pm\sqrt{80}}{10}=1\pm\dfrac2{\sqrt5}.$

But $f(1)=-4<0$ and $f(2)=1>0$ showing that there is a root between $1$ and $2$. This root can’t be $\tan^218^\circ$ as $0<\tan18^\circ<1$. It follows that $\tan^218^\circ$ is the lesser of the two possible roots, i.e.
$$\tan^218^\circ\ =\ 1-\frac2{\sqrt5}$$
and we are done. (Cool) (The other root turns out to be $\tan^254^\circ$.)