Find the polygon according to the relation given below

Question

If A, B, C and D are four consecutive vertices of a regular polygon we have to $\frac{1}{AB} = \frac{1}{AC}+\frac{1}{AD}$ . How many sides does the polygon have?

Solution by trigonometry.. $AB=2rsen(x)\\ AC=2rsen(2x)\\ AD=2rsen(3x)\\ \therefore \frac{1}{sen(x)}=\frac{1}{sen2x}+\frac{1}{sen3x}\implies \frac{1}{sen(x)}-\frac{1}{sen2x}=\frac{1}{sen3x}\\ \\sen(2x)=2sen(x)cos(x):\frac{2cos(x)-1}{sen(2x)}=\frac{1}{sen(3x)}\\ \therefore 2sen(3x)cos(x)-sen(3x)=sen(2x)\\ sen(A)cos(B)=\frac{{}sen(A+B)+sen(A-B)}{2}:\\ sen(4x)+sen(2x)-sen(3x)=sen(2x) \implies sen(4x)=sen(3x)\\ \therefore 4x+3x=180^o. \implies \boxed{x=\frac{180^o}{7}}.$

(Solução:LuizFuentes)

Would it be possible to solve by geometry?

Answer 1

Let $AB=x, AC=y$ and $AD=z$.

In the figure, $H$ is a point on $BD$ such that $DH=x$

Thus $\angle DHC= \frac{180^o-\alpha}{2}=90^o-\frac{\alpha}{2}$

$$\angle BHC=180^o-\left( 90^o-\frac{\alpha}{2}\right)=90^o+\frac{\alpha}{2} \tag{1}$$

Also $$\angle ABD=180^o-3 \alpha \tag{2}$$

Noting that $BD=AC=y \implies$ $$HB=y-x \tag{3}$$

From the given $$\frac{1}{AB}=\frac{1}{AC}+\frac{1}{AD}$$

$$\implies \frac{1}{x}=\frac{1}{y}+\frac{1}{z}$$

$$\implies \frac{1}{x}-\frac{1}{y}=\frac{1}{z}$$

$$\implies \frac{y-x}{xy}=\frac{1}{z}$$

$$\implies \frac{y-x}{x}=\frac{y}{z}$$

$$\implies \frac{HB}{BC}=\frac{BD}{DA}$$

$$\because \angle BGA= \angle HBC = \alpha$$

$$\therefore \Delta DBA \sim \Delta BHC$$

Hence $$\angle DBA = \angle BHC \tag{4}$$

Put $(1), (2)$ into $(4)$

$$180^0-3 \alpha= 90^0+\frac{\alpha}{2}$$

$$\alpha = \frac{180^o}{7} $$

Answer 2

$\square ABCD$ is an isosceles trapezoid. Let its diagonals meet at $X$, forming isosceles $\triangle XAD$, and let $\overline{CD'}$ create rhombus $\square ABCD'$ and isosceles $\triangle CDD'$.

As inscribed angles of the regular polygon's circumcircle subtending congruent chords, we have $$\angle CBD = \angle BCA = \angle BDC = \angle BDA \quad =: \theta$$ We readily conclude that $\angle CXD = \angle CD'D = 2\theta$, so that $\square XCDD'$ is cyclic. This, in turn, implies that $\angle XD'C$ (marked with a dot in the figure) also has measure $\theta$. Moreover, calculating in two ways the power of point $A$ relative to that quadrilateral's circumcircle, we have $$|AD'||AD|=|AX||AC| \tag1$$

Note that the above holds for any regular polygon. (Indeed, to help convey this universality, the figure is based on the $9$-gon instead of the target $7$-gon.) Now, we'll invoke the exercise's given relation, and manipulate it a bit in light of $\square ABCD'$ being a rhombus: $$\begin{align} \frac1{|AB|} = \frac1{|AC|}+\frac1{|AD|} &\qquad\to\qquad (\,|AD|-|AB|\,)|AC| = |AB||AD| \tag2 \\ &\qquad\to\qquad |D'D||AC| = |AD'||AD| \overset{(1)}{=} |AX||AC| \tag3 \\[8pt] &\qquad\to\qquad |D'D| = |AX| = |XD| \tag4 \end{align}$$ Therefore, $\triangle DXD'$ is isosceles (which doesn't look true above, indicating that we've broken universality) and has angle sum $180^\circ=\theta+3\theta+3\theta=7\theta$. We conclude that $A$, $B$, $C$, $D$ are vertices of a regular $7$-gon. $\square$