MHB Heptagon Challenge: Proving $\dfrac{1}{2}<\dfrac{S_Q+S_R}{S_P}<2-\sqrt{2}$

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The discussion focuses on proving the inequality $\frac{1}{2} < \frac{S_Q + S_R}{S_P} < 2 - \sqrt{2}$ for the areas of three regular heptagons, denoted as $S_P$, $S_Q$, and $S_R$. The heptagons are defined with specific side lengths, where $P_1P_2 = Q_1Q_3 = R_1R_4$. Participants analyze the geometric properties and relationships between the areas based on these dimensions. The proof involves comparing the areas derived from the side lengths and applying geometric inequalities. The conclusion emphasizes the validity of the established inequality for the areas of the heptagons.
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Let $P_1P_2P_3P_4P_5P_6P_7,\,Q_1Q_2Q_3Q_4Q_5Q_6Q_7,\,R_1R_2R_3R_4R_5R_6R_7$ be regular heptagons with areas $S_P,\,S_Q$ and $S_R$ respectively. Let $P_1P_2=Q_1Q_3=R_1R_4$. Prove that $\dfrac{1}{2}<\dfrac{S_Q+S_R}{S_P}<2-\sqrt{2}$
 
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\coordinate[label=left:$P_1$] (A) at (-1.96,-2.28);
\coordinate[label=left:$P_7$] (B) at (-3,0);
\coordinate[label=left:$P_6$] (C) at (-1.94,2.3);
\coordinate[label=above:$P_5$] (D) at (0.54,2.92);
\coordinate[label=right:$P_4$] (E) at (2.5,1.7);
\coordinate[label=right:$P_3$] (F) at (2.88,-0.9);
\coordinate[label=below:$P_2$] (G) at (0.7,-2.9);
\coordinate[label=below:$a$] (H) at (-0.4,-2.26);
\coordinate[label=below:$b$] (I) at (1.2,-0.86);
\coordinate[label=below:$c$] (J) at (0.7,0.6);
\draw (A) -- (B) -- (C) -- (D) -- (E) -- (F)--(G)--(A);
\draw (A) -- (D);
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\draw (D) -- (F);
[/TIKZ]
Let $P_1P_2=a,\,P_1P_3=b,\,P_1P_4=c$. By the Ptolomeus thoerem for the quadrangle $P_1P_3P_4P_5$ it follows that $ab+ac=bc$, i.e. $\dfrac{a}{b}+\dfrac{a}{c}=1$. Since $\triangle P_1P_2P_3\equiv \triangle Q_1Q_2Q_3$, then $\dfrac{Q_1Q_2}{Q_1Q_3}=\dfrac{a}{b}$ and hence $Q_1Q_2=\dfrac{a^2}{b}$.

Analogously $R_1R_2=\dfrac{a^2}{c}$. Therefore, $\dfrac{S_Q+S_R}{S_P}=\dfrac{a^2}{b}+\dfrac{a^2}{c}$. Then $\dfrac{a^2}{b}+\dfrac{a^2}{c}>\dfrac{1}{2}\left(\dfrac{a}{b}+\dfrac{a}{c}\right)^2=\dfrac{1}{2}$ (equality is not possible because $\dfrac{a}{b}\ne\dfrac{a}{c}$.

On the other hand

$\dfrac{a^2}{b^2}+\dfrac{a^2}{c^2}=\left(\dfrac{a}{b}+\dfrac{a}{c}\right)^2-\dfrac{2a^2}{bc}=1-\dfrac{2a^2}{bc}$---(1)

By the Sine theorem, we get

$\dfrac{a^2}{bc}=\dfrac{\sin^2 \dfrac{\pi}{7}}{\sin^2 \dfrac{\pi}{7}\sin^2 \dfrac{4\pi}{7}}=\dfrac{1}{4\cos\dfrac{2\pi}{7}\left(1+\cos\dfrac{2\pi}{7}\right)}$

Since $\cos\dfrac{2\pi}{7}<\cos \dfrac{\pi}{4}=\dfrac{\sqrt{2}}{2}$, then $\dfrac{a^2}{bc}>\dfrac{1}{4\dfrac{\sqrt{2}}{2}\left(1+\dfrac{\sqrt{2}}{2}\right)}=\sqrt{2}-1$. From here and from (1), we get the right hand side inequality of the problem.
 
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