*Posted by Mitchell in : potd , trackback
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Let $ABCDEF$ be a convex hexagon. $AD$ intersects $BE$ at $G$, $BE$ intersects $CF$ at $H$, and $AD$ intersects $CF$ at $I$. Additionally, $GF$ intersects $IE$ at $J$, $HD$ intersects $GC$ at $K$, and $IB$ intersects $HA$ at $L$. Let $[\mathcal{P}]$ denote the area of polygon $\mathcal{P}$. Given that \[\begin{align*} [GHI] &= 4 \\ [GDE] &= 5 \\ [HBC] &= 6 \\ [IFA] &= 7 \end{align*}\] find the minimum possible value of $[JFE] + [KDC] + [LBA] – [GKHLIJ]$.

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This diagram can be constructed by first constructing triangle $GHI$, then extending the sides to $A, B, C, D, E, F$. Therefore, it is easy to see that the ratios \[\begin{align*} a &= \frac{AI}{IG} \\ b &= \frac{BH}{HG} \\ c &= \frac{CH}{HI} \\ d &= \frac{DG}{GI} \\ e &= \frac{EG}{GH} \\ f &= \frac{FI}{IH} \end{align*}\] can take on any values, so long as they satisfy the relations \[\begin{align*} bc &= \frac{[HBC]}{[GHI]} = \frac{3}{2} \\ af &= \frac{[IFA]}{[GHI]} = \frac{7}{4} \\ de &= \frac{[GDE]}{[GHI]} = \frac{5}{4}. \end{align*}\] Then, note that$[JFE] + [KDC] + [LBA] – [GKHLIJ]$ is equal to \[([GAB] – [GIB] – [GHA]) + ([ICD] – [IHD] – [IGC])\]\[ + ([HEF] – [HGF] – [HIE]) + 2[GHI].\] But \[\begin{align*}[GAB] – [GIB] – [GHA] &= \frac{GA}{GI} \frac{GB}{GH} [GHI] – \frac{GB}{GH} [GHI] – \frac{GA}{GI} [GHI] \\ &= [GHI]\left((1 + a)(1+b) – (1+b) – (1+ a)\right) \\ &= 4(ab – 1).\end{align*}\] Similarly, $[ICD] – [IHD] – [IGC] = 4(cd – 1)$ and $([HEF] – [HGF] – [HIE]) = 4(ef – 1)$. Thus, we finally have that $[JFE] + [KDC] + [LBA] – [GKHLIJ]$ is $4(ab + cd + ef) – 4$. Using AM-GM on $ab + cd + ef$, we find that $ab + cd + ef $ is greater than or equal to $3 \sqrt[3]{abcdef} =3 \sqrt[3]{ \frac{5}{4} \frac{7}{4} \frac{3}{2}} =\frac{3\sqrt[3]{105}}{2\sqrt[3]{4}} $ with equality holding when $ab = cd = ef$, so the final answer is $4\left(\frac{3\sqrt[3]{105}}{2\sqrt[3]{4}}\right) – 4 = \boxed{3\sqrt[3]{210} – 4}.$

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