equal areas inside a pentagon (2013 HOMC Junior Q8 Senior Q8)

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equal areas inside a pentagon (2013 HOMC Junior Q8 Senior Q8)

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#1 h Jul 18, 2019, 3:14 PM • 1 Y

Y by Adventure10

Let $ABCDE$ be a convex pentagon and area of $\vartriangle ABC =$ area of $\vartriangle BCD =$ area of $\vartriangle CDE=$ area of $\vartriangle DEA =$ area of $\vartriangle EAB$ . Given that area of $\vartriangle ABCDE = 2$ . Evaluate the area of area of $\vartriangle ABC$ .

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#2 h Jan 9, 2024, 7:56 PM

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Given the regular pentagon $ABCDE\ :\ A(0,0),B(b,0),C(b+b\cos 72^{\circ},b\sin 72^{\circ}),D(\frac{b}{2},b(\sin 72^{\circ}+\sin 36^{\circ})\ ),E$ .

Area $=2 \cdot \frac{b}{2} \cdot b\sin 72^{\circ}+\frac{b}{2} \cdot b(\sin 72^{\circ}+\sin 36^{\circ})\ =\ 2$ ,
$b^{2}[\sin 72^{\circ}+\frac{1}{2}(\sin 72^{\circ}+\sin 36^{\circ})=2$ ,
$b^{2}[\sin 72^{\circ}+\sin 54^{\circ}\cos 18^{\circ}]=2$ ,
$b^{2}\sin 72^{\circ}(1+\sin 54^{\circ})=2$ ,
$b^{2}\sin 72^{\circ}=\frac{2(5-\sqrt{5})}{5}$ .

Area $\triangle ABC\ =\ \frac{1}{2}b^{2}\sin 72^{\circ}=\frac{5-\sqrt{5}}{5}$ .

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