Thứ Ba, 24 tháng 2, 2015

Golden Ratio in Equilateral Triangles

Triangles $ABC$ and $FGH$ are equilateral, with the circumcircle $(FGH)$ being the incircle of $\Delta ABC$ in such a manner that $F$ is on $AH,$ $G$ on $CF,$ and $H$ on $BG.$
golden ratio by Tran Quang Hung, construction #2
Then $\displaystyle\frac{FH}{AF}=\frac{GH}{BH}=\frac{FG}{CG}=\phi,$ the golden ratio.

Proof

Without loss of generality we may assume that the sides of $\Delta FGH$ equal $1,$ while those of $\Delta ABC$ equal $2.$ Let, say, $AF=x.$ Define $E$ (not shown) as the midpoint of $AB.$ Then, by the Power of a point theorem, $AF\cdot AH=AE^{2},$ i.e., $x(x+1)=1.$ Thus $x$ is the positive solution of the quadratic equation $x^{2}+x-1=0,$ i.e., $x=\frac{1}{2}(\sqrt{5}-1).$ Immediately
$\displaystyle\frac{FH}{AF}=\frac{1}{AF}=\frac{1+\sqrt{5}}{2}=\phi.$ 

Reference

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