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Question Number 200120 by Mingma last updated on 14/Nov/23

Answered by ajfour last updated on 14/Nov/23

Commented by ajfour last updated on 14/Nov/23

(s/(t+p))=(t/(s+q))=(q/p) (similarity of △s) ⇒ ((s+t)/(s+t+p+q))=(q/p) ..(i) such perimeters equal ⇒ q+p+p+q=s+t+q ⇒ s+t=2p+q subdtituting in (i) ((2p+q)/(3p+2q))=(q/p) say (p/q)=((AB)/(CD))=λ ⇒ ((2λ+1)/(3λ+2))=(1/λ) 2λ^2 −2λ−2=0 λ^2 −λ−1=0 λ=((1+(√5))/2) = ϕ

$$\frac{{s}}{{t}+{p}}=\frac{{t}}{{s}+{q}}=\frac{{q}}{{p}}\:\:\:\:\left({similarity}\:{of}\:\bigtriangleup{s}\right) \\ $$$$\Rightarrow\:\:\frac{{s}+{t}}{{s}+{t}+{p}+{q}}=\frac{{q}}{{p}}\:\:\:\:..\left({i}\right) \\ $$$${such}\:{perimeters}\:{equal}\:\Rightarrow \\ $$$${q}+{p}+{p}+\cancel{{q}}={s}+{t}+\cancel{{q}} \\ $$$$\Rightarrow\:\:{s}+{t}=\mathrm{2}{p}+{q} \\ $$$${subdtituting}\:{in}\:\left({i}\right) \\ $$$$\frac{\mathrm{2}{p}+{q}}{\mathrm{3}{p}+\mathrm{2}{q}}=\frac{{q}}{{p}} \\ $$$${say}\:\:\frac{{p}}{{q}}=\frac{{AB}}{{CD}}=\lambda \\ $$$$\Rightarrow\:\:\frac{\mathrm{2}\lambda+\mathrm{1}}{\mathrm{3}\lambda+\mathrm{2}}=\frac{\mathrm{1}}{\lambda} \\ $$$$\mathrm{2}\lambda^{\mathrm{2}} −\mathrm{2}\lambda−\mathrm{2}=\mathrm{0} \\ $$$$\lambda^{\mathrm{2}} −\lambda−\mathrm{1}=\mathrm{0} \\ $$$$\lambda=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:=\:\varphi \\ $$

Commented by Mingma last updated on 14/Nov/23

Very elegant, sir!

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