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
$$\frac{{s}}{{t}+{p}}=\frac{{t}}{{s}+{q}}=\frac{{q}}{{p}}\:\:\:\:\left({similarity}\:{of}\:\bigtriangleup{s}\right) \\ $$$$\Rightarrow\:\:\frac{{s}+{t}}{{s}+{t}+{p}+{q}}=\frac{{q}}{{p}}\:\:\:\:..\left({i}\right) \\ $$$${such}\:{perimeters}\:{equal}\:\Rightarrow \\ $$$${q}+{p}+{p}+\cancel{{q}}={s}+{t}+\cancel{{q}} \\ $$$$\Rightarrow\:\:{s}+{t}=\mathrm{2}{p}+{q} \\ $$$${subdtituting}\:{in}\:\left({i}\right) \\ $$$$\frac{\mathrm{2}{p}+{q}}{\mathrm{3}{p}+\mathrm{2}{q}}=\frac{{q}}{{p}} \\ $$$${say}\:\:\frac{{p}}{{q}}=\frac{{AB}}{{CD}}=\lambda \\ $$$$\Rightarrow\:\:\frac{\mathrm{2}\lambda+\mathrm{1}}{\mathrm{3}\lambda+\mathrm{2}}=\frac{\mathrm{1}}{\lambda} \\ $$$$\mathrm{2}\lambda^{\mathrm{2}} −\mathrm{2}\lambda−\mathrm{2}=\mathrm{0} \\ $$$$\lambda^{\mathrm{2}} −\lambda−\mathrm{1}=\mathrm{0} \\ $$$$\lambda=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:=\:\varphi \\ $$
Commented by Mingma last updated on 14/Nov/23
Very elegant, sir!