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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

$$\frac{s}{t+p}=\frac{t}{s+q}=\frac{q}{p}(similarityof△s)$$$$\Rightarrow \frac{s+t}{s+t+p+q}=\frac{q}{p}..\left(i\right)$$$$suchperimetersequal\Rightarrow$$$$q+p+p+\cancel{q}=s+t+\cancel{q}$$$$\Rightarrow s+t=2p+q$$$$subdtitutingin\left(i\right)$$$$\frac{2p+q}{3p+2q}=\frac{q}{p}$$$$say\frac{p}{q}=\frac{AB}{CD}=\lambda$$$$\Rightarrow \frac{2\lambda +1}{3\lambda +2}=\frac{1}{\lambda }$$$$2{\lambda }^2-2\lambda -2=0$$$${\lambda }^2-\lambda -1=0$$$$\lambda =\frac{1+\sqrt{5}}{2}=\phi$$

Commented by Mingma last updated on 14/Nov/23

Very elegant, sir!

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