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Question Number 191060 by Mingma last updated on 17/Apr/23

Answered by witcher3 last updated on 18/Apr/23

$$\mathrm{BS}=\mathrm{a}\frac{\mathrm{d}}{\mathrm{b}},\mathrm{BR}=\frac{\mathrm{d}}{\mathrm{b}}\mathrm{c}$$$$\mathrm{AP}=\mathrm{c}\frac{\mathrm{d}}{\mathrm{a}},\mathrm{AQ}=\mathrm{b}\frac{\mathrm{d}}{\mathrm{a}}$$$$\mathrm{CM}=\mathrm{a}.\frac{\mathrm{d}}{\mathrm{c}},\mathrm{NC}=\frac{\mathrm{d}}{\mathrm{c}}\mathrm{b}$$$$\mathrm{c}(1-\frac{\mathrm{d}}{\mathrm{a}})=\mathrm{BP},\mathrm{c}(1-\frac{\mathrm{d}}{\mathrm{b}})=\mathrm{RA}$$$$\mathrm{PR}=\mathrm{c}(\frac{\mathrm{d}}{\mathrm{a}}+\frac{\mathrm{d}}{\mathrm{b}}-1)$$$$\mathrm{CQ}=\mathrm{b}(1-\frac{\mathrm{d}}{\mathrm{a}}),\mathrm{NA}=\mathrm{b}(1-\frac{\mathrm{d}}{\mathrm{c}}),\mathrm{NQ}=\mathrm{b}(\frac{\mathrm{d}}{\mathrm{a}}+\frac{\mathrm{d}}{\mathrm{c}}-1)$$$${\mathrm{pp}}^{\prime }=\frac{\frac{\mathrm{d}}{\mathrm{a}}+\frac{\mathrm{d}}{\mathrm{b}}-1}{\frac{\mathrm{d}}{\mathrm{b}}}.\frac{\mathrm{ad}}{\mathrm{b}}==\frac{\mathrm{db}+\mathrm{da}-\mathrm{ab}}{\mathrm{b}}$$$${\mathrm{p}}^{\prime }\mathrm{q}=\frac{\mathrm{ad}}{\mathrm{c}}.\frac{\mathrm{b}(\frac{\mathrm{d}}{\mathrm{a}}+\frac{\mathrm{d}}{\mathrm{c}}-1)}{\frac{\mathrm{d}}{\mathrm{c}}\mathrm{b}}$$$$=\frac{\mathrm{dc}+\mathrm{da}-\mathrm{ac}}{\mathrm{c}}$$$$\mathrm{pq}=\mathrm{d}=\frac{\mathrm{dcb}+\mathrm{dab}-\mathrm{acb}+\mathrm{dcb}+\mathrm{dac}-\mathrm{acb}}{\mathrm{bc}}$$$$\mathrm{d}(\mathrm{cb}+\mathrm{ab}+\mathrm{cb})=2\mathrm{acb}$$$$\mathrm{d}=\frac{2\mathrm{acb}}{\mathrm{ab}+\mathrm{bc}+\mathrm{cb}}=\frac{2}{{\mathrm{a}}^{-}+{\mathrm{b}}^{-}+{\mathrm{c}}^{-}}$$$$$$

Commented by Mingma last updated on 18/Apr/23

Very nice work, sir!

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