Geometry Questions

Question Number 105769 by ajfour last updated on 31/Jul/20

Commented by ajfour last updated on 31/Jul/20

$${ACDB}\:{is}\:{a}\:{straight}\:{line}.\:{The} \\$$$${three}\:{coloured}\:{regions}\:{have}\:{equal} \\$$$${areas}.\:{Given}\:{OA}={p},\:{OB}={q}. \\$$$${Find}\:{coordinates}\:{of}\:{A}\:{and}\:{B}. \\$$

Answered by mr W last updated on 31/Jul/20

$${A}\left(\sqrt{{p}^{\mathrm{2}} −{h}^{\mathrm{2}} },{h}\right) \\$$$${B}\left({k},\sqrt{{q}^{\mathrm{2}} −{k}^{\mathrm{2}} }\right) \\$$$${eqn}.\:{of}\:{AB}: \\$$$$\frac{{y}−{h}}{\sqrt{{q}^{\mathrm{2}} −{k}^{\mathrm{2}} }−{h}}=\frac{{x}−\sqrt{{p}^{\mathrm{2}} −{h}^{\mathrm{2}} }}{{k}−\sqrt{{p}^{\mathrm{2}} −{h}^{\mathrm{2}} }} \\$$$${x}_{{C}} =\sqrt{{p}^{\mathrm{2}} −{h}^{\mathrm{2}} }−\frac{{h}\left({k}−\sqrt{{p}^{\mathrm{2}} −{h}^{\mathrm{2}} }\right)}{\sqrt{{q}^{\mathrm{2}} −{k}^{\mathrm{2}} }−{h}} \\$$$${y}_{{D}} ={h}−\frac{\left(\sqrt{{q}^{\mathrm{2}} −{k}^{\mathrm{2}} }−{h}\right)\sqrt{{p}^{\mathrm{2}} −{h}^{\mathrm{2}} }}{{k}−\sqrt{{p}^{\mathrm{2}} −{h}^{\mathrm{2}} }} \\$$$${A}_{{red}} ={A}_{{yellow}} \\$$$$\Rightarrow−{h}={y}_{{D}} ={h}−\frac{\left(\sqrt{{q}^{\mathrm{2}} −{k}^{\mathrm{2}} }−{h}\right)\sqrt{{p}^{\mathrm{2}} −{h}^{\mathrm{2}} }}{{k}−\sqrt{{p}^{\mathrm{2}} −{h}^{\mathrm{2}} }} \\$$$$\Rightarrow\frac{\left(\sqrt{{q}^{\mathrm{2}} −{k}^{\mathrm{2}} }−{h}\right)\sqrt{{p}^{\mathrm{2}} −{h}^{\mathrm{2}} }}{{k}−\sqrt{{p}^{\mathrm{2}} −{h}^{\mathrm{2}} }}=\mathrm{2}{h}\:\:\:...\left({i}\right) \\$$$${A}_{{blue}} ={A}_{{red}} \\$$$$\Rightarrow−{k}={x}_{{C}} =\sqrt{{p}^{\mathrm{2}} −{h}^{\mathrm{2}} }−\frac{{h}\left({k}−\sqrt{{p}^{\mathrm{2}} −{h}^{\mathrm{2}} }\right)}{\sqrt{{q}^{\mathrm{2}} −{k}^{\mathrm{2}} }−{h}} \\$$$$\Rightarrow\frac{{h}\left({k}−\sqrt{{p}^{\mathrm{2}} −{h}^{\mathrm{2}} }\right)}{\sqrt{{q}^{\mathrm{2}} −{k}^{\mathrm{2}} }−{h}}={k}+\sqrt{{p}^{\mathrm{2}} −{h}^{\mathrm{2}} }\:\:\:...\left({ii}\right) \\$$$$\left({i}\right)×\left({ii}\right): \\$$$$\sqrt{{p}^{\mathrm{2}} −{h}^{\mathrm{2}} }=\mathrm{2}\left({k}+\sqrt{{p}^{\mathrm{2}} −{h}^{\mathrm{2}} }\right) \\$$$$\Rightarrow{k}=−\frac{\sqrt{{p}^{\mathrm{2}} −{h}^{\mathrm{2}} }}{\mathrm{2}} \\$$$${from}\:\left({i}\right): \\$$$$\frac{{k}−\sqrt{{p}^{\mathrm{2}} −{h}^{\mathrm{2}} }}{\sqrt{{q}^{\mathrm{2}} −{k}^{\mathrm{2}} }−{h}}=\frac{\sqrt{{p}^{\mathrm{2}} −{h}^{\mathrm{2}} }}{\mathrm{2}{h}} \\$$$$\mathrm{4}{q}^{\mathrm{2}} −{p}^{\mathrm{2}} =\mathrm{15}{h}^{\mathrm{2}} \\$$$$\Rightarrow{h}=−\sqrt{\frac{\mathrm{4}{q}^{\mathrm{2}} −{p}^{\mathrm{2}} }{\mathrm{15}}} \\$$$$\Rightarrow{k}=−\sqrt{\frac{\mathrm{4}{p}^{\mathrm{2}} −{q}^{\mathrm{2}} }{\mathrm{15}}} \\$$

Commented by mr W last updated on 31/Jul/20

Commented by ajfour last updated on 01/Aug/20

$${Thanks}\:{Sir},\:{you}\:{spirit}\:{my}\:{level} \\$$$${up}! \\$$

Answered by ajfour last updated on 01/Aug/20

$${let}\:\:\:{A}\left({p}\mathrm{cos}\:\theta,\:−{p}\mathrm{sin}\:\theta\right) \\$$$$\:\:\:\:\:\:\:\:{B}\left(−{q}\mathrm{sin}\:\phi,\:{q}\mathrm{cos}\:\phi\right) \\$$$${let}\:\:{C}\left({c},\mathrm{0}\right)\:;\:\:{D}\left(\mathrm{0},{d}\right) \\$$$$\underset{−} {{slope}\:{of}\:{AB}} \\$$$$\:\:{m}=−\left(\frac{{q}\mathrm{cos}\:\phi+{p}\mathrm{sin}\:\theta}{{p}\mathrm{cos}\:\theta+{q}\mathrm{sin}\:\phi}\right)=−\frac{{d}}{{c}}\:\:\:...\left({i}\right) \\$$$$\:\:\:\:{p}\mathrm{sin}\:\theta={d}\:\:\:,\:\:\:{q}\mathrm{sin}\:\phi={c}\:\:\:\:\:...\left({ii}\right),\:\left({iii}\right), \\$$$$\&\:\:−{m}=\frac{{p}\mathrm{sin}\:\theta}{{p}\mathrm{cos}\:\theta−{c}}=\frac{{q}\mathrm{cos}\:\phi−{d}}{{q}\mathrm{sin}\:\phi}\:\:\:..\left({iv}\right) \\$$$$\Rightarrow\:\:\frac{\sqrt{{q}^{\mathrm{2}} −{c}^{\mathrm{2}} }+{d}}{\sqrt{{p}^{\mathrm{2}} −{d}^{\mathrm{2}} }+{c}}=\frac{{d}}{{c}}\:\:\:\:\:\:\:....\left({I}\right) \\$$$$\:\:\:\:\:\:\frac{{d}}{\sqrt{{p}^{\mathrm{2}} −{d}^{\mathrm{2}} }−{c}}\:=\:\frac{\sqrt{{q}^{\mathrm{2}} −{c}^{\mathrm{2}} }−{d}}{{c}}\:\:\:\:....\left({II}\right) \\$$$${from}\:\:\left({I}\right)\:\&\:\left({II}\right) \\$$$$\:\:\:\:\:\frac{{u}}{{v}}=\:\frac{\sqrt{{q}^{\mathrm{2}} −{c}^{\mathrm{2}} }−{d}}{\sqrt{{p}^{\mathrm{2}} −{d}^{\mathrm{2}} }−{c}}\:=\:\frac{{d}}{{c}} \\$$$$\&\:\:\:\:\:\:\:\:\:\:{uv}={cd} \\$$$$\Rightarrow\:\:\:\:{u}={d}\:,\:\:{v}={c} \\$$$$\Rightarrow\:\:\:\:{c}^{\mathrm{2}} +\mathrm{4}{d}^{\mathrm{2}} ={q}^{\mathrm{2}} \\$$$$\:\:\:\:\:\:\:\:\:\mathrm{4}{c}^{\mathrm{2}} +{d}^{\mathrm{2}} ={p}^{\mathrm{2}} \\$$$$\Rightarrow\:\:\:{c}=\sqrt{\frac{\mathrm{4}{p}^{\mathrm{2}} −{q}^{\mathrm{2}} }{\mathrm{15}}}\:,\:\:{d}=\sqrt{\frac{\mathrm{4}{q}^{\mathrm{2}} −{p}^{\mathrm{2}} }{\mathrm{15}}} \\$$$$\Rightarrow \\$$$$\:\:{A}\left(\sqrt{\frac{\mathrm{16}{p}^{\mathrm{2}} −\mathrm{4}{q}^{\mathrm{2}} }{\mathrm{15}}}\:,\:−\sqrt{\frac{\mathrm{4}{q}^{\mathrm{2}} −{p}^{\mathrm{2}} }{\mathrm{15}}}\:\right) \\$$$$\:{B}\left(−\sqrt{\frac{\mathrm{4}{p}^{\mathrm{2}} −{q}^{\mathrm{2}} }{\mathrm{15}}}\:,\:\sqrt{\frac{\mathrm{16}{q}^{\mathrm{2}} −\mathrm{4}{p}^{\mathrm{2}} }{\mathrm{15}}}\:\right)\:\bigstar \\$$$$\\$$