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Question-169539

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Question Number 169539 by Shrinava last updated on 02/May/22
Commented by mr W last updated on 02/May/22
(R/(OI))=(√(3/2)) ?
$$\frac{{R}}{{OI}}=\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}\:? \\ $$
Commented by Shrinava last updated on 02/May/22
Yes dear sir
$$\mathrm{Yes}\:\mathrm{dear}\:\mathrm{sir} \\ $$
Answered by mr W last updated on 02/May/22
sin A=2 sin (A/2) cos (A/2)=((2(√(abcd)))/(ad+bc)) sin B=2 sin (B/2) cos (B/2)=((2(√(abcd)))/(ab+cd)) 3 sin A sin B=1 3×((4abcd)/((ab+cd)(ad+bc)))=1 ((abcd)/((ab+cd)(ad+bc)))=(1/(12)) OI=R(√(1−((2rμ)/R))) r=((√(abcd))/(a+c)) R=(1/4)(√(((ab+cd)(ac+bd)(ad+bc))/(abcd))) μ=(√(((ab+cd)(ad+bc))/((a+c)^2 (ac+bd)))) 1−((2rμ)/R)=1−8×((√(abcd))/(a+c))(√((abcd)/((ab+cd)(ac+bd)(ad+bc))))(√(((ab+cd)(ad+bc))/((a+c)^2 (ac+bd)))) =1−((8abcd)/((a+c)^2 (ac+bd))) =1−((8abcd)/((a+c)(b+d)(ac+bd))) =1−((8abcd)/((ab+cd+ad+bc)(ac+bd))) =1−((4abcd)/((ab+cd)(ad+bc))) =1−4×(1/(12)) =(2/3) Ω=(R/(OI))=(1/( (√(1−((2rμ)/R)))))=(1/( (√(2/3))))=(√(3/2)) ✓
$$\mathrm{sin}\:{A}=\mathrm{2}\:\mathrm{sin}\:\frac{{A}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{A}}{\mathrm{2}}=\frac{\mathrm{2}\sqrt{{abcd}}}{{ad}+{bc}} \\ $$$$\mathrm{sin}\:{B}=\mathrm{2}\:\mathrm{sin}\:\frac{{B}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{B}}{\mathrm{2}}=\frac{\mathrm{2}\sqrt{{abcd}}}{{ab}+{cd}} \\ $$$$\mathrm{3}\:\mathrm{sin}\:{A}\:\mathrm{sin}\:{B}=\mathrm{1} \\ $$$$\mathrm{3}×\frac{\mathrm{4}{abcd}}{\left({ab}+{cd}\right)\left({ad}+{bc}\right)}=\mathrm{1} \\ $$$$\frac{{abcd}}{\left({ab}+{cd}\right)\left({ad}+{bc}\right)}=\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$ \\ $$$${OI}={R}\sqrt{\mathrm{1}−\frac{\mathrm{2}{r}\mu}{{R}}} \\ $$$${r}=\frac{\sqrt{{abcd}}}{{a}+{c}} \\ $$$${R}=\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\frac{\left({ab}+{cd}\right)\left({ac}+{bd}\right)\left({ad}+{bc}\right)}{{abcd}}} \\ $$$$\mu=\sqrt{\frac{\left({ab}+{cd}\right)\left({ad}+{bc}\right)}{\left({a}+{c}\right)^{\mathrm{2}} \left({ac}+{bd}\right)}} \\ $$$$\mathrm{1}−\frac{\mathrm{2}{r}\mu}{{R}}=\mathrm{1}−\mathrm{8}×\frac{\sqrt{{abcd}}}{{a}+{c}}\sqrt{\frac{{abcd}}{\left({ab}+{cd}\right)\left({ac}+{bd}\right)\left({ad}+{bc}\right)}}\sqrt{\frac{\left({ab}+{cd}\right)\left({ad}+{bc}\right)}{\left({a}+{c}\right)^{\mathrm{2}} \left({ac}+{bd}\right)}} \\ $$$$=\mathrm{1}−\frac{\mathrm{8}{abcd}}{\left({a}+{c}\right)^{\mathrm{2}} \left({ac}+{bd}\right)} \\ $$$$=\mathrm{1}−\frac{\mathrm{8}{abcd}}{\left({a}+{c}\right)\left({b}+{d}\right)\left({ac}+{bd}\right)} \\ $$$$=\mathrm{1}−\frac{\mathrm{8}{abcd}}{\left({ab}+{cd}+{ad}+{bc}\right)\left({ac}+{bd}\right)} \\ $$$$=\mathrm{1}−\frac{\mathrm{4}{abcd}}{\left({ab}+{cd}\right)\left({ad}+{bc}\right)} \\ $$$$=\mathrm{1}−\mathrm{4}×\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Omega=\frac{{R}}{{OI}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\frac{\mathrm{2}{r}\mu}{{R}}}}=\frac{\mathrm{1}}{\:\sqrt{\frac{\mathrm{2}}{\mathrm{3}}}}=\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}\:\checkmark \\ $$
Commented by Shrinava last updated on 03/May/22
Perfect dear sir thank you so much
$$\mathrm{Perfect}\:\mathrm{dear}\:\mathrm{sir}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$
Commented by mr W last updated on 03/May/22
you may find all the formulas in: https://en.m.wikipedia.org/wiki/Bicentric_quadrilateral
Commented by Shrinava last updated on 03/May/22
Thank you very much professor
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{professor} \\ $$

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