Question-169539

Question Number 169539 by Shrinava last updated on 02/May/22

Commented by mr W last updated on 02/May/22

\frac{R}{O I} = \sqrt{\frac{3}{2}} ?

Commented by Shrinava last updated on 02/May/22

Yes dear sir

Answered by mr W last updated on 02/May/22

\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2} = \frac{2 \sqrt{a b c d}}{a d + b c}

\sin B = 2 \sin \frac{B}{2} \cos \frac{B}{2} = \frac{2 \sqrt{a b c d}}{a b + c d}

3 \sin A \sin B = 1

3 \times \frac{4 a b c d}{(a b + c d) (a d + b c)} = 1

\frac{a b c d}{(a b + c d) (a d + b c)} = \frac{1}{12}

O I = R \sqrt{1 - \frac{2 r μ}{R}}

r = \frac{\sqrt{a b c d}}{a + c}

R = \frac{1}{4} \sqrt{\frac{(a b + c d) (a c + b d) (a d + b c)}{a b c d}}

μ = \sqrt{\frac{(a b + c d) (a d + b c)}{{(a + c)}^{2} (a c + b d)}}

1 - \frac{2 r μ}{R} = 1 - 8 \times \frac{\sqrt{a b c d}}{a + c} \sqrt{\frac{a b c d}{(a b + c d) (a c + b d) (a d + b c)}} \sqrt{\frac{(a b + c d) (a d + b c)}{{(a + c)}^{2} (a c + b d)}}

= 1 - \frac{8 a b c d}{{(a + c)}^{2} (a c + b d)}

= 1 - \frac{8 a b c d}{(a + c) (b + d) (a c + b d)}

= 1 - \frac{8 a b c d}{(a b + c d + a d + b c) (a c + b d)}

= 1 - \frac{4 a b c d}{(a b + c d) (a d + b c)}

= 1 - 4 \times \frac{1}{12}

= \frac{2}{3}

Ω = \frac{R}{O I} = \frac{1}{\sqrt{1 - \frac{2 r μ}{R}}} = \frac{1}{\sqrt{\frac{2}{3}}} = \sqrt{\frac{3}{2}} ✓

Commented by Shrinava last updated on 03/May/22

Perfect dear sir thank you so 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