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please......can anyone solve it for me......i m in a need of help..............

#MathProblem

Note by Ronak Pawar
7 years, 8 months ago

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Comments

Let XX be the point of contact of the smaller circle with the diameter, and let O1O_1 be the centre of the smaller circle. Let the angles ∠O1OC=θ\angle O_1OC = \theta and ∠O1AO=α\angle O_1AO = \alpha.

The straight line OO1OO_1 meets the larger circle at its point of tangency with the smaller circle, and hence OO1=1−rOO_1 = 1-r. Thus (1−r)sin⁡θ=r(1-r)\sin\theta = r, and hence r  =  sin⁡θ1+sin⁡θ r \; = \; \frac{\sin\theta}{1 + \sin\theta} Consequently AX  =  1+(1−r)cos⁡θ  =  1+cos⁡θ1+sin⁡θ  =  1+sin⁡θ+cos⁡θ1+sin⁡θ  =  2(1+cos⁡θ)1+sin⁡θ+cos⁡θ AX \; = \; 1 + (1-r)\cos\theta \; = \; 1 + \frac{\cos\theta}{1+\sin\theta} \; = \; \frac{1+\sin\theta+\cos\theta}{1+\sin\theta} \; = \; \frac{2(1+\cos\theta)}{1+\sin\theta+\cos\theta} the last identity being a trigonometric identity that is easy to prove. Hence AX  =  2(1−sin⁡θ1+sin⁡θ+cos⁡θ)  =  2(1−rAX)  =  2(1−tan⁡α) AX \; =\; 2\left(1 - \frac{\sin\theta}{1+\sin\theta+\cos\theta}\right) \; = \; 2\left(1 - \frac{r}{AX}\right) \; = \; 2(1-\tan\alpha) so that r  =  AXtan⁡α  =  2(tan⁡α−tan⁡2α) r \; = \; AX\tan\alpha \; = \; 2\big(\tan\alpha - \tan^2\alpha\big) Since ABAB and ACAC are both tangents to the smaller circle we have ∠BAO1=α\angle BAO_1 = \alpha, and so it follows that ∠BAC=2α\angle BAC = 2\alpha, and hence a=2cos⁡2αa = 2\cos2\alpha. Thus 2−a2+a=4sin⁡2α4cos⁡2α  =  tan⁡2α \frac{2-a}{2+a} = \frac{4\sin^2\alpha}{4\cos^2\alpha} \; = \; \tan^2\alpha and hence we deduce that r  =  2(2−a2+a−2−a2+a) r \; = \; 2\left(\sqrt{\frac{2-a}{2+a}} - \frac{2-a}{2+a}\right) as required.

Mark Hennings - 7 years, 8 months ago
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