simple geometry-

Geometry Level 4

Let $PQ$ and $RS$ be tangents at the extremities of diameter $PR$ of a circle of radius $r$ . if $PS$ and $RQ$ intersect at a point $X$ on the circumference of the circle, then $2r$ equals to -

\sqrt{PQ.RS}

\frac{PQ+RS}{2}

\frac{\sqrt{PQ^{2}+RS^{2}}}{2}

\frac{2PQ+RS}{PQ+RS}

2 solutions

Aniket Verma
Mar 3, 2015

$PR = 2r$

$\angle PXR=90^{o}$

Let $\angle XPR = \Theta$

therefore $\angle XRP= 90^{o} - \Theta$

so, $RS = 2rtan(\Theta)$ , and $PQ = 2rtan(90^{o} - \Theta)$ $=$ $2rcot(\Theta)$

therefore, $PQ\times RS = 2rtan(\Theta)\times 2rcot(\Theta)$

=> $PQ\times RS = (2r)^{2}$

hence, $2r = \sqrt{PQ\times RS}$

https://brilliant.org/discussions/thread/inmo-2017/

Ayush Pattnayak - 4 years, 5 months ago

Bhavesh Ahuja
Feb 4, 2015

Make a suitable diagram according to the question. You will see that angle PXR is 90 as diameter subtends 90 degrees on the circumference. Also angles QPR and SRP are 90 as radius is perpendicular to tangent. Now assume angle PQR as x, then angle XPQ will be 90-x, angle XPR= x, angle XRP= 90-x, angle XRS = x and angle XSR= 90-x. Therefore, triangles QPR and PRS are similar. QPR~PRS. so (QP/PR) = (PR/RS). So PR^2 = QP.RS

simple geometry-

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