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![[Post New]](/templates/default/images/icon_minipost_new.gif) 20 Mar 2007 21:23:33 IST
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A circle passes through the vertex A of an equilateral triangle ABC and is tangent to BC at its mid-point . Find the ratio in which the circle divides each of the sides AB and AC ???
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Umang |
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20 Mar 2007 21:33:54 IST
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hi is tne answer 3:4??
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20 Mar 2007 21:39:09 IST
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The options I hav dont think so !! anyways , tell me how u tried ???
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Umang |
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20 Mar 2007 21:46:08 IST
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pls explain how u did !~
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Umang |
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20 Mar 2007 21:49:04 IST
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i assumed B as origin.if D be mid point of BC then AD=2r.AD will be the diameter of circle because the radius through O is perpendicular to BC and in equi.triangle line joining AD is also perp.to BC.and we know the centre lies on AD.now D(a/2,0)where a is side length of triangle.A(a/2,a(root)3/2)from parametric eqn.2r=dist between these.r can be found out.now use the fact that AOX is congruent to AOY where X and Y are points of intersection with AB and AC.use cosine formula in any one of them to get AX or AY.thus u will obtain the ratio.im not sure if AD can be called diam.let me rethink.
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20 Mar 2007 21:50:55 IST
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ok i think i got my mistake.is the answer 1:3 or 3:1??
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20 Mar 2007 22:02:26 IST
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hey friend do tell me the answer if mines correct or not.ill find out the problem in my solution then.
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20 Mar 2007 22:02:52 IST
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hey friend do tell me the answer if mines correct or not.ill find out the problem in my solution then.
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