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| Find the Ratio (Posted on 2007-08-20) |
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I (radius=r) is the incircle of triangle ABC (right angled at B) and O (radius=R) is another circle which touches the extended side BC and AC. Find the ratio (R/r) if the point of intersection of circles I and O is midpoint of AC.
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Submitted by Praneeth
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Rating: 3.2500 (4 votes)
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Solution:
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3+2*√2
The tangents from an external point to the circle are equal. As Incircle touches hypotenuse at midpoint, ABC is isosceles right angledtriangle.
=> BC=AC/√2
Draw perpendiculars from I and O on to BC and points of intersection beP and Q respectively. Draw perpendicular from I to OQ to meet at R.
http://perplexus.info/show.php?pid=5713&cid=38269
Now BC=r+PC, AC=2*PC.
=> (r+PC)*√2=2*PC => PC=(√2 +1)*r --(1)
Consider triangle IOR
IR=2*PC
IO=r+R
OR=R-r(Assume R>r)
As it is aright angled triangle,
4*PC˛+(R-r)˛=(R+r)˛
=> PC = √(R*r)
Substitute this in eq(1)
√(R*r)=(√2 +1)*r
=> √(R/r)=√2 +1
=> R/r = 3+2*√2.
Other solutions are given in comments by Bractals and Charlie. |
Comments: (
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Subject |
Author |
Date |
 | Answer | K Sengupta | 2008-08-10 06:34:25 |
 | solution | Charlie | 2007-08-20 11:48:05 |
| Solution | Bractals | 2007-08-20 11:23:46 |
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