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![[Post New]](/templates/default/images/icon_minipost_new.gif) 8 Apr 2007 22:05:22 IST
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OOOPS SORRY FRIEND!!!!!!!!1Here is the corrected question;- If there are 'm' non concurrent and non parallel lines, then what is the maximum number of points which are equidistsnt from all the 'm' lines?
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Ken
From: UNITED STATES, Green Bay, Wisconsin
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11 Apr 2007 18:08:10 IST
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Its really a very nice question .. short of IQ quiz..
The set of points which are equidistant from two lines is the angle bisector b/w those two lines.. and there can be two angle bisector for each pair of two lines..
Now, a point will be equidistant from all the m lines, if it is the point of concurrence of the angle bisectors of every possible pair of two lines..
There will be mC2 pair of lines... and two bisectors for each set..
So, we can select one bisector for each pair of lines in "2 raised to the power mC2" ways...and this will be the required max no of points.. if all these selection of bisectors are concurrent.... got it??
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Sudeep Kumar
(B tech, IITd)
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11 Apr 2007 20:55:08 IST
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But Sir,I have got a question in T.M.H where m=3 and the answer is given as 4.Your explination is correct but I should do now?According to your explaination the answer should come " 2 raised to the power 3 (because 3C2 is 3)" which is 8.
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Ken
From: UNITED STATES, Green Bay, Wisconsin
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12 Apr 2007 20:22:43 IST
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Please post how it has been solved.. or tell me the page no. and name of book... i will recheck my solution and tell you if i have missed something or if there is some misprint in the book.... and thnx for pointing it out.
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Sudeep Kumar
(B tech, IITd)
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13 Apr 2007 09:19:14 IST
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Dear,
I also agree with Mr. Sudeep Kumar, Please check the solution given in your book and revert the same back to us so we could clarify your doubt and also find where we are missing the step.
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