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Absolutely Geometric (Posted on 2007-12-21) Difficulty: 3 of 5
In triangle PQR, the respective length of the sides PQ and PR are denoted by u and v while the length of the median PS is denoted by w. It is known that w is the geometric mean of u and v, and Angle QPR = 60o.

Determine |cos(Angle PQR) - cos(Angle QRP)|, where |x| denotes the absolute value of x.




See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

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Some Thoughts Look forward! | Comment 4 of 6 |

Hi!

I try to improve my solution and i find this :

Let note QR=s

From the cos theorem s^2=u^2+w^2-2uvcos(P)

but

|cos(Q)-cos(R)| = ((v^2-u^2+s^2)/v-(u^2-v^2+s^2)/u)/(2*s)

and after replacing and calculus i find that

|cos(Q)-cos(R)| = (1+cos(P))/sqrt(2)

and after some trigonometric function

|cos(Q)-cos(R)| = sqrt(2)(cos(P/2))^2

and for P=60 

|cos(Q)-cos(R)| = sqrt(9/8)=1.0606

 


  Posted by Chesca Ciprian on 2007-12-22 05:22:58
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