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| Find GM, Get One Side? (Posted on 2007-06-11) |
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In triangle PQR; QR = p, PR = q and PQ = r with Angle QPR = 2* Angle PQR. The length of all the three sides of the triangle are different.
Is it always true that: p2 = q (q + r)?.
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Submitted by K Sengupta
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Rating: 3.0000 (2 votes)
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Solution:
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Let the angles QPR, PQR and PRQ be denoted respectively by P, Q and R.
Then,
Sin P/p = Sin Q/q …..(i)
We also have,
Sin P = Sin 2Q = 2 *Cos Q* SinQ
Or, Sin P/Sin Q = 2*Cos Q
Or, p/q= 2*Cos Q …..(ii)
But, we know that:
Cos Q = (r^2 + p^2 - q^2)/(2rp)
Or, p/q = (r^2 + p^2 - q^2)/(2rp)
Or, 2*r*p^2 = q(r^2 + p^2 - q^2)
Or, (q-r)(p^2 – q(q+r)) = 0
Or, p^2 = q(q +r), since q=r is inadmissible as the three sides have distinct lengths.
Thus, it is always true that p^2 = q(q+r)
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Comments: (
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Subject |
Author |
Date |
 | Sol | Praneeth Yalavarthi | 2007-07-09 12:20:12 |
 | re: Solution | JayDeeKay | 2007-06-13 13:10:29 |
| Solution | hoodat | 2007-06-11 13:48:24 |
| Solution | Bractals | 2007-06-11 12:49:11 |
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