4I got b as answer
1..IF P,Q,R MAY BE PERIMETER ,CIRCUMRADIUS AND INRADIUS OF AN ARBITARY TRIANGLE THE WHICH OF THE FOLLOWING MAY NOT BE ALWAYS TRUE--
(A)P>Q+R
(B) P≤Q+R
(C)P/6<Q+R<6P (MULTI ANSWER)
-
UP 0 DOWN 0 0 14
14 Answers
1i got b and c......cudnt come up wid a as yet ....d ans was given as a,b,c...
1ques is asking which may not be true but soumik u have proved a,b I think u have made a mistake sumwhere....wat is actually Jenson inequality....i have proved p/6<R+r....a/2R<1 , b/2R<1 , c/2R<1 so p/2R<3 or p/6<R or p/6<R+r but cudnt proof R+r<6p....so c cud be the ans...
1assume a triangle of sides 100, 101 & 200 ( it gives b & c)
assume a triangle of sides 100 , 100 & 1 (it gives a )
so abc all are correct
11avra, so that gives p/6<R+r for all triangles. How come then c be the ans?
11After delivering utter non-sense in my last post, let me close this thread finally with something that should clear all dbts.....
sinA+sinB+sinC=4\prod{cos\frac{A}{2}}
From Jenson's \sum{cos\frac{A}{2}}\le \frac{3\sqrt{3}}{2}
From A-M-GM 4\underbrace{\prod{cos\frac{A}{2}}}_x \le \frac{2\sqrt{3}}{2}
Now x can be made as close to zero as we wish - but not 0.....so effectively range turns out to be \left(0,\frac{3\sqrt{3}}{2} \right]
So this function being continuous can take any value within its range - let it take the value \frac{1}{a}
0<a<6, so that gives 2R= \frac{P}{4x}<6P\Rightarrow R<3P\Rightarrow \boxed{R+r<6P}
The Equilateral traingle confirms case b).
f(P,R,r)=(R+r-P)
Also
f(P,R,r)=f(A,B)=\frac{1}{2[sinA+sinB+sin(A+B)]}}+\frac{2sin\frac{A}{2}sin\frac{B}{2}cos\left(\frac{A+B}{2} \right)}{sinA+sinB+sin\left(\frac{A+B}{2} \right)}-1
Numerator of the function is
1+4sin\frac{A}{2}sin\frac{B}{2}cos\left(\frac{A+B}{2} \right)-2\left(sinA+sinB+sin(A+B) \right) - Denominator is always >0
Define a new function G(A,B)=sinA+sinB+sin(A+B)-\frac{1}{2}
G(0,0)<0 & G\left( \frac{\pi}{2},\frac{\pi}{2}\right)>0
So from Mean-Value-Theorem, there exists (A',B') in the open interval \left(0,\frac{\pi}{2} \right) for which G(A',B')=0.....
G(A',B')=0 implies f(A,B)>0
Which confirms a).
Thus answers are (a,b).