996(3)We minimize the number of pairs by making sure no two boxes differ by more than oneball; one can easily check that the boxes must each contain [n/k] balls
5911st one :P
let a=rcosα
b=rsinα
thus r2=a2+b2 or r = √a2+b2
so asinx + bcosx = r(cosαsinx + sinαcosx)
= rsin(α+x)
So max(asinx + bcosx) = r
=√a2+b2
proved
2305(3)
Well here's a more elaborate solution solution to number 3.
(Just in case u didn't get Swarna's solution).
Let the n balls be put into the k boxes with n1 balls in the 1st box,n2 balls in the 2nd box,.......,nk balls in the kth box.
Total number of pairs =
\dpi{200} \fn_phv \sum_{i=0}^{k}\binom{n_{i}}{2}
(since there are pC2 pairs in a box containing p balls)
Now this is the term we want to minimize.
If ni - nj > 2 for some i j, then moving one ball from i to j decreases the number of pairs in the same box .
This is because before total no. of pairs were:
\dpi{200} \fn_phv \binom{n_{i}}{2}+\binom{n_{j}}{2}
After shifting total no. of pairs=
\dpi{200} \fn_phv \binom{n_{i}-1}{2}+\binom{n_{j}+1}{2}
On checking the 2 we find that on shifting the total no. of pairs just decreased by ni - nj -1.
Hence we reduce the size of larger ones and increase the size of smaller ones till they hit their average(i.e [n/k])
Hence the boxes must contain
[n/k] or [n/k+1] balls.
Good now go for the other problems.
996
Swarna Kamal Dhyawala
·2013-03-03 09:35:25
To avoid misunderstanding between angle A and area, I write Area.
From the law of cosines we know:
-a^2 + b^2 + c^2 = 2bc cosA
= 2bc sinA * cosA/sinA
= 4Area * cotA
We may find similar formulas with cotB and cotC. Adding these gives
a^2 + b^2 + c^2 = 4Area * (cotA + cotB + cotC)
Now we get
S = (a^2+b^2+c^2)/(4Area)
then
S = cotA + cotB + cotC
we know this identity of triangle
tanA + tanB + tanC = tanA tanB tanC
by division through by tanA tanB tanC
cotB cotC + cotA cotC + cotA cotB = 1.
By squaring both sides of S = cotA + cotB + cotC we find
S^2 = (cotA)^2 + (cotB)^2 + (cotC)^2 + 2.
Now we know that
(cotA - cotB)^2 + (cotB - cotC)^2 + (cotC - cotA)^2 >= 0
and thus
2((cotA)^2 + (cotB)^2 + (cotC)^2)
- 2(cotB cotC + cotA cotC + cotA cotB) >=0
or
2(S^2 - 2) - 2 >= 0
2S^2 - 6 >= 0
S^2 - 3 >= 0
and we find that S >= sqrt(3) (as it must be positive), which is
exactly what we were looking for.
792.
16A2=(a+b+c)(b+c-a)(c+a-b)(b+a-c).............hero's formula≤(a+b+c)((a+b+c)/3)3
..........(AM-GM inequality)
→4A≤(a+b+c)23√3≤√3a2+b2+c23(AM-GM inequality)......... hence proved
