1no ,it's not the answer .
for 2p it's p(p+1) and for 2p-1 it's p2
I'm getting really confused.
How many triangles can be constructed whose largest side is
(a) 2p
(b) 2p -1
also find the number if 2p =10 .i.ie the lenght of largest side is 10 .
i.e. a≤b≤c =10 .
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18 Answers
62ok srinath...
try this..
Case : 2p
let us take one side (the largest one) Fix this as side "AB"
Let the 2nd largest side have length "X"
The smallest side will have length "Y"
X can go from 2p to p+1 (if it becomes 2p, then it will no longer be possible to have X+Y>2p
So X is 2p then Y can take values 1,2,..... 2p (2p values)
X is 2p-1, they Y can take values 2,3... 2p-1 (2p-2 values)
x is p+1 y can take value p, p+1 only (2 value)
actually for X=k, y can take values 2p+1-k, 2p+2-k.,....... k
these are 2k-2p values
the summation will go from k=P+1 to 2P
this will be 2+4+.... 2p = 2(1+2+.....p) = p(p+1)
62ok ok.. so u want distinct traingles... :)
lets see take it from there then :)
1how come , we can always rotate the triangle no , so only two are possible . right?
(2,2,2) and (2,2,1) or any other combo of 2,2,1 yields onyl one triangle?
62ok.. lets see..
for n=2, it will have triangle
(2,2,2)
(1,2,2)
(2,1,2)
(2,2,1)
so for p=1.. the answer should be 4... but 2p is not working.. (neither is my answer :D
so i will check this again... this is a wonderful question :)
162draw the graph..
the answer is very simple
n-2C2 (if i am not mistaken...!) tell me if i am.. i will post the full solution if u dont want to try drawing the graph of the above graphs...
1this is integral values we wnt ,so 1≤x≤2p ,same for y .
and also , x+y >2p , fine.
then ,what should I do? our sir said use counters. I didn't understnd him.
62take the 3 sides be (x,y,2p)
now try with 0<x<2p, 0<y<2p and x+y>2p
same with the second case!...
This is similar almost .. for 2nd case..
just try dude.. u need to find integral points in a square of side 2p and 2p-1 such blah blah blah...