Let tn be the no. of triangles with integral sides out of the side lengths {1,2,3,....n} . Then t20-t19 = ?
11I'm getting 204 as the ans.....now i should also ask, what's wrong with my method?
1Its easy
t19 is a subset as discussed of t20
so all the solution triangles will be those with side 20
now a,b,20 be the triangle
so a+b>20
a+b=21 there are 9 solutions (2,19),(3,18),....(10,11)
a+b=22 2 to 10 (11,11) is not allowed 8 solution
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on adding till a+b=37 which has one solution so total 9+2(8+7+6+5+4+3+2+1) which equals 81
1I never said I got it........... I just saw the answer key....that's all.
1JEET Topper...............
The correct answer in 81 ... :-P
the answer keys of KVPY are out but not the solutions >>>
11yes 153 is its correct answer and its quite easy .
simply try to find a pattern in how the triangles are being formed.
1but Anirudh (17,18,19) is not included in t20 - t19. Nishant sir please help
1Nishant sir that is what i have done. isnt my counting correct?
@anirudh this is because a,b,c are symmetric and the triangles with sides for example (3,4,5) and (5,4,3),etc are the same
62See the trick is that t19 is a subset of t20
so atleast one side should be 20
Now count the cases :)
62oh sorry guys i was solving the problem with the sum of sides :(
Will post the full solution soon :P
1bhaiya you solution should be correct . I also don't remember the options fully.
1i wanted to ask in fibonacci solutions why can't we keep a=11 and b>9 . and so on.
nishant bhaiya . i think you have misread the question .
for t19 isn't 17,18,19 a valid solution .
62Let me try and calculate t19 .. after which, i hpe you guys find the value of t20 and thus complete the solution..
look at the largest side of the triangle
That side cannot be greater than 9 (because sum of 2 sides is greater than the 3rd)
No fo triangles with largest side as 9 =5 (smallest side goes from 1, 2, 3, 4, 5)
No fo triangles with largest side as 8 =3 (smallest side goes from 3, 4, 5)
No fo triangles with largest side as 7 =2 (smallest side goes from 5, 6)
The largest side cannot be 6
hence the number of distinct triangles of sum of sides 19 will be 10...
If you want to find the total number, multiply the number of scalenes by 3 and isoceles triangles by 2 (there iwll be no equilateral triangles)
1i don't think i saw 45 as one of the options but the method looks to be right....may be 45 was given...i don't remember
1soumik i'm getting 45
let the sides be a,b,c
t19 = all triangles formed from the set {1,2,3,4,...,19}
t20 = all triangles formed from the set {1,2,3,4,...,19,20}
so t20 - t19 excludes all triangles formed with sides from the set{1,2,3,...19}
so t20 - t19 is the set of all triangles that can be formed from the set {1,2,3,4,...19} with one side 20
NOTE: the above set doesnt include 20 as all triangles have disinct sides
let c=20
so all triangles are the solution of the inequality a+b>20 with a,bε{1,2,3,4,...19}
manually setting a=1 we get b>19 which is not possible
setting a=2, b>18 so 1 solution
setting a=3, b>17 so 2 solutions
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setting a=10, b>10 so 9 solutions
adding we get total triangles= 45
11Suppose there are 3 blanks, viz _,_,_....which are to be filled with integers....
If 1 of them is filled by 1, then for any 'n', there is only 1 possibility, i.e. only the set (1,n,n) can define the triangle effectively.
Since 'n' varies from 1 to 20 - so there are 20 triangles.......Now when 1 of them is occupied by 2, then (n-1) can also be an option, so there are (4.19) triangles possible.....
So effectively, the total number of triangles are 20+(4.19)+(9.18)+.... (64.13)
Similarly when n=19 ,
the total triangles are 19+(4.18)+(9.17)+....(64.12)....Subtracting them, we have 1+4+....64, which is easy......My logic is okay, but 1 trouble is calculation mistakes - someone pls check it out!