PRMO 2017 question 20

 What is the number of triples (a, b, c) of positive integers such that (i) a<b<c< 10 and (ii) a, b, c, 10 form the sides of a quadrilateral?

Solution:
4 sides of a rectangle have the property that sum of any 3 sides is larger than the 4th side. Similar to what we have in a triangle. And it can be generalized to any polygon. So:

a + b + c > 10
a + b + 10 > c Which will anyway hold since c <= 9
a + c + 10 > b Which will anyway hold since b <= 8
c + b + 10 > a Which will anyway hold since a <= 7

So only the first one is worth examining.
Now a,b,c have to be integers between 1 to 9.
So we have 9C3 ways of choosing a,b,c.
Out of which we have to remove those triplets which violate the first condition.

9C3 = 9.8.7/3.2 = 84
Now list down triplets which violate 
a + b + c > 10
i.e. check for a,b,c for which
a + b + c <= 10

Those are 11 triplets: 
1,2,3
1,2,4
1,2,5
1,2,6
1,2,7
1,3,4
1,3,5
1,3,6
1,4,5
2,3,4
2,3,5

So answer 84 - 11 = 73.

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