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The number of triangles with each side having integral length and the longest side is of 11 units is:-? MY ATTEMPT:- I applied following constraints: $12 \leq {a+b} \leq 22$ $a,b \geq 1$ I made different cases that a+b =12, a+b=13 and so on. My answer came was 160 but it is not the correct one.

Ruchit Vithani
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  • Yah. That means sum of two sides that is a+b should be greater than or equal to 12. Because third side is 11 and all sides are integral. – Ruchit Vithani Jul 13 '17 at 05:31
  • @Ruchit Vithani: Check your counts. Note that $(a,b)=(1,11)$ gives the same triangle as $(a,b) = (11,1)$, so for the purposes of the count, you can assume $a \le b$. – quasi Jul 13 '17 at 05:40
  • Why are you specifying that a+b <= 22? That is not a condition. The only condition is that $a \le 11; b \le 11$. You can not, for instance $a = 13$ and $b =2$ but only because you are told the longest side is $11$ so $a \le 11$. Other wise there would be no limit to the sides and there are none to the sum of the sides at all. – fleablood Jul 13 '17 at 06:55

2 Answers2

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Your conditions are not quite correct. There is no stipulation on what $a + b$ is the only stipulation is that $11$ is the longest sides.

We need $a \le b \le 11$ so we don't count multiple instances the same and so that the maximum side(s) is $11$.

We need $a + b > 11$ or $a + b \ge 12$ to satisfy the (non-trivial) triangle inequality.

So the answer is $\sum\limits_{a=1}^{11}\sum\limits_{b=\max(a,12-a)}^{11}1$

$= \sum\limits_{a=1}^{11}[12- \max(a,12-a)]$

$= \sum\limits_{a=1}^{11}\min(12-a, a)$

$= 1 + 2 + 3 + 4+ 5 + 6 +5 + 4 + 3+2+1 = 21 + 15 = 36$.

I.E. $(a,b) = $

$(1,11)$

$(2,10)..(2, 11)$

$(3,9)....(3,11)$.

.....

$(5,7)....(5,11)$.

$(6,6)....(6,11)$.

$(7,7)....(7,11)$.

......

$(10,11)...(11,11)$

$(11,11)$

fleablood
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Forget the previous answer
Let us assume $a\geq b$ so if $a=11$ the b has 11 (1-11) possibilities
if $a=10$ b has 9 possibilites (2-10). $a=9$ we have 7 possibilities(3-9)... and so on
so the total is: 11+9+7...1=36 (sum of first n odd numbers in n^2)

Vishnu N
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