IPMAT Indore 2024 · Quantitative Ability · Question 11

Step 1 — state the conditions. We need integers a\le b\le c with

a+b+c=15 \text{and}a+b>c \left( \text{triangle inequality}\right)

Step 2 — bound the largest side. From a+b>c and a+b=15-c:

15-c>c \Rightarrow c<7.5 \Rightarrow c\le 7

Also c\ge ⌈15/3⌉=5, since c is the largest of three numbers summing to 15.

Step 3 — list by c.

Total: 1+2+4=7

The seven triangles are \left( 5,5,5 \right), \left( 3,6,6 \right), \left( 4,5,6 \right), \left( 1,7,7 \right), \left( 2,6,7 \right), \left( 3,5,7 \right), \left( 4,4,7 \right).

Answer: 7

Shortcut. Use the standard formula. For an odd perimeter p, the number of incongruent integer-sided triangles is the nearest integer to

\frac{{\left( p+3 \right)}^{2}}{48}

Here \frac{{18}^{2}}{48}=\frac{324}{48}=6.75\to 7. If you’d rather not memorise it, the “largest side \le ⌊\left( p-1 \right)/2⌋=7” bound cuts the listing to three short rows.