Permutations & Combinations
Permutation and Combination
star_batch_jee_advanced_2025
Grade None
The number of isosceles triangles with integer sides if no side exceeding 2008 is:
$(1004)^2$ if equal sides do not exceed 1004
$2(1004)^2$ if equal sides exceed 1004
$3(1004)^2$ if equal sides have any length $\leq$ 2008
$(2008)^2$ if equal sides have any length $\leq$ 2008
Step-by-Step Solution
Key Concept: The triangle inequality $b < 2a$ creates two regimes based on whether $2a$ exceeds the maximum side length 2008, and these regimes contribute $(1004)^2$ and $2(1004)^2$ respectively.
For an isosceles triangle with equal sides $a$ and base $b$, the triangle inequality requires $a + a > b$, so $b < 2a$. With constraint $a, b \leq 2008$: when equal sides $a \leq 1004$, base $b$ can be $1, 2, \ldots, 2a-1$ giving $2a-1$ triangles per $a$, totaling $(1004)^2$. When $1004 < a \leq 2008$, base $b$ can be $1, 2, \ldots, 2008$ (since $2a > 2008$) giving $2008$ triangles per $a$, totaling $1004 \times 2008 = 2(1004)^2$. Combined total: $(1004)^2 + 2(1004)^2 = 3(1004)^2$.
Correct Answer: 1,2,3