Question

The number of isosceles triangles with integer sides if no side exceeds 2018 is

• A. $(1009{)}^2$ if equal sides do not exceed 1009
  
• B. $2(1009{)}^2$ if equal sides exceed 1009
  
• C. $3(1009{)}^2$ if equal sides have any length $\le 2018$
  
• D. $(2018{)}^2$ if equal sides have any length $\le 2018$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $(1009{)}^2$ if equal sides do not exceed 1009

B $2(1009{)}^2$ if equal sides exceed 1009

C $3(1009{)}^2$ if equal sides have any length $\le 2018$


If the sides are a,a,b then the triangle forms only when 2a>b .So for any $a\in N$, b can change from 1 to 2a-1 ,where $a\le 1009\Rightarrow$ no.of such triangles =1+3+5+---+(2(1009)-1)$=(1009{)}^2$
And if $1010\le a\le 2018,$
But a has 1009 possibilities hence
No.of triangles $=1009\times 2018$
$=2(1009{)}^2$
$\therefore$ Total no.of isosceles triangles $=3(1009{)}^2$