2
$\begingroup$

Let the sum of the three sides of a triangle be $100,$ and all the sides are positive integers length, how many possible isosceles triangles are there?

$\endgroup$
  • $\begingroup$ Hint: count solutions to the (in)equations $2a+b=100$ and $2a \geq b$. $\endgroup$ – Daniel Franke Aug 3 '13 at 2:36
  • $\begingroup$ @DanielFranke $2a>b$. $\endgroup$ – S.B. Aug 3 '13 at 2:37
5
$\begingroup$

Hint: We can certainly do $(49,49,2)$, $(48,48,4)$, $(47,47,6)$ and so on for a while. But the sum of the two equal sides cannot be less than or equal to the third side. That should tell you where we need to stop.

$\endgroup$
2
$\begingroup$

Hint: Let the side length $=x$, then we know the base length $=100-2x$. Then set up your inequality(s).

$\endgroup$
0
$\begingroup$

x=2*a+100-2*a if (100-2*a)

2*34+32=100 2*35+30=100 2*36+28=100 2*37+26=100 2*38+24=100 2*39+22=100 2*40+20=100 2*41+18=100 2*42+16=100 2*43+14=100 2*44+12=100 2*45+10=100 2*46+8=100 2*47+6=100 2*48+4=100 2*49+2=100 number of solution: 16

$\endgroup$
  • $\begingroup$ How will the side lengths $1,1,98$ form a triangle? $\endgroup$ – ronno Aug 4 '13 at 10:42
  • $\begingroup$ thx , i forgot something $\endgroup$ – user52413 Aug 4 '13 at 11:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.