Oh, that 3 sided polygon!

Geometry Level 3

How many isosceles triangles having a perimeter of $2015$ units, and an obtuse angle, with integer-side lengths are possible?

The answer is 87.

1 solution

Xuming Liang
Jan 17, 2016

On the isosceles triangle, label the congruent sides $a$ and the base $b$ . Since only the apex angle can be obtuse, postive integers $a,b$ are subjected to the following:

$2a+b=2015\\ 2a>b \\ 2a^2<b^2$

the second equation comes from the triangle inequality, while the third is the criterium for obtuse angle.

The first two equations give $a\ge 504$ .

The first and last equations give $a<\frac {2015}{2+\sqrt {2}}\implies a\le 590$ .

Hence $504\le a\le 590\implies$ there are $\boxed {87}$ values for $a$ , each of which corresponds to a unique triangle.

Maybe you can elaborate because many people wouldn't be able to understand. You can add that if one angle is obtuse then cosine of that angle is negative which gives the 1st inequality. We get 2nd inequality from value of cosine is always greater than or equal to -1.

Kushagra Sahni - 5 years, 5 months ago

Oh, that 3 sided polygon!

The answer is 87.

1 solution

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