So I need help with this problem:

How many different non-congruent isosceles triangles can be formed by connecting three of the dots in a 4x4 square array of dots like the one shown below?

My work, is that I tried drawing different isoceles triangles and came up with the answer 9. I inputted it and it was incorrect.

So I went to look to see if there was an original post about this question, which I found, and I realized that I had the same exact answer as this post: https://web2.0calc.com/questions/please-help_50163.

Can somebody help? Thanks!

Guest Dec 22, 2019

edited by
Guest
Dec 22, 2019

#1**+1 **

This is an AoPS problem, isn't it?

I got 9 as well. But it turns out the answer is 11:

This is the solution if you want to see their reasoning:

We can construct segments of length \( 1, \sqrt 2, \sqrt 5, \sqrt {10}, 2, 2\sqrt 2, \sqrt {13}, 3, 3\sqrt 2\) (these can be obtained systematically by considering all lengths of segments from the top left dot to other dots). The following results can be obtained either by inspection or analysis of slope.

For each of \( 1, \sqrt 5, \sqrt {13}, 3\), there are zero isosceles triangles having it as its base length.

For \(\sqrt{10}\) there is one.

For each of \(2\sqrt 2, 3\sqrt 2\) there are two.

For each of \( 2, \sqrt 2\) there are three.

This gives a total of 11 non-congruent isosceles triangles.

Guest Dec 22, 2019