Twelve points are spaced evenly around a circle, lettered from A to L. Let N be the total number of isosceles triangles, including equilateral triangles, that can be constructed from three of these points. A different orientation of the same lengths counts as a different triangle, because a different combination of points form the vertices. What is the value of N?

A. 48
B. 52
C. 60
D. 72
E. 120

refer to this topic https://gmatclub.com/forum/twelve-point ... l#p1812685

Regards

[quote="Carcass"]There are 12 points at EQUAL interval..
So the isosceles triangle can have EQUAL sides of length.
Let's work on the central vertex.

A as the vertex between EQUAL sides..
The EQUAL sides can be 1 unit, two unit, 3 unit, 4 unit or 5 unit...
6 unit will be the diameter so not possible.
So A as central vertex can make 5 isosceles triangles.

Thus 12 vertices from A to L can make 12*5=60 such isosceles triangles.

However 4 units results in an equilateral triangle.
So we are counting the same triangle THREE times.
So actual different triangles with 4 unit as length = 12/3=4..
So we have 12-4=8 extra triangles counted..


hello expert please explain
"However 4 units results in an equilateral triangle.
So we are counting the same triangle THREE times.
So actual different triangles with 4 unit as length = 12/3=4..
So we have 12-4=8 extra triangles counted "

I might be missing something but how does when know that a unit 4 will create a equilateral triangle

I don't even understand this solution. Please a detailed explanation would be good.

The solution provided above is from a quant moderator on Gmat Club

The image also is another explanation provided by an expert.

Will see if GreenLight will step in. I do no not know how to help more sir

I am sorry

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