Twelve points are spaced evenly around a circle, lettered

AAPL wrote:Magoosh

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

OA B.

Let's draw a circle with 12 equally spaced points, named A through L.

Originating from the point A, make all the possible isosceles triangles: ∆ABL, ∆ACK, ∆ADJ, ∆AEI, and ∆AFH

So, there are 5 isosceles triangles with vertex A. Since there are 12 points, we have 12*5 = 60 isosceles triangles

Note that ∆AEI is an equilateral triangle, thus, is counted thrice: once taking A as the vertex; once taking E as the vertex; and once taking I as the vertex, thus, we must remove 2 equilateral triangles.

Since 3 points make 2 extra equilateral triangles, 12 points will make (12/3)*2 = 8 equilateral triangles.

Thus, the actual number of isosceles triangles, incl. equilateral triangles = 60 – 8 = 52

The correct answer: B

Hope this helps!

-Jay
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