# Problem of the Day

Thirteen distinct points lie in a plane with exactly of the points lying on the same line and the other points non-collinear. If 276 unique triangles can be drawn using these points as vertices, what is the value of ?

A. 2

B. 3

C. 4

D. 5

E. 7

Reveal Answer

### Solution

[latexpage]

If all of the points were non-collinear, there would be $_{13}C_3=286$ unique triangles. However, any group of three points chosen that are collinear will fail to make a triangle. $286-{_{n}}C_{3}=276$. Solving this equation gives $n=5$.

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