2017 AIME I Problems/Problem 1

Problem 1

Fifteen distinct points are designated on $\triangle ABC$: the 3 vertices $A$, $B$, and $C$; $3$ other points on side $\overline{AB}$; $4$ other points on side $\overline{BC}$; and $5$ other points on side $\overline{CA}$. Find the number of triangles with positive area whose vertices are among these $15$ points.

Solution 1

Every triangle is uniquely determined by 3 points. There are $\binom{15}{3}=455$ ways to choose 3 points, but that counts the degenerate triangles formed by choosing three points on a line. There are $\binom{5}{3}$ invalid cases on segment $AB$, $\binom{6}{3}$ invalid cases on segment $BC$, and $\binom{7}{3}$ invalid cases on segment $CA$ for a total of $65$ invalid cases. The answer is thus $455-65=\boxed{390}$.

Solution 2

We simply choose $3$ points from $15$ to begin with since a triangle consists of $3$ points and there are $15$ points total. That gives us $\binom{15}{3}$. Now, we need to subtract of the degenerate triangles. These are just the triangles that are collinear. We can count the number of degenerate triangles on each side of $ABC$, then subtract is from $\binom{15}{3}$. For $\overline{AB}$, we have $\binom{5}{3}$ degenerate triangles (dont forget $A$ and $B$), for $\overline{BC}$, we have $\binom{6}{3}$ degenerate triangles, and for $\overline{AC}$, we have $\binom{7}{3}$ degenerate triangles. So, our final answer is $\binom{15}{3} - \binom{7}{3} - \binom{6}{3} - \binom{5}{3} = \boxed{390}$

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Video Solution

https://youtu.be/BiiKzctXDJg ~Shreyas S


See Also

2017 AIME I (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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