AIME 2017 I · 第 1 题

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 remove the degenerate triangles from this total. The degenerate triangles are just the triangles that are collinear. We can count the number of degenerate triangles on each side of $ABC$, then subtract it from $\binom{15}{3}$. For $\overline{AB}$, we have $\binom{5}{3}$ degenerate triangles (don't forget $A$ and $B$). Then, 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 $\binom{15}{3} - \binom{7}{3} - \binom{6}{3} - \binom{5}{3} = \boxed{390}$

-jb2015007 and cnwyshi (minor edits like grammar)

Solution 3 (Casework Ba\$h)

We will use casework based on the number of common vertices to solve this problem.

Case 1: No common vertices

There are two subcases for this solution, being three distinct sides, and two vertices on a side. This is because, for a triangle to have area 0, all three points must be on one side, so this scenario is possible.

First, to calculate the amount of triangles where all three vertices are on 3 distinct sides of a triangle, one just uses multiplication, for $5\times4\times3=60$.

The other case is a bit trickier, as there are three possible sides where the two common vertices could be; AC, AB and BC.

If the two vertices are on segment AC, where there are 5 points to choose from, the last point must be on a different segment. These can be combined in any way, so there are $(5, 2) \times 7$ ways to choose this case, or $70$.

If the two vertices are on segment AB, there are 3 points to choose two from, and 9 other points. This yields $(3, 2) \times 9 = 3 \times 9 = 27$ possibilities.

Finally, if the two vertices are on BC, there are 4 points to choose two from, and 8 others. This yields $(4, 2) \times 8 = 6 \times 8 = 48$ possibilities.

Adding all of the zero vertice cases together results in $60 + 70 + 27 + 48 = 205$ non-common-vertice solutions.

Case 2: One Common Vertice

There are three cases to work with here; one where the vertice is A, one where it is B, and one where it is C.

If the common vertice is A, and all other vertices are on distinct line segments (AB, BC, or CA), there is guaranteed to be a line segment where a vertice does not occur. This means that there are three subcases for this type, relative to the large case of a vertice on A: a vertice not being on AB (meaning vertices on BC and AC), a vertice not being on BC (meaning vertices on AC and AB) and a vertice not on AC (meaning vertices on AB and BC).

For these subcases, only multiplying the number of points on each line suffices, so for this case for A specifically, we have $(3\times5)+(4\times5)+(3\times4)=15+20+12=47$ ways.

This is true for all one-vertice distinct line segment cases (because of symmetry), so this case can be replicated three times, for $47 \times 3 = 141$ cases of 1-vertice distinct line segment.

However, it is also possible that two points are on the same line segment, which must be directly opposite to the vertice (because no 3 points can be on the same line). This means that there must be 2 dots on that line segment opposite the vertice. The line segments opposite vertice A, B, and C are BC, AC and AB respectively, which have 4, 5, and 3 dots respectively. This problem asks for us to choose 2 from four, five, and 3 and add them up, which gets $(4,2)+(5,2)+(3,2)=6+10+3=19$ cases.

In total, for one vertice cases, we have $141+19=160$ cases. This brings the aggregate number of cases to $205+160=365$ cases.

Case 3: Two Common Vertices

The only way a triangle could not have an area of greater than 0 is if all three of its vertices are on the same line. Two are already on the same line, so the last point cannot be on the same line as the other two. This is also the two vertice case, so the final point cannot be a vertice. Out of the 12 non-vertice points, 3, 4, or 5 of them will be unavailable, whether if B, A, or C is chosen, respectively. This gives $9+8+7=24$ cases for two-vertice solutions.

Adding this to the aggregate gives $365+24=389$ cases.

Case 4: Three Common Vertices

By trivial, this is the original triangle, which is still a case. This gives 1 three-vertice solution.

This brings the total number to $390$ cases, which is the final solution. $\boxed{390}$.

Video Solution

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