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18 Mar 2026, 10:41
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Hi everyone,
I noticed a recurring confusion (including mine) regarding counting triangles in polygons, especially in DS questions.
There are TWO very different questions that often get mixed up:
1) Triangulation of an n-sided polygon
This asks: Into how many triangles can the polygon be split WITHOUT overlap?
Rule:
Number of triangles = n − 2
Examples:
• Square (4 sides) → 2 triangles
• Pentagon (5 sides) → 3 triangles
2) Total number of triangles formed by choosing vertices
This asks: How many different triangles can be formed using the polygon’s vertices?
Rule:
Number of triangles = C(n,3)
Example:
Pentagon:
C(5,3) = 10 triangles
Important DS trap:
• n − 2 counts decomposition triangles
• C(n,3) counts ALL possible triangles (many overlap)
Similarly:
• Number of quadrilaterals = C(n,4)
→ A pentagon has C(5,4) = 5 quadrilaterals
This distinction helped me avoid wrong conclusions in DS where the question asks for “total number” vs “minimum decomposition”.
Hope this helps others avoid the same trap.
Happy to hear if anyone has seen this tested in a tricky way.
I noticed a recurring confusion (including mine) regarding counting triangles in polygons, especially in DS questions.
There are TWO very different questions that often get mixed up:
1) Triangulation of an n-sided polygon
This asks: Into how many triangles can the polygon be split WITHOUT overlap?
Rule:
Number of triangles = n − 2
Examples:
• Square (4 sides) → 2 triangles
• Pentagon (5 sides) → 3 triangles
2) Total number of triangles formed by choosing vertices
This asks: How many different triangles can be formed using the polygon’s vertices?
Rule:
Number of triangles = C(n,3)
Example:
Pentagon:
C(5,3) = 10 triangles
Important DS trap:
• n − 2 counts decomposition triangles
• C(n,3) counts ALL possible triangles (many overlap)
Similarly:
• Number of quadrilaterals = C(n,4)
→ A pentagon has C(5,4) = 5 quadrilaterals
This distinction helped me avoid wrong conclusions in DS where the question asks for “total number” vs “minimum decomposition”.
Hope this helps others avoid the same trap.
Happy to hear if anyone has seen this tested in a tricky way.
18 Mar 2026, 23:51
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This is a solid breakdown and the distinction really does matter more than people realize when they're under test conditions.
Your two cases are exactly right:
- n - 2 = minimum triangulations (no overlaps, partition/decomposition mode)
- C(n,3) = total possible triangles using any 3 vertices (many overlap)
One thing worth adding: C(n,3) assumes no three vertices are collinear. For a convex polygon this is always true, but the GMAT sometimes describes "points on a circle" or other configurations — always check that the problem specifies convex. Almost every GMAT problem does, but it's worth having in the back of your head.
The DS trap you named is real. Questions that go something like "how many triangles can be formed from the vertices of a polygon with n sides" will use C(n,3), while "how many non-overlapping triangles can it be divided into" uses n-2. The word "formed" vs "divided into" is the signal. I've gotten burned on this before precisely because I saw "triangles" and went on autopilot with n-2.
One extension that's also worth knowing: for quadrilaterals, C(n,4) counts all overlapping ones. For a hexagon (6 vertices), C(6,4) = 15 quadrilaterals total, but only 3 non-overlapping quadrilateral decompositions. Same logic, bigger numbers.
The "total number" vs "minimum decomposition" language is the thing to lock in. Once you have that reflex, these DS questions become pretty mechanical.
Your two cases are exactly right:
- n - 2 = minimum triangulations (no overlaps, partition/decomposition mode)
- C(n,3) = total possible triangles using any 3 vertices (many overlap)
One thing worth adding: C(n,3) assumes no three vertices are collinear. For a convex polygon this is always true, but the GMAT sometimes describes "points on a circle" or other configurations — always check that the problem specifies convex. Almost every GMAT problem does, but it's worth having in the back of your head.
The DS trap you named is real. Questions that go something like "how many triangles can be formed from the vertices of a polygon with n sides" will use C(n,3), while "how many non-overlapping triangles can it be divided into" uses n-2. The word "formed" vs "divided into" is the signal. I've gotten burned on this before precisely because I saw "triangles" and went on autopilot with n-2.
One extension that's also worth knowing: for quadrilaterals, C(n,4) counts all overlapping ones. For a hexagon (6 vertices), C(6,4) = 15 quadrilaterals total, but only 3 non-overlapping quadrilateral decompositions. Same logic, bigger numbers.
The "total number" vs "minimum decomposition" language is the thing to lock in. Once you have that reflex, these DS questions become pretty mechanical.
19 Mar 2026, 10:52
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Great addition, thank you Edskore!
Edskore
