Reducing Sides in Convex Polygons

A convex polygon is always equivalent to a polygon with fewer sides that has the same area.

This principle is based on the fact that the number of sides in a polygon can be reduced through decomposition and rearrangement.

Of course, the triangle is an exception to this rule, as a polygon with only two sides does not exist. By definition, a polygon must have at least three sides.

This concept is particularly useful for demonstrating that in plane geometry, changing the shape of a polygon does not alter its area.

Note: This can be applied to both convex and concave polygons. However, it is much simpler and more intuitive for convex polygons. In concave polygons, the process might require more complicated cuts and rearrangements to manage indentations.

How to Construct an Equivalent Polygon with One Less Side

There are various ways to reduce the number of sides.

For example, by decomposing the polygon into smaller polygons and reassembling it into a different shape.

One of the simplest methods involves using the diagonals of the polygon.

For instance, consider any polygon (regular or irregular). Here, we use a hexagon as an example.

an example of a hexagon

Draw a diagonal between two points with an intermediate point.

For example, the diagonal line between A and C with B as the intermediate point.

the diagonal between two points

Translate the diagonal line so that it passes through the intermediate point B.

the translation of the diagonal

Finally, extend one of the segments ending at A or C until it intersects the line.

For example, extend the segment AF until it reaches point A'.

extend one of the sides adjacent to A or C

Then, draw a segment A'C connecting point A' with point C, eliminating vertex B from the polygon.

the new side A'C

The final result is a polygon A'CDEF with five sides, a pentagon.

This new pentagon is equivalent to the initial hexagon, meaning it has the same area.

the pentagon

Proof: To demonstrate this, observe the triangles AA'C and ABC.
triangles AA'C and ABC
These two triangles have the same base (AC) and the same height (AA'≅BB'). Therefore, according to the congruence theorem of triangles, they are equivalent triangles, meaning they have the same area.
the two triangles are equivalent
At this point, it is enough to observe that the hexagon is composed of the polygon ACDEF plus the triangle ABC. $$ ABCDEF \doteq ACDEF + ABC $$ The pentagon A'CDEF is composed of the polygon ACDEF plus the triangle AA'C. $$ A'CDEF \doteq ACDEF + AA'C $$ Knowing that triangles ABC and AA'C are equivalent, it follows that both the hexagon and the pentagon are equicomposed polygons, composed of the same equivalent polygons. $$ ABCDEF \doteq ACDEF + ABC \doteq ACDEF + AA'C \doteq A'CDEF $$ $$ ABCDEF \doteq A'CDEF $$ Therefore, the hexagon ABCDEF and the pentagon A'CDEF are equivalent.

By repeating the process, I can further reduce the pentagon A'CDEF into an equivalent quadrilateral C'DEF (4 sides).

the reduction from pentagon to quadrilateral

Finally, I can reduce the quadrilateral C'DEF into an equivalent triangle C'D'F (3 sides).

an equivalent triangle

From this, I deduce that every convex polygon can be reduced to an equivalent triangle.

Note: The method I described for reducing the number of sides using diagonals is not a standard method in Euclidean geometry. In practice, changing the number of sides of a polygon while maintaining the same area requires decomposition and rearrangement techniques that are not as straightforward as simply drawing or shifting diagonals.

Observations

Some observations and notes:

  • Concave Polygons
    In the case of concave polygons, reducing the number of sides while maintaining the area is not always possible in Euclidean geometry.

And so on.

 
 

Please feel free to point out any errors or typos, or share suggestions to improve these notes. English isn't my first language, so if you notice any mistakes, let me know, and I'll be sure to fix them.

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