Well-Known Fact ala Euclid: If two polygons are congruent, they must have the same area and the same perimeter.
Question A: If two polygons have the same area and the same perimeter, do they have to be congruent?
Note: Think of a trapezoid (e.g. red pattern block)...how could you cut it with one straight line and reassemble it to show that the answer to Question A is "No"?
Question B: If two triangles have the same area and the same perimeter, do they have to be congruent?
Source: M. Barabash's "A Non-Visual Counterexample in Elementary Geometry," College Mathematics Journal, Vov. 2005
Hint: For the trapezoid, drop a perpendicular from a top vertex to the base and flip the "cut-off triangle. A new quadrilateral is formed but its perimeter and area remain the same.
For the triangle case, think about the green equilaterial triangle....and an old "string technique" used by Descartes to draw an ellipse.
Solution Commentary: Marita Barabash suggests:
Isn't this a great argument? Refer to the full article where a fixed-area argument is also given.
- Start with equilateral triangle ABC (with area S) with base AB of length a
- Form new isosceles triangle AB'C' (with area S') by reducing base AB' to a-2x and making sides AC' and B'C' equal with length a+x
- WLOG, assume S>S' (the argument can easily be adjusted for the other direction as well)
- On triangle ABC, stretch a string of length 2a, starting at vertex A, through vertex C, and ending in vertex B...place a pencil point inside the string at vertex C
- As you move the pencil point (keeping the string taut), you form an infinite number of non-isosceles triangles with constant perimeter 3a but with areas decreasing from S to 0. One of the triangles formed must have an area equal to S' (and have same perimeter), but yet is not congruent to triangle AB'C'
And a side note: This proof provides a great opportunity to introduce students to the idea that the moving pencil traces 1/4 the perimeter of an ellipse...and a chance to explore WHY!