Solutions (and a couple of
additional questions): Return to my Puzzle pages
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© Copyright 2002, Jim Loy
Solutions (and a couple of
additional questions):
1. Is it possible to divide an equilateral triangle into two equal (same area) triangles, which are not congruent to each other? Of course not, the diagram shows the only way to dissect an equilateral triangle into two equal triangles, and they are congruent.
2. Divide an equilateral triangle into
three equal triangles, none of which is congruent to any of the others. The
diagram on the right (or some equivalent reflection or rotation) shows the only
two ways.
3. Divide an equilateral triangle into
four equal triangles, none of which is congruent to any of the others. On the
left are six ways to do it. I have labeled two of them A and B, to help with
the solution of question #4. Let me pose a fifth question: Question #5 - Is
there any other way to divide our equilateral triangle into four
equal triangles with none of them being congruent? The answer to this question
is below.
4. Divide an equilateral triangle into
five equal triangles, none of which is congruent to any of the others. On the
right are two ways, similar to A and B in the solution #3. Of course, there are
many other solutions.
5. Here is a seventh solution to question
#3. I think that there is no eighth solution.
Let me ask a question #6: Do the two methods in questions 3 and 4, labeled A and B, always work? In other words, can we divide an equilateral triangle into n equal triangles, using those two methods, no two triangles being congruent? The answer to that is below.
6. Yes, both methods work for all
integers n. The only difficulty is proving that no two triangles are
congruent. You might want to think about that. Examining method A, we see that
the smallest angle of each triangle (labeled a, b, c, . . . on the right)
follows this relationship: a<b<c<d<e< . . . In other words, none
of the smallest angles is equal, and the triangles are not congruent. The proof
may or may not be easy?
Examining method B, we see that two
triangles (or more) are congruent only when one of the sides is vertical (first
diagram on the left), or if one of the triangles is isosceles (second diagram
on the left). I think the Pythagorean theorem will show that the colored
triangle in each diagram cannot have the correct area.
Return to the original puzzle.