A certain triangle ABC can be divided into 5 triangles each similar to ABC.
What are the angles A,B,C and how is the triangle divided?
If there is a line of the division coming out of a corner, say corner C then it must be true that two of the other angles add up to C. If A+B=C and since for any triangle A+B+C=180 we have
C + C = 180
C = 90
Which is not surprising since the altitude from the right angle always divides a right triangle into two similar right triangles. Any four such divisions of the original triangle would give us the 5 similar triangles sought.
We have two further cases to contend with:
1) A division line coming out of a corner with A+A=C
2) No division lines coming out of corners.
Case 1) seems to always lead to isosceles triangles. If we force the triangle to be isosceles with A=B then we have 4A=180 or A=45 with is a right triangle already covered above.
This is not fully resolved but I dont think it can yield 5 triangles in any other way.
Case 2) Seems able to yield divisions into other numbers besides 5 for ANY triangle. Connect the midpoints of the sides to get 4. Connect two sides with a segment parallel to the third side but only 1/3 the way up then make a little zig-zag to create a total of 6 triangles. 1/4 of the way can make 8 triangles. So we can create any even and increase any by 3. Thus ANY triangle can be split into any number of similar triangles except 2, 3, or 5.
Since the parallel sides are a must, this case will never work.
Posted by Jer
on 2011-09-06 13:01:30