Wednesday, May 11, 2016

2 Fun 3-4-5 Pythagorean Triples

I'm going to warehouse these problems from @five_triangles here. I really like how they both show constructions for a 3-4-5  Pythagorean Triple. My plan is to keep looking for other variants on this theme. This is almost enough material to make a followup day after the Pythagorean Theorem although it still needs a few slightly easier problems to go with it.


Examine triangle DCO.
  • DC is R + r
  • DO is 2R - r
  • OC is R
Apply the Pythagorean Theorem:  $$ (R+r)^2 = (2R - r)^2 + R^2$$
After simplifying this reduces to $$3Rr = 2R^2$$ and then assuming  R is positive $$ r = \frac{2}{3} R$$ Or put another way: r is 1/3 of the side and R is 1/2 of the side. Plug those values back into the original triangle and voila you find the 3-4-5.


This construction is a bit trickier. 
  1. Note the congruent segments in the two big triangles
  2. Angle chase to find the 2 sets of similar triangles and the right angles.
  3. From the first set AFC and AFE and the given hypotenuse lengths You can immediately see the areas are in a ratio of 4:1 as is BF:EF since they share an altitude AF.
  4. Let H be the intersection of GC and the line from B. This is a right angle from the initial angle chase.  This second set EFG and EBH are also similar.   For EBH you know have a right triangle with a hypotenuse of 5x and one side of 4x from step 3 where x is the length of EF. The Pythagorean theorem immediately says EH must be 3x. 
  5. Its easy to do the 3:1 scaling if you want to find the area of EFG from there.

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