# Application of the Pythagorean Theorem

#### andyimm

Greetings,

I was hoping you could help me prove the following:

For the case of all 3, 4, 5 special right triangles, the relationship of:

a/3+b=c

is always true.

Or can you find a counterexample? I believe it to be true always but am looking for a formal proof.

It is much easier to work with than:

a^2+b^2=c^2

Please prove me right Thanks much,

Andy

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#### Plato

MHF Helper
Greetings,
I was hoping you could help me prove the following:
For the case of all 3, 4, 5 special right triangles, the relationship of:
a/3+b=c
is always true.
Or can you find a counterexample? I believe it to be true always but am looking for a formal proof.
It is much easier to work with than:
a^2+b^2=c^2
Well it all depends upon what you mean by 3, 4, 5 special right triangles.
A $$\displaystyle 5,~12,~13$$ is clearly a right triangle that lacks that property
BUT does $$\displaystyle 3k,~4k,~5k,~k\in\mathbb{Z}^+$$ form right triangle with that property?

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#### ChipB

MHF Helper
I think what the OP is asking is to show that for any triplet a = 3k, b=4k, c = 5k it is true that a/3 + b = c.

Simple enough: Start by adding a/3 + b:

a/3+b = 3k/3 + 4k = 5k = c

Hence a/3+b =c.

As Plato points out this does not work for other examples of Pythagorean triples, such as (5, 12, 13). You might be tempted to think (a, c,b,c)= (5, 12, 13) has the property a/5+b = c, and hence the denominator of the 'a' term is always the value of 'a' for the base Pythagorean triplet - but that only works if the base triple has the property c = b+1, such as for the triplets (5, 12, 13), or (7, 24, 25), etc. It does not work if the base Pythagorean triplet has the property c = b+2, such as for the triplets (8, 15, 17), or (12, 35, 37), etc.

• 1 person