All of us know the way to calculate the hypotenuse of a right triangle: Using the Pythagorean Theorem.
I came up with a substitute to this. Let the shortest leg of the right triangle be '$a$' units, and the relatively bigger leg be '$b$' units, where $a < b < \text{Hypotenuse}$.
My theorem is that in such a case, there will always be a variable '$x$' such that:
$a^2 – x^2 =2bx$.
As for the applications of this statement, we can find the hypotenuse of the triangle by finding the positive root of this equation, and then added to the biggest leg to obtain the value of the hypotenuse. I derived a few basic conclusions from The Dickson's method of generating Pythagorean Triples, to prove this.
Let us give it a try:
Let $a = 5$, $b = 12$.
$a^2-x^2=2bx$.
$25-x^2=24x$
$x^2+24x-25=0$
$x=1,-5$.
Therefore hypotenuse $= 12+1=13$, which satisfies Pythagorean Theorem.
This satisfies only the triangles which have unequal sides.
Obviously, we don't need a substitute for isosceles right triangles since given the legs '$a$', the hypotenuse is $a\sqrt{2}$.
My questions are:
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Can the difficulty level of application of this theorem be compared to that of Pythagoras? If so, in the general case, is this difficult or easy to apply?
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Are there any more possible practical applications of this theorem (like in geometric proofs)?
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(Please forgive my imperiousness) Can I get this published somewhere?
I would like a detailed answer on how better or more tedious this is than the Pythagorean Theorem, and one practical example (like a geometric proof) involving this theorem.
Thanks,
Sandeep
Best Answer
Your equation: $$ a^2-x^2=2bx\Longrightarrow x^2+2bx-a^2=0 $$ So, $$ x=-b\pm\sqrt{b^2+a^2}, $$ and $x_2=-b+\sqrt{a^2+b^2}$ is a positive root. By your statement, $$ c=b+x_2=\sqrt{a^2+b^2}. $$ And what? If you want to find hypotenuse, you need evaluate square root in any way.