Substitute for finding the hypotenuse of a right triangle?

Question

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:

1. 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?

2. Are there any more possible practical applications of this theorem (like in geometric proofs)?

3. (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

Answer 1

1. Yes, in the general case it is applicable but difficult to apply as compared to Pythagorean theorem. Because the Pythagorean theorem is straight forward to apply while your theorem involves root finding of quadratic equation which is not simple in each case.
2. It can be applied in simple cases & geometric proofs but in general cases it may create lengthy calculations.
3. Yes, but for publishing this article, first you will have to submit manuscript for review by peers of journal if it fulfills the standards of particular journal, it can be published. Because journals have their own rules/terms/standards to publish articles if some reject this article then some other may accept this up-to the defined level.

Answer 2

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.