# Substitute for finding the hypotenuse of a right triangle?

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

• So, instead of the hypotenuse h, you prefer to look at h - b, where b is the longer of the two legs. Not harder than finding h, clearly. But of course you could also look at h + b or 3h - 79.8b or a + b + h and so on. To make it interesting, I'd say you had to find a strong geometric theorem that used your expression. – lulu Jul 22 '15 at 11:54
• If I'm right, that's what I'm asking... Is there any practical purpose of this theorem other than being able to find the hypotenuse? – Sandeep Jul 22 '15 at 12:00
• Well, I don't see one but of course there might be something. As a rule, though, I'd start with the Geometry (or maybe with the Number Theory) instead of just algebraically manipulating the definitions. You might, for example, take a look at the form a + b - h . That one has Geometry in it! It is the diameter of the inscribed circle. – lulu Jul 22 '15 at 12:52

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.
• But the Pythagorean Theorem will make us Ind the squares of two numbers, adding them, and finding their positive square root. Wouldn't both of these be approximately the same difficulty? – Sandeep Jul 22 '15 at 11:54
• Alright, but finding the roots of a quadratic equation is more difficult than simply taking sq. root of sum of squares of two known numbers i.e. legs of right triangle. – Harish Chandra Rajpoot Jul 22 '15 at 13:24
• You stated this can be applied in simple proofs. Can you give one case where it makes things easy and one case where it creates complications? – Sandeep Jul 25 '15 at 9:25
• Alright, this can be used to find out the hypotenuse if legs are known, one of unknown legs if hypotenuse & other leg are known & other acute angles in a triangle instead of using Pythagorean theorem – Harish Chandra Rajpoot Jul 25 '15 at 9:31

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.

• Isn't factoring the equation better than using the general formula? That way, you don't need to evaluate square root. And it always ends up with the equation being factorisable. – Sandeep Jul 26 '15 at 16:49
• @Sandeep, factoring? Ok, how do you factor $a^2 - x^2 - 2bx = 0$ in a general case? – Michael Galuza Jul 26 '15 at 16:51
• But 'a' and 'b' are given. And speaking from a statistical point of view, it is always factorisable. – Sandeep Jul 26 '15 at 16:53
• @Sandeep, yes, it is. By finding roots ))) Anyway, your method is correct, but useless. If I want to find hypotenuse, I evaluate square root. All I need is Pythagorean Theorem. Continue your studying and research! – Michael Galuza Jul 26 '15 at 16:58
• Thanks a lot for your 'valuable' advice. Just as you prefer the Pythagorean Theorem, I prefer my own. Let's not carry on the argument any further. As a second thought, you might guess for who the bounty is going to be... – Sandeep Jul 26 '15 at 17:14