I Tried Creating a New Math Theorem: A Weighted Fourth-Power Distance Identity

We all know famous theorems like the Pythagorean theorem.

a^2+b^2=c^2

It is simple, beautiful, and powerful.

That made me wonder:

Can we create a small but real mathematical theorem today?

Not just a random formula.

A theorem should have:

a clear statement,
a proof,
examples,
and a reason why it is interesting.

So I explored a geometry idea involving points on a circle and distances from any point in the plane.

The result is a small theorem I am calling:

Shaswat’s Weighted Moment-Circle Theorem

This theorem is about a hidden symmetry in the sum of fourth powers of distances.

The Main Idea

Imagine some points placed on a circle.

Now take any point P anywhere in the plane.

From P, measure the distance to every point on the circle.

Usually, these distances depend on the direction of P.

But under a special balance condition, something interesting happens:

The weighted sum of the fourth powers of those distances depends only on how far P is from the center of the circle.

It does not depend on the direction of P.

That means the value is rotationally symmetric.

The Theorem

Let the points be:

A_1,A_2,\ldots,A_n

These points lie on a circle of radius R, centered at O.

Assign positive weights:

w_1,w_2,\ldots,w_n

Let the total weight be:

W=w_1+w_2+\cdots+w_n

Represent each point A_i using a complex number u_i, with the center O as the origin.

So every point satisfies:

|u_i|=R

Now suppose the weighted first and second moments vanish:

\sum_{i=1}^{n} w_i u_i = 0

and

\sum_{i=1}^{n} w_i u_i^2 = 0

Then, for any point P in the plane with OP = ρ, we have:

\sum_{i=1}^{n} w_i PA_i^4 W(R^4+4R^2\rho^2+\rho^4)

In Simple Words

If the points on the circle are balanced strongly enough, then:

\sum w_i PA_i^4

depends only on the distance of P from the center.

It does not depend on where around the circle P is located.

So if two points P and Q are the same distance from the center O, then:

\sum w_i PA_i^4 \sum w_i QA_i^4

That is the hidden symmetry.

Why This Is Interesting

For regular polygons, distance identities like this are already known.

For example, if the points are the vertices of a regular polygon, the symmetry is expected.

But this theorem explains the phenomenon using moment conditions.

Instead of saying:

The points must form a regular polygon.

We say:

The weighted first and second complex moments must vanish.

That gives a more flexible condition.

It can include some non-regular weighted arrangements too.

Proof

Let the complex coordinate of point P be p.

Since OP = ρ, we have:

|p|=\rho

Each point A_i has complex coordinate u_i, and since all A_i lie on the circle of radius R:

|u_i|=R

The squared distance between P and A_i is:

PA_i^2 = |p-u_i|^2

Expanding this:

|p-u_i|^2 |p|^2+|u_i|^2-p\overline{u_i}-\overline{p}u_i

Since |p| = ρ and |u_i| = R, we get:

PA_i^2 \rho^2+R^2-p\overline{u_i}-\overline{p}u_i

Let:

T=\rho^2+R^2

and

Q_i=p\overline{u_i}+\overline{p}u_i

Then:

PA_i^2=T-Q_i

Therefore:

PA_i^4=(T-Q_i)^2

Now sum with weights:

\sum w_iPA_i^4 \sum w_i(T-Q_i)^2

Expanding:

\sum w_iPA_i^4 \sum w_iT^2 2T\sum w_iQ_i + \sum w_iQ_i^2

Since T is constant:

\sum w_iT^2 = WT^2

Now look at the middle term:

\sum w_iQ_i \sum w_i(p\overline{u_i}+\overline{p}u_i)

So:

\sum w_iQ_i p\sum w_i\overline{u_i} + \overline{p}\sum w_i u_i

By assumption:

\sum w_i u_i=0

Taking conjugates gives:

\sum w_i\overline{u_i}=0

Therefore:

\sum w_iQ_i=0

So the middle term disappears.

Now calculate the last term:

Q_i^2= (p\overline{u_i}+\overline{p}u_i)^2

Expanding:

Q_i^2 p^2\overline{u_i}^{\,2} + 2|p|^2|u_i|^2 + \overline{p}^{\,2}u_i^2

Now sum with weights:

\sum w_iQ_i^2 p^2\sum w_i\overline{u_i}^{\,2} + 2|p|^2\sum w_i|u_i|^2 + \overline{p}^{\,2}\sum w_i u_i^2

By the second assumption:

\sum w_i u_i^2=0

Taking conjugates:

\sum w_i\overline{u_i}^{\,2}=0

Also:

|p|^2=\rho^2

and

|u_i|^2=R^2

So:

\sum w_i|u_i|^2 \sum w_iR^2 WR^2

Therefore:

\sum w_iQ_i^2 2\rho^2WR^2

Now substitute back:

\sum w_iPA_i^4 WT^2+2\rho^2WR^2

Since:

T=\rho^2+R^2

we get:

\sum w_iPA_i^4 W(\rho^2+R^2)^2+2W\rho^2R^2

Expanding:

\sum w_iPA_i^4 W(\rho^4+2\rho^2R^2+R^4+2\rho^2R^2)

So:

\sum w_iPA_i^4 W(R^4+4R^2\rho^2+\rho^4)

Hence proved.

Example: Regular Triangle

Take an equilateral triangle on a circle of radius R.

Let all weights be equal to 1.

Because of symmetry:

u_1+u_2+u_3=0

and

u_1^2+u_2^2+u_3^2=0

So the theorem applies.

Here:

W=3

Therefore, for any point P with OP = ρ:

PA_1^4+PA_2^4+PA_3^4 3(R^4+4R^2\rho^2+\rho^4)

So if P moves around a circle centered at O, the value stays constant.

Only the distance from the center matters.

Example: Square

For a square centered at O, again the first and second moments vanish.

So for four vertices:

W=4

Hence:

PA_1^4+PA_2^4+PA_3^4+PA_4^4 4(R^4+4R^2\rho^2+\rho^4)

This is another clean version of the identity.

Locus Interpretation

Suppose we want all points P such that:

\sum w_iPA_i^4=C

Using the theorem:

W(R^4+4R^2\rho^2+\rho^4)=C

This only depends on ρ, which is the distance of P from the center.

So the locus is a circle centered at O, if a real solution exists.

That is beautiful because a complicated-looking distance equation becomes a simple circular locus.

What Makes This Theorem Useful?

This theorem gives a compact way to understand fourth-power distance sums.

It connects:

geometry,
complex numbers,
weighted averages,
symmetry,
and moments.

The result also gives a simple test:

\sum w_i u_i=0

and

\sum w_i u_i^2=0

If both conditions hold, then the fourth-power distance sum becomes radial.

That means the expression becomes much easier to analyze.

Is This Completely New?

I want to be careful here.

Many formulas about distances to regular polygons and points on a circle already exist.

So I am not claiming that every part of this is historically new.

A safer and more honest claim is:

This is a weighted moment-based formulation of a fourth-power distance identity on the circle.

The regular polygon case is known, but this moment-condition version gives a nice generalization and a clean proof.

Before publishing this as a formal research paper, it should be checked by a mathematics professor or someone experienced in geometry/algebra.

But as a mathematical note, learning experiment, or blog post, it is a valid and interesting result.

Final Theorem Again

Let A_i be points on a circle of radius R, represented by complex numbers u_i, and let w_i > 0.

If:

\sum w_i u_i=0

and

\sum w_i u_i^2=0

then for any point P with OP = ρ:

\sum w_iPA_i^4 W(R^4+4R^2\rho^2+\rho^4)

where:

W=\sum w_i

That is the theorem.

Closing Thought

The beautiful thing about mathematics is that even simple objects like a circle can hide deep symmetry.

Sometimes, creating a theorem is not about finding something extremely complicated.

It is about noticing a pattern, stating it clearly, proving it, and connecting it to existing ideas.

This was my attempt at doing that.

Thanks for reading.

I Tried Creating a New Math Theorem: A Weighted Fourth-Power Distance Identity

I Tried Creating a New Math Theorem: A Weighted Fourth-Power Distance Identity

Shaswat’s Weighted Moment-Circle Theorem

The Main Idea

The Theorem

In Simple Words

Why This Is Interesting

Proof

Example: Regular Triangle

Example: Square

Locus Interpretation

What Makes This Theorem Useful?

Is This Completely New?

Final Theorem Again

Closing Thought

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