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.

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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.

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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:

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

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.