Geometrical Proof

Here is an interesting geometry problem-

Let \(A_1\), \(A_2 \ \cdots \ A_n\) be the vertices of a regular \(n\)-sided polygon \(P\). Let \(X\) be a random point on the incircle of \(P\).

Prove that \(\displaystyle\sum_{i=1}^{n} \overline{A_iX}^2=nR^2\left[1+\cos^2\left(\dfrac{\pi}{n}\right)\right]\).


Note : Here, \(R\) denotes the circumradius of \(P\).

Note by Pratik Shastri
3 years, 11 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

@KARAN SHEKHAWAT Here is the solution :

Let \(O\) represent the center of the polygon.

Now, \[\begin{align}\vec{XA_i}&=\vec{XO}+\vec{OA_i}\\ |\vec{XA_i}|^2&=|\vec{XO}|^2+|\vec{OA_i}|^2+2(\vec{XO} \cdot \vec{OA_i})\\ \sum_{i=1}^{n} |\vec{XA_i}|^2&=n(r^2+R^2)+2\sum_{i=1}^{n}(\vec{XO} \cdot \vec{OA_i})\\ &=n(r^2+R^2)+2\vec{XO} \cdot \left(\sum_{i=1}^{n} \vec{OA_i}\right)\end{align}\] Since \(\displaystyle\sum_{i=1}^{n} \vec{OA_i}=0\), \[\sum_{i=1}^{n} |\vec{XA_i}|^2=\sum_{i=1}^{n} \overline{A_iX}^2=n(R^2+r^2)\]

All that's left to do is to express \(r\) in terms of \(R\). That's a matter of simple trigonometry.

Note 1 : \(r\) is the inradius of \(P\).

Note 2 : \(\displaystyle\sum_{i=1}^{n}(\vec{XO} \cdot \vec{OA_i})=\vec{XO} \cdot \left(\sum_{i=1}^{n} \vec{OA_i}\right)\) due to the distributivity of the dot product over vector addition.

Pratik Shastri - 3 years, 10 months ago

Log in to reply

Awesome ! Thanks a lot ! It's is much better than complex number tecnique atleast for me :) You are genius !

Karan Shekhawat - 3 years, 10 months ago

Log in to reply

Can we use particular case of polygon in which vertex are n'th roots of of unity To prove it ?

Karan Shekhawat - 3 years, 10 months ago

Log in to reply

You could do that. We are to prove the general result which can also be, to my knowledge, proven that way.

Pratik Shastri - 3 years, 10 months ago

Log in to reply

I'm get stuck in that ! If you Don't mind Can you please Upload its solution So that we can learn from it ?

Thanks !

Karan Shekhawat - 3 years, 10 months ago

Log in to reply

@Karan Shekhawat Anyways I will upload it within a day or two.

Pratik Shastri - 3 years, 10 months ago

Log in to reply

@Karan Shekhawat Well, I didn't solve it using complex numbers, I used vectors.

Pratik Shastri - 3 years, 10 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...