What is n???

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Rushil

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Rushil

#1 h Oct 4, 2005, 12:04 AM • 2 Y

Y by Adventure10, Mango247

Let $A_1, A_2, \ldots, A_n$ be an $n$ -sided regular polygon. If $\frac{1}{A_1 A_2} = \frac{1}{A_1 A_3} + \frac{1}{A_1A_4}$ , find $n$ .

This post has been edited 1 time. Last edited by Rushil, Dec 3, 2005, 2:48 PM

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shobber

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shobber

#2 h Nov 26, 2005, 2:49 PM • 2 Y

Y by Adventure10, Mango247

$n$ should be $7$ . We can get $A_1A_4=A_1A_5$ from the equation.

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varun

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varun

#3 h Dec 1, 2005, 6:14 AM • 2 Y

Y by Adventure10, Mango247

How do you get it.

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shobber

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shobber

#4 h Dec 1, 2005, 11:26 AM • 2 Y

Y by Adventure10, Mango247

varun wrote:

How do you get it.

Ptolemy. Use the fact that $A_1A_2=A_2A_3$ and $A_1A_3=A_2A_4$ and $A_1A_4=A_2A_5$ and so on...

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shyong

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shyong

#5 h Dec 1, 2005, 11:49 AM • 2 Y

Y by Adventure10, Mango247

So what is $A_1$ , $A_2$ anyway ?

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varun

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varun

#6 h Dec 1, 2005, 11:53 AM • 2 Y

Y by Adventure10, Mango247

shobber wrote:

varun wrote:

How do you get it.

Ptolemy. Use the fact that $A_1A_2=A_2A_3$ and $A_1A_3=A_2A_4$ and $A_1A_4=A_2A_5$ and so on...

I want to know how you prove $A_1A_4=A_1A_5$

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FMako

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FMako

#7 h Dec 2, 2005, 3:06 AM • 2 Y

Y by Adventure10, Mango247

shobber wrote:

varun wrote:

How do you get it.

Ptolemy. Use the fact that $A_1A_2=A_2A_3$ and $A_1A_3=A_2A_4$ and $A_1A_4=A_2A_5$ and so on...

You are assuming that the $n$ -gon is regular.

$A_xA_y = z$ , where $z$ is the distance between the points $A_x$ and $A_y$ .

Correct me, if I am wrong.

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varun

270 posts

varun

#8 h Dec 2, 2005, 3:22 AM • 1 Y

Y by Adventure10

FMako wrote:

shobber wrote:

varun wrote:

How do you get it.

Ptolemy. Use the fact that $A_1A_2=A_2A_3$ and $A_1A_3=A_2A_4$ and $A_1A_4=A_2A_5$ and so on...

You are assuming that the $n$ -gon is regular.

$A_xA_y = z$ , where $z$ is the distance between the points $A_x$ and $A_y$ .

Correct me, if I am wrong.

You are correct, but still no satisfactory answer has been provided.

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anjani_kumarsingh

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anjani_kumarsingh

#9 h Dec 2, 2005, 9:26 AM • 1 Y

Y by Adventure10

Are $A_1 , A_2 ,....$ the vertices of a regular polygon. If so the the problem becomes very easy.
Rushil, can we use Calculus in solving the problems at RMO.

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varun

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varun

#10 h Dec 2, 2005, 9:39 AM • 1 Y

Y by Adventure10

yes they are vertices.
But, I don't think use of calculus is allowed.
anyway you asked Rushil.

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Rushil

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Rushil

#11 h Dec 2, 2005, 12:03 PM • 2 Y

Y by Adventure10, Mango247

I dont really know whether Calculus is allowed at RMO - the local RMO coordinator said it was OK ( to a 'limited' extenet) but one of my friends says its use is not fully credited even though it is allowed!

So the best answer is : Calculus is allowed but you wont get full marks on solving the whole question using Calculus!

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leepakhin

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leepakhin

#12 h Dec 2, 2005, 2:46 PM • 2 Y

Y by Adventure10, Mango247

I think the problem is true only when the polygon is regular.
As we all know, a regular heptagon satisfy the conditions. Let the vertices be $A_1$ to $A_7$ . If we add as many points as we like on the side of $A_1A_4$ not containing $A_2$ and $A_3$ , the equation $\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$ is still valid. Therefore the polyogn must be regular.

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varun

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varun

#13 h Dec 2, 2005, 3:41 PM • 2 Y

Y by Adventure10, Mango247

I know the answer is $7$ .
This is a problem of Indian National Mathematics Olympiad.
You have to prove why it is satisfied for $n=7$

I did it with one method. Wanted to know other solutions and so I've asked it.

Please prove it.

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leepakhin

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leepakhin

#14 h Dec 3, 2005, 12:03 PM • 2 Y

Y by Adventure10, Mango247

Until now, nobody has ever posted their full solutions so I am posting mine.

(The polygon should be regular).
Obviously the polygon has at least 5 sides. Since it is regular, it is cyclic.
Let $A_1A_2=A_2A_3=A_3A_4=A_4A_5=x$ , $A_1A_3=y$ , $A_1A_4=z$ .
$\frac{1}{x}=\frac{1}{y}+\frac{1}{z}$
$\Rightarrow yz=zx+xy$
By Ptolemy's Theorem (applying to quadrilateral $A_1A_3A_4A_5$ ),
$A_1A_5\cdot x+xy=A_3A_5\cdot z=yz=zx+xy$
$\Rightarrow A_1A_5=z=A_1A_4$
Therefore, $A_1$ lies on the perpendicular bisector of $A_4A_5$ which passes the centre of the circle.
$\therefore$ The polygon has $7$ sides.

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Rushil

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Rushil

#15 h Dec 3, 2005, 2:31 PM • 2 Y

Y by Adventure10, Mango247

Of course there's the 'long, dirty and algebraic solution' using the expression for sides and the lengths involved in terms of angles they subtend at centre of polygon! Of course, that's for the regular polygon! I'll check what the problem says - regular or any polygon!?!?!?!?

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Arne

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Arne

#16 h Dec 3, 2005, 2:39 PM • 2 Y

Y by Adventure10, Mango247

The problem can't possibly be true for ANY polygon. I'm SURE they meant a regular polygon.

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Rushil

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Rushil

#17 h Dec 3, 2005, 2:47 PM • 2 Y

Y by Adventure10, Mango247

Ya! I feel the same way! I think I'll change the statement now! If someone else finds the result true for any polygon, or the correct verified statement , please tell!!

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shobber

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shobber

#18 h Dec 3, 2005, 3:03 PM • 2 Y

Y by Adventure10, Mango247

I just checked one of my book. The $\Longleftarrow$ part was a problem of 21st All USSR Math Olympiad.

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xirti

607 posts

xirti

#19 h Dec 27, 2005, 4:11 PM • 1 Y

Y by Adventure10

Try this one:

Find all regular polygons $A_1A_2..A_n$ such that $\frac{1}{A_1A_2}=\sum_{k=2}^p \frac{1}{A_1A_k}$ for some p.

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sayantanchakraborty

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sayantanchakraborty

#20 h Dec 23, 2013, 9:15 AM • 2 Y

Y by Adventure10, Mango247

WLOG let us assume that the circumradius of the regular n gon be 1.
Let the angle subtended by each side onto the center be $\theta$
Then applying the law of sines to triangles $OA_1A_2,OA_1A_3,OA_1A_4$ we get
$A_1A_2=2sin(\theta/2)$
$A_1A_3=2sin(\theta)$
$A_1A_4=2sin(3\theta/2)$

Plugging these back in the equation gives
$cosec(\theta/2)=cosec(\theta)+cosec(3\theta/2)$
Simple manipulations of this yeild $\theta=2\pi/7$ whence $n=7$ .

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nik9796

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nik9796

#21 h Dec 24, 2013, 5:59 PM • 2 Y

Y by Adventure10, Mango247

It can be solved using complex numbers.
Let Ao be the centre of the regular polygon which gives us Ar=Ao+|A1-Ao|e^2i(r-1)(pi)/n
Now |Ar-A1|=|A1-Ao|2sin2(r-1)(pi)/n
Putting this in given identity and simplifying gives sin(3pi/n)=sin(4pi/n) which gives n=7

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nik9796

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nik9796

#22 h Dec 24, 2013, 6:03 PM • 2 Y

Y by Adventure10, Mango247

It can be solved using complex numbers.
Let Ao be the centre of the regular polygon which gives us Ar=Ao+|A1-Ao|e^2i(r-1)(pi)/n
Now |Ar-A1|=|A1-Ao|2sin2(r-1)(pi)/n
Putting this in given identity and simplifying gives sin(3pi/n)=sin(4pi/n) which gives n=7

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srramanujan

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srramanujan

#23 h Apr 29, 2021, 1:22 AM

Y by

nik9796 wrote:

It can be solved using complex numbers.
Let $A_o$ be the centre of the regular polygon which gives us $A_r=A_o+|A_1-A_o|e^{\frac{2i(r-1)(pi)}{n}}$
Now $|A_r-A_1|=|A_1-A_o|2\sin \frac{2(r-1)(pi)}{n}$
Putting this in given identity and simplifying gives $\sin\frac{ 3pi}{n}=\sin \frac{4pi}{n}$ which gives $n=7$

This post has been edited 3 times. Last edited by srramanujan, Apr 29, 2021, 1:25 AM
Reason: edit

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srramanujan

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srramanujan

#24 h Apr 29, 2021, 1:48 AM

Y by

Number of diagonals in $n-gon$ also tells the number of quadrilaterals,right?
So starting with minimal $n=4$ , this is not a possibility because then $A_1A_2=A_1A_4$ which we lead $\dfrac{1}{A_1A_3}=0$ which is pretty inappropriate.

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