Geometry

**Drexel28** · November 30th 2009, 08:20 PM

Hello, everyone. Of late I have been kind of interested in geometry. Here are two that I found interesting. They aren't particularly difficult but their result is nice.

Problem 1: Let $P_n$ be a regular $n$ -gon of side length $a$ . Prove that $\text{Area of }P_n=a^2\frac{n}{4}\cot\left(\frac{\pi}{n}\right)$

Spoiler:

Problem 2: Prove that no equilateral triangle lying in $\mathbb{R}^2$ may have all it's vertices on lattice point. Do this without Pick's theorem.

Good luck!

**Unbeatable0** · November 30th 2009, 09:18 PM

What about a stronger claim for your second problem?
Prove that no equilateral triangle lying in

may have all it's vertices on rational points (x and y rational).
This can be made even stronger, but only the wording of the stronger
argument will hint about a way to prove it

.

**Drexel28** · November 30th 2009, 09:20 PM

Originally Posted by Unbeatable0

What about a stronger claim for your second problem?
Prove that no equilateral triangle lying in

may have all it's vertices on rational points (x and y rational).
This can be made even stronger, but only the wording of the stronger
argument will hint about a way to prove it

.

Yes, the argument that I came up with at least furnishes a solution for this result. The wording, as you alluded to though, suggests the solution. Thus, I gave a weaker question in hopes of maybe disguising the answer.

**aman_cc** · December 1st 2009, 06:51 AM

Originally Posted by Drexel28

Yes, the argument that I came up with at least furnishes a solution for this result. The wording, as you alluded to though, suggests the solution. Thus, I gave a weaker question in hopes of maybe disguising the answer.

Without loss of generality (in the context of this problem), three points of the equilateral triangle are:

1. $(0,0)$
2. $(acos\theta,asin\theta)$
3. $(\frac{a}{2}(cos\theta-\sqrt3sin\theta),\frac{a}{2}(sin\theta-\sqrt3cos\theta))$

Here $a$ - side and $\theta$ angle the side makes with +ve x axiz

So if $acos\theta$ and $asin\theta)$ are rational, point #3 can't have rational co-ordinates.

Is this correct?

What was the more generic/stronger statement you guys were referring to?

**Drexel28** · December 1st 2009, 12:37 PM

Originally Posted by aman_cc

Without loss of generality (in the context of this problem), three points of the equilateral triangle are:

1. $(0,0)$
2. $(acos\theta,asin\theta)$
3. $(\frac{a}{2}(cos\theta-\sqrt3sin\theta),\frac{a}{2}(sin\theta-\sqrt3cos\theta))$

Here $a$ - side and $\theta$ angle the side makes with +ve x axiz

So if $acos\theta$ and $asin\theta)$ are rational, point #3 can't have rational co-ordinates.

Is this correct?

What was the more generic/stronger statement you guys were referring to?

I don't quite follow your argument friend. I don't doubt it, I just am a little confused.

P.S. What I was referring to was that my argument proved thta no equilateral triangle may have all rational coordinates. Also, once you have shown exactly what you mean in the above perhaps you would like to give a non-trigonometric argument.

**aman_cc** · December 1st 2009, 09:54 PM

Originally Posted by Drexel28

I don't quite follow your argument friend. I don't doubt it, I just am a little confused.

P.S. What I was referring to was that my argument proved thta no equilateral triangle may have all rational coordinates. Also, once you have shown exactly what you mean in the above perhaps you would like to give a non-trigonometric argument.

Ok. Let try to explain

Let (x1,y1), (x2,y2), (x3,y3) be an equilateral triangle will all rational points.
Surely then A=(0,0), B=(x2-x1,y2-y1), C=(x3-x1,y3-y1) is also an equilateral triangle will all rational points. Thus one point is at the origin.

So let length of a side of this triangle be 'a'. And $\theta$ be the angle which the side (meeting at A) you encounter first when you trace a ray from +ve x-axis in the anti-clockwise direction make with +ve x-axis. Angle the other side which meets at A makes is $\alpha=\theta + \pi/3$

So the co-ordinates of A,B,C are:
$(0,0)$
$(acos\theta,asin\theta)$
$(acos\alpha,asin\alpha)$

For point 3, substitute for $/alpha$ and expand in terms of $/theta$ . You will get the result I wrote earlier.
Now it is clear that if point 2 is rational then point 3 can't be rational.

Hence the result.

Convincing?
What's your approach please?

**Drexel28** · December 2nd 2009, 05:38 AM

Originally Posted by aman_cc

Ok. Let try to explain

Let (x1,y1), (x2,y2), (x3,y3) be an equilateral triangle will all rational points.
Surely then A=(0,0), B=(x2-x1,y2-y1), C=(x3-x1,y3-y1) is also an equilateral triangle will all rational points. Thus one point is at the origin.

So let length of a side of this triangle be 'a'. And $\theta$ be the angle which the side (meeting at A) you encounter first when you trace a ray from +ve x-axis in the anti-clockwise direction make with +ve x-axis. Angle the other side which meets at A makes is $\alpha=\theta + \pi/3$

So the co-ordinates of A,B,C are:
$(0,0)$
$(acos\theta,asin\theta)$
$(acos\alpha,asin\alpha)$

For point 3, substitute for $/alpha$ and expand in terms of $/theta$ . You will get the result I wrote earlier.
Now it is clear that if point 2 is rational then point 3 can't be rational.

Hence the result.

Convincing?

I understand now.

What's your approach please?

Spoiler:

**aman_cc** · December 2nd 2009, 06:04 AM

Originally Posted by Drexel28

I understand now.

Spoiler:

Thanks. This is essentially argument I gave. Just that I think putting it in trigonometry form made my solution look (luckily with no obvious intent to do that

) smaller and without requiring all the effort you had to do.

**Drexel28** · December 2nd 2009, 06:19 AM

Originally Posted by aman_cc

Thanks. This is essentially argument I gave. Just that I think putting it in trigonometry form made my solution look (luckily with no obvious intent to do that

) smaller and without requiring all the effort you had to do.

Yes, I do agree that yours was shorter. Also, I appreciate that you don't feel this is a competition to see who has the best solution. I did mine that way not because it wasn't easier to do trig, but because I found it slightly more "pretty". The concepts were more fluid than citing trig. Your solution, as I said, sure does have the advantage of compactness though

.

P.S. I cam up with three other solutions to this besides mine and yours if you are interested...two are geometric..one is not.

**aman_cc** · December 2nd 2009, 06:23 AM

Originally Posted by Drexel28

Yes, I do agree that yours was shorter. Also, I appreciate that you don't feel this is a competition to see who has the best solution. I did mine that way not because it wasn't easier to do trig, but because I found it slightly more "pretty". The concepts were more fluid than citing trig. Your solution, as I said, sure does have the advantage of compactness though

.

P.S. I cam up with three other solutions to this besides mine and yours if you are interested...two are geometric..one is not.

Sure !! I would love to read them. (PS: I have sent across my email to you as a pvt msg)

**Drexel28** · December 2nd 2009, 06:33 AM

Originally Posted by aman_cc

Sure !! I would love to read them. (PS: I have sent across my email to you as a pvt msg)

Maybe an idea would suffice for each and anyone who wishes can fill in the gaps.

1. Pick's theorem Pick's theorem - Wikipedia, the free encyclopedia tells us particularly that if the triangle in question had all lattice points that it's area would be rational. This is a problem though because....

2. We can use the basic ideas of finding the area of an equaliteral triangle in the plane to show that, similar to Pick's Theorem, the area must be rational.

3. Number theory! Having all sides lattice points creates a contradiction number theoretically.

**aman_cc** · December 2nd 2009, 07:00 AM

Originally Posted by Drexel28

Maybe an idea would suffice for each and anyone who wishes can fill in the gaps.

1. Pick's theorem Pick's theorem - Wikipedia, the free encyclopedia tells us particularly that if the triangle in question had all lattice points that it's area would be rational. This is a problem though because....

2. We can use the basic ideas of finding the area of an equaliteral triangle in the plane to show that, similar to Pick's Theorem, the area must be rational.

3. Number theory! Having all sides lattice points creates a contradiction number theoretically.

Logic for 1 and 2 -
Area is rational - Just use the std formula in co-ordinate geometry.
Also let side = a. So, its easy to see $a^2$ is rational. But area is also $(\sqrt3/4)a^2$ , which can't be rational. Hence proved

For (3) - Any hints on the direction to take plz?

**Drexel28** · December 2nd 2009, 07:05 AM

Originally Posted by aman_cc

Logic for 1 and 2 -
Area is rational - Just use the std formula in co-ordinate geometry.
Also let side = a. So, its easy to see $a^2$ is rational. But area is also $(\sqrt3/4)a^2$ , which can't be rational. Hence proved

For (3) - Any hints on the direction to take plz?

Consider that the squares of the distances between the three points must all be equal.

**Unbeatable0** · December 2nd 2009, 08:12 AM

Your solutions are ineed interesting.

And here, if you'd like, is the stronger argument I was referring to
(in spoiler tags because it does imply a way for a proof)

Note: This is not originally mine. It was taken from an Israeli math forum.

Spoiler:

**Drexel28** · December 2nd 2009, 08:19 AM

Originally Posted by Unbeatable0

Your solutions are ineed interesting.

And here, if you'd like, is the stronger argument I was referring to
(in spoiler tags because it does imply a way for a proof)

Note: This is not originally mine. It was taken from an Israeli math forum.

Spoiler:

Question:

Spoiler:

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