1. ## Geometry

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:
See if you can do it multiple ways: induction, calculus, or plain old geometry

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!

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 .

3. Originally Posted by Unbeatable0
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.

4. 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?

5. 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.

6. 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?

7. 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.

Spoiler:
I tried to do this from a purely coordinate geometric stand point. WLOG (like you said) assume that one of the vertices of our triangle is the orgin $O$. Thus we are really examining $\triangle OBC$, where $B,C$ are the other verticies. We may assume that $B=\left(2a,2b\right)$ for some $a,b\in\mathbb{Q}$. From here we break this into two cases, namely: $a,b\ne0$, or $a\text{ or }b=0$. Clearly both cannot be zero since the origin and $B$ are distinct points.

Case 1, $a\text{ or }b=0$: It is clear that we assume WLOG that $a=0$ and $b>0$ for all other cases are analgous. Thus this triangle will have $B=(0,2b)=(0,c)$ and $C=(x,y)$ for some $x,y$. Clearly seeing though that the length of $\overline{OB}$ is $|c|=c$ (since they are both on the y-axis) we may conclude that the length of $\overline{OC},\overline{BC}$ is as well. Realizing though that the lengths of $\overline{OC},\overline{BC}$ are given independently by the distance equations $d\left(O,C\right)=\sqrt{x^2+y^2},d\left(B,C\right) =\sqrt{x^2+(y-c)^2}$ and appealing once again to the equilaterallness of the triangle we may conclude that, in fact, $d\left(O,C\right)=\sqrt{x^2+y^2}=\sqrt{x^2+(y-c)^2}=d\left(B,C\right)$. Solving this equation yields $y=\frac{c}{2}$. As stated earlier though $d\left(O,B\right)=d\left(O,C\right)=d\left(B,C\rig ht)=c$ and focusing mainly on $d\left(O,C\right)$ we see that $\sqrt{x^2+y^2}=\sqrt{x^2+\frac{c^2}{4}}=c\implies x=\pm\frac{\sqrt{3}c^2}{4}$ which is clearly irrational for any rational $c$.

Case 2, $a,b\ne0$: Now we have that $B=(2a,2b)$ where neither $a$ or $b$ is zero. Constructing the midpoint $M$ of $\overline{OB}$ gives us that $M=(a,b)$ and that the length of the segment $\overline{OM}$ is $d\left(O,M\right)=\sqrt{a^2+b^2}$. Now, form $\overline{MC}$ to be the median connecting $M$ and $C$. If we consider the two segments $\overline{OB},\overline{MC}$ to be the restrictions of the lines $\ell_1,\ell_2$ respectively we can see that $\text{slope of }\overline{OB}=\frac{b}{a}$ and since $\overline{OB}\perp\overline{MC}$ we may conclue that $\text{slope of }\overline{MC}=\frac{-a}{b}$ (now you see why we needed $a,b\ne0$). Letting $C=(x,y)$ for some $x,y$ again and realizing that $M,C$ both lie on $\ell_2$ we may conclude that $C=\left(a+\Delta x,b+\Delta y\right)=\left(a+ \Delta x,b+\frac{-a}{b}\Delta x\right)\quad\color{red}\star$ where $\Delta x$ represents the change in $x$ between the points $M$ and $C$. Appealing to this we see that $\text{length of }\overline{MC}=\sqrt{\left(a+\Delta x-a\right)^2+\left(b+\frac{-a}{b}\Delta x-b\right)^2}=\left|\Delta x\right|\sqrt{1+\frac{a^2}{b^2}}$. Remembering though that forming the median $\overline{MC}$ bisected $\angle OBC$ we see that the new triangle formed, $\triangle OMC$ is a $30-60-90$ triangle. In particular, this tells us that $\text{length of }\overline{MC}=\sqrt{3}\cdot\text{length of }\overline{OB}=\sqrt{3}\cdot\sqrt{a^2+b^2}$. Equating our two lengths of $\overline{MC}$ gives us $\left|\Delta x\right|\sqrt{1+\frac{a^2}{b^2}}=\sqrt{3}\sqrt{a^2 +b^2}\implies \left|\Delta x\right|\sqrt{a^2+b^2}=\sqrt{3}\cdot b\cdot\sqrt{a^2+b^2}$ and since $a,b\ne0\implies\sqrt{a^2+b^2}\ne0$ we may conclude that $\left|\Delta x\right|=\sqrt{3}b\implies \Delta x=\pm\sqrt{3}b$. Thus, substitution into $\color{red}\star$ gives us that $C=\left(a\pm\sqrt{3}b,b\mp\sqrt{3}a\right)$ which is clearly irrational since $a,b\in\mathbb{Q}$.

The conclusion follows. $\square$

8. Originally Posted by Drexel28
I understand now.

Spoiler:
I tried to do this from a purely coordinate geometric stand point. WLOG (like you said) assume that one of the vertices of our triangle is the orgin $O$. Thus we are really examining $\triangle OBC$, where $B,C$ are the other verticies. We may assume that $B=\left(2a,2b\right)$ for some $a,b\in\mathbb{Q}$. From here we break this into two cases, namely: $a,b\ne0$, or $a\text{ or }b=0$. Clearly both cannot be zero since the origin and $B$ are distinct points.

Case 1, $a\text{ or }b=0$: It is clear that we assume WLOG that $a=0$ and $b>0$ for all other cases are analgous. Thus this triangle will have $B=(0,2b)=(0,c)$ and $C=(x,y)$ for some $x,y$. Clearly seeing though that the length of $\overline{OB}$ is $|c|=c$ (since they are both on the y-axis) we may conclude that the length of $\overline{OC},\overline{BC}$ is as well. Realizing though that the lengths of $\overline{OC},\overline{BC}$ are given independently by the distance equations $d\left(O,C\right)=\sqrt{x^2+y^2},d\left(B,C\right) =\sqrt{x^2+(y-c)^2}$ and appealing once again to the equilaterallness of the triangle we may conclude that, in fact, $d\left(O,C\right)=\sqrt{x^2+y^2}=\sqrt{x^2+(y-c)^2}=d\left(B,C\right)$. Solving this equation yields $y=\frac{c}{2}$. As stated earlier though $d\left(O,B\right)=d\left(O,C\right)=d\left(B,C\rig ht)=c$ and focusing mainly on $d\left(O,C\right)$ we see that $\sqrt{x^2+y^2}=\sqrt{x^2+\frac{c^2}{4}}=c\implies x=\pm\frac{\sqrt{3}c^2}{4}$ which is clearly irrational for any rational $c$.

Case 2, $a,b\ne0$: Now we have that $B=(2a,2b)$ where neither $a$ or $b$ is zero. Constructing the midpoint $M$ of $\overline{OB}$ gives us that $M=(a,b)$ and that the length of the segment $\overline{OM}$ is $d\left(O,M\right)=\sqrt{a^2+b^2}$. Now, form $\overline{MC}$ to be the median connecting $M$ and $C$. If we consider the two segments $\overline{OB},\overline{MC}$ to be the restrictions of the lines $\ell_1,\ell_2$ respectively we can see that $\text{slope of }\overline{OB}=\frac{b}{a}$ and since $\overline{OB}\perp\overline{MC}$ we may conclue that $\text{slope of }\overline{MC}=\frac{-a}{b}$ (now you see why we needed $a,b\ne0$). Letting $C=(x,y)$ for some $x,y$ again and realizing that $M,C$ both lie on $\ell_2$ we may conclude that $C=\left(a+\Delta x,b+\Delta y\right)=\left(a+ \Delta x,b+\frac{-a}{b}\Delta x\right)\quad\color{red}\star$ where $\Delta x$ represents the change in $x$ between the points $M$ and $C$. Appealing to this we see that $\text{length of }\overline{MC}=\sqrt{\left(a+\Delta x-a\right)^2+\left(b+\frac{-a}{b}\Delta x-b\right)^2}=\left|\Delta x\right|\sqrt{1+\frac{a^2}{b^2}}$. Remembering though that forming the median $\overline{MC}$ bisected $\angle OBC$ we see that the new triangle formed, $\triangle OMC$ is a $30-60-90$ triangle. In particular, this tells us that $\text{length of }\overline{MC}=\sqrt{3}\cdot\text{length of }\overline{OB}=\sqrt{3}\cdot\sqrt{a^2+b^2}$. Equating our two lengths of $\overline{MC}$ gives us $\left|\Delta x\right|\sqrt{1+\frac{a^2}{b^2}}=\sqrt{3}\sqrt{a^2 +b^2}\implies \left|\Delta x\right|\sqrt{a^2+b^2}=\sqrt{3}\cdot b\cdot\sqrt{a^2+b^2}$ and since $a,b\ne0\implies\sqrt{a^2+b^2}\ne0$ we may conclude that $\left|\Delta x\right|=\sqrt{3}b\implies \Delta x=\pm\sqrt{3}b$. Thus, substitution into $\color{red}\star$ gives us that $C=\left(a\pm\sqrt{3}b,b\mp\sqrt{3}a\right)$ which is clearly irrational since $a,b\in\mathbb{Q}$.

The conclusion follows. $\square$

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.

9. 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.

10. 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)

11. 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.

12. 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?

13. 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.

14. 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:

Prove that in any triangle in $\mathbb{R}^2$ whose vertices are in $\mathbb{Q}^2$, the tangents
of the angles are rational.

15. Originally Posted by Unbeatable0
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:

Prove that in any triangle lying in $\mathbb{Q}^2$, the tangents
of the angles are rational.
Question:

Spoiler:
What do you mean "lying in $\mathbb{Q}^2$? Do you mean that the verticies of the triangle are rational points or that literally every point of the triangle is rational? Because I have a problem witht the second one...

Page 1 of 2 12 Last