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    MHF Contributor Drexel28's Avatar
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    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!
    Last edited by Drexel28; November 30th 2009 at 10:20 PM.
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    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 .
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Unbeatable0 View Post
    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.
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    Quote Originally Posted by Drexel28 View Post
    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?
    Last edited by aman_cc; December 1st 2009 at 08:02 AM.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by aman_cc View Post
    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.
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    Quote Originally Posted by Drexel28 View Post
    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?
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by aman_cc View Post
    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:
    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

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    Quote Originally Posted by Drexel28 View Post
    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.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by aman_cc View Post
    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.
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    Quote Originally Posted by Drexel28 View Post
    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)
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by aman_cc View Post
    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.
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    Quote Originally Posted by Drexel28 View Post
    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?
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    Quote Originally Posted by aman_cc View Post
    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.
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    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.
    Last edited by Unbeatable0; December 4th 2009 at 02:35 AM. Reason: Reworded
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Unbeatable0 View Post
    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 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...
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