Mathematics / Mathematik / Matemática
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Winfried Weber Group moderatorThe company name is only visible to registered members.beuatiful formular for an equilateral triangle
Helo,
I have the following situation
we have an equilateral triangle ABC, each side has length d:=l(AB)=l-(BC)=l(CA) ("gleichseitiges Dreieck ABC")
we take an optional point P on the plane and we know:
a:= l(PA)
b:=l(PB)
c:=l(PC)
Then we have the following beautiful symmetric formular:
3*(a^4+b^4+c^4+d^4) = (a^2+b^2+c^2+d^2)^2
That means: With a, b and c, the length d of the equilateral triangle can be calculated.
But has anybody an idea how to show this formular?
It looks a little bit like the formular of Pythagoras (a^2+b^2=c^2), and I think you have to add some "helping points" Q, R, ... so you get similar triangles or triangles with angles 90°, 60° and 30°, but how many points, where exactly, how?
Best regards,
Winfried
- 08 Jan 2007, 09:21 am
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Stanislava SimeonovaThe company name is only visible to registered members.Re: beuatiful formular for an equilateral triangle
Hello!
I believe you could solve this one without additional points, but instead using areas of triangles.
For example, I took the point P on the "left side" of AC, so it's basically outside the triangle, in this case you know that
S(ABP)+S(BPC)=S(ABC)+S(ACP) , where S is the area.
There is a formula that says that the area of any triangle is the square root of the following p*(p-a)*(p-b)*(p-c), where p is half the perimeter of the triangle, or in other words p=(a+b+c)/2, where a,b and c are the sides of any given triangle.
Thus, if you use that formula and the corelation between the areas of the triangles, which btw you probably need to turn to the power of 2 on both sides, because of this square root, you will have an equation only with d,a,b and c.
That's my idea, but I didn't have time yet to check out whether the equation is too hard to be worked on :)
Best Regards,
Stany
- 09 Jan 2007, 08:51 am
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Winfried Weber Group moderatorThe company name is only visible to registered members.
Re^2: beautiful formular for an equilateral triangle
Hello Stany,
thanks for the answer!
For example, I took the point P on the "left side" of AC, so it's basically outside the triangle, in this case you know that
S(ABP)+S(BPC)=S(ABC)+S(ACP) , where S is the area.
EDIT: Of course you are right, it is clear.
EDIT: there is the following simular formular:
for every common triangle ABC with sides a,b,c we have for the area A:= S(ABC):
16* A^2 = (a^2+b^2+c^2)^2 - 2*(a^4+b^4+c^4)
Best Regards,
Winfried
This post was modified on 11 Jan 2007 at 01:01 pm.- 10 Jan 2007, 2:56 pm
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Winfried Weber Group moderatorThe company name is only visible to registered members.
Re^2: beuatiful formular for an equilateral triangle
I believe you could solve this one without additional points, but instead using areas of triangles.
For example, I took the point P on the "left side" of AC, so it's basically outside the triangle, in this case you know that
S(ABP)+S(BPC)=S(ABC)+S(ACP) , where S is the area.
ok
There is a formula that says that the area of any triangle is the square root of the following p*(p-a)*(p-b)*(p-c), where p is half the perimeter of the triangle, or in other words p=(a+b+c)/2, where a,b and c are the sides of any given triangle.
ok,
I found the the s a m e formular ("without using p")
For every common triangle ABC with sides a,b,c we have for the area A:= S(ABC):
16* A^2 = (a^2+b^2+c^2)^2 - 2*(a^4+b^4+c^4)
1.) Can somebody give me an hint for a proof of this formular?
2.) What is the relationship the the other formular with the point and the equilaterla triangle?
(a^2+b^2+c^2+d^2)^2 = 3 * (a^4+b^4+c^4+d^4)
Best Regards,
Winfried
- 12 Jan 2007, 10:03 am
