Let P be the centroid of the equilateral triangle T with vertices ( x 1 , y 1 ) , ( x 2 , y 2 ) , ( x 3 , y 3 ) on the Cartesian coordinate plane. It is known that 1 8 7 2 = x 1 + x 2 + x 3 = x 1 x 2 + x 2 x 3 + x 3 x 1 . T is then rotated 3 5 ° clockwise such that the new vertices are ( x 1 ′ , y 1 ′ ) , ( x 2 ′ , y 2 ′ ) , ( x 3 ′ , y 3 ′ ) and the centroid is still at P .
Find the value of ( x 1 ′ + x 2 ′ + x 3 ′ ) ( x 1 ′ x 2 ′ + x 2 ′ x 3 ′ + x 3 ′ x 1 ′ ) .
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Yep, this is the more direct approach. I just thought it was really cool that the roots of a cubic can be linked to this.
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Yes, indeed, it is. We can find this cubic with trigonometry as well, of course; it comes out to be ( x − x ˉ ) 3 − 4 3 r 2 ( x − x ˉ ) − 4 r 3 cos ( 3 θ ) where r is the radius of the circumcircle and θ is the phase angle of the vertices relative to the centroid. We can see that the rotation affects the constant term only.
The value of x 1 ′ + x 2 ′ + x 3 ′ and x 1 ′ x 2 ′ + x 2 ′ x 3 ′ + x 3 ′ x 1 ′ will remain the same regardless of how much you rotate T . This is a consequence of François Viète's trigonomic solutions for roots to a cubic polynomial.
Viète found that, given a cubic polynomial a x 3 + b x 2 + c x + d , if one shifts the polynomial to the right by 3 a b and defines:
p q = 3 a 2 3 a c − b 2 = 2 7 a 3 2 b 3 − 9 a b c + 2 7 a 2 d
Then the polynomial reduces to x 3 + p x + q . Using the trignometric identity 4 cos 3 x − 3 cos x − cos ( 3 x ) = 0 , if the polynomial has 3 real roots, the roots to this second polynomial can be expressed as:
x i = 2 − 3 p cos ( 3 1 arccos ( 2 p 3 q p − 3 ) − 3 2 π i ) , i = 1 , 2 , 3
Geometrically, the roots of a x 3 + b x 2 + c x + d will be the x-coordinates of an equilateral triangle with a circumradius of 2 − 3 p , a centroid with x-coordinate 3 a b and angle of rotation of 3 1 cos − 1 ( 2 p 3 q − p 3 ) . Details of the derivation can be found here
We can define x 1 , x 2 , x 3 as the roots of the cubic x 3 − 1 8 x 2 + 7 2 x + d , which can be seen by Vieta's Formulas, where d is a value whereby the roots are real, and have x 1 , x 2 , x 3 be the x-coordinates of an equilateral triangle T , whose contruction is described above. Since the circumradius and the x-coordinate of the centroid is independant of d , upon changing d , T will simply rotate, keeping the circumradius and the x-coordinate of the centroid constant. As such, upon rotating the triangle, only d changes, keeping b = − 1 8 and c = 7 2 constant. So x 1 ′ , x 2 ′ , x 3 ′ are the roots to the cubic x 3 − 1 8 x 2 + 7 2 x + d ′ , keeping the value of x 1 ′ + x 2 ′ + x 3 ′ and x 1 ′ x 2 ′ + x 2 ′ x 3 ′ + x 3 ′ x 1 ′ constant by Vieta's Formulas.
Here is a link to an interactive online visual. The value of d is restrained by requiring the roots to be real, however, this does not limit the amount T can rotate because the triangle is not uniquely determined by x i . By reflecting T about the y-axis, this keeps keeping the x-coordinates of its vertices unchanged.
A fascinating problem; I really enjoy thinking about it. Thank you for posting!
I think it is misleading to say that "the value of d is restrained by requiring the roots to be real, limiting the amount T can rotate." While the value of d is certainly constrained, that does not limit the amount you can rotate; the problem is rather that the triangle is not uniquely determined by the x k .
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Oh yes, that is what I tried to convey at the second sentence starting with "however". I had a terrible time trying to phrase it though. I'll try again.
EDIT: I've tried again and stolen some phrases from you
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It is a simple exercise in trigonometry to show that x 1 x 2 + x 1 x 3 + x 2 x 3 = 3 x ˉ 2 − 4 a 2 for the vertices of any equilateral triangle, where ( x ˉ , y ˉ ) is the centroid and a is the side length. Since T preserves both x ˉ and a , the answer is 1 2 × 7 2 = 1 2 9 6 .