Equilateral Triangle and Algebra

Geometry Level 4

Let P P be the centroid of the equilateral triangle T T with vertices ( x 1 , y 1 ) , ( x 2 , y 2 ) , ( x 3 , y 3 ) (x_1,y_1), (x_2,y_2), (x_3,y_3) on the Cartesian coordinate plane. It is known that 18 = x 1 + x 2 + x 3 72 = x 1 x 2 + x 2 x 3 + x 3 x 1 . \begin{aligned} 18 &= x_1+x_2+x_3 \\ 72&= x_1x_2 + x_2x_3 + x_3x_1. \end{aligned} T T is then rotated 35 ° 35° clockwise such that the new vertices are ( x 1 , y 1 ) , ( x 2 , y 2 ) , ( x 3 , y 3 ) (x_1',y_1'), (x_2',y_2'), (x_3',y_3') and the centroid is still at P . P.

Find the value of ( x 1 + x 2 + x 3 ) ( x 1 x 2 + x 2 x 3 + x 3 x 1 ) . \left( x_1'+x_2'+x_3' \right)\left( x_1'x_2' + x_2'x_3' + x_3'x_1' \right).


The answer is 1296.

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2 solutions

Otto Bretscher
Nov 10, 2018

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 a 2 4 x_1 x_2+x_1 x_3+x_2 x_3=3\bar{x}^2-\frac{a^2}{4} for the vertices of any equilateral triangle, where ( x ˉ , y ˉ ) (\bar{x},\bar{y}) is the centroid and a a is the side length. Since T T preserves both x ˉ \bar{x} and a a , the answer is 12 × 72 = 1296 12\times 72=\boxed{1296} .

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.

Julian Poon - 2 years, 7 months ago

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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 3 r 2 4 ( x x ˉ ) r 3 4 cos ( 3 θ ) (x-\bar{x})^3-\frac{3r^2}{4}(x-\bar{x})-\frac{r^3}{4}\cos(3\theta) where r r is the radius of the circumcircle and θ \theta is the phase angle of the vertices relative to the centroid. We can see that the rotation affects the constant term only.

Otto Bretscher - 2 years, 7 months ago
Julian Poon
Nov 9, 2018

The value of x 1 + x 2 + x 3 x_1'+x_2'+x_3' and x 1 x 2 + x 2 x 3 + x 3 x 1 x_1'x_2' + x_2'x_3' + x_3'x_1' will remain the same regardless of how much you rotate T 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 ax^3+bx^2+cx+d , if one shifts the polynomial to the right by b 3 a \frac{b}{3a} and defines:

p = 3 a c b 2 3 a 2 q = 2 b 3 9 a b c + 27 a 2 d 27 a 3 \begin{aligned} p&=\frac{3ac-b^2}{3a^2} \\ q&=\frac{2b^3-9abc+27a^2d}{27a^3} \end{aligned}

Then the polynomial reduces to x 3 + p x + q x^3 + px + q . Using the trignometric identity 4 cos 3 x 3 cos x cos ( 3 x ) = 0 4 \cos^3 x - 3\cos x - \cos(3x) = 0 , if the polynomial has 3 real roots, the roots to this second polynomial can be expressed as:

x i = 2 p 3 cos ( 1 3 arccos ( 3 q 2 p 3 p ) 2 π i 3 ) , i = 1 , 2 , 3 x_i = 2\sqrt{-\frac{p}{3}}\cos\left( \frac{1}{3} \arccos \left( \frac{3q}{2p} \sqrt{\frac{-3}{p}} \right) - \frac{2 \pi i}{3} \right), \quad i=1,2,3

Geometrically, the roots of a x 3 + b x 2 + c x + d ax^3 + bx^2 + cx + d will be the x-coordinates of an equilateral triangle with a circumradius of 2 p 3 \displaystyle 2\sqrt{-\frac{p}{3}} , a centroid with x-coordinate b 3 a \displaystyle \frac{b}{3a} and angle of rotation of 1 3 cos 1 ( 3 q 2 p 3 p ) \displaystyle \frac{1}{3}\cos^{-1}\left(\frac{3q}{2p}\sqrt{-\frac{3}{p}}\right) . Details of the derivation can be found here

We can define x 1 , x 2 , x 3 x_1, x_2, x_3 as the roots of the cubic x 3 18 x 2 + 72 x + d x^3 - 18x^2 + 72x + d , which can be seen by Vieta's Formulas, where d d is a value whereby the roots are real, and have x 1 , x 2 , x 3 x_1, x_2, x_3 be the x-coordinates of an equilateral triangle T T , whose contruction is described above. Since the circumradius and the x-coordinate of the centroid is independant of d d , upon changing d d , T T will simply rotate, keeping the circumradius and the x-coordinate of the centroid constant. As such, upon rotating the triangle, only d d changes, keeping b = 18 b = -18 and c = 72 c=72 constant. So x 1 , x 2 , x 3 x_1', x_2', x_3' are the roots to the cubic x 3 18 x 2 + 72 x + d x^3 - 18x^2 + 72x + d' , keeping the value of x 1 + x 2 + x 3 x_1'+x_2'+x_3' and x 1 x 2 + x 2 x 3 + x 3 x 1 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 d is restrained by requiring the roots to be real, however, this does not limit the amount T T can rotate because the triangle is not uniquely determined by x i x_i . By reflecting T 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 d is restrained by requiring the roots to be real, limiting the amount T T can rotate." While the value of d 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 x_k .

Otto Bretscher - 2 years, 7 months ago

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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

Julian Poon - 2 years, 7 months ago

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