Is there an easier way to solve a "Find the locus" problem?Note: I am not...

Baladdaa9

Baladdaa9

Answered

2022-07-21

Is there an easier way to solve a "Find the locus" problem?
Note: I am not concerned with the accuracy of my solution so you don't need to redo any of my calculations.
Original question: Suppose ABC is an equilateral triangular lamina of side length unity, resting in two-dimensions. If A and B were constrained to move on the x- and y-axis respectively, then what is the locus of the centre of the triangle?
My solution: Let the vertices of Δ A B C be A = ( x 0 , 0 ), B = ( 0 , y 0 ) and C = ( x , y ). Let the centre of the triangle be D = ( X , Y ). We wish to find the locus of D under the constraints:
x 0 2 + y 0 2 = 0 ( x x 0 ) 2 + y 2 = 0 x 2 + ( y y 0 ) 2 = 0
Since D is the centre of the triangle, we know
X = x + x 0 3 and Y = y + y 0 3
Parametrizing: Let θ be the angle that the edge AB makes with the x-axis. We use (theta) as our parameter of choice to write, x 0 = cos θ y 0 = sin θ
for θ [ 0 , 2 π )
This implies, ( x cos θ ) 2 + y 2 = 0
We can parametrize this with another parameter ϕ [ 0 , 2 π ) as
x = cos θ + cos ϕ y = sin ϕ
Plugging these back into third equation gives
( cos θ + cos ϕ ) 2 + ( sin ϕ sin θ ) 2 = 1
After lots of algebraic manipulations...
X = 1 2 cos θ ± 1 2 3 sin θ Y = 1 2 sin θ ± 1 2 3 cos θ

Answer & Explanation

Jaylene Tyler

Jaylene Tyler

Expert

2022-07-22Added 10 answers

Step 1
It is probably easier to embed the construction in the complex plane and set A = sin θ , B = i cos θ to have C = ( A B ) e ± i π / 3 + B = e i θ ± i π / 3 + B
so that: 3 D = A + B + C = e i θ + e i θ ± i π / 3 + i cos θ ..
Step 2
That leads to the same equations for the ellipses.
If three vertices of a tetrahedron lie on the x-, y- and z- axis we may assume they are A(a,0,0), B(0,b,0), C(0,0,c) and a 2 + b 2 = a 2 + c 2 = b 2 + c 2 must hold, so | a | = | b | = | c | and there are just 16 possible locations for D and the centroid G of ABCD.

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