Converging Points

Geometry Level 2

{ A n + 1 = α A n + ( 1 − α ) B n B n + 1 = α B n + ( 1 − α ) C n C n + 1 = α C n + ( 1 − α ) A n \large {\left\{\begin{matrix}A_{n+1}=\alpha A_n+(1-\alpha)B_n \\ B_{n+1}=\alpha B_n+(1-\alpha)C_n \\ C_{n+1}=\alpha C_n+(1-\alpha)A_n\end{matrix}\right.}

Let A 1 , B 1 , C 1 A_1,B_1,C_1 be three distinct non-collinear points on the coordinate plane.
Also A n , B n , C n A_n,B_n,C_n satisfy the recurrence relation above ( 0 < α < 1 0<\alpha<1 ).

Then lim ⁡ n → ∞ A n \displaystyle\lim_{n\to\infty}A_n is the __________ \text{\_\_\_\_\_\_\_\_\_\_} of the triangle A 1 B 1 C 1 A_1B_1C_1 .

Orthocenter A n = A 1 A_n=A_1 Incenter Centroid Circumcenter

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展豪 張 by 展豪 張 , 22, Hong Kong

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2 solutions

Eric Chan
Mar 25, 2016

Expanding the equations given, we get :

A n + 1 + B n + 1 + C n + 1 = A n + B n + C n A_{n+1}+B_{n+1}+C_{n+1} = A_n+B_n+C_n

i.e. sum of coordinates are always the same regardless of n n .

⟹ \implies centroid are the same.(which is the centre of mass)

8 Helpful 0 Interesting 1 Brilliant 0 Confused

Of course, we have to show that the limit exists ...

Calvin Lin Staff - 5 years, 2 months ago

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It can be shown by considering the sequence: { a n } = A n B n \{a_n\}=A_nB_n which is a GS with 0 < r < 1 0<r<1 ...
OMG I forgot to specify 0 < α < 1 0<\alpha<1 ... edited, thanks!

展豪 張 - 5 years, 2 months ago

Cool solution!

展豪 張 - 5 years, 2 months ago

is there a way to prove that the average of the coordinates give you the centroid?

Willia Chang - 4 years, 11 months ago
Caique Harger
Mar 23, 2016

I do not nos how to actually prove it matematically but as an bn and cn Grow the sabe way as n Grows, it only could be the centroid. AM I wrong?

1 Helpful 2 Interesting 0 Brilliant 0 Confused

Writing in matrix form:
[ A n + 1 B n + 1 C n + 1 ] = [ α 1 − α 0 0 α 1 − α 1 − α 0 α ] [ A n B n C n ] \begin{bmatrix}A_{n+1} \\ B_{n+1} \\ C_{n+1} \end{bmatrix}=\begin{bmatrix}\alpha & 1-\alpha & 0 \\ 0 & \alpha & 1-\alpha \\ 1-\alpha & 0 & \alpha \end{bmatrix} \begin{bmatrix}A_n \\ B_n \\ C_n \end{bmatrix}
or
[ A n + 1 B n + 1 C n + 1 ] = [ α 1 − α 0 0 α 1 − α 1 − α 0 α ] n [ A 1 B 1 C 1 ] \begin{bmatrix}A_{n+1} \\ B_{n+1} \\ C_{n+1} \end{bmatrix}=\begin{bmatrix}\alpha & 1-\alpha & 0 \\ 0 & \alpha & 1-\alpha \\ 1-\alpha & 0 & \alpha \end{bmatrix}^n \begin{bmatrix}A_1 \\ B_1 \\ C_1 \end{bmatrix}
Now consider the 3 3 by 3 3 matrix as a matrix of Markov Chain.
By symmetry, lim ⁡ n → ∞ [ α 1 − α 0 0 α 1 − α 1 − α 0 α ] n = [ 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 ] \displaystyle \lim_{n\to\infty}\begin{bmatrix}\alpha & 1-\alpha & 0 \\ 0 & \alpha & 1-\alpha \\ 1-\alpha & 0 & \alpha \end{bmatrix}^n=\begin{bmatrix}\frac 13 & \frac 13 & \frac 13 \\ \frac 13 & \frac 13 & \frac 13 \\ \frac 13 & \frac 13 & \frac 13 \end{bmatrix}
So A ∞ = B ∞ = C ∞ = 1 3 ( A 1 + B 1 + C 1 ) = A_\infty=B_\infty=C_\infty=\frac 13 (A_1+B_1+C_1)= centroid of triangle A 1 B 1 C 1 A_1B_1C_1 .

展豪 張 - 5 years, 2 months ago

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