Squares $ BCA_{1}A_{2}$ , $ CAB_{1}B_{2}$ , $ ABC_{1}C_{2}$ are outwardly drawn on sides of triangle $ \triangle ABC$. If $ AB_{1}A'C_{2}$ , $ BC_{1}B'A_{2}$ , $ CA_{1}C'B_{2}$ are parallelograms then prove that: (i) Lines $ BC$ and $ AA'$ are orthogonal. (ii)Triangles $ \triangle ABC$ and $ \triangle A'B'C'$ have common centroid
Problem
Source: Federation of Bosnia, 1. Grades 2008.
Tags: geometry, parallelogram, analytic geometry, vector, complex numbers
balan razvan
23.04.2008 21:57
who is $ A',B',C'$?
delegat
23.04.2008 23:04
Points $ A'$ , $ B'$ and $ C'$ are defined by parallelograms mentioned above.
balan razvan
23.04.2008 23:13
delegat wrote: (i) Lines $ BC$ and $ AA'$ coincide and lines $ BC$ and $ AA'$ are orthogonal. ?? $ BC$ and $ AA'$ ,$ BC$ and $ AA'$?
delegat
23.04.2008 23:39
Your last comment is OK In my problem sheet it is written exactly as I wrote I will check where s mistake.
ch1n353ch3s54a1l
25.04.2008 07:24
delegat wrote: Squares $ BCA_{1}A_{2}$ , $ CAB_{1}B_{2}$ , $ ABC_{1}C_{2}$ are outwardly drawn on sides of triangle $ \triangle ABC$. If $ AB_{1}A'C_{2}$ , $ BC_{1}B'A_{2}$ , $ CA_{1}C'B_{2}$ are parallelograms then prove that: (i) Lines $ BC$ and $ AA'$ coincide and lines $ BC$ and $ AA'$ are orthogonal. (ii)Triangles $ \triangle ABC$ and $ \triangle A'B'C'$ have common centroid I hope these work...
Two lines are perpendicular iff $ \frac{z_1-z_2}{z_3-z_4}\in i\mathbb{R^*}$ where $ z_1$ and $ z_2$ are the coordinates of two points of one line, and $ z_3$ and $ z_4$ are the coordinates of two points of the other line.
Let $ A$ be the origin, and $ BC$ is parallel to the real axis. We need $ \frac{a'-a}{b-c}\in i\mathbb{R^*}$, but since $ BC$ is parallel to the real axis, they have the same imaginary part, so $ b-c\in \mathbb{R}$.
Now $ AA'=a'=(b_1-a)+(a'-b_1)$. $ b_1=ic$ and $ c_2=(a'-b_1)=i^3b$. So $ a'=ic+i^3b=ic-ib=i(c-b)\in i\mathbb{R}$, so $ AA'$ and $ BC$ are perpendicular.
From above, $ a'=i(c-b)$. Also $ b'=b+i(b-(c-b))$ and $ c'=c+i((c-b)-c)$ and the centroid of $ A'B'C'$ is $ \frac{b+c}{3}$ and the centroid of $ ABC$ is $ \frac{b+c}{2}$ as well...
me@home
13.08.2008 19:00
By its construction $ A'-A = i(B-A)+(-i)(C-A)$. Thus $ A'-A=i(B-C)$ , which implies part1. For part2, notice that \[ \sum_{cyc}A'-A = i \left(\sum_{cyc}B-C\right)=0\]So the two sets of points have the same vector sum and therefore equal centroid.
jayme
24.10.2014 09:58
Dear Mathlinkers, you can see http://jl.ayme.pagesperso-orange.fr/ p. 65-67 for (1), p. 75-76 for (2) Sincerely Jean-Louis