5

Let $ABC$ be a triangle inscribled inside circle $(O)$ . M is a point inside the triangle $ABC$ ($M \notin BC,CA,AB$)

$AM,BM,CM$ meets $(O)$ again at $A',B',C'$ respectively. Midperpendicular of $MA', MB', MC'$ meet $BC,CA,AB$ at $A_1,B_1,C_1$ respectively.

1.Prove that $A_1,B_1,C_1$ is collinear

2.$M'$ is symmetric to M through $A_1B_1C_1$. Prove that $M'\in (O)$

Please give me some hint

Thanksenter image description here

Angela Pretorius
  • 10,200
septimus
  • 163
  • This doesn't look easy. Before I take a crack at it, would it be acceptable to translate the problem into $\mathbb{R}^2$ or the complex plane and try to solve it algebraically, or are you looking more for a pure geometric proof? – Will Nelson May 24 '13 at 20:11
  • Fun fact: The locus of midpoints of segments $MP$, with $P$ on the given circle, is itself a circle; the new circle's radius is half that of the given one, and its center is the midpoint of $OM$. With all the midpoints in the figure, this would seem to be a relevant fact, but I haven't determined how to use it. – Blue May 24 '13 at 20:57

1 Answers1

2

Here's a vector approach.

With a view to invoking Menelaus' Theorem, we write $$A_1 = a B + (1-a) C \qquad B_1 = b C + (1-b) A \qquad C_1 = c A + (1-c) B$$ and we hope to show $$\frac{a}{1-a} \cdot \frac{b}{1-b} \cdot \frac{c}{1-c} = -1 \qquad (*)$$

Let $A^{\prime\prime} = M + \frac{1}{2}\frac{|A^\prime M|}{|AM|}(M-A)$ be the midpoint of $MA^\prime$. Since $\overrightarrow{A^{\prime\prime}A_1} \perp \overrightarrow{AM}$, we have $$( A_1 - A^{\prime\prime} )\cdot( M - A ) = 0$$ so that $$( \; a B + ( 1 - a ) C \; )\cdot( M - A ) = A^{\prime\prime}\cdot(M-A) = M\cdot(M-A) + \frac{1}{2}|AM||A^\prime M|$$ Thus, $$a \; ( \; B - C \; )\cdot( M - A ) = ( M - C )\cdot(M-A) + \frac{1}{2}|AM||A^\prime M|$$ and $$-(1-a) \; (\; B - C \; )\cdot( M - A ) = ( M - B )\cdot( M - A ) + \frac{1}{2}|AM||A^\prime M|$$ whence $$\frac{a}{1-a} = - \frac{2\;\overrightarrow{CM}\cdot\overrightarrow{AM}+ |AM||A^\prime M|}{2\;\overrightarrow{AM}\cdot\overrightarrow{BM}+|AM||A^\prime M|}$$

Now, recall the Power of a Point theorem, which says that $M$ divides chords through it into sub-segments whose lengths have a constant product, denoted here as $p$:

$$|AM||A^\prime M| = |BM||B^\prime M| = |CM||C^\prime M| = p$$

This guarantees that the left-hand side of $(*)$ expands to a circular product: $$ \left(- \frac{2\;\overrightarrow{CM}\cdot\overrightarrow{AM}+ p}{2\;\overrightarrow{AM}\cdot\overrightarrow{BM}+p}\right) \left(- \frac{2\;\overrightarrow{AM}\cdot\overrightarrow{BM}+ p}{2\;\overrightarrow{BM}\cdot\overrightarrow{CM}+p}\right) \left( - \frac{2\;\overrightarrow{BM}\cdot\overrightarrow{CM}+ p}{2\;\overrightarrow{CM}\cdot\overrightarrow{AM}+p}\right) $$ which reduces to $-1$; according to Menelaus, $A_1$, $B_1$, $C_1$ are indeed collinear.

For now, I'll leave proof that $M^\prime$ lies on the circle as homework.

Blue
  • 75,673
  • here is a easier solution http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=535534 – septimus Jun 04 '13 at 03:38
  • 1
    @septimus: Oh, you wanted an easy solution. Why didn't you say so? :) (I believe my brute-force argument can be streamlined a bit, but I can certainly appreciate a solution that incorporates anti-Steiner points and radical axes. Thanks for the link.) I'm not sure of the protocol when you find a solution off-site, but I suppose you should post (with credit and link) your favorite approach from AoPS as an answer here so that you can officially accept it; otherwise, the StackExchange "Community" robot will keep bumping your question into the main queue to generate more answers. – Blue Jun 04 '13 at 05:06