Given the triangle $ABC$ we consider the points $X,Y,Z$ such that the triangles $ABZ,BCX,CAZ$ are equilateral, and they don't have intersection with $ABC$. Let $B'$ be the midpoint of $BC$, $N'$ the midpoint of $CY$, and $M,N$ the midpoints of $AZ,CX$, respectively. Prove that $B'N' \bot MN$.
Problem
Source: Argentina IMO 2005 TST, problem 3
Tags: ratio, geometry, geometric transformation, rotation, power of a point, radical axis, geometry proposed
grobber
28.04.2005 04:06
Cute one . Consider the transformation we get by composing the spiral similarity cetnered at $B$ of angle $\frac\pi 6$ and ratio $\frac{\sqrt 3}2$ with the rotation of angle $\frac \pi 3$ centered at $C$ (the rotation comes first, then the spiral similarity). This transformation is again a spiral similarity of angle $\frac \pi 2$. The image of $Y$ is $M$, and the image of $B$ is $N$, which means that the image of the line $BY$ is $MN$, which shows that $BY\perp MN$. Now, since $BY\|B'N'$, we get the conclusion. P.S. the above even gives the ratio of the lengths of the two segments.
RaMlaF
28.04.2005 04:14
wow! your solutions is Great! Another nice idea is to use the fact that $MN$ is parallel to the line between the centers of $ABZ,BCX$. Hence by radical axis, and since the Fermat point of $ABC$ belongs to $BY$ and to the intersection of the circles $ABZ,BCX$, we get $MN\bot BY$ and the conclusion follows.
salamitrosky
15.05.2005 01:18
jojo, excelent, es una muy interesante solución la de grobber. Una mínima generalización del problema es: Given the triangle ABC we consider the points A', B', C' such that the triangles BCA', CAB', ABC' are equilateral an externals to ABC. Let Ca, Ba, Ab, Cb, Bc y Ac be the midpoints of BA', A'C, CB', B'A, AC' y C'B. The intersection of BaBc and CaCb is X, the intersection of AbAc and CaCb is Y, and the intersection of BaBc and AbAc is Z. Prove that XYZ is homotetic with the triangle NaNbNc (centers of BCA', CAB', ABC'), with center F and reason 1/2. (perdón por mi inglés, is terrible).
Charlesww
15.05.2005 17:29
pretty easy I think, the rotation (CCA :pi:/3 2) of BY from C is AX, and the rotation (CA :pi:/6 2/ :sqrt: 3) of MN from B is AX, too. So we get it.
Binomial-theorem
03.01.2013 13:38
3500th post Complex number bash ftw.
Hints from AoPS127) Let $A$ be the origin. What must be true about $(m-n)/(n'-b')$ if and only if $\overline{MN} \perp \overline{B'N'}$?
85) If you're stuck with a ratio that you'd like to show is inamginary, here's a hint. If $\frac{z_1}{z_2}=\frac{z_3}{z_4}=k$, then what does $\frac{z_1b+z_3c}{z_2b+z_4c}$ equal?
Let $A$ be the origin
We desire: $\frac{m-n}{b'-n'}$ being imaginary.
We know: $b-a=\zeta(z-a)\implies b=\zeta*z$
Also, $x-c=\zeta(b-c)$ and $y=\zeta(c)$.
Now we have a HIGHLY computationally intense problem so let's try it.
$m=\frac{z+a}{2}=\frac{\frac{b}{\zeta}+a}{2}$. Next, $n=\frac{x+c}{2}=\frac{2c+\zeta(b-c)}{2}$. Also, $b'=\frac{b+c}{2}$ and $n'=\frac{c+y}{2}=\frac{c+\zeta(c)}{2}$.
$\frac{m-n}{b'-n'}=\frac{\frac{b}{\zeta}-2c-\zeta(b-c)}{b-\zeta*c}$ which we need to be imaginary.
This is the same thing as $\frac{\frac{b}{\zeta}-2c-\zeta(b)+\zeta*c}{b-\zeta*c}$
This is also the same as $\frac{b-\zeta^2(b)-2\zeta*c+\zeta^2*c}{b\zeta-\zeta^2*c}$
Therefore, we get $\frac{b(1-\zeta^2)+c(\zeta^2-2\zeta)}{b\zeta+c(-\zeta^2)}$
We now prove $\frac{1-\zeta^2}{\zeta}=\frac{\zeta^2-2\zeta}{-\zeta^2}$.
This is crucial in the next part of the problem.
$\frac{1-\zeta^2}{\zeta}=\frac{\zeta-2}{-\zeta}$ or $1-\zeta^2=-\zeta+2$ Therefore, we get $\zeta^2-\zeta+1=0$ which is true by the definition of $\zeta$.
Yay, lemma proven.
Find $\frac{bz_1+cz_3}{bz_2+cz_4}$ where $\frac{z_1}{z_2}=\frac{z_3}{z_4}$.
We note that letting this value be $k$, we get $z_1=kz_2$.
Therefore, we have $\frac{bkz_2+ckz_4}{bz_2+cz_4}=k$.
Therefore, we must now prove that $\frac{1-\zeta^2}{\zeta}$ is imaginary.
Since $z$ is a primitive 6th root of unity, we get $z=e^{\frac{\pi}{3}i}$ or $z=e^{\frac{5\pi}{3}i}$.
Therefore, $z=\frac{\sqrt3}{2}+\frac{i}{2}$ or $z=\frac{1}{2}-\frac{i\sqrt3}{2}$.
In the first case, we get $z^2=e^{\frac{2\pi}{3}i}$ or therefore it is the same as $\frac{-1}{2}+\frac{\sqrt3}{2}i$
We then get $\frac{\frac32-\frac{\sqrt3}{2}i}{\frac{\sqrt3}{2}+\frac{i}{2}}$
$= \frac{3-\sqrt3*i}{\sqrt3+i}=\frac{(3-\sqrt3*i)(\sqrt3+i)}{4}=\frac{3-4\sqrt{3}i-3}{4}=-\sqrt{3}i$.
In the second case, we get $z^2=e^{\frac{4\pi}{3}i}=-\frac{1}{2}-\frac{\sqrt3}{2}i$.
Also, we get $\frac{1-z^2}{z}=\frac{3+\sqrt3i}{1-\sqrt3i}=\frac{(3+\sqrt3i)(1+\sqrt3i)}{4}=\frac{4\sqrt3i}{4}=\sqrt3i$.
We are finally done. $\Box$.
sayantanchakraborty
28.02.2014 20:22
We denote the affix of each uppercase letter by a lowercase letter.Then by rotation formula we have, $x=c+(b-c)\epsilon$ $y=a+(c-a)\epsilon$ $z=b+(a-b)\epsilon$ Thus $n=c+\frac{(b-c)\epsilon}{2}$ $n'=\frac{(c+a)+(c-a)\epsilon}{2}$ $m=\frac{(a+b)+(a-b)\epsilon}{2}$ $b'=\frac{b+c}{2}$ It is just calculation to see that $\frac{m-n}{b'-n'}=i\mathbb{R}$ for some nonzero real $\mathbb{R}$. Note:Here $\epsilon$ represents $cos60^{\circ}+isin60^{\circ}$. Bye. Maths is the doctor of science... Sayantan..
jayme
01.03.2014 10:59
Dear Mathlinkers, for a more general proof, you can see http://perso.orange.fr/jl.ayme vol. 16 p. 43-44 Sincerely Jean-Louis