Three visions of a fact

Leo Giugiuc has posted a problem at the CutTheKnotMath facebook page with credits to Petru Puiu Braica, Romania. The problem appeared under the caption "Three visions of a fact" which I use as the title of this page:

Consider distinct complex numbers $x,$ $y,$ $z$ such that $|x|=|y|=|z|\ne |x+y+z|$ and $|x-y|\ne |x-z|.$ Let $k$ be a real number, $k\ne 0,$ and $k\ne 1;$ define $\displaystyle w=\frac{y-kz}{1-k}.$

Prove that there is a complex number $o$ for which


if and only if $|x-y|=|x-w|.$

Vision 1

For a triangle with the circumcenter at the origin, the sum of the vertices coincides with the orthocenter. Accordingly, we introduce $h=x+y+z.$ The points $x,y,x,h$ form an orthocentric system. In particular, $y$ is the orthocenter of $\Delta xzh.$ Reflections of the orthocenter in the sides of a triangle lie on the circumcircle of that triangle. In particular, reflection of $y$ in $xh$ lies on $(xzh).$ In addition that point, say $w,$ is collinear with $y$ and $z$ (because $yz\perp xh)$ and is, therefore, their linear combination. For the reflection $w$ of $y$ in $xh,$ $xy=xw.$

19 June 2015, Created with GeoGebra

Vision 2

Given three circles of the same radius through point $H,$ denote their other common points $X,$ $Y,$ and $Z.$ Then $XH\perp YZ.$ For this reason, the line $YZ$ intersects the two circles $(XY)$ and $(XZ)$ in symmetric points relative to $XH.$ In the applet below, if $H_x$ is the intersection of $XH$ with $YZ$ and $V$ and $W$ the other two points of intersection of the two circles with $YZ$ then $YH_x = H_xW$ and $VH_x = H_xZ,$ implying, in particular, that $XY=XW.$

19 June 2015, Created with GeoGebra

Vision 3

This vision is algebraic, using complex numbers.

We consider in complex plane the points $A = x,$ $B = y$ and $C = z.$ Obviously $ABC$ is a triangle and and $O = 0$ is its circumcenter, hence, by Sylvester's theorem, $H=x+y+z$ is its orthocenter.

By the hypothesis, $\Delta ABC$ is not right angled, $BC\ni D = w$ and $D\ne B,$ $D\ne C.$ We have to prove that the points $A,$ $H,$ $C,$ $D$ are concyclic iff $AB = AD.$

Without loss of generality, we choose $A=i,$ $B=-b,$ and $C=c,$ $b=\cot B,$ $c=\cot C.$ Then $b,c\ne 0,$ $b\ne c,$ $bc\ne 1,$ $H=bci,$ and $D=a,$ where $a\ne -b$ and $a\ne c.$ The circumcircle $(AHC)$ cuts the line $BC$ in at most two distinct points. Since one of them is $C,$ it is enough to show that the other is $D'=b.$ We have

$\displaystyle\begin{align} \frac{H-A}{H-C}:\frac{D'-A}{D'-C} &= \frac{bci-i}{bci-c}\cdot\frac{b-c}{b-i}\\ &=\frac{(bc-1)(b-c)}{c}\cdot\frac{i}{i(b+i)(b-i)}\\ &=\frac{(bc-1)(b-c)}{c(b^{2}+1)}. \end{align}$

Hence, the cross-ratio $\displaystyle\frac{H-A}{H-C}:\frac{D'-A}{D'-C}$ is a real number, meaning that $A,H,C,D'$ are concyclic. This exactly means that $A,H,C,D'$ are concyclic iff $AB=AD.$


The third vision is a solution by Denisa Maria Rostogol, Romania, 10th grade; it was communicated to me in private correspondence by Leo Giugiuc under whose tutelage it was obtained.

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