A Problem with Two Similar Rectangles

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Using complex numbers, let $A=i,\,$ $B=-u,\,$ $C=v,\,$ with $u=\cot B\,$ and $v=\cot C.\,$ We have $\displaystyle \frac{E-C}{A-C}=-ki,\,$ implying $E=c+k+cki.\,$ Similarly, $D=-b-k+bki.\,$ From Here, $BE\,$ is defined by $-ckx+(b+c+k)y=bck\,$ and $CD\,$ by $BKX+(b+c+k)y=bck.\,$ Since $AH,\,$ the altitude, is defined by $x=0,\,$ suffice it to show that

$\left|\begin{array}{ccc}\,1&0&0\\-ck&b+c+k&bck\\bk&b+c+k&bck\end{array}\right|=0,$

which is obvious.

Let $h=AH.\,$ Then $BH=\sqrt{c^2-h^2}\,$ and $CH=\sqrt{b^2-h^2}.\,$ Drop a perpendicular from $E\,$ to $BC.\,$ BY $AAA\,$ test, triangles $AHC\,$ and $CFE\,$ are similar. Thus $EF=k\sqrt{b^2-h^2}\,$ and $CF=kh.$

Triangles $BHX\,$ and $BFE\,$ are similar. Thus,

$\displaystyle HX=\frac{EF\cdot BH}{BF}=\frac{k\sqrt{b^2-h^2}\sqrt{c^2-h^2}}{kh+\sqrt{b^2-h^2}+\sqrt{c^2-h^2}}.$

This expression was derived using the blue rectangle and is completely symmetric in $b\,$ and $c.\,$ Thus, a similarly derived expression using the orange rectangle results in the same expression. Hence, $BE\,$ and $CD\,$ intersect $AH\,$ at the same point.

The problem strongly reminds of what is known as Vecten's configuration, with the difference that squares on the sides of a triangle have been replaced with similar rectangles. The configuration has many interesting properties. The proof of the one at hand is practically the same as before (see F. G.-M., Exercices de Géométrie, Éditions Jacques Gabay, sixiéme édition, 1991, p 225).

Drop perpendiculars from $B\,$ and $C\,$ to $CD\,$ and $BE,\,$ respectively.Let $K\,$ and $L\,$ be the points of intersection. We'll prove that $BK\,$ and $CL\,$ meet on the extension of $AH.\,$ If that point is denoted $J,\,$ then this will make $BK,\,$ $CL\,$ and $AH\,$ the altitudes in $\Delta BCJ,\,$ from which the required statement follows immediately.

To start with, let $J\,$ be the intersection of $BK\,$ and $AH.\,$ Consider triangles $BCD\,$ and $ABJ.\,$ By simple angle chasing, $\angle BDC=\angle ABJ\,$ and $\angle CBD=\angle BAJ.\,$ Thus the two triangles are similar; in particular,

$\displaystyle AJ=BC\cdot\frac{AB}{BD}=BC\cdot\frac{AC}{CE},$

due to the similarity of the rectangles. This means that $J\,$ also serves as the intersection of $CL\,$ and $AH.\,$ The proof is complete.

The statement is a direct consequences of Jacobi's theorem:

In 2001, Floor van Lamoen has introduced the notion of friendship among triangle centers. In particular, in Vecten configuration, the extension of the altitude through the common vertex of $\Delta ABC\,$ is the median in $\Delta AGF,\,$ which is one of the flank triangles. This property also holds for the present configuration:

This is left as an exercise.

The problem by Kadir Altintas has been kindly posted by Leo Giugiuc at the CutTheKnotMath facebook page, along with his solution (Solution 1). Solution 2 is by Amit Itagi; Solution 4 is by Daniel Liu and, independently, by Kostas Vittas.