# Similar Parallelograms on Sides of a Triangle

### Problem ### Solution

We use barycentric coordinates and Conway's notations. Let $S\,$ be twice the area of $\Delta ABC.\,$ Call $\lambda=\cot\angle CBB_a,\,$ $\mu=\cot\angle BCB_a.\,$ Then we have point $B_a=(-a^2:S_C+S\mu:S_B+S\lambda).\,$ Since the parallelograms are similar, we have the same angles for the other sides and we can calculate the points $C_b=(S_B+S\lambda:-b^2:S_A+S\mu)\,$ and $A_c=(S_B+S\mu:S_A+S\lambda:-c^2).$

Now we calculate the points

\begin{align} A' &= (b^2+c^2+S(\lambda+\mu):-b^2:-c^2),\\ B' &= (-a^2:a^2+c^2+S(\lambda+\mu):-c^2),\\ C' &= (-a^2:-b^2:a^2+b^2+S(\lambda+\mu)) \end{align}

and the intersection point of $A'X,\,$ $B'Y,\,$ and $C'Z:\,$

$Q=(b^2+c^2+(\lambda+\mu)S:c^2+a^2+(\lambda+\mu)S:a^2+b^2+(\lambda+\mu)S).$

This point $Q\,$ divides the segment $GK\,$ in the ratio

$\displaystyle \frac{GQ}{QK}=-\frac{a^2+b^2+c^2}{3(a^2+b^2+c^2+S(\lambda+\mu))}.$

The proof holds for the parallelograms all drawn towards the interior of the triangle as well as in the exterior, because the angles involved could be considered signed: ### Acknowledgment

The problem has been originally proposed by Kadir Altintas and posted by Takis Chronopoulos at the Οι Ρομαντικοι της Γεωμετριας (Romantics of Geometry) facebook group: It was later upgraded to the case of similar and similarly oriented parallelograms. Kostas Theodoros Rekoumis has observed that the problem of concurrence is solved in a one step reference to Desargues' theorem even when the parallelograms are not necessarily similar. The above extension of the problem and the solution are due to Francisco Javier García Capitán. Copyright © 1996-2018 Alexander Bogomolny

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