Se P è un punto arbitrario sul lato BC del triangolo ABC, come posso tracciare un segmento Q attraverso il triangolo che sia parallelo a BC e sottenda un angolo retto in P?
La risposta... Non cliccate qui se non ci avete pensato! Poi è troppo tardi!
Un cerchio di diametro QR passa per P.
Da un punto qualsiasi S su AB si tracci il segmento ST parallelo a BC e il segmento AP.
Il semicerchio di diametro ST incrocia AP in L e il triangolo SLT è retto.
Il punto Q è individuato dalla parallela a SL passante per P.
I Rudi Mathematici (Rodolfo Clerico / Rudy d’Alembert, Piero Fabbri / Piotr Rezierovic Silverbrahms, Francesca Ortenzio / Alice Riddle) sono autori della omonima e-zine di matematica ricreativa, pubblicata in rete dal 1999.
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valter for Fields Medal.
Io tentato tre tipi di costruzione con GeoGebra simulando l’uso della sola riga e compasso; le propongo …anche se, forse, non sono così “Quick”:
https://drive.google.com/file/d/1Dhw_0BcEjJh79kLxh5donuIoj6ts48h_/view?usp=sharing, https://drive.google.com/file/d/1ZtqVI4Yu2hwBFS3yKHrM-wmRbtmqvkjr/view?usp=sharing
https://drive.google.com/file/d/1ZtqVI4Yu2hwBFS3yKHrM-wmRbtmqvkjr/view?usp=sharing
https://drive.google.com/file/d/1cnu1l51N51SzvFZ9eb31ua8u_wsxSD4W/view?usp=sharing