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An extension of the Fermat's problem.
Virgil Nicula   0
Oct 25, 2015
PP20. Let a rectangle $ABCD$ with $\left\{\begin{array}{ccc} AB & = & a\\\\ BC & = & b\end{array}\right\|$ and a circle $w$ with the diameter $[AB]$ . For a mobile point $M\in w$ so

that the $AB$ doesn't separate $M$ , $C$ let $\left\{\begin{array}{ccc} MA & = & x\\\\ MB & = & y\end{array}\right\|$ and the intersections $G$ , $F$ of $AB$ with $MC$ , $MD$ respectively.

Prove that $\boxed{AG^2+BF^2=AB^2+\left(a^2-2b^2\right)\left(\frac {xy}{xy+ab}\right)^2}\ (*)$ . Remark. For $a=b\sqrt 2$ obtain the Fermat's problem.
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Virgil Nicula
Oct 25, 2015
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An extension of the Fermat's problem.
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Virgil Nicula
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Virgil Nicula
#1 h  Oct 25, 2015, 9:07 AM • 2 Y
Y by Adventure10, Mango247
PP20. Let a rectangle $ABCD$ with $\left\{\begin{array}{ccc} AB & = & a\\\\ BC & = & b\end{array}\right\|$ and a circle $w$ with the diameter $[AB]$ . For a mobile point $M\in w$ so

that the $AB$ doesn't separate $M$ , $C$ let $\left\{\begin{array}{ccc} MA & = & x\\\\ MB & = & y\end{array}\right\|$ and the intersections $G$ , $F$ of $AB$ with $MC$ , $MD$ respectively.

Prove that $\boxed{AG^2+BF^2=AB^2+\left(a^2-2b^2\right)\left(\frac {xy}{xy+ab}\right)^2}\ (*)$ . Remark. For $a=b\sqrt 2$ obtain the Fermat's problem.
This post has been edited 3 times. Last edited by Virgil Nicula, Oct 25, 2015, 9:21 AM
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