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RE: [obm-l] 2 questoes do IME
Sauda,c~oes,
Mais esclarecimentos da 2a. questão. Agora
parece que podemos parar e dar o problema
como resolvido. Uma figura no pdf da versao 9
do material do IME seria legal. :))
ii) IME 1985/1986 (6a questao, item (b))
>Determine o lugar geometrico dos centros dos
>circulos que cortam dois circulos exteriores,
>de centros O1 e O2 e raios respectivamente
>iguais a R1 e R2, em pontos diametralmente opostos.
Dear Luis,
here is the solution for your second problem.
Let R be the radius of a circle, with center P, intersecting the
circles C1 and C2 in antipodal points.
We have
R^2 = R1^2 + O1P^2 = R2^2 + O2P^2
or O1P^2 - O2P^2 = R2^2 - R1^2
So P lies on a perpendicular to O1O2
For the radical axis of C1 and C2 we have
O1P^2 - O2P^2 = R1^2 - R2^2
So the locus and the radical axis lie symmetrically wrt the midpoint
of O1O2
In Dutch we call this line the "antimachtlijn" translated in English
as "antiradical axis".
I know there is another name in English but I can't remember it.
If I remember well it already appeared in Hyacinthos but I couldn't
find it.
Kind regards
Eric
[]'s
Luis
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