(The above Q isn't too hard, just something to get this thread rolling again .)
Find the maximal possible angle that a circle centred at the origin can meet the ellipse x^2/a^2 + y^2/b^2 = 1.
(The above Q isn't too hard, just something to get this thread rolling again .)
Update 2018
This marathon will continue into 2018.
2014 HSC
2015 Gap Year
2016-2018 UOW Bachelor of Maths/Maths Advanced.
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If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
Also, a new problem in the meantime of a different flavour:
An ellipse is inscribed in the triangle ABC, with points of contact X,Y,Z on the triangle sides opposite A,B,C respectively.
Prove that AX,BY,CZ are concurrent. (That is, prove that there exists a point that all three of these lines pass through.)
I don't think that is quite correct, but it is close. (Perhaps an erroneous trig manipulation?)
As a test, if , then we expect the maximal angle of contact to be close to , because the circle and the ellipse are basically identical. Also, the maximal angle should never exceed .
Last edited by Paradoxica; 13 Oct 2017 at 3:34 PM.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
My meaning was that "angle of contact" means "acute angle of contact", otherwise the two circles that meet the ellipse tangentially give you the maximal angle (pi) and the problem is boring.
Your new answer seems correct with absolute values.
My answer (including the a-dependence) was the equivalent expression
Haha not even bashy, once you spot you can parallel project / affine transform the circle to an ellipse, there are no calculations to do, you just need to prove that the cevians meeting the incircle contact points are concurrent.
This is within the scope of HSC geometry. Alternatively, at this stage the concurrence is trivial from Ceva's theorem, which isn't hard to prove itself.
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