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Please help me. I didn't get any idea to solve this challenging problem:

Suppose $A, B, C$ are three non-collinear points corresponding to complex numbers $$z_0 = ai, \;z_1 = \frac{1}{2}+bi,\; z_2 = 1+ci$$ ($a, b$ and $c$ being real numbers), respectively. Prove that the curve $$z = z_0 \cos^4 t + 2 z_1 \cos^2 t \cdot \sin^2 t + z_2 \sin^4 t \qquad (t \in \mathbb R)$$ shares a single common point with the line bisecting $AB$ and parallel to $AC$ and $\Delta ABC$, and find this point.

Dustan Levenstein
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Aryabhatta
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    Well, there's an obvious place to start: if you need to discuss the line "bisecting $AB$ and parallel to $AC$ in $\triangle ABC$", you probably need to find that line first. Have you been able to do so? – Nick Peterson Mar 29 '17 at 15:57
  • Pretty sure the complex numbers are just being used as vectors in this problem. – Dustan Levenstein Mar 29 '17 at 16:01
  • Please take the time to type in important parts of your question instead of including them in an image. Images are not searchable and are inaccessible to people using screen readers. You can’t expect us to take our time to help you if you’re not going to take your own time to do this. – amd Mar 29 '17 at 19:41
  • I've decided to convert the image to text this time, but you should take @amd's advice in the future. – Dustan Levenstein Mar 30 '17 at 18:25

1 Answers1

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  • Note that "the line bisecting $AB$ and parallel to $AC$" is simply the line that connects the bisectors of $AB$ and $BC$.
  • Change variable to $u=\sin^2 t$.
  • As @Dustan Levenstein remarks, the complex numbers here are just being used as a real vector space; their multiplicative structure is irrelevant.
  • Note that the coefficients of $z_0$, $z_1$, $z_2$ sum to $1$, so you can add the same vector to all three points and the curve will translate similarly.
  • Thus, by an appropriate affine transformation you can switch to a nicer coordinate system where the coordinates are $A(0,0)$, $B(1,2)$, $C(2,0)$, without changing the expression for the curve.
  • You should now be able to prove that the $y$-coordinate reaches $1$ exactly once while $u$ goes from $0$ to $1$. This shows that the curve meets the line in a single point.
  • By symmetry, this point can only be the point halfway between the midpoints of $AB$ and $BC$.
  • This allows to you find the point in the original complex coordinates.
hmakholm left over Monica
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