Isl 2002 g3 (complex this time)

by Wolstenholme, Aug 4, 2014, 4:43 AM

The circle $S$ has centre $O$ , and $BC$ is a diameter of $S$ . Let $A$ be a point of $S$ such that $\angle AOB<120{{}^\circ}$ . Let $D$ be the midpoint of the arc $AB$ which does not contain $C$ . The line through $O$ parallel to $DA$ meets the line $AC$ at $I$ . The perpendicular bisector of $OA$ meets $S$ at $E$ and at $F$ . Prove that $I$ is the incentre of the triangle $CEF.$

Proof:

WLOG $S$ is the unit circle. WLOG let $A, B, C, D, I, E, F$ have complex coordinates $a^2, -1, 1, d, x, e, f$ respectively.

Clearly $d = -ia$ . Since $OD \perp AB$ by definition and since $AI \perp AB$ since $BC$ is a diameter of $S$ we have that quadrilateral $ADOI$ is a parallelogram and so we have that $x = a^2 + ai$ .

Now, note that $E, F \in S$ and that the projections of either point onto $AO$ has complex coordinate $\frac{a^2}{2}$ . This implies that $e, f$ are the roots of the equation $z^2 - a^2z + a^4 = 0$ . Therefore by Vieta's formulas $e + f = a^2$ and $ef = a^4$ .

Now since the circumcircle of $CEF$ is $S$ it suffices to show that $x = -\sqrt{ef} - \sqrt{c}(\sqrt{e} + \sqrt{f})$ . Choosing the appropriate sign for the square roots given that $\angle{AOB} < 120$ we have that $\sqrt{ef} = -a^2$ and since $(\sqrt{e} + \sqrt{f})^2 = e + f + 2\sqrt{ef} = -a^2$ we have that $\sqrt{c}(\sqrt{e} + \sqrt{f}) = -ai$ so $-\sqrt{ef} - \sqrt{c}(\sqrt{e} + \sqrt{f}) = a^2 + ai$ as desired.

This is literally the worst solution possible but I felt that I had to post it for its ridiculousness. In fact, you can easily prove that $ADOI$ is a parallelogram with solely complex numbers (it requires solving a simple linear system of two equations).

This post has been edited 1 time. Last edited by Wolstenholme, Aug 5, 2014, 12:03 AM

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glad to have found a fellow chipotle lover <3
by nukelauncher, Aug 13, 2020, 6:40 AM
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by fukano_2, Jun 14, 2020, 6:24 AM
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by AopsUser101, Jan 29, 2020, 8:27 PM
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by mathleticguyyy, Jun 5, 2019, 8:26 PM
Hi. Nice Blog!
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by songssari, Jun 12, 2016, 8:21 AM
shouts make blogs happier
by briantix, Mar 18, 2016, 9:57 PM
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by mishka1980, Sep 12, 2015, 10:33 PM
This is late, but where is the ARML results post?
by donot, Aug 31, 2015, 11:07 PM
"I am Sam"
"Sam I am"
by mathwizard888, Aug 12, 2015, 9:13 PM
HW $\textcolor{white}{}$
by Eugenis, Apr 20, 2015, 10:10 PM
Uh-oh ARML practice is Thursday... I should start the homework.
by nosaj, Apr 20, 2015, 12:34 AM
Yes I am Sam, and Chebyshev polynomials aren't trivial, although they do make some problems trivial
by Wolstenholme, Apr 15, 2015, 10:00 PM
How are Chebyshev Polynomials trivial?
by nosaj, Apr 13, 2015, 4:10 AM
Are you Sam?
by Eugenis, Apr 4, 2015, 2:05 AM
@Brian: yes, yes I did #whoneedsalgskillz?

@gauss1181; hey!
by Wolstenholme, Mar 1, 2015, 11:25 PM
hello!!!
by gauss1181, Nov 27, 2014, 12:19 AM
Hi Wolstenholme did you actually use calc on that tstst problem
by briantix, Aug 2, 2014, 12:25 AM

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Isl 2002 g3 (complex this time)

by Wolstenholme, Aug 4, 2014, 4:43 AM

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