ELMO SL 2013 G7

by Wolstenholme, Aug 1, 2014, 9:28 PM

Let $ABC$ be a triangle inscribed in circle $\omega$ , and let the medians from $B$ and $C$ intersect $\omega$ at $D$ and $E$ respectively. Let $O_1$ be the center of the circle through $D$ tangent to $AC$ at $C$ , and let $O_2$ be the center of the circle through $E$ tangent to $AB$ at $B$ . Prove that $O_1, O_2,$ and the nine-point center of $ABC$ are collinear.

Proof:

Let $H$ be the orthocenter of $\triangle{ABC}$ . I shall show that $O_{1}HO_{2}O$ is a parallelogram, which will immediately imply the desired result. We proceed with a complex number bash. Assume WLOG that the circumcircle of $\triangle{ABC}$ is the unit circle and denote the complex coordinates of $A, B, C$ by $a, b, c$ respectively.

Denote the complex coordinate of $O_2$ by $x$ . Then since $O_{2}B \perp AB$ we have that $\frac{x - b}{\overline{x} - \overline{b}} = -\frac{b - a}{\overline{b} - \overline{a}} = ab$ so $x$ satisfies the equation $x = ab\overline{x} + b - a$ .

Now, denote the complex coordinates of $E$ by $e$ . Since $E$ is on the $C$ -median of $\triangle{ABC}$ we have that $\frac{e - c}{\overline{e} - \overline{c}} = \frac{a + b - 2c}{\overline{a} + \overline{b} - 2\overline{c}} = \frac{(a + b - 2c)abc}{bc + ac - 2ab}$ so $e$ satisfies the equation $e = \frac{(a + b - 2c)abc}{bc + ac - 2ab}\overline{e} + \frac{(a + b)(c^2 - ab)}{bc + ac - 2ab}$

But $e$ lies on the unit circle so $\overline{e} = \frac{1}{e}$ and so, after plugging in, we find that $e$ satisfies: $e^2 - \frac{(a + b)(c^2 - ab)}{bc + ac - 2ab}e - \frac{(a + b - 2c)abc}{bc + ac - 2ab} = 0$ . Since $c$ also is a root of this quadratic, by Vieta's formulas we find that $e = \frac{(2c - a - b)ab}{bc + ac - 2ab}$ .

Now let $M$ denote the midpoint of segment $BE$ and let $m$ be its complex coordinate. Then we easily find that $m = \frac{b + e}{2} = \frac{b(3ac + bc - 3ab - a^2)}{2(bc + ac - 2ab)}$ . Since $O_{2}M \perp BE$ we have that $\frac{x - m}{\overline{x} - \overline{m}} = -\frac{b - e}{\overline{b} - \overline{e}}$ . But since $E \in \omega$ we have that $\overline{e} = \frac{1}{e}$ so $\frac{x - m}{\overline{x} - \overline{m}} = be = \frac{(2c - a - b)ab^2}{bc + ac - 2ab}$ .

Therefore $x = \frac{(2c - a - b)ab^2}{bc + ac - 2ab}\overline{x} + m - \overline{m} \cdot \frac{(2c - a - b)ab^2}{bc + ac - 2ab}$ . But we can compute that $\overline{m} = \frac{3ab + a^2 - 3ac - bc}{2ab(a + b - 2c)}$ so $\overline{m} \cdot \frac{(2c - a - b)ab^2}{bc + ac - 2ab} = -\frac{b(3ab + a^2 - 3ac - bc)}{2(bc + ac - 2ab)} = m$ and so $x = \frac{(2c - a - b)ab^2}{bc + ac - 2ab}\overline{x}$ .

Plugging this into our original equation $x = ab\overline{x} + b - a$ , we find that $\frac{(2c - a - b)ab^2}{bc + ac - 2ab}\overline{x} = ab\overline{x} + b - a$ and solving for $\overline{x}$ we obtain $\overline{x} = \frac{ac + bc - 2ab}{ab(c - b)}$ . Letting $O_1$ have complex coordinate $y$ we similarly find that $\overline{y} = \frac{ab + cb - 2ac}{ac(b - c)}$ .

Now since $O$ has complex coordinate $0$ and $H$ has complex coordinate $a + b + c$ it suffices to show that $\overline{x} + \overline{y} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ . But $\overline{x} + \overline{y} = \frac{ac + bc - 2ab}{ab(c - b)} + \frac{ab + cb - 2ac}{ac(b - c)} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ as desired so we are done.

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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 mathleticguyyy, Jun 5, 2019, 8:26 PM
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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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ELMO SL 2013 G7

by Wolstenholme, Aug 1, 2014, 9:28 PM

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