IMO 2013 #3 Revisited

by Wolstenholme, Oct 28, 2014, 3:40 AM

So I feel bad about my earlier solution... it was ugly and took TOO LONG. I actually did the complex bash without even looking at synthetic stuff because I had heard that it was a massive challenge to do so.

But, since the complex bash was so awful, I revisited the problem and came up with this simple hybrid proof:

Actually, using a key synthetic observation, we can massively simplify the complex bash. Let $O_1$ be the circumcenter of $\triangle{A_1B_1C_1}$ . Assume WLOG that $O_1$ is on arc $BC$ not containing $A$ of the circumcircle of $\triangle{ABC}$ and that $BC \ge CA$ . Then we have that $O_1A_1 = O_1B_1$ and it is clear that $A_1B = AB_1 = \frac{BC + AC - AB}{2}$ . Moreover since angles $\angle{O_1BA_1}$ and $\angle{O_1AB_1}$ both subtend arc $O_1C$ of the circumcircle of $\triangle{ABC}$ they are equal so we have that $\triangle{OB_1A} \cong \triangle{OA_1B}$ so $O_1A = O_1B$ so $O_1$ is the midpoint of arc $AB$ containing $C$ .

Now we can proceed with the complex bash. Assume WLOG that the circumcircle of $\triangle{ABC}$ is the unit circle. Let $A, B, C, A_1, B_1, C_1, O_1$ have coordinates $a^2, b^2, c^2, a_1, b_1, c_1, o_1$ . Then the center of the $A$ -excircle has coordinate $-bc + ab + ac$ so we find that $a_1 = \frac{1}{2}\left(b^2 + c^2 + ab + ac - \frac{bc^2}{a} - \frac{b^2c}{a}\right)$ . Similarly $b_1 = \frac{1}{2}\left(a^2 + c^2 + ba + bc - \frac{ac^2}{b} - \frac{a^2c}{b}\right)$ and $c_1 = \frac{1}{2}\left(a^2 + b^2 + ac +bc - \frac{ab^2}{c} - \frac{a^2b}{c}\right)$ . We know from our synthetic observation that $o_1 = ab$ .

Now note that $2(a_1 - o_1) = \frac{ab^2 + ac^2 - a^2b + a^2c - bc^2 - b^2c}{a}$ and that $2(c_1 - o_1) = \frac{a^2c - 2abc + b^2c + ac^2 + bc^2 - ab^2 - a^2b}{c}$ . Therefore $4|a_1 - o_1| = 4(a_1 - o_1)(\overline{a_1} - \overline{o_1}) = \frac{(ab^2 + ac^2 - a^2b - bc^2 + a^2c - b^2c)(ab^2 + ac^2 - a^2b - b^2c - a^2c + b^2c)}{a^2b^2c^2}$ . We also find $4|c_1 - o_1| = 4(c_1 - o_1)(\overline{c_1} - \overline{o_1}) = \frac{(a^2c - 2abc + b^2c + ac^2 + bc^2 - ab^2 - a^2b)(a^2c - 2abc + b^2c - ac^2 - bc^2 + ab^2 + a^2b)}{a^2b^2c^2}$ . These expressions should clearly be equal, and writing them as differences of squares we obtain $(c^2 - ab)^2(a - b)^2 - c^2(a - b)^2(a + b)^2 = c^2(a - b)^4 - (c^2 - ab)^2(a + b)^2$ and rearranging yields $2(c^2 - ab)^2(a^2 + b^2) = 2c^2(a - b)^2(a^2 + b^2)$ which factors as $(a^2 + b^2)(c^2 - a^2)(c^2 - b^2) = 0$ . It's clear that $c^2 - a^2$ and $c^2 - b^2$ are nonzero (or else $C = A$ or $C = B$ which is impossible) so $a^2 + b^2 = 0$ . This implies $A$ and $B$ are antipodes on the circumcircle of $\triangle{ABC}$ which implies that $\angle{ACB} = 90$ as desired.

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without even looking at synthetic stuff because I had heard that it was a massive challenge to do so.

this is actually really really false (you just need to draw a good diagram)

by NewAlbionAcademy, Oct 28, 2014, 5:36 AM

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No, I meant I had heard that the complex bash was a challenge - so I wanted to do it

by Wolstenholme, Oct 28, 2014, 2:35 PM

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No, I meant I had heard that the complex bash was a challenge - so I wanted to do it

by Wolstenholme, Oct 28, 2014, 2:35 PM

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oh oops darn

by NewAlbionAcademy, Oct 28, 2014, 11:11 PM

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glad to have found a fellow chipotle lover <3
by nukelauncher, Aug 13, 2020, 6:40 AM
the random chinese tst problem is the only thing I read, but I'll assume your blog is nice and give you a shout even though you probably never use aops anymoer
by fukano_2, Jun 14, 2020, 6:24 AM
wolstenholme - op
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Hi. Nice Blog!
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shouts make blogs happier
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You were just featured on AoPS's facebook page.
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?

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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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IMO 2013 #3 Revisited

by Wolstenholme, Oct 28, 2014, 3:40 AM

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