IMO 2013 #3

by Wolstenholme, Oct 27, 2014, 7:28 PM

Let the excircle of triangle $ABC$ opposite the vertex $A$ be tangent to the side $BC$ at the point $A_1$ . Define the points $B_1$ on $CA$ and $C_1$ on $AB$ analogously, using the excircles opposite $B$ and $C$ , respectively. Suppose that the circumcentre of triangle $A_1B_1C_1$ lies on the circumcircle of triangle $ABC$ . Prove that triangle $ABC$ is right-angled.

Proof:

Get ready guys... this is not going to be fun...

We proceed with complex numbers. Assume WLOG that the circumcircle of $\triangle{ABC}$ is the unit circle. Let $A = a^2, B = b^2, C = c^2, A_1 = a_1, B_1 = b_1, C_1 = c_1$ . 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 $c_1 = \frac{1}{2}\left(a^2 + b^2 + ac +bc - \frac{ab^2}{c} - \frac{a^2b}{c}\right)$ .

Now we can compute $a_2 - a_1 = \frac{(a - b)(a + b)(ab - ac - bc - c^2)}{2ab}$ and $a_3 - a_1 = \frac{(a - c)(a + c)(ac - ab - bc - b^2)}{2ac}$ . Similarly $\overline{a_2 - a_1} = \frac{(a - b)(a + b)(ab + ac + bc - c^2)}{2a^2b^2c^2}$ and $\overline{a_3 - a_1} = \frac{(a - c)(a + c)(ab + ac + bc - b^2)}{2a^2b^2c^2}$ and $\overline{a_2 - a_3} = \frac{(b - c)(b + c)(ab + ac + bc - a^2)}{2a^2b^2c^2}$ .

It is well-known that the circumcenter $O_1$ of $\triangle{A_1B_1C_1}$ has coordinate $o_1 = a_1 + \frac{(a_2 - a_1)(a_3 - a_1)(\overline{a_2 - a_3})}{(\overline{a_2 - a_1})(a_3 - a_1) - (\overline{a_3 - a_1})(a_2 - a_1)}$ . We can calculate the numerator of the fraction on the right to be $\frac{(b + c)(b - c)(a - b)(a - c)(a + b)(a + c)(ab - ac - bc - c^2)(ac - ab - bc - b^2)(ab + ac + bc - a^2)}{8a^4b^3c^3}$ and we can calculate its denominator as $\frac{(a - b)(a + b)^2(a - c)(a + c)^2(b - c)(b + c)^2}{4a^3b^3c^3}$ so $o_1 = a_1 + \frac{(ab - ac - bc - c^2)(ac - ab - bc - b^2)(ab + ac + bc - a^2)}{2a(a + b)(b + c)(c + a)}$ $= \boxed{\frac{a^3(b^2 + c^2) + b^3(c^2 + a^2) + c^3(a^2 + b^2) + 2abc(ab + bc + ca)}{(a + b)(b + c)(c + a)}}$

Now, note that $O_1$ is on the circumcircle of $\triangle{ABC} \Longleftrightarrow o_1\overline{o_1} = 1 \Longleftrightarrow$ $\left(\sum_{\text{sym}}(a^3b^2 + a^2b^2c)\right)\left(\sum_{\text{sym}}(a^3b + a^2bc)\right) = abc(a + b)^2(b + c)^2(c + a)^2$ . Now, we want to factor out $(a^2 + b^2)(b^2 + c^2)(c^2 + a^2)$ and it is easy to check that the remaining term should have degree $3$ , be symmetric (so we can make it out of the three-variable symmetric polynomials), and have $10$ terms. There are not a lot of options for this cubic factor, and so through plugging in small numbers for $a, b, c$ and/or guessing right, we have that the above equation factors as $(a^2 + b^2)(b^2 + c^2)(c^2 + a^2)((a + b + c)(ab + ac + bc) + abc) = 0$ .

Now I will show that $(a + b + c)(ab + ac + bc) + abc \ne 0$ . Assuming the contrary we find that $|a + b + c| = (a + b + c)(\overline{a} + \overline{b} + \overline{c}) = \frac{(ab + bc + ca)(a + b + c)}{abc} = -1 < 0$ , contradiction. Therefore two of $a^2, b^2, c^2$ sum to 0 which implies that the line between a pair of $A, B, C$ is a diameter of the circumcircle of $\triangle{ABC}$ which implies the desired result.

This post has been edited 1 time. Last edited by Wolstenholme, Oct 27, 2014, 8:44 PM

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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
wolstenholme - op
by AopsUser101, Jan 29, 2020, 8:27 PM
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Hi. Nice Blog!
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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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IMO 2013 #3

by Wolstenholme, Oct 27, 2014, 7:28 PM

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