According to http://www.mathlinks.ro/Forum/viewtopic.php?t=20406 , we have $\frac{MN}{OI}=\frac{a}{R}$, where O is the circumcenter and I is the incenter of triangle ABC. In other words, $\frac{MN}{a}=\frac{OI}{R}$. Since $OI=\sqrt{R^2-2Rr}$, this becomes $\frac{MN}{a}=\frac{\sqrt{R^2-2Rr}}{R}$. Thus we have $\frac{MN}{BC}=\frac{MN}{a}=\frac{\sqrt{R^2-2Rr}}{R}=\sqrt{\frac{R^2-2Rr}{R^2}}=\sqrt{1-2\frac{r}{R}}$. darij

It's amazing that this configuration was throughly discussed 1.5 years ago in Crux with Mayhem (October 2003 issue, solution to problem 2776). Guess the problems just keep getting recycled, don't they. Anyways, on the contest, I came up with a brute-force solution, expressing everything in terms of $a,b,c$. It worked, but definitely not as nicely.

billzhao wrote: It's amazing that this configuration was throughly discussed 1.5 years ago in Crux with Mayhem (October 2003 issue, solution to problem 2776). Guess the problems just keep getting recycled, don't they. Anyways, on the contest, I came up with a brute-force solution, expressing everything in terms of $a,b,c$. It worked, but definitely not as nicely. billzhao, the most amazing is this: http://www.mathlinks.ro/Forum/viewtopic.php?t=30539 This appears in Moldova MO 2005! Anyway, here is my solution. Let the perendicular from I and O to AB be F',F respectively. E',E are defined similarly. Now by computing out the length of FF',EE',AM,AN, the answer is trivial by Euler Theorem. and, at last, how well did the Canadians perform in APMO?

billzhao wrote: Anyways, on the contest, I came up with a brute-force solution, expressing everything in terms of $a,b,c$. It worked, but definitely not as nicely. That's also what I did in the contest... Though it is not nice at all, it took me not much time to solve this problem. mecrazywong wrote: billzhao, the most amazing is this: http://www.mathlinks.ro/Forum/viewtopic.php?t=30539 This appears in Moldova MO 2005! Unbelievable!

Let A' , B' , C' are excenters of triangle ABC Easy to see that angle AC'N = B'A'C' - 2*CC'A' = O'C'I , were O' is circumcenter of A'B'C' AC'BIN is cyclic X is Miquel point of lines BA , CN , MN , BC Not hard to prove that X is midpoint of B'C' and triangle XMN ~ XBC , so XN/XC = MN/BC C'O' intersect IA at point V VC'I ~ AC'N and C'O'/O'V =C'X/XA , so O'C'I~XC'N XN/CX = XN/C'X = IO'/C'O' = sqrt(R*R-2*R*r)/R . done

Denote $BC=a$, $AC=b$ and $AB=c$. Let the incenter of the triangle be $I$ and let the point at which the incircle is tangent to $AB$ be $D$. Now note that \[a=2R\cdot \sin{A} = 2R \cdot \sin{\frac{A}{2}} \cdot \cos{\frac{A}{2}}\] and by trigonometric ratios in triangle $\triangle{AID}$, \[AD=r\cdot \frac{\cos{\frac{A}{2}}}{\sin{\frac{A}{2}}}.\] Combining these ratios with the fact that $AD=\frac{b+c-a}{2}$ yields that \[\frac{b+c-a}{a}=\frac{2AD}{a}=\frac{r}{R\cdot \sin^2{\frac{A}{2}}}.\] By cosine law and the fact that $1-\cos{A}=2\sin^2{\frac{A}{2}}$, it follows that \[\frac{MN^2}{a^2}=\frac{(b-a)^2 + (c-a)^2 -2(c-a)(b-a) \cos{A}}{a^2}\] \[\frac{MN^2}{a^2}=\frac{(b^2 + c^2 - 2bc\cos{A})-a(1-\cos{A})(b+c-a)}{a^2}\] \[\frac{MN^2}{a^2}=1-\frac{(1-\cos{A})(b+c-a)}{a}\] \[\frac{MN^2}{a^2}=1-\frac{2\sin^2{\frac{A}{2}}(b+c-a)}{a}\] \[\frac{MN^2}{a^2}=1-\frac{2r}{R}.\] Therefore \[\frac{MN}{BC}=\sqrt{1-\frac{2r}{R}}.\]

Let the midpoint of arc $BAC$ of $\odot ABC$ be $R$. Then rotation about $O$ with rotation angle $\angle BAC$ carries $M \to N$ and $B \to C$. Thus $$\frac{MN^2}{BC^2}=\frac{RM^2}{RB^2}$$ Applying cosine formula and extended sine formula, $$\frac{RM^2}{RB^2}=\frac{(2R\cos{\frac{A}{2}})^2+a^2-2\cdot2R\cos{\frac{A}{2}}\cdot a \cdot \cos({B+\frac{A}{2}}}){(2R\cos{\frac{A}{2}})^2}$$$$= 1+\frac{(\sin A)^2}{(\cos{\frac{A}{2}})^2} - 4\sin{\frac{A}{2}}\sin({B+\frac{A}{2}})$$$$=1-4\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}}$$ Now the remaining part is easy.

I knew that in the case when angle $A= 30^0$, $MN^2$ becomes equal to $R^2-2Rr$. Therefore the ratio becomes equal to $\sqrt{1-\frac{2r}{R}}$. First express $\sqrt{1-\frac{2r}{R}}$ in terms of $a,b$ and $c$. Then apply cosine rule on triangles $AMN$ and $ABC$ to get the value of $MN^2$ in terms of $a,b,c$. Now just expand both the sides and they become equal. Its really long but does the work.

We will use barycentric coordinates with reference triangle $\triangle ABC$. By the problem conditions, $M=\left(\frac{a}{c},\frac{c-a}{c},0\right)$ and $N=\left(\frac{a}{b},0,\frac{b-a}{b}\right)$. Then, $\overrightarrow{MN}=\left\langle \frac{a(b-c)}{bc},\frac{c-a}{c},\frac{a-b}{b}\right\rangle$, and thus $$|MN|^2=-a^2\left(\frac{(c-a)(a-b)}{bc}\right)-b^2\left(\frac{a(b-c)(a-b)}{b^2c}\right)-c^2\left(\frac{a(b-c)(c-a)}{bc^2}\right)$$$$=\frac{a^2(a-b)(a-c)-ab(a-b)(b-c)+ac(a-b)(b-c)}{bc}.$$From here, it is clear that $$\frac{|MN|^2}{|BC|^2}=\frac{a(a-b)(a-c)-b(a-b)(b-c)+c(a-c)(b-c)}{abc}$$$$=\frac{a^3+b^3+c^3-a^2b-ab^2-b^2c-bc^2-c^2a-ca^2+3abc}{abc}$$$$=\frac{abc-(a+b-c)(a-b+c)(-a+b+c)}{abc}.$$Because $$[ABC]=\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}=\frac{rs}{2},$$it follows that $$r=\frac{1}{2}\sqrt{\frac{(a+b-c)(a-b+c)(-a+b+c)}{a+b+c}}.$$Similarly, because $$[ABC]=\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}=\frac{abc}{4R},$$we can rewrite $R$ as $$\frac{abc}{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}.$$As a result, $$\frac{r}{R}=\frac{(a+b-c)(a-b+c)(-a+b+c)}{2abc},$$and $\frac{MN}{BC}=\sqrt{1-\frac{2r}{R}}$.

We use barycentrics on $\triangle ABC.$ Notice $M=(a:c-a:0)$ and $N=(a:0:b-a)$ so $\overrightarrow{MN}=\left(\frac{a(b-c)}{bc},\frac{c-a}{c},\frac{a-b}{b}\right).$ Hence, \begin{align*}|MN|^2&=-a^2\cdot\frac{c-a}{c}\cdot\frac{a-b}{b}-b^2\cdot\frac{a(b-c)}{bc}\cdot\frac{a-b}{b}-c^2\cdot\frac{a(b-c)}{bc}\cdot\frac{c-a}{c}\\&=-\frac{a((-a+b+c)(a-b+c)(a+b-c)-abc)}{bc}\\&=\frac{a}{bc}\cdot(abc-8(s-a)(s-b)(s-c))\end{align*}and $MN/BC=\sqrt{1-\frac{8(s-a)(s-b)(s-c)}{abc}}.$ Notice $$\sqrt{s(s-a)(s-b)(s-c)}=[ABC]=\frac{abc}{4R}=sr$$so \begin{align*}\frac{2r}{R}&=\frac{2\sqrt{(s-a)(s-b)(s-c)/s}}{\frac{abc}{4\sqrt{s(s-a)(s-b)(s-c)}}}\\&=\frac{8(s-a)(s-b)(s-c)}{abc}.\end{align*}Therefore, $MN/MC=\sqrt{1-2r/R}.$