### 2012 IMO SL #G6

**Problem: **Let $ABC$ be a triangle with circumcenter $O$ and incenter $I$. The points $D,E$ and $F$ on the sides $BC,CA$ and $AB$ respectively are such that $BD+BF=CA$ and $CD+CE=AB$. The circumcircles of the triangles $BFD$ and $CDE$ intersect at $P \neq D$. Prove that $OP=OI$.

**Solution 1: **Let circle with radius $OI$ intersect $BI, CI$ at $S, T \ne I$ respectively. Let $H$ be the foot of $S$ to $AB$ and let $BI$ meet $(ABC)$ at $M \ne A.$ It's known that if any circle passing through $S,B$ intersects $BC,BA$ at $D',F'$ respectively then $BD'+BF' = 2BH.$ But$$BH = BS \cdot \cos \left(\frac{1}{2} \angle ABC \right) = IM \cdot \cos \left(\frac{1}{2} \angle ABC \right) = MA \cdot \cos \left(\frac{1}{2} \angle ABC \right) = \frac{1}{2}AC$$so in fact $BFSD$ cyclic, similarly $CDTE$ cyclic. Then$$\angle TPS = 360^\circ - \angle TPD - \angle SPD = \frac{1}{2} \angle ABC + \frac{1}{2} \angle ACB = \angle TIS$$so $P$ lies on $(TIS)$ centered at $O,$ done. $\square$

**Solution 2: **Let $\Gamma$ be the circumcircle of $ABC$, and let $\omega_a = (AEF)$, $\omega_b = (BFD)$, $\omega_c = (CDE)$. We'll show that $\text{Pow}(P, \Gamma) = \text{Pow}(I, \Gamma)$ with barycentric coordinates. Let $BD = p$ and $CD = q$, with $p + q = a$. Then we have $BF = b - p$ and $AF = c - b + p$, while $CE = c - q$ and $AE = b - c + q$. Now because $u$ is the power of $A$ to the circle $[u, v, w]$ we have $\omega_a = [0, c(b - p), b(c - q)]$, and $\omega_b = [c(c - b + p), 0, aq]$ and $\omega_c = [b(b - c + q), ap, 0]$. By Miquel's Theorem $P$ is on all three circles, so $P$ is the intersection of the radical axes of $(\omega_a, \omega_b)$ and $(\omega_a, \omega_c)$. This means it satisfies\[-c(c - b + p)x + c(b - p)y + (b(c - q) - aq)z = 0\]and its symmetric counterpart\[-b(b - c + q)x + (c(b - p) - ap)y + b(c - q)z = 0.\]On the other hand, we have \[\text{Pow}(P, \Gamma) = \text{Pow}(P, \Gamma) - \text{Pow}(P, \omega_a) = -c(b - p)y - b(c - q)z\]if $P$ is homogenized, while\[\text{Pow}(I, \Gamma) = -2Rr = -\frac{abc}{2K}\cdot \frac{K}{s} = -\frac{abc}{a + b + c},\]so we want to show\[\frac{c(b - p)y + b(c - q)z}{x + y + z} = \frac{abc}{a + b + c}.\]This rearranges to\[-abcx + c(b^2 + bc - p(a + b + c))y + b(c^2 + bc - q(a + b + c))z = 0.\]But this equation is $b$ times the first equation plus $c$ times the second, so since $P$ satisfies both of those equations, it must satisfy this one as well. $\square$

**Solution 3: **Let $\omega$ be the circle centered at $O$ with radius $OI$. Notice that $AF+AE=BD+DC=BC$, hence the problem is symmetric about $A,B,C$. Let $A_1,A_2$ be the second intersection of $AI$ with $\omega$ and $(AFE)$ respectively. Let $M$ be the projection of $O$ on $AA'$, then $$AA_1=2AM-AI=2R\sin\angle OAI-4R\sin\frac{B}{2}\sin\frac{C}{2}=2R\sin\frac{B-C}{2}-4R\sin\frac{B}{2}\sin\frac{C}{2}=2R\sin\frac{A}{2}$$Meanwhile, by Ptolemy's theorem, $$AA_2=\frac{(AF+AE)\cdot AF}{FE}=BC\cdot\frac{AF}{FE}=\frac{2R\sin A\sin\frac{A}{2}}{\sin A}=2R\sin\frac{A}{2}=AA_1$$Hence $A_1=A_2$, denote this common point by $A'$, define $B'$ and $C'$ symmetrically. We finish the problem by directly applying Miquel's theorem to triangle $BIC$ and points $B',C',D$. $\square$

very creative solution 3

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