7 pages trigo bash go brrr

Here is a short, mostly synthetic, solution.

Here's a even shorter solution. Define $BI \cap DE=Y$ and $CI \cap DF=X$, by Iran Lemma they both lie in $(AIEF)$. Redefine $(DXY) \cap AD=K$, finally before we go with the claims note $EF$ is also diameter in $(AIEF)$ and $\angle EDF=45$. Claim 1: $DXYP$ is cyclic. Proof: Note $\angle AXF=\angle AYE=\angle EDF=\angle XAY=45$, thus $AXDY$ is a parallelogram , also by Reim's theorem we get $XY \parallel BC$ thus in fact $DPXY$ is a Isosceles Trapezoid. Claim 2: $P,K,M$ are colinear. Proof: Note that by 3TL lemma in $\triangle DEF$ we get that $MX,MY$ are tangent to $(DPXY)$, now by cross ratio: $$-1=(X, Y; AD \cap XY, \infty_{XY}) \overset{D}{=} (X, Y; K, P) \implies K,P,M \; \text{colinear!}$$Finishing:, It is now trivial to end by noting $(DPK), (AEF)$ are symetric w.r.t. $XY$ thus done!.

MathLuis wrote: Here's a even shorter solution. Define $BI \cap DE=Y$ and $CI \cap DF=X$, by Iran Lemma they both lie in $(AIEF)$. Redefine $(DXY) \cap AD=K$, finally before we go with the claims note $EF$ is also diameter in $(AIEF)$ and $\angle EDF=45$. Claim 1: $DXYP$ is cyclic. Proof: Note $\angle AXF=\angle AYE=\angle EDF=\angle XAY=45$, thus $AXDY$ is a parallelogram , also by Reim's theorem we get $XY \parallel BC$ thus in fact $DPXY$ is a Isosceles Trapezoid. Claim 2: $P,K,M$ are colinear. Proof: Note that by 3TL lemma in $\triangle DEF$ we get that $MX,MY$ are tangent to $(DPXY)$, now by cross ratio: $$-1=(X, Y; AD \cap XY, \infty_{XY}) \overset{D}{=} (X, Y; K, P) \implies K,P,M \; \text{colinear!}$$Finishing:, It is now trivial to end by noting $(DPK), (AEF)$ are symetric w.r.t. $XY$ thus done!. Nice! What is Iran Lemma? Why is this called that?

Here's the solution I wrote during the contest. Obviously $AEIF$ is a square. We must prove that $\frac{PD}{\sin{\angle PKD}} = AI$. Now we can ignore the entire picture, and just focus on the quadrilateral $AIPD$. The only constraint left is $\frac{AI}{ID} = \sqrt{2}$. A good way to use it is by noticing a pair of similar triangles: $$\frac {AI}{ID} = \sqrt{2} = \frac{ID}{IM} \implies \triangle IDM \sim \triangle IAD $$$\implies \angle IAD = \angle MDI = \angle APM \implies \angle PKD = \angle PAI$. So the condition is equivalent to $\frac{PD}{\sin{\angle PAI}} = AI$, which is easy to see after drawing the perpendicular from $I$ to $AP$.

@Fatemeh06 you can check either the american triangle center config handout by @i3435 at the Gaussian Curvature Website, or check EGMO (i think its also mentioned there), its a lemma on the configuration of the problem Iran TST 2009/9, thus the name.

Many of my friends bashed this, but here is a nice synthetic solution I came up with during the contest: Let $W$ be the orthocenter of $DEF$. We will prove the following claims: (1) $AW=AF=AE$ (2) $W \in AP$ (3) $W\in (PDK)$ For (1), since $\angle EWF = 180^{\circ}-\angle EDF = 135^{\circ} = 180^{\circ}-\frac{\angle FAE}{2}$ and $AE=AF$, we have that $W$ lies on the circle centered at $A$ passing through $E$ and $F$. For (2), we have $\angle WAE = 2\angle WFE = 2(90^{\circ} - \angle DEF) = \angle ABC$, so $W\in AP$. Finally, we will prove (3). For this we could erase all points except $D,E,F,A,M,P,K$, where $P$ is defined as the foot of the perpendicular from $A$ to the $D$-tangent to $DEF$. Now let $K'$ be the point on $AD$, such that $\angle DK'W=90^{\circ}$. We will prove $M,K',P$ are collinear. Let $E_1$ and $F_1$ be the feet of the altitudes from $E,F$ in triangle $DEF$. So now $P,K',W,E_1,F_1$ lie on the same circle (the one of diameter $DW$). Let $DA$ intersect $DEF$ for the second time at $L$. Then since $DA$ is the symmedian of $DEF$, $(E,F,L,D)=-1$. Projecting though $D$ onto the circle through $P,K',W,E_1,F_1$ gives us this is equivalent to $(DE_1,DF_1,DK',DP)=-1$, so $(E_1,F_1,K',P)=-1$. So $K'$ lies on the symmedian of $E_1F_1P$, which is $PM$. So $K'\equiv K$, so (3) is proved. Finally, to conclude the solution, from (1) $AW=AE=DI$ and from (2) $AW\perp BC$, so $IDWA$ is a parallelogram. Hence $AI=WD$. From (3) the diameter of $(PKD)$ is $WD$ and the diameter of $(AEF)$ is $AI=EF$, so $AI=WD$. $\square$

Obviously $M$ is the midpoint of $AI$. $IM*IA=IE^2=ID^2$, so $\angle IDM = \angle IAD$. Because $M$ lies on the perpendicular bisector to $PD$, $\angle IDM=\angle APM$. So $\angle APM = \angle IAD$, and so $\angle PKD = \angle PAI$. Now let $I'$ be the projection of $I$ onto $AP$. Then $\angle I'MI=2\angle PAI=2\angle PKD$, and $I'I=PD$, so if we translate $I'I$ by vector $I'P$ $M$ would become the circumcenter of $PKD$, and so the radius is $MI$, which is also the radius of the circumcircle of $AFE$ because $M$ is its center

Let $EF \cap BC=T$. Angle chasing and PoP gives that the triangles $PKD$ and $DTM$ are similar. Rest is easy. Their radii' ratio is equal to $ \frac{PD}{MD}$. So it is enough to prove that $ \frac{MI}{R}= \frac{PD}{MD}$, where $R$ is the radius of $(DTM)$. Obviously $(AMPT)$ is cyclic. Then $\frac{MI}{R}=2\sin (\angle ITM)$. And of course, $\angle ITM= \angle MTA=\angle MPA= \angle PMD/2$. Since $PMD$ is isosceles, we have $ \frac{PD}{MD}=2\sin (\angle ITM)$. Done!