Dear mathlinkers, a reference Sharygin I., Problema II 137, Problemas de geometria, Editions Mir (1986) 95. Sincerely Jean-Louis

In fact, if $AB<AC$, then: $DE^2-DF^2=AC^2-AB^2=IE^2-IF^2$. Hence the result. [Edited]

I had no idea what "karno"'s theorem was. Turns out it's Carnot's Theorem: http://en.wikipedia.org/wiki/Carnot's_theorem Thanks for teaching me something I didn't know!

It's very easy for #2. Here my solution: Let $ O $ be the circumcenter of $ ABC $ and $ M $ be the midpoint of $ AI $.Then we have $ p(E,(M))-p(E,(O))=p(F,(M))-p(F,(O))=-(p-b)(p-c) $, so we are done.

My solution: Let $ O $ be the circumcenter of $ \triangle ABC $ . Let $ I_a, I_b, I_c $ be the excenter of $ \triangle ABC $ and $ T=I_bE \cap I_cF $ . From the condition $ \Longrightarrow E, F $ is the tangent point of $ (I_b), (I_c) $ with $ AB, AC $, respectively , so $ A, E, F, T $ lie on a circle with diameter $ AT $ . ... $ ( \star ) $ Since $ I, O $ is the orthocenter, nine point center of $ \triangle I_aI_bI_c $, respectively , so the reflection $ I' $ of $ I $ in $ O $ is the circumcenter of $ \triangle I_aI_bI_c $ , hence from $ I'I_b \parallel TI_c, I'I_c \parallel TI_b $ and $ I'I_b=I'I_c $ we get $ I'I_bTI_c $ is a rhombus . Since $ AT, AI' $ are symmetry WRT $ I_bI_c $ , so we get $ AT, AI' $ are isogonal conjugate of $ \angle BAC $ , hence combine with $ (\star ) $ we get $ AI' \perp EF $ . ie. $ DI \perp EF $ ( $ \because AIDI' $ is a parallelogram ) Q.E.D

Here is a solution that uses $0$ creativity but I need less than $5$ minutes to come up with.After a simple angle chase we deduce it to prove that $<BDI=<AEF$ and $<AFE=<CDI$,now apply sine rule in $BDI$ and$CDI$ and then $M$,$N$ be points where incircle touches $AB,AC$ and now apply sine rule $BIM$ and $CIN$ and we are finished(these is not trigo bashing cause this is just sine rule and it can easily be avoided).

2003 Silk Road P2?

Sardor wrote: It's very easy for #2. Here my solution: Let $ O $ be the circumcenter of $ ABC $ and $ M $ be the midpoint of $ AI $.Then we have $ p(E,(M))-p(E,(O))=p(F,(M))-p(F,(O))=-(p-b)(p-c) $, so we are done. Very nice solution Sardor!

Let $M$ and $N$ be the feet of the perpendiculars from $I$ to $AB$ and $AC$, respectively. It suffices to prove that $ED^2+FI^2=DF^2+EI^2$. But $ED^2+FI^2=DF^2+EI^2\Leftrightarrow (EB^2+BD^2)+FI^2=(FC^2+CD^2)+EI^2\Leftrightarrow BD^2+FI^2=CD^2+EI^2\Leftrightarrow BD^2+(FN^2+NI^2)=CD^2+(EM^2+MI^2)\Leftrightarrow BD^2+FN^2=CD^2+EM^2\Leftrightarrow BD^2+AB^2=CD^2+AC^2$ , which is correct. Done.

BarisKoyuncu wrote: Let $M$ and $N$ be the feet of the perpendiculars from $I$ to $AB$ and $AC$, respectively. It suffices to prove that $ED^2+FI^2=DF^2+EI^2$. But $ED^2+FI^2=DF^2+EI^2\Leftrightarrow (EB^2+BD^2)+FI^2=(FC^2+CD^2)+EI^2\Leftrightarrow BD^2+FI^2=CD^2+EI^2\Leftrightarrow BD^2+(FN^2+NI^2)=CD^2+(EM^2+MI^2)\Leftrightarrow BD^2+FN^2=CD^2+EM^2\Leftrightarrow BD^2+AB^2=CD^2+AC^2$ , which is correct. Done. BarisKoyuncu wrote: Let $M$ and $N$ be the feet of the perpendiculars from $I$ to $AB$ and $AC$, respectively. It suffices to prove that $ED^2+FI^2=DF^2+EI^2$. But $ED^2+FI^2=DF^2+EI^2\Leftrightarrow (EB^2+BD^2)+FI^2=(FC^2+CD^2)+EI^2\Leftrightarrow BD^2+FI^2=CD^2+EI^2\Leftrightarrow BD^2+(FN^2+NI^2)=CD^2+(EM^2+MI^2)\Leftrightarrow BD^2+FN^2=CD^2+EM^2\Leftrightarrow BD^2+AB^2=CD^2+AC^2$ , which is correct. Done. This is a nice solution!! I didn't think about using Pythagoras theorem for the 90-degree angle.