This is a particular case of a problem recently discussed. See http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=535534 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=532462

by perspectivity, please help me! We have that a triangle $A_1B_1C_1$ such that construct by the lines - perpendicular bisector of the $OA',OB',OC'$, here $A_1$ is intersection point of the perpendicular bisector $OB'$ and $OC'$ and similar that ... For (i) we need prove that $ABC$ and $A_1B_1C_1$ have a line perspectivity. So we prove that they have a center perspectivity. But, i haven't proof. Please help me, in this idea!

But how about second part ???

My solution: Let $ H $ be the orthocenter of $ \triangle ABC $ . Let $ D $ be the intersection of $ AH $ and $ (ABC) $ . Let $ K $ be the Anti-steiner point of $ OH $ (so-called Kiepert focus) of $ \triangle ABC $ . Let $ O_a, O_b, O_c $ be the reflection of $ O $ in $ BC, CA, AB $ , respectively . Easy to see $ P_A, P_B, P_C $ is the center of $ (OO_aA'), (OO_bB'), (OO_cC') $ . Since $ D $ is the reflection of $ H $ in $ BC $ , so $ O_a, D, K $ are collinear . Since $ \angle OO_aK=\angle ADK=\angle OA'K $ , so $ K, O_a, A', O $ are concyclic . ie. $ K $ lie on $ (OO_aA') $ . Similarly, we can prove $ (OO_bB') $ and $ (OO_cC') $ pass through $ K $ , so we get $ (OO_aA'), (OO_bB'), (OO_cC') $ are coaxial. ie. $ P_A, P_B, P_C $ are collinear at $ l $ Since $ l $ is the perpendicular bisector of $ OK $ , so the distance between $ O $ and $ l $ is $ \frac {R}{2} $ . Q.E.D

Let $\triangle A_1B_1C_1$ be the anti-pedal triangle of $\triangle ABC$. a)We'll prove it polar dual. Given a triangle $\triangle ABC$ ,let $D,E,F$ be the intouch point of the incircle with sides $BC,AC,AB$.Let $D',E',F'$ be the antipodes of $D,E,F$ and let $D''$ be the result of $\mathcal{H}_{I,2}(D')$.Prove that $AD",BE",CF"$ are concurrent. This however is immediate by Sondat's theorem as $\triangle D"E"F"$ and $\triangle ABC$ are orthologic in $I$. b) It suffices to show that the previous concurrency point wrt $\triangle A_1B_1C_1$ lies on $(O,2R)$,or after homothety $\mathcal{H}_{O,\frac{1}{2}}$ that the lines connecting $A'$,midpoint of $A_1O$ and analogously for others intersect on $\odot O$.But the previous line intersect in Feurbach point of $\triangle A_1B_1C_1$ and hence we're done.$\blacksquare$

nikolapavlovic wrote: Let $\triangle A_1B_1C_1$ be the anti-pedal triangle of $\triangle ABC$. a)We'll prove it polar dual. Given a triangle $\triangle ABC$ ,let $D,E,F$ be the intouch point of the incircle with sides $BC,AC,AB$.Let $D',E',F'$ be the antipodes of $D,E,F$ and let $D''$ be the result of $\mathcal{H}_{I,2}(D')$.Prove that $AD",BE",CF"$ are concurrent. This however is immediate by Sondat's theorem as $\triangle D"E"F"$ and $\triangle ABC$ are orthologic in $I$. b) It suffices to show that the previous concurrency point wrt $\triangle A_1B_1C_1$ lies on $(O,2R)$,or after homothety $\mathcal{H}_{O,\frac{1}{2}}$ that the lines connecting $A'$,midpoint of $A_1O$ and analogously for others intersect on $\odot O$.But the previous line intersect in Feurbach point of $\triangle A_1B_1C_1$ and hence we're done.$\blacksquare$ what is the proof of this concurrency over Feurbach point?

will use complex numbers in order to solve the problem! we take the circumcircle as the unit circle, we can obtain $P_a$ as it's one perpendicular bisector of $OA'$ and on side $BC$ as $P_a=\frac{a(\sum ab)}{a^2-bc}$ and similarly $P_B=\frac{b(\sum ab)}{b^2-ac}$, $P_c=\frac{c(\sum ab)}{c^2-ba}$ for the first part we should prove: $\frac{P_a-P_b}{P_a-P_c} \in R \implies \frac{P_a-P_b}{P_a-P_c} =\frac{(b-a)(c^2-ab)}{(c-a)(b^2-ac)}$ which is obviously real! and for the second part we prove the altitude of $\triangle OP_aP_b$ WRT $O$ is equal to $\frac{1}{2}$ we know the area of $\triangle XYZ$ is equal to: $S_{xyz} = \frac{i}{4}.\left|\begin{array}{clr} x & y & z\\ \overline{x} & \overline{y} & \overline{z} \\ 1 & 1 & 1 \end{array}\right|$ which whould implies us to prove : $\frac{i}{4}.\left|\begin{array}{clr} p_a & p_b & 0\\ \overline{p_a} & \overline{p_b} & 0 \\ 1 & 1 & 1 \end{array}\right| = \frac{1}{4}. |p_a-p_b|$ and because we know both parts are real so we can square both sides and it will end up with result desired !