Let $ABC$ be triangle with circumcircle $(O)$. Let $AO$ cut $(O)$ again at $A'$. Perpendicular bisector of $OA'$ cut $BC$ at $P_A$. $P_B,P_C$ define similarly. Prove that : I) Point $P_A,P_B,P_C$ are collinear. II ) Prove that the distance of $O$ from this line is equal to $\frac {R}{2}$ where $R$ is the radius of the circumcircle.
Problem
Source: Iran Third Round 2013 - Geometry Exam - Problem 5
Tags: geometry, circumcircle, geometric transformation, reflection, perpendicular bisector, geometry unsolved
08.09.2013 04:43
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
17.09.2013 15:39
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!
24.09.2013 22:25
But how about second part ???
21.10.2014 14:41
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
08.03.2017 21:39
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.$\blacksquare$
02.03.2018 15:37
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?
30.07.2018 05:16
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 !