P is a point in the plane of ABC

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Source: India tst 2006 p4

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#1 h Jun 27, 2012, 4:50 AM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle and let $P$ be a point in the plane of $ABC$ that is inside the region of the angle $BAC$ but outside triangle $ABC$ .

(a) Prove that any two of the following statements imply the third.

(i) the circumcentre of triangle $PBC$ lies on the ray $\stackrel{\to}{PA}$ .

(ii) the circumcentre of triangle $CPA$ lies on the ray $\stackrel{\to}{PB}$ .

(iii) the circumcentre of triangle $APB$ lies on the ray $\stackrel{\to}{PC}$ .

(b) Prove that if the conditions in (a) hold, then the circumcentres of triangles $BPC,CPA$ and $APB$ lie on the circumcircle of triangle $ABC$ .

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#2 h Jul 3, 2012, 1:47 AM • 2 Y

Y by Adventure10, Mango247

Part (a). Let $O_A$ denote the circumcenter of $PBC$ , and define $O_B$ and $O_C$ similarly. Let $P_A$ be the reflection of $P$ through $O_A$ , and define $P_B$ and $P_C$ similarly. We will prove that (i) and (ii) imply (iii); the proof for the other two cases is entirely similar.

Assume (i) and (ii) hold, so that $O_A\in PA$ and $O_B\in PB$ . Since $O_A$ is the intersection of the perpendicular bisectors of $PB$ and $PC$ , $P_AC\perp PC$ and $P_AB\perp PB$ . Similarly, $P_BA\perp PA$ and $P_BC\perp PC$ . Therefore $P_BA$ , $P_AB$ , and $PC$ are altitudes of triangle $P_AP_BP$ , so they concur at the orthocenter $H$ . But if $J$ is $H$ under a homothety with ratio $\frac12$ about $P$ , it is clear that $J$ is the intersection of the perpendicular bisectors of $PB$ and $PA$ , so $J=O_C$ . Since $H$ lies on $PC$ , so does $O_C$ , as desired.

Part (b). Preserving the notation defined above, note that $P_A$ is the orthocenter of $P_BP_CP$ since $P_C$ is the orthocenter of $P_AP_BP$ as deduced in the second paragraph. Therefore the circumcircle of $ABC$ is the nine-point circle of triangle $O_CO_BP$ because $A,B,C$ are the feet of the altitudes. Now $O_C$ and $O_B$ are midpoints of sides $P_CP$ and $P_BP$ , and $O_A$ is midpoint of $P_AP$ , and it is known that the nine-point circle passes through these three points. Therefore the circumcenters lie on circle $ABC$ , as desired.

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