Write a full solution.

Let \(H\) be orthocenter of \(\triangle ABC\) and point \(I,J,K\) be a midpoint between each vertex \(A,B,C\) and \(H\) respectively. If \(P\) is a point on circumcircle of \(\triangle ABC\) not on the vertex \(A,B,C\), and \(M\) be the midpoint between \(P,H\). Prove that \(I,J,K,M\) lie on the same circle.

Let the tangents of circumcircle of \(\triangle ABC\) at point \(B,C\) intersect at point \(D\). Prove that \(\overline{AD}\) is symmedian line of \(ABC\).

Let \(P,Q\) be 2 points that are isogonal conjugate to each other in \(\triangle ABC\). If \(\overline{PP_{1}}\) and \(\overline{QQ_{1}}\) is perpendicular to \(\overline{BC}\) at point \(P_{1}\) and \(Q_{1}\) respectively, \(\overline{PP_{2}}\) and \(\overline{QQ_{2}}\) is perpendicular to \(\overline{CA}\) at point \(P_{2}\) and \(Q_{2}\) respectively, and \(\overline{PP_{3}}\) and \(\overline{QQ_{3}}\) is perpendicular to \(\overline{AB}\) at point \(P_{3}\) and \(Q_{3}\) respectively. Prove that \(P_{1},P_{2},P_{3},Q_{1},Q_{2},Q_{3}\) lie on the same circle, and the center of that circle lies in the midpoint of \(P\) and \(Q\).

Let \(I,N,H\) be the center of incircle, nine-point circle, and orthocenter of \(\triangle ABC\) respectively. Construct \(\overline{ID}, \overline{NM}\) perpendicular to \(\overline{BC}\) at point \(D,M\) respectively. If \(\overline{AH}\) intersect circumcircle of \(\triangle ABC\) at point \(K\) such that \(Y\) is a midpoint of \(\overline{AK}\). Prove that \(|ID - NM| = \displaystyle \left |r - \displaystyle \frac{AY}{2}\right|\) where \(r\) is an inradius of \(\triangle ABC\).

Let \(U\) be a foot of altitude of \(\triangle ABC\) from point \(A\). If \(U',U''\) are reflection of \(U\) by \(\overline{CA},\overline{AB}\) respectively, such that \(\overline{U'U''}\) intersect \(\overline{CA},\overline{AB}\) at point \(V,W\) respectively. Prove that \(\overline{BV},\overline{CW}\) is perpendicular to \(\overline{CA},\overline{AB}\) respectively.

This note is a part of Thailand Math POSN 2nd round 2015.

## Comments

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TopNewestThe first question can be solved using homothety. The points A,B,C and P lie on the circumcircle. So consider a homothetic transformation of the circumcircle about orthocenter and shrink the circle to half of its radius. The mid points of H and A,B,C, P will lie on this circle. This is the nine point circle of the triangle. – Pranav Rao · 1 year, 9 months ago

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– Shrihari B · 1 year, 9 months ago

Hi Pranav I do I know you ? I know a Pranav Rao and he is from mumbai too. This is shrihari.Log in to reply