Geometry (Thailand Math POSN 2nd round)

Write a full solution.

  1. Let HH be orthocenter of ABC\triangle ABC and point I,J,KI,J,K be a midpoint between each vertex A,B,CA,B,C and HH respectively. If PP is a point on circumcircle of ABC\triangle ABC not on the vertex A,B,CA,B,C, and MM be the midpoint between P,HP,H. Prove that I,J,K,MI,J,K,M lie on the same circle.

  2. Let the tangents of circumcircle of ABC\triangle ABC at point B,CB,C intersect at point DD. Prove that AD\overline{AD} is symmedian line of ABCABC.

  3. Let P,QP,Q be 2 points that are isogonal conjugate to each other in ABC\triangle ABC. If PP1\overline{PP_{1}} and QQ1\overline{QQ_{1}} is perpendicular to BC\overline{BC} at point P1P_{1} and Q1Q_{1} respectively, PP2\overline{PP_{2}} and QQ2\overline{QQ_{2}} is perpendicular to CA\overline{CA} at point P2P_{2} and Q2Q_{2} respectively, and PP3\overline{PP_{3}} and QQ3\overline{QQ_{3}} is perpendicular to AB\overline{AB} at point P3P_{3} and Q3Q_{3} respectively. Prove that P1,P2,P3,Q1,Q2,Q3P_{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 PP and QQ.

  4. Let I,N,HI,N,H be the center of incircle, nine-point circle, and orthocenter of ABC\triangle ABC respectively. Construct ID,NM\overline{ID}, \overline{NM} perpendicular to BC\overline{BC} at point D,MD,M respectively. If AH\overline{AH} intersect circumcircle of ABC\triangle ABC at point KK such that YY is a midpoint of AK\overline{AK}. Prove that IDNM=rAY2|ID - NM| = \displaystyle \left |r - \displaystyle \frac{AY}{2}\right| where rr is an inradius of ABC\triangle ABC.

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

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

Note by Samuraiwarm Tsunayoshi
4 years, 7 months ago

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The 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 - 3 years, 11 months ago

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

Shrihari B - 3 years, 10 months ago

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