MathDB

g9

Part of Indonesia MO Shortlist - geometry

Problems(4)

DE + EF + FD >= 1/2 (AB + BC + CA) if DC + CE = EA + AF = FB + BD

Source: Indonesia INAMO Shortlist 2008 G9

8/25/2021
Given a triangle ABCABC, the points DD, EE, and FF lie on the sides BCBC, CACA, and ABAB, respectively, are such that DC+CE=EA+AF=FB+BD.DC + CE = EA + AF = FB + BD. Prove that DE+EF+FD≥12(AB+BC+CA).DE + EF + FD \ge \frac12 (AB + BC + CA).
geometryGeometric Inequalitiesperimetersemiperimeter
[sum A_kB_k/2A_kB_k] where 2008 // lines divide ABC in 2008 equal areas

Source: Indonesia INAMO Shortlist 2009 G9 https://artofproblemsolving.com/community/c1101409_

12/11/2021
Given triangle ABCABC. Let A1B1A_1B_1, A2B2A_2B_2,... ..., A2008B2008A_{2008}B_{2008} be 20082008 lines parallel to ABAB which divide triangle ABCABC into 20092009 equal areas. Calculate the value of ⌊A1B12A2B2+A1B12A3B3+...+A1B12A2008B2008⌋ \left\lfloor \frac{A_1B_1}{2A_2B_2} + \frac{A_1B_1}{2A_3B_3} + ... + \frac{A_1B_1}{2A_{2008}B_{2008}} \right\rfloor
floor functionequal areasgeometry
<MKN=90^o wanted, midpoint and 2 arc midpoints, intersecting circles

Source: Indonesia INAMO Shortlist 2010 G9

8/27/2021
Given two circles Γ1\Gamma_1 and Γ2\Gamma_2 which intersect at points AA and BB. A line through AA intersects Γ1\Gamma_1 and Γ2\Gamma_2 at points CC and DD, respectively. Let MM be the midpoint of arc BCBC in Γ1\Gamma_1 ,which does not contains AA, and NN is the midpoint of the arc BDBD in Γ2\Gamma_2, which does not contain AA. If KK is the midpoint of CDCD, prove that ∠MKN=90o.\angle MKN = 90^o.
right anglearc midpointcirclesgeometry
angle bisector wanted, # ABCD, cyclic BCED, EF = EG = EC

Source: Indonesia INAMO Shortlist 2017 G9 https://artofproblemsolving.com/community/c1101409_indonesia_shortlist__geometry

11/15/2021
It is known that ABCDABCD is a parallelogram. The point EE is taken so that BCEDBCED is a cyclic quadrilateral. Let ℓ\ell be a line that passes through AA, intersects the segment DCDC at point FF and intersects the extension of the line BCBC at GG. Given EF=EG=ECEF = EG = EC. Prove that ℓ\ell is the bisector of the angle ∠BAD\angle BAD.
geometryangle bisectorcyclic quadrilateralparallelogramequal segments