MathDB

2021 Iranian Geometry Olympiad

Part of Iranian Geometry Olympiad

Subcontests

(5)
5
3

<A_1A_2A_3 +< A_2A_3A_4 +... + < A_{2021}A_1A_2 = 360^o

Let A1,A2,...,A2021A_1, A_2, . . . , A_{2021} be 20212021 points on the plane, no three collinear and ∠A1A2A3+∠A2A3A4+...+∠A2021A1A2=360o,\angle A_1A_2A_3 + \angle A_2A_3A_4 +... + \angle A_{2021}A_1A_2 = 360^o, in which by the angle ∠Ai−1AiAi+1\angle A_{i-1}A_iA_{i+1} we mean the one which is less than 180o180^o (assume that A2022=A1A_{2022} =A_1 and A0=A2021A_0 = A_{2021}). Prove that some of these angles will add up to 90o90^o.
Proposed by Morteza Saghafian - Iran

Igo 2021 intermediate p5

Consider a convex pentagon ABCDEABCDE and a variable point XX on its side CDCD. Suppose that points K,LK, L lie on the segment AXAX such that AB=BKAB = BK and AE=ELAE = EL and that the circumcircles of triangles CXKCXK and DXLDXL intersect for the second time at YY . As XX varies, prove that all such lines XYXY pass through a fixed point, or they are all parallel. Proposed by Josef Tkadlec - Czech Republic

AD is Euler line of triangle IKL

Given a triangle ABCABC with incenter II. The incircle of triangle ABCABC is tangent to BCBC at DD. Let PP and QQ be points on the side BC such that ∠PAB=∠BCA\angle PAB = \angle BCA and ∠QAC=∠ABC\angle QAC = \angle ABC, respectively. Let KK and LL be the incenter of triangles ABPABP and ACQACQ, respectively. Prove that ADAD is the Euler line of triangle IKLIKL.
Proposed by Le Viet An, Vietnam
4
3

concentric circumcircles wanted, starting with isosceles trapezoid

In isosceles trapezoid ABCDABCD (AB∥CDAB \parallel CD) points EE and FF lie on the segment CDCD in such a way that D,E,FD, E, F and CC are in that order and DE=CFDE = CF. Let XX and YY be the reflection of EE and CC with respect to ADAD and AFAF. Prove that circumcircles of triangles ADFADF and BXYBXY are concentric.
Proposed by Iman Maghsoudi - Iran

Igo 2021 intermediate p4

Let ABCABC be a scalene acute-angled triangle with its incenter II and circumcircle Γ\Gamma. Line AIAI intersects Γ\Gamma for the second time at MM. Let NN be the midpoint of BCBC and TT be the point on Γ\Gamma such that IN⊥MTIN \perp MT. Finally, let PP and QQ be the intersection points of TBTB and TCTC, respectively, with the line perpendicular to AIAI at II. Show that PB=CQPB = CQ. Proposed by Patrik Bak - Slovakia

any circle passing through two points contains at least 673 of others

20212021 points on the plane in the convex position, no three collinear and no four concyclic, are given. Prove that there exist two of them such that every circle passing through these two points contains at least 673673 of the other points in its interior. (A finite set of points on the plane are in convex position if the points are the vertices of a convex polygon.)
3
3

angles of lines wanted, heart by 3 semicircles constructed

As shown in the following figure, a heart is a shape consist of three semicircles with diameters ABAB, BCBC and ACAC such that BB is midpoint of the segment ACAC. A heart ω\omega is given. Call a pair (P,P′)(P, P') bisector if PP and P′P' lie on ω\omega and bisect its perimeter. Let (P,P′)(P, P') and (Q,Q′)(Q,Q') be bisector pairs. Tangents at points P,P′,QP, P', Q, and Q′Q' to ω\omega construct a convex quadrilateral XYZTXYZT. If the quadrilateral XYZTXYZT is inscribed in a circle, find the angle between lines PP′PP' and QQ′QQ'. https://cdn.artofproblemsolving.com/attachments/3/c/8216889594bbb504372d8cddfac73b9f56e74c.png
Proposed by Mahdi Etesamifard - Iran

Igo intermediate p3

Given a convex quadrilateral ABCDABCD with AB=BCAB = BC and ∠ABD=∠BCD=90\angle ABD = \angle BCD = 90.Let point EE be the intersection of diagonals ACAC and BDBD. Point FF lies on the side ADAD such that AFFD=CEEA\frac{AF}{F D}=\frac{CE}{EA}.. Circle ω\omega with diameter DFDF and the circumcircle of triangle ABFABF intersect for the second time at point KK. Point LL is the second intersection of EFEF and ω\omega. Prove that the line KLKL passes through the midpoint of CECE. Proposed by Mahdi Etesamifard and Amir Parsa Hosseini - Iran

proving tangency and collinearity

Consider a triangle ABCABC with altitudes AD,BEAD, BE, and CFCF, and orthocenter HH. Let the perpendicular line from HH to EFEF intersects EF,ABEF, AB and ACAC at P,TP, T and LL, respectively. Point KK lies on the side BCBC such that BD=KCBD=KC. Let ω\omega be a circle that passes through HH and PP, that is tangent to AHAH. Prove that circumcircle of triangle ATLATL and ω\omega are tangent, and KHKH passes through the tangency point.
2
3

diagonal of inscribed KLMN in square ABCD is // to side of ABCD having 1/2 area

Points K,L,M,NK, L, M, N lie on the sides AB,BC,CD,DAAB, BC, CD, DA of a square ABCDABCD, respectively, such that the area of KLMNKLMN is equal to one half of the area of ABCDABCD. Prove that some diagonal of KLMNKLMN is parallel to some side of ABCDABCD.
Proposed by Josef Tkadlec - Czech Republic

IGO 2021 P2

Let ABCDABCD be a parallelogram. Points E,FE, F lie on the sides AB,CDAB, CD respectively, such that ∠EDC=∠FBC\angle EDC = \angle FBC and ∠ECD=∠FAD\angle ECD = \angle FAD. Prove that AB≥2BCAB \geq 2BC.
Proposed by Pouria Mahmoudkhan Shirazi - Iran

two circles intersecting, prove that three lines are concurrent

Two circles Γ1\Gamma_1 and Γ2\Gamma_2 meet at two distinct points AA and BB. A line passing through AA meets Γ1\Gamma_1 and Γ2\Gamma_2 again at CC and DD respectively, such that AA lies between CC and DD. The tangent at AA to Γ2\Gamma_2 meets Γ1\Gamma_1 again at EE. Let FF be a point on Γ2\Gamma_2 such that FF and AA lie on different sides of BDBD, and 2∠AFC=∠ABC2\angle AFC=\angle ABC. Prove that the tangent at FF to Γ2\Gamma_2, and lines BDBD and CECE are concurrent.
1
3

a shape with at least two reflection symmetries from 4 given shapes

With putting the four shapes drawn in the following figure together make a shape with at least two reflection symmetries. https://cdn.artofproblemsolving.com/attachments/6/0/8ace983d3d9b5c7f93b03c505430e1d2d189fd.png
Proposed by Mahdi Etesamifard - Iran

IGO 2021 P1

Let ABCABC be a triangle with AB=ACAB = AC. Let HH be the orthocenter of ABCABC. Point EE is the midpoint of ACAC and point DD lies on the side BCBC such that 3CD=BC3CD = BC. Prove that BE⊥HDBE \perp HD.
Proposed by Tran Quang Hung - Vietnam

Proving ∠BHF=90

Acute-angled triangle ABCABC with circumcircle ω\omega is given. Let DD be the midpoint of ACAC, EE be the foot of altitude from AA to BCBC, and FF be the intersection point of ABAB and DEDE. Point HH lies on the arc BCBC of ω\omega (the one that does not contain AA) such that ∠BHE=∠ABC\angle BHE=\angle ABC. Prove that ∠BHF=90∘\angle BHF=90^\circ.