Angles

141 results

Croatian Mathematical Olympiad 2010 Problem 1-3

Neka je DD točka na stranici AC\overline{AC} trokuta ABCABC. Neka su EE i FF točke na dužinama BD\overline{BD} i BC\overline{BC} redom, takve da je BAE=CAF\measuredangle BAE = \measuredangle CAF. Neka su PP i QQ točke na dužinama BC\overline{BC} i BD\overline{BD} redom, takve da je EPCDEP \parallel CD i FQCDFQ \parallel CD. Dokaži da je BAP=CAQ\measuredangle BAP = \measuredangle CAQ.

Croatian Mathematical Olympiad 2010 Problem M-3

Unutar trokuta ABCABC dana je točka PP takva da je

ABP=PCA=13(ABC+BCA).\measuredangle ABP = \measuredangle PCA = \frac{1}{3} (\measuredangle ABC + \measuredangle BCA).

Dokaži da je ABAC+PB=ACAB+PC\frac{|AB|}{|AC| + |PB|} = \frac{|AC|}{|AB| + |PC|}.

Croatian Mathematical Olympiad 2011 Problem 2-3

Na polukružnici s promjerom AB\overline{AB} dane su točke KK i LL. Simetrala dužine AB\overline{AB} siječe dužinu KL\overline{KL} u točki UU i pritom su točke AA i KK s jedne strane te simetrale, a BB i LL s druge. Neka je NN nožište okomice iz sjecišta pravaca AKAK i BLBL na pravac ABAB, a VV točka na pravcu KLKL takva da je VAU=VBU\measuredangle VAU = \measuredangle VBU.

Dokaži da su pravci NVNV i KLKL međusobno okomiti.

Croatian Mathematical Olympiad 2011 Problem I-3

Neka je kk upisana kružnica šiljastokutnog trokuta ABCABC sa središtem u točki II, a kck_c pripisana kružnica istog trokuta nasuprot kuta BCA\angle BCA. Ako je točka DD diralište stranice AB\overline{AB} i kružnice kck_c, a točka SS sjecište pravca DIDI s kružnicom kck_c (različito od točke DD), dokaži da je pravac DIDI simetrala kuta ASB\angle ASB.

Croatian Mathematical Olympiad 2011 Problem M-3

Unutar šiljastokutnog trokuta ABCABC dana je točka SS takva da je SAB=SBC=SCA\measuredangle SAB = \measuredangle SBC = \measuredangle SCA. Pravci ASAS, BSBS, CSCS sijeku redom kružnice opisane trokutima SBCSBC, SCASCA, SABSAB u točkama A1A_1, B1B_1, C1C_1. Dokaži nejednakost P(A1CB)+P(B1AC)+P(C1BA)3P(ABC).P(A_1CB) + P(B_1AC) + P(C_1BA) \geq 3P(ABC).

Croatian Mathematical Olympiad 2013 Problem 2-3

U trokutu ABCABC kut pri vrhu BB iznosi 120°120°. Neka su A1,B1,C1A_1, B_1, C_1 redom točke na stranicama BC\overline{BC}, CA\overline{CA}, AB\overline{AB}, takve da su AA1,BB1,CC1AA_1, BB_1, CC_1 simetrale kutova trokuta ABCABC. Odredi kut A1B1C1\measuredangle A_1B_1C_1.

Croatian Mathematical Olympiad 2014 Problem 1-3

Dan je trokut ABCABC u kojem je AB>AC|AB| > |AC|. Neka je PP polovište stranice BC\overline{BC}, a SS točka u kojoj simetrala kuta BAC\measuredangle BAC sijeće tu stranicu. Paralela s pravcem ASAS kroz točku PP sijeće pravce ABAB i ACAC redom u točkama XX i YY. Neka je ZZ točka takva da je YY polovište dužine XZ\overline{XZ} te neka se pravci BYBY i CZCZ sijeku u točki DD.

Dokaži da je simetrala kuta BDC\measuredangle BDC paralelna s pravcem ASAS.

Croatian Mathematical Olympiad 2015 Problem I-3

U četverokutu ABCDABCD je DAB=110\measuredangle DAB = 110^\circ, ABC=50\measuredangle ABC = 50^\circ, BCD=70\measuredangle BCD = 70^\circ. Neka su MM i NN polovišta dužina AB\overline{AB} i CD\overline{CD} redom. Za točku PP na dužini MN\overline{MN} vrijedi AM:CN=MP:NP|AM| : |CN| = |MP| : |NP| i AP=CP|AP| = |CP|. Odredi veličinu APC\measuredangle APC.

Croatian Mathematical Olympiad 2016 Problem I-3

Točka OO je središte kružnice opisane šiljastokutnom trokutu ABCABC. Točke EE i FF redom su odabrane na dužinama OB\overline{OB} i OC\overline{OC} tako da je BE=OF|BE| = |OF|. Ako su MM i NN redom polovišta kružnih lukova EOA^\widehat{EOA} i AOF^\widehat{AOF}, dokaži da je ENO+OMF=2BAC\measuredangle ENO + \measuredangle OMF = 2\measuredangle BAC.

Croatian Mathematical Olympiad 2017 Problem 1-3

U trokutu ABCABC vrijedi AB<BC|AB| < |BC|. Točka II je središte kružnice upisane tom trokutu. Neka je MM polovište stranice AC\overline{AC}, a NN polovište luka AC^\widehat{AC} opisane kružnice tog trokuta koji sadrži točku BB. Dokaži da je

IMA=INB.\measuredangle IMA = \measuredangle INB.

Croatian Mathematical Olympiad 2020 Problem 2-3

Dana je kružnica promjera AB\overline{AB}. Na toj kružnici, s različitih strana pravca ABAB, nalaze se točke CC i DD takve da vrijedi AC<BC|AC| < |BC| i AC<AD|AC| < |AD|. Točka PP pripada dužini BC\overline{BC} te vrijedi CAP=ABC\measuredangle CAP = \measuredangle ABC. Okomica iz točke CC na pravac ABAB siječe pravac BDBD u točki QQ. Pravci PQPQ i ADAD sijeku se u točki RR, a pravci PQPQ i CDCD u točki TT.

Ako je AR=RQ|AR| = |RQ|, dokaži da su pravci ATAT i PQPQ međusobno okomiti.

Croatian Mathematical Olympiad 2022 Problem I-3

Neka je ABCABC raznostraničan šiljastokutan trokut. Točka NN je polovište duljeg luka BC^\widehat{BC} kružnice opisane trokutu ABCABC. Neka je kk kružnica promjera BC\overline{BC}.

Simetrala kuta BAC\measuredangle BAC siječe kružnicu kk u točkama DD i EE, a D1D_1 i E1E_1 su točke takve da su DD1\overline{DD_1} i EE1\overline{EE_1} promjeri kružnice kk.

Dokaži da polovište dužine BC\overline{BC} pripada kružnici opisanoj trokutu NE1D1NE_1D_1.

Croatian Mathematical Olympiad 2025 Problem 1-1

Neka je MM polovište stranice AB\overline{AB} trokuta ABCABC u kojem je BC>AC|BC| > |AC|, te neka je NN nožište okomice iz točke AA na dužinu CM\overline{CM}. Neka je PP točka na pravcu ANAN takva da je PBPB okomito na CBCB.

Ako vrijedi CPB=CBA\measuredangle CPB = \measuredangle CBA, dokaži da je BAC=90°\measuredangle BAC = 90°.

International Mathematical Olympiad 1960 Problem 7

An isosceles trapezoid with bases aa and cc and altitude hh is given.

(a) On the axis of symmetry of this trapezoid, find all points PP such that both legs of the trapezoid subtend right angles at PP.

(b) Calculate the distance of PP from either base.

(c) Determine under what conditions such points PP actually exist. (Discuss various cases that might arise.)

International Mathematical Olympiad 1975 Problem 3

On the sides of an arbitrary triangle ABCABC, triangles ABR,BCP,CAQABR, BCP, CAQ are constructed externally with CBP=CAQ=45°\angle CBP = \angle CAQ = 45°, BCP=ACQ=30°\angle BCP = \angle ACQ = 30°, ABR=BAR=15°\angle ABR = \angle BAR = 15°. Prove that QRP=90°\angle QRP = 90° and QR=RPQR = RP.

International Mathematical Olympiad 1983 Problem 2

Let AA be one of the two distinct points of intersection of two unequal coplanar circles C1C_1 and C2C_2 with centers O1O_1 and O2O_2, respectively. One of the common tangents to the circles touches C1C_1 at P1P_1 and C2C_2 at P2P_2, while the other touches C1C_1 at Q1Q_1 and C2C_2 at Q2Q_2. Let M1M_1 be the midpoint of P1Q1P_1Q_1, and M2M_2 be the midpoint of P2Q2P_2Q_2. Prove that O1AO2=M1AM2\angle O_1AO_2 = \angle M_1AM_2.

International Mathematical Olympiad 1986 Problem 2

A triangle A1A2A3A_{1}A_{2}A_{3} and a point P0P_{0} are given in the plane. We define As=As3A_{s} = A_{s-3} for all s4s \geq 4. We construct a set of points P1,P2,P3,P_{1}, P_{2}, P_{3}, \ldots, such that Pk+1P_{k+1} is the image of PkP_{k} under a rotation with center Ak+1A_{k+1} through angle 120120^{\circ} clockwise (for k=0,1,2,k = 0, 1, 2, \ldots). Prove that if P1986=P0P_{1986} = P_{0}, then the triangle A1A2A3A_{1}A_{2}A_{3} is equilateral.

International Mathematical Olympiad 1989 Problem 2

In an acute-angled triangle ABCABC the internal bisector of angle AA meets the circumcircle of the triangle again at A1A_1. Points B1B_1 and C1C_1 are defined similarly. Let A0A_0 be the point of intersection of the line AA1AA_1 with the external bisectors of angles BB and CC. Points B0B_0 and C0C_0 are defined similarly. Prove that:

(i) The area of the triangle A0B0C0A_0B_0C_0 is twice the area of the hexagon AC1BA1CB1AC_1BA_1CB_1.

(ii) The area of the triangle A0B0C0A_0B_0C_0 is at least four times the area of the triangle ABCABC.

International Mathematical Olympiad 1993 Problem 2

Let DD be a point inside acute triangle ABCABC such that ADB=ACB+π/2\angle ADB = \angle ACB + \pi/2 and ACBD=ADBCAC \cdot BD = AD \cdot BC.

(a) Calculate the ratio (ABCD)/(ACBD)(AB \cdot CD)/(AC \cdot BD).

(b) Prove that the tangents at CC to the circumcircles of ACD\triangle ACD and BCD\triangle BCD are perpendicular.

International Mathematical Olympiad 1995 Problem 5

Let ABCDEFABCDEF be a convex hexagon with AB=BC=CDAB = BC = CD and DE=EF=FADE = EF = FA, such that BCD=EFA=π/3\angle BCD = \angle EFA = \pi/3. Suppose GG and HH are points in the interior of the hexagon such that AGB=DHE=2π/3\angle AGB = \angle DHE = 2\pi/3. Prove that AG+GB+GH+DH+HECFAG + GB + GH + DH + HE \geq CF.

International Mathematical Olympiad 1997 Problem 2

The angle at AA is the smallest angle of triangle ABCABC. The points BB and CC divide the circumcircle of the triangle into two arcs. Let UU be an interior point of the arc between BB and CC which does not contain AA. The perpendicular bisectors of ABAB and ACAC meet the line AUAU at VV and WW, respectively. The lines BVBV and CWCW meet at TT. Show that

AU=TB+TC.AU = TB + TC.

International Mathematical Olympiad 2001 Problem 5

In a triangle ABCABC, let APAP bisect BAC\angle BAC, with PP on BCBC, and let BQBQ bisect ABC\angle ABC, with QQ on CACA.

It is known that BAC=60\angle BAC = 60^{\circ} and that AB+BP=AQ+QBAB + BP = AQ + QB.

What are the possible angles of triangle ABCABC?

International Mathematical Olympiad 2004 Problem 1

Let ABCABC be an acute-angled triangle with ABACAB \neq AC. The circle with diameter BCBC intersects the sides ABAB and ACAC at MM and NN respectively. Denote by OO the midpoint of the side BCBC. The bisectors of the angles BAC\angle BAC and MON\angle MON intersect at RR. Prove that the circumcircles of the triangles BMRBMR and CNRCNR have a common point lying on the side BCBC.

International Mathematical Olympiad 2004 Problem 5

In a convex quadrilateral ABCDABCD the diagonal BDBD does not bisect the angles ABC\angle ABC and CDA\angle CDA. The point PP lies inside ABCDABCD and satisfies PBC=DBA and PDC=BDA.\angle PBC = \angle DBA \text{ and } \angle PDC = \angle BDA.

Prove that ABCDABCD is a cyclic quadrilateral if and only if AP=CPAP = CP.

International Mathematical Olympiad 2007 Problem 2

Consider five points A,B,C,DA, B, C, D and EE such that ABCDABCD is a parallelogram and BCEDBCED is a cyclic quadrilateral. Let \ell be a line passing through AA. Suppose that \ell intersects the interior of the segment DCDC at FF and intersects line BCBC at GG. Suppose also that EF=EG=ECEF = EG = EC. Prove that \ell is the bisector of angle DABDAB.

International Mathematical Olympiad 2010 Problem 2

Let II be the incentre of triangle ABCABC and let Γ\Gamma be its circumcircle. Let the line AIAI intersect Γ\Gamma again at DD. Let EE be a point on the arc BDC^\widehat{BDC} and FF a point on the side BCBC such that BAF=CAE<12BAC.\angle BAF = \angle CAE < \frac{1}{2}\angle BAC. Finally, let GG be the midpoint of the segment IFIF. Prove that the lines DGDG and EIEI intersect on Γ\Gamma.

International Mathematical Olympiad 2014 Problem 4

Points PP and QQ lie on side BCBC of acute-angled triangle ABCABC so that PAB=BCA\angle PAB = \angle BCA and CAQ=ABC\angle CAQ = \angle ABC. Points MM and NN lie on lines APAP and AQAQ, respectively, such that PP is the midpoint of AMAM, and QQ is the midpoint of ANAN. Prove that lines BMBM and CNCN intersect on the circumcircle of triangle ABCABC.

International Mathematical Olympiad 2018 Problem 6

A convex quadrilateral ABCDABCD satisfies ABCD=BCDAAB \cdot CD = BC \cdot DA. Point XX lies inside ABCDABCD so that XAB=XCDandXBC=XDA.\angle XAB = \angle XCD \quad \text{and} \quad \angle XBC = \angle XDA. Prove that BXA+DXC=180°\angle BXA + \angle DXC = 180°.

International Mathematical Olympiad 2019 Problem 2

In triangle ABCABC, point A1A_1 lies on side BCBC and point B1B_1 lies on side ACAC. Let PP and QQ be points on segments AA1AA_1 and BB1BB_1, respectively, such that PQPQ is parallel to ABAB. Let P1P_1 be a point on line PB1PB_1, such that B1B_1 lies strictly between PP and P1P_1, and PP1C=BAC\angle PP_1C = \angle BAC. Similarly, let Q1Q_1 be a point on line QA1QA_1, such that A1A_1 lies strictly between QQ and Q1Q_1, and CQ1Q=CBA\angle CQ_1Q = \angle CBA.

Prove that points PP, QQ, P1P_1, and Q1Q_1 are concyclic.

International Mathematical Olympiad 2020 Problem 1

Consider the convex quadrilateral ABCDABCD. The point PP is in the interior of ABCDABCD. The following ratio equalities hold: PAD:PBA:DPA=1:2:3=CBP:BAP:BPC.\angle PAD : \angle PBA : \angle DPA = 1 : 2 : 3 = \angle CBP : \angle BAP : \angle BPC.

Prove that the following three lines meet in a point: the internal bisectors of angles ADP\angle ADP and PCB\angle PCB and the perpendicular bisector of segment ABAB.

International Mathematical Olympiad 2021 Problem 3

Let DD be an interior point of the acute triangle ABCABC with AB>ACAB > AC so that DAB=CAD\angle DAB = \angle CAD. The point EE on the segment ACAC satisfies ADE=BCD\angle ADE = \angle BCD, the point FF on the segment ABAB satisfies FDA=DBC\angle FDA = \angle DBC, and the point XX on the line ACAC satisfies CX=BXCX = BX. Let O1O_1 and O2O_2 be the circumcentres of the triangles ADCADC and EXDEXD, respectively. Prove that the lines BCBC, EFEF, and O1O2O_1O_2 are concurrent.

International Mathematical Olympiad 2022 Problem 4

Let ABCDEABCDE be a convex pentagon such that BC=DEBC = DE. Assume that there is a point TT inside ABCDEABCDE with TB=TDTB = TD, TC=TETC = TE and ABT=TEA\angle ABT = \angle TEA. Let line ABAB intersect lines CDCD and CTCT at points PP and QQ, respectively. Assume that the points P,B,A,QP, B, A, Q occur on their line in that order. Let line AEAE intersect lines CDCD and DTDT at points RR and SS, respectively. Assume that the points R,E,A,SR, E, A, S occur on their line in that order. Prove that the points P,S,Q,RP, S, Q, R lie on a circle.

International Mathematical Olympiad 2024 Problem 4

Let ABCABC be a triangle with AB<AC<BCAB < AC < BC. Let the incentre and incircle of triangle ABCABC be II and ω\omega, respectively. Let XX be the point on line BCBC different from CC such that the line through XX parallel to ACAC is tangent to ω\omega. Similarly, let YY be the point on line BCBC different from BB such that the line through YY parallel to ABAB is tangent to ω\omega. Let AIAI intersect the circumcircle of triangle ABCABC again at PAP \neq A. Let KK and LL be the midpoints of ACAC and ABAB, respectively.

Prove that KIL+YPX=180\angle KIL + \angle YPX = 180^{\circ}.

Middle European Mathematical Olympiad 2009 Problem I-3

Let ABCDABCD be a convex quadrilateral such that ABAB and CDCD are not parallel and AB=CDAB = CD. The midpoints of the diagonals ACAC and BDBD are EE and FF. The line EFEF meets segments ABAB and CDCD at GG and HH, respectively. Show that AGH=DHG\measuredangle AGH = \measuredangle DHG.

Middle European Mathematical Olympiad 2009 Problem T-6

Suppose that ABCDABCD is a cyclic quadrilateral and CD=DACD = DA. Points EE and FF belong to the segments ABAB and BCBC respectively, and ADC=2EDF\measuredangle ADC = 2\measuredangle EDF. Segments DKDK and DMDM are height and median of the triangle DEFDEF, respectively. LL is the point symmetric to KK with respect to MM. Prove that the lines DMDM and BLBL are parallel.

Middle European Mathematical Olympiad 2013 Problem I-3

Let ABCABC be an isosceles triangle with AC=BCAC = BC. Let NN be a point inside the triangle such that 2ANB=180°+ACB2\angle ANB = 180° + \angle ACB. Let DD be the intersection of the line BNBN and the line parallel to ANAN that passes through CC. Let PP be the intersection of the angle bisectors of the angles CANCAN and ABNABN.

Show that the lines DPDP and ANAN are perpendicular.