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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 2019 Problem 2-3

Na stranici AB\overline{AB} tetivnog četverokuta ABCDABCD postoji točka XX sa svojstvom da dijagonala BD\overline{BD} raspolavlja dužinu CX\overline{CX}, a dijagonala AC\overline{AC} raspolavlja dužinu DX\overline{DX}.

Koliki je najmanji mogući omjer AB:CD|AB| : |CD| u takvom četverokutu?

Croatian Mathematical Olympiad 2021 Problem 1-3

Neka je ABCDABCD konveksni četverokut u kojem je B>90°\measuredangle B > 90°, D>90°\measuredangle D > 90° te A=C\measuredangle A = \measuredangle C. Neka su EE i FF redom točke osnosimetrične točki AA u odnosu na pravce BCBC i CDCD. Neka dužine AE\overline{AE} i AF\overline{AF} sijeku pravac BDBD redom u točkama KK i LL.

Dokaži da se kružnice opisane trokutima BKEBKE i FLDFLD diraju.

Croatian Mathematical Olympiad 2021 Problem I-3

Dan je konveksan četverokut ABCDABCD čije se dijagonale sijeku u točki PP. Neka su XX i YY točke odabrane tako da četverokuti ABPXABPX, CDXPCDXP, BCPYBCPY i DAYPDAYP budu tetivni. Pravci ABAB i CDCD sijeku se u točki QQ, pravci BCBC i DADA u točki RR, a pravci XRXR i YQYQ u točki ZZ. Dokaži da točke XX, YY, ZZ i PP pripadaju istoj kružnici.

Croatian Mathematical Olympiad 2021 Problem M-3

Neka je ABCDABCD konveksni četverokut čije se dijagonale sijeku u točki EE. Pravci ABAB i CDCD sijeku se u točki PP, a pravci ADAD i BCBC u točki QQ. Neka je XX sjecište kružnica opisanih trokutima EBCEBC i EDAEDA različito od EE, a YY sjecište kružnica opisanih trokutima EABEAB i ECDECD različito od EE. Konačno, neka je WW sjecište kružnica opisanih trokutima PBCPBC i PDAPDA različito od PP. Dokaži da su trokuti WQYWQY i WXPWXP slični.

Croatian Mathematical Olympiad 2023 Problem I-3

Neka je ABCDABCD tetivni četverokut. Neka su MM i NN redom polovišta dužina BC\overline{BC} i AD\overline{AD}. Pretpostavimo da točke Q,A,B,PQ, A, B, P leže na pravcu u tom poretku, da je ACAC tangenta opisane kružnice trokuta ADQADQ te da je BDBD tangenta opisane kružnice trokuta BCPBCP. Dokaži da se pravac CDCD, tangenta opisane kružnice trokuta ANQANQ u točki AA i tangenta opisane kružnice trokuta BMPBMP u točki BB sijeku u jednoj točki.

Croatian Mathematical Olympiad 2025 Problem 3-2

Neka je ABCDABCD tetivan četverokut takav da je AB=AD|AB| = |AD|. Točke MM i NN nalaze se redom na stranicama BC\overline{BC} i CD\overline{CD} i pritom je BM+DN=MN|BM| + |DN| = |MN|. Dokaži da središte opisane kružnice trokuta AMNAMN pripada pravcu ACAC.

International Mathematical Olympiad 1989 Problem 4

Let ABCDABCD be a convex quadrilateral such that the sides ABAB, ADAD, BCBC satisfy AB=AD+BCAB = AD + BC. There exists a point PP inside the quadrilateral at a distance hh from the line CDCD such that AP=h+ADAP = h + AD and BP=h+BCBP = h + BC. Show that:

1h1AD+1BC.\frac{1}{\sqrt{h}} \geq \frac{1}{\sqrt{AD}} + \frac{1}{\sqrt{BC}}.

International Mathematical Olympiad 1998 Problem 1

In the convex quadrilateral ABCDABCD, the diagonals ACAC and BDBD are perpendicular and the opposite sides ABAB and DCDC are not parallel. Suppose that the point PP, where the perpendicular bisectors of ABAB and DCDC meet, is inside ABCDABCD. Prove that ABCDABCD is a cyclic quadrilateral if and only if the triangles ABPABP and CDPCDP have equal areas.

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 2005 Problem 5

Let ABCDABCD be a fixed convex quadrilateral with BC=DABC = DA and BCBC not parallel with DADA. Let two variable points EE and FF lie of the sides BCBC and DADA, respectively and satisfy BE=DFBE = DF. The lines ACAC and BDBD meet at PP, the lines BDBD and EFEF meet at QQ, the lines EFEF and ACAC meet at RR.

Prove that the circumcircles of the triangles PQRPQR, as EE and FF vary, have a common point other than PP.

International Mathematical Olympiad 2008 Problem 6

Let ABCDABCD be a convex quadrilateral with BABC|BA| \neq |BC|. Denote the incircles of triangles ABCABC and ADCADC by ω1\omega_1 and ω2\omega_2 respectively. Suppose that there exists a circle ω\omega tangent to the ray BABA beyond AA and to the ray BCBC beyond CC, which is also tangent to the lines ADAD and CDCD. Prove that the common external tangents of ω1\omega_1 and ω2\omega_2 intersect on ω\omega.

International Mathematical Olympiad 2014 Problem 3

Convex quadrilateral ABCDABCD has ABC=CDA=90°\angle ABC = \angle CDA = 90°. Point HH is the foot of the perpendicular from AA to BDBD. Points SS and TT lie on sides ABAB and ADAD, respectively, such that HH lies inside triangle SCTSCT and CHSCSB=90°,THCDTC=90°.\angle CHS - \angle CSB = 90°, \quad \angle THC - \angle DTC = 90°.

Prove that line BDBD is tangent to the circumcircle of triangle TSHTSH.

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 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 4

Let Γ\Gamma be a circle with centre II, and ABCDABCD a convex quadrilateral such that each of the segments ABAB, BCBC, CDCD and DADA is tangent to Γ\Gamma. Let Ω\Omega be the circumcircle of the triangle AICAIC. The extension of BABA beyond AA meets Ω\Omega at XX, and the extension of BCBC beyond CC meets Ω\Omega at ZZ. The extensions of ADAD and CDCD beyond DD meet Ω\Omega at YY and TT, respectively. Prove that AD+DT+TX+XA=CD+DY+YZ+ZC.AD + DT + TX + XA = CD + DY + YZ + ZC.

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 2022 Problem T-6

Let ABCDABCD be a convex quadrilateral such that AC=BDAC = BD and the sides ABAB and CDCD are not parallel. Let PP be the intersection point of the diagonals ACAC and BDBD. Points EE and FF lie, respectively, on segments BPBP and APAP such that PC=PEPC = PE and PD=PFPD = PF. Prove that the circumcircle of the triangle determined by the lines ABAB, CDCD and EFEF is tangent to the circumcircle of the triangle ABPABP.

Middle European Mathematical Olympiad 2023 Problem T-5

We are given a convex quadrilateral ABCDABCD whose angles are not right. Assume there are points P,Q,R,SP, Q, R, S on its sides AB,BC,CD,DAAB, BC, CD, DA, respectively, such that PSBDPS \parallel BD, SQBCSQ \perp BC, PRCDPR \perp CD. Furthermore, assume that the lines PR,SQPR, SQ, and ACAC are concurrent. Prove that the points P,Q,R,SP, Q, R, S are concyclic.

Grade 9 2009 Problem 2

Zadan je konveksan četverokut ABCDABCD koji nije paralelogram. Neka pravac koji prolazi kroz polovišta dijagonala četverokuta siječe stranice AB\overline{AB} i CD\overline{CD} redom u točkama MM i NN. Dokaži da trokuti ABNABN i CDMCDM imaju jednake površine.

Grade 9 2011 Problem 4

Dan je tetivni četverokut ABCDABCD. Simetrala dužine BC\overline{BC} siječe dužinu AB\overline{AB} u točki EE. Kružnica koja prolazi točkom EE, vrhom CC i polovištem FF stranice BC\overline{BC} siječe dužinu CD\overline{CD} u točki GG. Dokaži da su pravci ADAD i FGFG međusobno okomiti.

Grade 10 1996 Problem 3

Neka je A1A2A3A4A_1A_2A_3A_4 konveksan četverokut, SS sjecište njegovih dijagonala. Označimo sa sks_k površinu trokuta AkSAk+1A_kSA_{k+1}, (A5=A1A_5 = A_1), k=1,2,3,4k = 1, 2, 3, 4. Dokažite da je s22=s1s3i2s4=s1+s3s_2^2 = s_1 s_3 \quad \text{i} \quad 2 s_4 = s_1 + s_3 ako i samo ako je A1A2A3A4A_1A_2A_3A_4 paralelogram.

Grade 10 2008 Problem 4

Dan je četverokut ABCDABCD s kutovima α=60\alpha = 60^\circ, β=90\beta = 90^\circ, γ=120\gamma = 120^\circ. Dijagonale AC\overline{AC} i BD\overline{BD} sijeku se u točki SS, pri čemu je 2BS=SD=2d2|BS| = |SD| = 2d. Iz polovišta PP dijagonale AC\overline{AC} spuštena je okomica PM\overline{PM} na dijagonalu BD\overline{BD}, a iz točke SS okomica SN\overline{SN} na PB\overline{PB}.

Dokaži:

(a) MS=NS=d2|MS| = |NS| = \dfrac{d}{2};

(b) AD=DC|AD| = |DC|;

(c) P(ABCD)=9d22P(ABCD) = \dfrac{9d^2}{2}.

Grade 10 2009 Problem 2

Dan je četverokut ABCDABCD. Opisana kružnica trokuta ABCABC siječe stranice CD\overline{CD} i DA\overline{DA} redom u točkama PP i QQ, a opisana kružnica trokuta CDACDA stranice AB\overline{AB} i BC\overline{BC} redom u točkama RR i SS. Pravci BPBP i BQBQ sijeku pravac RSRS redom u točkama MM i NN. Dokaži da točke MM, NN, PP i QQ leže na istoj kružnici.

Grade 10 2026 Problem 6

Neka je ABCDABCD konveksan četverokut takav da je AB=4|AB| = 4, BC=7|BC| = 7, AD=5|AD| = 5, CBA=90|\measuredangle CBA| = 90^\circ, te su kutovi ADC\measuredangle ADC i DCB\measuredangle DCB šiljasti i međusobno sukladni. Odredi duljinu dužine CD\overline{CD}.

Grade 10 2024 Problem 4

Polukrug promjera PQ\overline{PQ} upisan je u pravokutnik ABCDABCD i dira njegove stranice AB\overline{AB} i AD\overline{AD}. Pritom se točka PP nalazi na stranici BC\overline{BC}, a točka QQ na stranici CD\overline{CD}. Ako je BP=2|BP| = 2 i DQ=1|DQ| = 1, odredi PQ|PQ|.

Grade 11 2016 Problem 1

U konveksnom četverokutu ABCDABCD vrijedi AD=CD|AD| = |CD| i ADC=90°\measuredangle ADC = 90°. Ako je AB=a|AB| = a, BC=b|BC| = b, BD=d|BD| = d, ABC=β\measuredangle ABC = \beta, dokaži da vrijedi 2d2=a2+b2+2absinβ.2d^2 = a^2 + b^2 + 2ab \sin \beta.

Grade 11 2015 Problem 4

U konveksnom četverokutu ABCDABCD vrijedi BAD=50°\measuredangle BAD = 50°, ADB=80°\measuredangle ADB = 80° i ACB=40°\measuredangle ACB = 40°.

Ako je DBC=30°+BDC\measuredangle DBC = 30° + \measuredangle BDC, izračunaj BDC\measuredangle BDC.

Grade 11 2018 Problem 4

U četverokutu ABCDABCD je DBC=DCB=50°\measuredangle DBC = \measuredangle DCB = 50° i DAB=ABC=BDC\measuredangle DAB = \measuredangle ABC = \measuredangle BDC. Dokaži da je ACBDAC \perp BD.

Grade 11 2026 Problem 4

U četverokutu ABCDABCD vrijedi ABC=90°|\measuredangle ABC| = 90°, BCD=120°|\measuredangle BCD| = 120° i CDA=90°|\measuredangle CDA| = 90°. Neka je MM sjecište dijagonala AC\overline{AC} i BD\overline{BD}. Ako je BM=1|BM| = 1 i MD=2|MD| = 2, odredi površinu četverokuta ABCDABCD.

Grade 12 2005 Problem 4

Neka je ABCDABCD konveksni četverokut i neka su PP i QQ redom točke na njegovim stranicama BC\overline{BC} i CD\overline{CD} takve da je BAP=DAQ\measuredangle BAP = \measuredangle DAQ. Dokažite da trokuti ABPABP i ADQADQ imaju jednake površine ako i samo ako je spojnica njihovih ortocentara okomita na pravac ACAC.

Grade 12 2014 Problem 5

Neka je ABCDABCD konveksni četverokut takav da vrijedi

BAD=90,BAC=2BDCiDBA+DCB=180.\measuredangle BAD = 90^\circ, \quad \measuredangle BAC = 2\measuredangle BDC \quad \text{i} \quad \measuredangle DBA + \measuredangle DCB = 180^\circ.

Odredi mjeru kuta DBA\measuredangle DBA.

Grade 12 2016 Problem 3

U konveksnom četverokutu ABCDABCD vrijedi BAC=48°,CAD=66°,CBD=DBA.\measuredangle BAC = 48°, \quad \measuredangle CAD = 66°, \quad \measuredangle CBD = \measuredangle DBA. Odredi kut BDC\measuredangle BDC.