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[Solutions] Sharygin Geometry Mathematical Olympiad 2021 (Final Round)

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Grade 8

  1. Let $ABCD$ be a convex quadrilateral. The circumcenter and the incenter of triangle $ABC$ coincide with the incenter and the circumcenter of triangle $ADC$ respectively. It is known that $AB = 1$. Find the remaining sidelengths and the angles of $ABCD$.
  2. Three parallel lines $\ell_a$, $\ell_b$, $\ell_c$ pass through the vertices of triangle $ABC$. A line $a$ is the reflection of altitude $AH_a$ about $\ell_a$. Lines $b, c$ are defined similarly. Prove that $a, b, c$ are concurrent.
  3. Three cockroaches run along a circle in the same direction. They start simultaneously from a point $S$. Cockroach $A$ runs twice as slow than $B$, and thee times as slow than $C$. Points $X$, $Y$ on segment $SC$ are such that $SX = XY =YC$. The lines $AX$ and $BY$ meet at point $Z$. Find the locus of centroids of triangles $ZAB$.
  4. Let $A_1$ and $C_1$ be the feet of altitudes $AH$ and $CH$ of an acute-angled triangle $ABC$. Points $A_2$ and $C_2$ are the reflections of $A_1$ and $C_1$ about $AC$. Prove that the distance between the circumcenters of triangles $C_2HA_1$ and $C_1HA_2$ equals $AC$.
  5. Points $A_1$, $A_2$, $A_3$, $A_4$ are not concyclic, the same for points $B_1$, $B_2$, $B_3$, $B_4$. For all $i$, $j$, $k$ the circumradii of triangles $A_iA_jA_k$ and $B_iB_jB_k$ are equal. Can we assert that $A_iA_j=B_iB_j$ for all $i, j$'?
  6. Let $ABC$ be an acute-angled triangle. Point $P$ is such that $AP = AB$ and $PB\parallel AC$. Point $Q$ is such that $AQ = AC$ and $CQ\parallel AB$. Segments $CP$ and $BQ$ meet at point $X$. Prove that the circumcenter of triangle $ABC$ lies on the circle $(PXQ)$.
  7. Let $ABCDE$ be a convex pentagon such that angles $CAB$, $BCA$, $ECD$, $DEC$ and $AEC$ are equal. Prove that $CE$ bisects $BD$.
  8. Does there exist a convex polygon such that all its sidelengths are equal and all triangle formed by its vertices are obtuse-angled?

Grade 9

  1. Three cevians concur at a point lying inside a triangle. The feet of these cevians divide the sides into six segments, and the lengths of these segments form (in some order) a geometric progression. Prove that the lengths of the cevians also form a geometric progression.
  2. A cyclic pentagon is given. Prove that the ratio of its area to the sum of the diagonals is not greater than the quarter of the circumradius.
  3. Let $ABC$ be an acute-angled scalene triangle and $T$ be a point inside it such that $\angle ATB = \angle BTC = 120^\circ$. A circle centered at point $E$ passes through the midpoints of the sides of $ABC$. For $B$, $T$, $E$ collinear, find angle $ABC$.
  4. Define the distance between two triangles to be the closest distance between two vertices, one from each triangle. Is it possible to draw five triangles in the plane such that for any two of them, their distance equals the sum of their circumradii?
  5. Let $O$ be the clrcumcenter of triangle $ABC$. Points $X$ and $Y$ on side $BC$ are such that $AX = BX$ and $AY = CY$. Prove that the circumcircle of triangle $AXY$ passes through the circumceuters of triangles $AOB$ and $AOC$.
  6. The diagonals of trapezoid $ABCD$ ($BC\parallel AD$) meet at point $O$. Points $M$ and $N$ lie on the segments $BC$ and $AD$ respectively. The tangent to the circle $AMC$ at $C$ meets the ray $NB$ at point $P$; the tangent to the circle $BND$ at $D$ meets the ray $MA$ at point $R$. Prove that $\angle BOP =\angle AOR$.
  7. Three sidelines of on acute-angled triangle are drawn on the plane. Fyodor wants to draw the altitudes of this triangle using a ruler and a compass. Ivan obstructs him using an eraser. For each move Fyodor may draw one line through two markeed points or one circle centered at a marked point and passing through another marked point. After this Fyodor may mark an arbitrary number of points (the common points of drawn lines, arbitrary points on the drawn lines or arbitrary points on the plane). For each move Ivan erases at most three of marked point. (Fyodor may not use the erased points in his constructions but he may mark them for the second time). They move by turns, Fydors begins. Initially no points are marked. Can Fyodor draw the altitudes?
  8. A quadrilateral $ABCD$ is circumscribed around a circle $\omega$ centered at $I$. Lines $AC$ and $BD$ meet at point $P$, lines $AB$ and $CD$ meet at point $£$, lines $AD$ and $BC$ meet at point $F$. Point $K$ on the circumcircle of triangle $E1F$ is such that $\angle IKP = 90^\circ$. The ray $PK$ meets $\omega$ at point $Q$. Prove that the circumcircle of triangle $EQF$ touches $\omega$.

Grade 10

  1. Let $CH$ be an altitude of right-angled triangle $ABC$ ($\angle C = 90^\circ$), $HA_1$, $HB_1$ be the bisectors of angles $CHB$, $AHC$ respectively, and $E, F$ be the midpoints of $HB_1$ and $HA_1$ respectively. Prove that the lines $AE$ and $BF$ meet on the bisector of angle $ACB$.
  2. Let $ABC$ be a scalene triangle, and $A_0$, $B_0$, $C_0$ be the midpoints of $BC$, $CA$, $AB$ respectively. The bisector of angle $C$ meets $A_0C_0$ and $B_0C_0$ at points $B_1$ and $A_1$ respectively. Prove that the lines $AB_1$, $BA_1$ and $A_0B_0$ concur.
  3. The bisector of angle $A$ of triangle $ABC$ ($AB > AC$) meets its circumcircle at point $P$. The perpendicular to $AC$ from $C$ meets the bisector of angle $A$ at point $K$. A cừcle with center $P$ and radius $PK$ meets the minor arc $PA$ of the circumcircle at point $D$. Prove that the quadrilateral $ABDC$ is circumscribed.
  4. Can a triangle be a development of a quadrangular pyramid?
  5. A secant meets one circle at points $A_1$, $B_1$։, this secant meets a second circle at points $A_2$, $B_2$. Another secant meets the first circle at points $C_1$, $D_1$ and meets the second circle at points $C_2$, $D_2$. Prove that point $A_1C_1 \cap  B_2D_2$, $A_1C_1 \cap  A_2C_2$, $A_2C_2  \cap B_1D_1$, $B_2D_2 \cap B_1D_1$ lie on a circle coaxial with two given circles.
  6. The lateral sidelines $AB$ and $CD$ of trapezoid $ABCD$ meet at point $S$. The bisector of angle $ASC$ meets the bases of the trapezoid at points $K$ and $L$ ($K$ lies inside segment $SL$). Point $X$ is chosen on segment $SK$, and point $Y$ is selected on the extension of $SL$ beyond $L$ such a way that $\angle AXC - \angle AYC = \angle ASC$. Prove that $\angle BXD - \angle BYD = \angle  BSD$.
  7. Let $I$ be the incenter of a right-angled triangle $ABC$, and $M$ be the midpoint of hypothenuse $AB$. The tangent to the circumcircle of $ABC$ at $C$ meets the line passing through $I$ and parallel to $AB$ at point $P$. Let $H$ be the orthocenter of triangle $PAB$. Prove that lines $CH$ and $PM$ meet at the incircle of triangle $ABC$.
  8. On the attraction "Merry parking", the auto has only two position of a steering wheel: "right", and "strongly right". So the auto can move along an arc with radius $r_1$ or $r_2$. The auto started from a point $A$ to the Nord, it covered the distance $\ell$ and rotated to the angle $a < 2\pi$. Find the locus of its possible endpoints.

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Name

Abel,5,Albania,2,AMM,2,Amsterdam,4,An Giang,47,Andrew Wiles,1,Anh,2,APMO,21,Austria (Áo),1,Ba Lan,1,Bà Rịa Vũng Tàu,77,Bắc Bộ,2,Bắc Giang,62,Bắc Kạn,4,Bạc Liêu,19,Bắc Ninh,54,Bắc Trung Bộ,3,Bài Toán Hay,5,Balkan,41,Baltic Way,32,BAMO,1,Bất Đẳng Thức,69,Bến Tre,73,Benelux,16,Bình Định,65,Bình Dương,39,Bình Phước,52,Bình Thuận,42,Birch,1,BMO,41,Booklet,12,Bosnia Herzegovina,3,BoxMath,3,Brazil,2,British,16,Bùi Đắc Hiên,1,Bùi Thị Thiện Mỹ,1,Bùi Văn Tuyên,1,Bùi Xuân Diệu,1,Bulgaria,6,Buôn Ma Thuột,2,BxMO,15,Cà Mau,22,Cần Thơ,28,Canada,40,Cao Bằng,12,Cao Quang Minh,1,Câu Chuyện Toán Học,43,Caucasus,3,CGMO,11,China - Trung Quốc,25,Chọn Đội Tuyển,532,Chu Tuấn Anh,1,Chuyên Đề,125,Chuyên SPHCM,7,Chuyên SPHN,30,Chuyên Trần Hưng Đạo,3,Collection,8,College Mathematic,1,Concours,1,Cono Sur,1,Contest,675,Correspondence,1,Cosmin Poahata,1,Crux,2,Czech-Polish-Slovak,28,Đà Nẵng,50,Đa Thức,2,Đại Số,20,Đắk Lắk,76,Đắk Nông,15,Danube,7,Đào Thái Hiệp,1,ĐBSCL,2,Đề Thi,1,Đề Thi HSG,2272,Đề Thi JMO,1,DHBB,32,Điện Biên,15,Định Lý,1,Định Lý Beaty,1,Đỗ Hữu Đức Thịnh,1,Do Thái,3,Doãn Quang Tiến,5,Đoàn Quỳnh,1,Đoàn Văn Trung,1,Đồng Nai,65,Đồng Tháp,64,Du Hiền Vinh,1,Đức,1,Dương Quỳnh Châu,1,Dương Tú,1,Duyên Hải Bắc Bộ,32,E-Book,31,EGMO,30,ELMO,19,EMC,11,Epsilon,1,Estonian,5,Euler,1,Evan Chen,1,Fermat,3,Finland,4,Forum Of Geometry,2,Furstenberg,1,G. Polya,3,Gặp Gỡ Toán Học,30,Gauss,1,GDTX,3,Geometry,14,GGTH,30,Gia Lai,40,Gia Viễn,2,Giải Tích Hàm,1,Giới hạn,2,Goldbach,1,Hà Giang,5,Hà Lan,1,Hà Nam,45,Hà Nội,256,Hà Tĩnh,92,Hà Trung Kiên,1,Hải Dương,71,Hải Phòng,57,Hậu Giang,14,Hélènne Esnault,1,Hilbert,2,Hình Học,33,HKUST,7,Hòa Bình,33,Hoài Nhơn,1,Hoàng Bá Minh,1,Hoàng Minh Quân,1,Hodge,1,Hojoo Lee,2,HOMC,5,HongKong,8,HSG 10,126,HSG 10 2010-2011,4,HSG 10 2011-2012,7,HSG 10 2012-2013,8,HSG 10 2013-2014,7,HSG 10 2014-2015,6,HSG 10 2015-2016,2,HSG 10 2016-2017,8,HSG 10 2017-2018,4,HSG 10 2018-2019,4,HSG 10 2019-2020,7,HSG 10 2020-2021,3,HSG 10 2021-2022,4,HSG 10 2022-2023,11,HSG 10 2023-2024,1,HSG 10 Bà Rịa Vũng Tàu,2,HSG 10 Bắc Giang,1,HSG 10 Bạc Liêu,2,HSG 10 Bình Định,1,HSG 10 Bình Dương,1,HSG 10 Bình Thuận,4,HSG 10 Chuyên SPHN,5,HSG 10 Đắk Lắk,2,HSG 10 Đồng Nai,4,HSG 10 Gia Lai,2,HSG 10 Hà Nam,4,HSG 10 Hà Tĩnh,15,HSG 10 Hải Dương,10,HSG 10 KHTN,9,HSG 10 Nghệ An,1,HSG 10 Ninh Thuận,1,HSG 10 Phú Yên,2,HSG 10 PTNK,10,HSG 10 Quảng Nam,1,HSG 10 Quảng Trị,2,HSG 10 Thái Nguyên,9,HSG 10 Vĩnh Phúc,14,HSG 1015-2016,3,HSG 11,135,HSG 11 2009-2010,1,HSG 11 2010-2011,6,HSG 11 2011-2012,10,HSG 11 2012-2013,9,HSG 11 2013-2014,7,HSG 11 2014-2015,10,HSG 11 2015-2016,6,HSG 11 2016-2017,8,HSG 11 2017-2018,7,HSG 11 2018-2019,8,HSG 11 2019-2020,5,HSG 11 2020-2021,8,HSG 11 2021-2022,4,HSG 11 2022-2023,7,HSG 11 2023-2024,1,HSG 11 An Giang,2,HSG 11 Bà Rịa Vũng Tàu,1,HSG 11 Bắc Giang,4,HSG 11 Bạc Liêu,3,HSG 11 Bắc Ninh,2,HSG 11 Bình Định,12,HSG 11 Bình Dương,3,HSG 11 Bình Thuận,1,HSG 11 Cà Mau,1,HSG 11 Đà Nẵng,9,HSG 11 Đồng Nai,1,HSG 11 Hà Nam,2,HSG 11 Hà Tĩnh,12,HSG 11 Hải Phòng,1,HSG 11 Kiên Giang,4,HSG 11 Lạng Sơn,11,HSG 11 Nghệ An,6,HSG 11 Ninh Bình,2,HSG 11 Quảng Bình,12,HSG 11 Quảng Nam,1,HSG 11 Quảng Ngãi,9,HSG 11 Quảng Trị,3,HSG 11 Sóc Trăng,1,HSG 11 Thái Nguyên,8,HSG 11 Thanh Hóa,3,HSG 11 Trà Vinh,1,HSG 11 Tuyên Quang,1,HSG 11 Vĩnh Long,3,HSG 11 Vĩnh Phúc,11,HSG 12,674,HSG 12 2009-2010,2,HSG 12 2010-2011,39,HSG 12 2011-2012,44,HSG 12 2012-2013,58,HSG 12 2013-2014,53,HSG 12 2014-2015,44,HSG 12 2015-2016,37,HSG 12 2016-2017,46,HSG 12 2017-2018,55,HSG 12 2018-2019,43,HSG 12 2019-2020,43,HSG 12 2020-2021,52,HSG 12 2021-2022,35,HSG 12 2022-2023,42,HSG 12 2023-2024,28,HSG 12 2023-2041,1,HSG 12 2024-2025,1,HSG 12 An Giang,9,HSG 12 Bà Rịa Vũng Tàu,13,HSG 12 Bắc Giang,18,HSG 12 Bạc Liêu,4,HSG 12 Bắc Ninh,13,HSG 12 Bến Tre,20,HSG 12 Bình Định,17,HSG 12 Bình Dương,8,HSG 12 Bình Phước,9,HSG 12 Bình Thuận,8,HSG 12 Cà Mau,7,HSG 12 Cần Thơ,7,HSG 12 Cao Bằng,5,HSG 12 Chuyên SPHN,11,HSG 12 Đà Nẵng,3,HSG 12 Đắk Lắk,21,HSG 12 Đắk Nông,1,HSG 12 Điện Biên,3,HSG 12 Đồng Nai,21,HSG 12 Đồng Tháp,18,HSG 12 Gia Lai,14,HSG 12 Hà Nam,5,HSG 12 Hà Nội,17,HSG 12 Hà Tĩnh,16,HSG 12 Hải Dương,16,HSG 12 Hải Phòng,20,HSG 12 Hậu Giang,4,HSG 12 Hòa Bình,10,HSG 12 Hưng Yên,10,HSG 12 Khánh Hòa,4,HSG 12 KHTN,26,HSG 12 Kiên Giang,12,HSG 12 Kon Tum,3,HSG 12 Lai Châu,4,HSG 12 Lâm Đồng,11,HSG 12 Lạng Sơn,8,HSG 12 Lào Cai,17,HSG 12 Long An,18,HSG 12 Nam Định,7,HSG 12 Nghệ An,13,HSG 12 Ninh Bình,12,HSG 12 Ninh Thuận,7,HSG 12 Phú Thọ,18,HSG 12 Phú Yên,13,HSG 12 Quảng Bình,14,HSG 12 Quảng Nam,12,HSG 12 Quảng Ngãi,7,HSG 12 Quảng Ninh,20,HSG 12 Quảng Trị,10,HSG 12 Sóc Trăng,4,HSG 12 Sơn La,5,HSG 12 Tây Ninh,6,HSG 12 Thái Bình,11,HSG 12 Thái Nguyên,13,HSG 12 Thanh Hóa,17,HSG 12 Thừa Thiên Huế,19,HSG 12 Tiền Giang,3,HSG 12 TPHCM,13,HSG 12 Tuyên Quang,3,HSG 12 Vĩnh Long,7,HSG 12 Vĩnh Phúc,20,HSG 12 Yên Bái,6,HSG 9,573,HSG 9 2009-2010,1,HSG 9 2010-2011,21,HSG 9 2011-2012,42,HSG 9 2012-2013,41,HSG 9 2013-2014,35,HSG 9 2014-2015,41,HSG 9 2015-2016,38,HSG 9 2016-2017,42,HSG 9 2017-2018,45,HSG 9 2018-2019,41,HSG 9 2019-2020,18,HSG 9 2020-2021,50,HSG 9 2021-2022,53,HSG 9 2022-2023,55,HSG 9 2023-2024,15,HSG 9 An Giang,9,HSG 9 Bà Rịa Vũng Tàu,8,HSG 9 Bắc Giang,14,HSG 9 Bắc Kạn,1,HSG 9 Bạc Liêu,1,HSG 9 Bắc Ninh,12,HSG 9 Bến Tre,9,HSG 9 Bình Định,11,HSG 9 Bình Dương,7,HSG 9 Bình Phước,13,HSG 9 Bình Thuận,5,HSG 9 Cà Mau,2,HSG 9 Cần Thơ,4,HSG 9 Cao Bằng,2,HSG 9 Đà Nẵng,11,HSG 9 Đắk Lắk,12,HSG 9 Đắk Nông,3,HSG 9 Điện Biên,5,HSG 9 Đồng Nai,8,HSG 9 Đồng Tháp,10,HSG 9 Gia Lai,9,HSG 9 Hà Giang,4,HSG 9 Hà Nam,10,HSG 9 Hà Nội,15,HSG 9 Hà Tĩnh,13,HSG 9 Hải Dương,16,HSG 9 Hải Phòng,8,HSG 9 Hậu Giang,6,HSG 9 Hòa Bình,4,HSG 9 Hưng Yên,11,HSG 9 Khánh Hòa,6,HSG 9 Kiên Giang,16,HSG 9 Kon Tum,9,HSG 9 Lai Châu,2,HSG 9 Lâm Đồng,14,HSG 9 Lạng Sơn,10,HSG 9 Lào Cai,4,HSG 9 Long An,10,HSG 9 Nam Định,9,HSG 9 Nghệ An,21,HSG 9 Ninh Bình,14,HSG 9 Ninh Thuận,4,HSG 9 Phú Thọ,13,HSG 9 Phú Yên,9,HSG 9 Quảng Bình,14,HSG 9 Quảng Nam,12,HSG 9 Quảng Ngãi,13,HSG 9 Quảng Ninh,17,HSG 9 Quảng Trị,10,HSG 9 Sóc Trăng,9,HSG 9 Sơn La,5,HSG 9 Tây Ninh,16,HSG 9 Thái Bình,11,HSG 9 Thái Nguyên,5,HSG 9 Thanh Hóa,12,HSG 9 Thừa Thiên Huế,9,HSG 9 Tiền Giang,7,HSG 9 TPHCM,11,HSG 9 Trà Vinh,2,HSG 9 Tuyên Quang,6,HSG 9 Vĩnh Long,12,HSG 9 Vĩnh Phúc,12,HSG 9 Yên Bái,5,HSG Cấp Trường,80,HSG Quốc Gia,113,HSG Quốc Tế,16,Hứa Lâm Phong,1,Hứa Thuần Phỏng,1,Hùng Vương,2,Hưng Yên,44,Huỳnh Kim Linh,1,Hy Lạp,1,IMC,26,IMO,58,IMT,2,IMU,2,India - Ấn Độ,47,Inequality,13,InMC,1,International,349,Iran,13,Jakob,1,JBMO,41,Jewish,1,Journal,30,Junior,38,K2pi,1,Kazakhstan,1,Khánh Hòa,30,KHTN,64,Kiên Giang,74,Kon Tum,24,Korea - Hàn Quốc,5,Kvant,2,Kỷ Yếu,46,Lai Châu,12,Lâm Đồng,48,Lăng Hồng Nguyệt Anh,1,Lạng Sơn,37,Langlands,1,Lào Cai,35,Lê Hải Châu,1,Lê Hải Khôi,1,Lê Hoành Phò,4,Lê Hồng Phong,5,Lê Khánh Sỹ,3,Lê Minh Cường,1,Lê Phúc Lữ,1,Lê Phương,1,Lê Viết Hải,1,Lê Việt Hưng,2,Leibniz,1,Long An,52,Lớp 10 Chuyên,709,Lớp 10 Không Chuyên,355,Lớp 11,1,Lục Ngạn,1,Lượng giác,1,Lưu Giang Nam,2,Lưu Lý Tưởng,1,Macedonian,1,Malaysia,1,Margulis,2,Mark Levi,1,Mathematical Excalibur,1,Mathematical Reflections,1,Mathematics Magazine,1,Mathematics Today,1,Mathley,1,MathLinks,1,MathProblems Journal,1,Mathscope,8,MathsVN,5,MathVN,1,MEMO,13,Menelaus,1,Metropolises,4,Mexico,1,MIC,1,Michael Atiyah,1,Michael Guillen,1,Mochizuki,1,Moldova,1,Moscow,1,MYM,25,MYTS,4,Nam Định,46,Nam Phi,1,National,276,Nesbitt,1,Newton,4,Nghệ An,73,Ngô Bảo Châu,2,Ngô Việt Hải,1,Ngọc Huyền,2,Nguyễn Anh Tuyến,1,Nguyễn Bá Đang,1,Nguyễn Đình Thi,1,Nguyễn Đức Tấn,1,Nguyễn Đức Thắng,1,Nguyễn Duy Khương,1,Nguyễn Duy Tùng,1,Nguyễn Hữu Điển,3,Nguyễn Minh Hà,1,Nguyễn Minh Tuấn,9,Nguyễn Nhất Huy,1,Nguyễn Phan Tài Vương,1,Nguyễn Phú Khánh,1,Nguyễn Phúc Tăng,2,Nguyễn Quản Bá Hồng,1,Nguyễn Quang Sơn,1,Nguyễn Song Thiên Long,1,Nguyễn Tài Chung,5,Nguyễn Tăng Vũ,1,Nguyễn Tất Thu,1,Nguyễn Thúc Vũ Hoàng,1,Nguyễn Trung Tuấn,8,Nguyễn Tuấn Anh,2,Nguyễn Văn Huyện,3,Nguyễn Văn Mậu,25,Nguyễn Văn Nho,1,Nguyễn Văn Quý,2,Nguyễn Văn Thông,1,Nguyễn Việt Anh,1,Nguyễn Vũ Lương,2,Nhật Bản,4,Nhóm $\LaTeX$,4,Nhóm Toán,1,Ninh Bình,61,Ninh Thuận,26,Nội Suy Lagrange,2,Nội Suy Newton,1,Nordic,21,Olympiad Corner,1,Olympiad Preliminary,2,Olympic 10,136,Olympic 10/3,6,Olympic 10/3 Đắk Lắk,6,Olympic 11,124,Olympic 12,52,Olympic 23/3,2,Olympic 24/3,10,Olympic 24/3 Quảng Nam,10,Olympic 27/4,24,Olympic 30/4,61,Olympic KHTN,8,Olympic Sinh Viên,78,Olympic Tháng 4,12,Olympic Toán,348,Olympic Toán Sơ Cấp,3,Ôn Thi 10,2,PAMO,1,Phạm Đình Đồng,1,Phạm Đức Tài,1,Phạm Huy Hoàng,1,Pham Kim Hung,3,Phạm Quốc Sang,2,Phan Huy Khải,1,Phan Quang Đạt,1,Phan Thành Nam,1,Pháp,2,Philippines,8,Phú Thọ,32,Phú Yên,43,Phùng Hồ Hải,1,Phương Trình Hàm,11,Phương Trình Pythagoras,1,Pi,1,Polish,32,Problems,1,PT-HPT,14,PTNK,65,Putnam,27,Quảng Bình,65,Quảng Nam,58,Quảng Ngãi,50,Quảng Ninh,61,Quảng Trị,42,Quỹ Tích,1,Riemann,1,RMM,14,RMO,24,Romania,38,Romanian Mathematical,1,Russia,1,Sách Thường Thức Toán,7,Sách Toán,70,Sách Toán Cao Học,1,Sách Toán THCS,7,Saudi Arabia - Ả Rập Xê Út,9,Scholze,1,Serbia,17,Sharygin,28,Shortlists,56,Simon Singh,1,Singapore,1,Số Học - Tổ Hợp,28,Sóc Trăng,37,Sơn La,22,Spain,8,Star Education,1,Stars of Mathematics,11,Swinnerton-Dyer,1,Talent Search,1,Tăng Hải Tuân,2,Tạp Chí,17,Tập San,3,Tây Ban Nha,1,Tây Ninh,37,Thái Bình,45,Thái Nguyên,61,Thái Vân,2,Thanh Hóa,69,THCS,2,Thổ Nhĩ Kỳ,5,Thomas J. 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MOlympiad.NET: [Solutions] Sharygin Geometry Mathematical Olympiad 2021 (Final Round)
[Solutions] Sharygin Geometry Mathematical Olympiad 2021 (Final Round)
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