MathDB
compare angles between altitude-median (2012 Kyiv City MO Round2 10.4)

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8/2/2020
In the triangle ABCABC with sides BC>AC>ABBC> AC> AB the angles between altiude and median drawn from one vertex are considered. Find out at which vertex this angle is the largest of the three.
(Rozhkova Maria)
geometryanglesgeometric inequality
concurrent or //, // chords of inters. circles (2012 Kyiv City MO Round2 11.4)

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8/2/2020
The circles w1{{w} _ {1}} and w2{{w} _ {2}} intersect at points PP and QQ. Let ABAB and CDCD be parallel diameters of circles w1{ {w} _ {1}} and w2{{w} _ {2}} , respectively. In this case, none of the points A,B,C,DA, B, C, D coincides with either PP or QQ, and the points lie on the circles in the following order: A,B,P,QA, B, P, Q on the circle w1{{w} _ {1} } and C,D,P,QC, D, P, Q on the circle w2{{w} _ {2}} . The lines APAP and BQBQ intersect at the point XX, and the lines CPCP and DQDQ intersect at the point Y,XYY, X \ne Y. Prove that all lines XYXY for different diameters ABAB and CDCD pass through the same point or are all parallel.
(Serdyuk Nazar)
geometryparallelconcurrentcircles
concurrency, <APE =<BAC, < CQF =< BCA (2013 Kyiv City MO Round2 11.4)

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8/5/2020
Let H H be the intersection point of the altitudes AP AP and CQ CQ of the acute-angled triangle ABC ABC . On its median BM BM marked points E E and F F so that APE=BAC \angle APE = \angle BAC and CQF=BCA \angle CQF = \angle BCA , and the point E E lies inside the triangle APB APB , and the point F F lies inside the triangle CQB CQB . Prove that the lines AE AE , CF CF and BH BH intersect at one point.
(Vyacheslav Yasinsky)
geometryequal anglesconcurrencyconcurrent
collinear symmetrics wrt midpoints , equilateral (2015 Kyiv City MO Round2 11.2)

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9/4/2020
The line passing through the center of the equilateral triangle ABC ABC intersects the lines AB AB , BC BC and CA CA at the points C1 {{C} _ {1}} , A1 {{A} _ {1}} and B1 {{B} _ {1}} , respectively. Let A2 {{A} _ {2}} be a point that is symmetric A1 {{A} _ {1}} with respect to the midpoint of BC BC ; the points B2 {{B} _ {2}} and C2 {{C} _ {2}} are defined similarly. Prove that the points A2 {{A} _ {2}} , B2 {{B} _ {2}} and C2 {{C} _ {2}} lie on the same line tangent to the inscribed circle of the triangle ABC ABC .
(Serdyuk Nazar)
geometryEquilateralcollinearSymmetric
concurrceny wanted, 3 circles related (2014 Kyiv City MO Round2 10.4 11.3)

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8/16/2020
Three circles are constructed for the triangle ABCABC : the circle wA{{w} _ {A}} passes through the vertices BB and CC and intersects the sides ABAB and AC AC at points A1{{A} _ {1}} and A2{{A} _ {2}} respectively, the circle wB{{w} _ {B}} passes through the vertices AA and CC and intersects the sides BABA and BCBC at the points B1{{B} _ {1}} and B2{{B} _ {2}} , wC{{w} _ {C}} passes through the vertices AA and BB and intersects the sides CACA and CBCB at the points C1{{C} _ {1}} and C2{{C} _ {2}} . Let A1A2B1B2=C{{A} _ {1}} {{A} _ {2}} \cap {{B} _ {1}} {{B} _ {2}} = {C} ', A1A2C1C2=B{{A} _ {1}} {{A} _ {2}} \cap {{C} _ {1}} {{C} _ {2}} = {B} ' ta B1B2C1C2=A{ {B} _ {1}} {{B} _ {2}} \cap {{C} _ {1}} {{C} _ {2}} = {A} ' is Prove that the perpendiculars, which are omitted from the points A,B,C{A} ', \, \, {B}', \, \, {C} ' to the lines BCBC , CACA and ABAB respectively intersect at one point.
(Rudenko Alexander)
concurrentgeometrycircles
square construction given 4 collinear (2016 Kyiv City MO Round2 10.2 11.2 )

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9/7/2020
On the horizontal line from left to right are the points P,Q,R,SP, \, \, Q, \, \, R, \, \, S. Construct a square ABCDABCD, for which on the line ADAD lies lies the point PP, on the line BCBC lies the point QQ, on the line ABAB lies the point RR, on the line CDCD lies the point SS .
geometrycollinearconstructionsquare
EF=diameter of circumcircle of AO_1O_2 (2017 Kyiv City MO Round2 10.3)

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9/10/2020
Circles w1w_1 and w2w_2 with centers at points O1O_1 and O2O_2 respectively, intersect at points AA and BB. A line passing through point BB, intersects the circles w1w_1 and w2w_2 at points CC and DD other than BB. Tangents to the circles w1w_1 and w2w_2 at points CC and DD intersect at point EE. Line EAEA intersects the circumscribed circle ww of triangle AO1O2AO_1O_2 at point FF. Prove that the length of the segment is EFEF is equal to the diameter of the circle ww.
(Vovchenko V., Plotnikov M.)
geometrycircumcirclediametercircles
ABTN is cyclic iff AB = AK, KT = KC,CN = BK (2017 Kyiv City MO Round2 11.2)

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9/10/2020
The median CMCM is drawn in the triangle ABCABC intersecting bisector angle BLBL at point OO. Ray AOAO intersects side BCBC at point KK, beyond point KK draw the segment KT=KCKT = KC. On the ray BCBC beyond point CC draw a segment CN=BKCN = BK. Prove that is a quadrilateral ABTNABTN is cyclic if and only if AB=AKAB = AK.
(Vladislav Yurashev)
geometrycyclic quadrilateralConcyclicequal segments
circumcircle of HXY equidistant from B,C (2018 Kyiv City MO Round2 10.3)

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9/14/2020
In the acute triangle ABCABC the orthocenter HH and the center of the circumscribed circle OO were noted. The line AOAO intersects the side BCBC at the point DD. A perpendicular drawn to the side BCBC at the point DD intersects the heights from the vertices BB and CC of the triangle ABCABC at the points XX and YY respectively. Prove that the center of the circumscribed circle ΔHXY\Delta HXY is equidistant from the points BB and CC.
(Danilo Hilko)
geometrycircumcircleequal segments
<TSC=<BAC wanted, parallelogram, circumcircle (2020 Kyiv City MO Round2 11.2)

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9/21/2020
A point PP was chosen on the smaller arc BCBC of the circumcircle of the acute-angled triangle ABCABC. Points RR and SS on the sidesAB AB and ACAC are respectively selected so that CPRSCPRS is a parallelogram. Point TT on the arc ACAC of the circumscribed circle of ABC\vartriangle ABC such that BTCPBT \parallel CP. Prove that TSC=BAC\angle TSC = \angle BAC.
(Anton Trygub)
geometryparallelogramcircumcircleequal angles
right angle wanted, AB=BC, circumcircle related (2018 Kyiv City MO Round2 11.2)

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9/14/2020
In the quadrilateral ABCDABCD , AB=BCAB = BC , the point KK is the midpoint of the side CDCD , the rays BKBK and ADAD intersect at the point MM , the circumscribed circle ΔABM \Delta ABM intersects the line ACAC for the second time at the point PP . Prove that BKP=90\angle BKP = 90 {} ^ \circ .
(Anton Trygub)
circumcirclegeometryright angle
<T_A T_B T_C= 90^o - 1/2 <ABC if OI//AC (2019 Kyiv City MO Round2 10.3)

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9/18/2020
Denote in the triangle ABCABC by TA,TB,TCT_A,T_B,T_C the touch points of the exscribed circles of ABC\vartriangle ABC, tangent to sides BC,ACBC, AC and ABAB respectively. Let OO be the center of the circumcircle of ABC\vartriangle ABC, and II is the center of it's inscribed circle. It is known that OIACOI\parallel AC. Prove that TATBTC=90o12ABC\angle T_A T_B T_C= 90^o - \frac12 \angle ABC.
(Anton Trygub)
geometryexcentersexcenterangles
AR = QR wanted inside a regular pentagon (2019 Kyiv City MO Round2 10.3.1)

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9/18/2020
Let ABCDEABCDE be a regular pentagon with center MM. Point PMP \ne M is selected on segment MDMD. The circumscribed circle of triangle ABPABP intersects the line AEAE for second time at point QQ, and a line that is perpendicular to the CDCD and passes through PP, for second time at the point RR. Prove that AR=QRAR = QR.
geometryequal segmentspentagonregular pentagon
circumcenter lies on midline of other triangle (2019 Kyiv City MO Round2 11.3)

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9/19/2020
The line \ell is perpendicular to the side ACAC of the acute triangle ABCABC and intersects this side at point KK, and the circumcribed circle ABC\vartriangle ABC at points PP and TT (point P on the other side of line ACAC, as the vertex BB). Denote by P1P_1 and T1T_1 - the projections of the points PP and TT on line ABAB, with the vertices A,BA, B belong to the segment P1T1P_1T_1. Prove that the center of the circumscribed circle of the P1KT1\vartriangle P_1KT_1 lies on a line containing the midline ABC\vartriangle ABC, which is parallel to the side ACAC.
(Anton Trygub)
geometryCircumcentermidline
centroid is incenter of other triangle, (2019 Kyiv City MO Round2 11.3.1)

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9/19/2020
It is known that in the triangle ABCABC the smallest side is BCBC. Let X,Y,KX, Y, K and LL - points on the sides AB,ACAB, AC and on the rays CB,BCCB, BC, respectively, are such that BX=BK=BC=CY=CLBX = BK = BC =CY =CL. The line KXKX intersects the line LYLY at the point MM. Prove that the intersection point of the medians KLM\vartriangle KLM coincides with the center of the inscribed circle ABC\vartriangle ABC.
geometryincenterCentroidequal segments
<PBM+<CBM=<PCA,<BM=90^o, <ABC+<APC=180^o (2020 Kyiv City MO Round2 10.2)

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9/21/2020
Let MM be the midpoint of the side ACAC of triangle ABCABC. Inside BMC\vartriangle BMC was found a point PP such that BMP=90o\angle BMP = 90^o, ABC+APC=180o\angle ABC+ \angle APC =180^o. Prove that PBM+CBM=PCA\angle PBM + \angle CBM = \angle PCA.
(Anton Trygub)
geometryanglesright ange
product of radii, common tangents to circles (2021 Kyiv City MO Round2 11.3.1)

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2/15/2021
Two circles k1k_1 and k2k_2 with radii r1r_1 and r2r_2 have no common points. The lineAB AB is a common internal tangent, and the line CDCD is a common external tangent to these circles, where A,Ck1A, C \in k_1 and B,Dk2B, D \in k_2. Knowing that AB=12AB=12 and CD=16CD =16, find the value of the product r1r2r_1r_2.
geometryTangentscircles
concyclic, <OAD+<OBC= <ODA + <OCB = 90^o (2021 Kyiv City MO Round2 10.4)

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2/14/2021
Inside the quadrilateral ABCDABCD marked a point OO such that OAD+OBC=ODA+OCB=90o\angle OAD+ \angle OBC = \angle ODA + \angle OCB = 90^o. Prove that the centers of the circumscribed circles around triangles OADOAD and OBCOBC as well as the midpoints of the sides ABAB and CDCD lie on one circle.
(Anton Trygub)
geometryConcyclicangles
OK// AB wanted in isosceles trapezoid (2021 Kyiv City MO Round2 10.4.1)

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2/15/2021
Let ABCDABCD be an isosceles trapezoid, AD=BCAD=BC, ABCDAB \parallel CD. The diagonals of the trapezoid intersect at the point OO, and the point MM is the midpoint of the side ADAD. The circle circumscribed around the triangle BCMBCM intersects the side ADAD at the point KK. Prove that OKABOK \parallel AB.
geometrytrapezoidisoscelesparallel
circle tangent to incircle, <BKA=45^o (2021 Kyiv City MO Round2 11.3)

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2/15/2021
In the triangle ABCABC, the altitude BHBH and the angle bisector BLBL are drawn, the inscribed circle ww touches the side of the ACAC at the point KK. It is known that BKA=45o\angle BKA = 45^o. Prove that the circle with diameter HLHL touches the circle ww.
(Anton Trygub)
geometrytangent circles
Slowly growing product

Source: Kyiv City MO 2021 Round 1, Problem 9.3

12/21/2023
Let an=1+2n2n31n4a_n = 1 + \frac{2}{n} - \frac{2}{n^3} - \frac{1}{n^4}. For which smallest positive integer nn does the value of Pn=a2a3a4anP_n = a_2a_3a_4 \ldots a_n exceed 100100?
algebra
Magic rectangle (almost)

Source: Kyiv City MO 2021 Round 1, Problem 7.4

12/21/2023
A rectangle 3×53 \times 5 is divided into 1515 1×11 \times 1 cells. The middle 33 cells that have no common points with the border of the rectangle are deleted. Is it possible to put in the remaining 1212 cells numbers 1,2,,121, 2, \ldots, 12 in some order, so that the sums of the numbers in the cells along each of the four sides of the rectangle are equal?
Proposed by Mariia Rozhkova
permutationsrectangleMagic squares
We balling

Source: Kyiv City MO 2021 Round 1, Problem 7.1

12/21/2023
Mom brought Andriy and Olesya 44 balls with the numbers 1,2,31, 2, 3 and 44 written on them (one on each ball). She held 22 balls in each hand and did not know which numbers were written on the balls in each hand. The mother asked Andriy to take a ball with a higher number from each hand, and then to keep the ball with the lower number from the two balls he took. After that, she asked Olesya to take two other balls, and out of these two, keep the ball with the higher number. Does the mother know with certainty, which child has the ball with the higher number?
Proposed by Bogdan Rublov
combinatoricsgameballs
Cutting rectangle game

Source: Kyiv City MO 2021 Round 1, Problem 7.2

12/21/2023
Andriy and Olesya take turns (Andriy starts) in a 2×12 \times 1 rectangle, drawing horizontal segments of length 22 or vertical segments of length 11, as shown in the figure below.
https://i.ibb.co/qWqWxgh/Kyiv-MO-2021-Round-1-7-2.png
After each move, the value PP is calculated - the total perimeter of all small rectangles that are formed (i.e., those inside which no other segment passes). The winner is the one after whose move PP is divisible by 20212021 for the first time. Who has a winning strategy?
Proposed by Bogdan Rublov
gamecombinatoricscuttinggeometryrectangle
Small factoring

Source: Kyiv City MO 2021 Round 1, Problem 7.3

12/21/2023
Petryk factored the number 106=100000010^6 = 1000000 as a product of 77 distinct positive integers. Among all such factorings, find the one in which the largest of these 77 factors is the smallest possible.
Proposed by Bogdan Rublov
Factoringnumber theory