MathDB
KL=LB, perpendiculars in right triangle (Kyiv City Olympiad 2006 8.3)

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6/28/2020
On the legs AC,BCAC, BC of a right triangle △ABC\vartriangle ABC select points MM and NN, respectively, so that ∠MBC=∠NAC\angle MBC = \angle NAC. The perpendiculars from points MM and CC on the line ANAN intersect ABAB at points KK and LL, respectively. Prove that KL=LBKL=LB.
(O. Clurman)
geometryright triangleequal segmentsperpendicularllllllllllllllllllllllllllllll
<FXE inside regular hecagon (Kyiv City Olympiad 2005 8.5 9.5)

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6/27/2020
Let ABCDEFABCDEF be a regular hexagon. On the line AFAF mark the point XXso that ∠DCX=45o \angle DCX = 45^o . Find the value of the angle FXEFXE.
(Vyacheslav Yasinsky)
geometryangleshexagonRegular
concurrency in a parallelogram (Kyiv City Olympiad 2006 9.4)

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6/28/2020
On the sides ABAB and CDCD of the parallelogram ABCDABCD mark points EE and FF, respectively. On the diagonals ACAC and BDBD chose the points MM and NN so that EM∥BDEM\parallel BD and FN∥ACFN\parallel AC. Prove that the lines AF,DEAF, DE and MNMN intersect at one point.
(B. Rublev)
geometryparallelogramconcurrencyconcurrent
triangle construction, circumcircle, incircle (Kyiv City Olympiad 2007 9.3)

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6/29/2020
On a straight line 44 points are successively set , A,P,Q,WA, P, Q,W , which are the points of intersection of the bisector ALAL of the triangle ABCABC with the circumscribed and inscribed circle. Knowing only these points, construct a triangle ABCABC .
geometrycircumcircleincircleconstruction
angles wanted, 2 triangles, trapezoid (Kyiv City Olympiad 2008 8.4)

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6/30/2020
There are two triangles ABCABC and BKLBKL on the plane so that the segment AKAK is divided into three equal parts by the point of intersection of the medians △ABC\vartriangle ABC and the point of intersection of the bisectors △BKL \vartriangle BKL (AKAK - median △ABC \vartriangle ABC, KAKA - bisector △BKL\vartriangle BKL ) and quadrilateral KALCKALC is trapezoid. Find the angles of the triangle BKLBKL.
(Bogdan Rublev)
geometrytrapezoidTrianglesangles
<BFL=?AF=LC<1/2AC, AB^2+BC^2=AL^2+LC^2 (Kyiv City Olympiad 2008 9.5)

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6/30/2020
In the triangle ABCABC on the side ACAC the points FF and LL are selected so that AF=LC<12ACAF = LC <\frac{1}{2} AC. Find the angle ∠FBL \angle FBL if AB2+BC2=AL2+LC2A {{B} ^ {2}} + B {{C} ^ {2}} = A {{L} ^ {2}} + L {{C } ^ {2}}
(Zhidkov Sergey)
geometryangles
orthocenter, perpendicular chords, AP = 2PB (Kyiv City Olympiad 2009 8.5 9.3)

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7/2/2020
A chord ABAB is drawn in the circle, on which the point PP is selected in such a way that AP=2PBAP = 2PB. The chord DEDE is perpendicular to the chord ABAB and passes through the point PP. Prove that the midpoint of the segment APAP is the orthocener of the triangle AEDAED.
geometryChordsorthocentercircle
TH bisects BC, circumcircle of acute related (2010 Kyiv City MO 9.4 )

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7/13/2020
In an acute-angled triangle ABCABC, the point OO is the center of the circumcircle, CHCH is the height of the triangle, and the point TT is the foot of the perpendicular dropped from the vertex CC on the line AOAO. Prove that the line THTH passes through the midpoint of the side BCBC .
geometrycircumcirclebisects segment
(MB- MS)(NC-NS) &lt;= 0, midpoints (Kyiv City Olympiad 2010 8.5)

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7/2/2020
In an acute-angled triangle ABCABC, the points MM and NN are the midpoints of the sides ABAB and ACAC, respectively. For an arbitrary point SS lying on the side of BCBC prove that the condition holds (MB−MS)(NC−NS)≤0(MB- MS)(NC-NS) \le 0
geometrymidpointsgeometric inequality
angle chasing, OD = BD = 1/3 BC (Kyiv City Olympiad 2010 8.4 9.6)

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7/2/2020
Point OO is the center of the circumcircle of the acute triangle ABCABC. The line AOAO intersects the side BCBC at point DD so that OD=BD=1/3BCOD = BD = 1/3 BC . Find the angles of the triangle ABCABC. Justify the answer.
geometryanglesCircumcenter
&lt;DKL=&lt;CLK if &lt;BMN=&lt;MNC, midpoints of cyclic (2011 Kyiv City MO 9.4)

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7/14/2020
Let ABCDABCD be an inscribed quadrilateral. Denote the midpoints of the sides AB,BC,CDAB, BC, CD and DADA through M,L,NM, L, N and KK, respectively. It turned out that ∠BMN=∠MNC\angle BM N = \angle MNC. Prove that: i) ∠DKL=∠CLK\angle DKL = \angle CLK. ii) in the quadrilateral ABCDABCD there is a pair of parallel sides.
geometryCyclicequal anglesmidpoints
AD = CD wanted, tangents to circumcircle (2011 Kyiv City MO 9.4.1)

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7/14/2020
The triangle ABCABC is inscribed in a circle. At points AA and BB are tangents to this circle, which intersect at point TT. A line drawn through the point TT parallel to the side ACAC intersects the side BCBC at the point DD. Prove that AD=CDAD = CD.
geometrycircumcircleTangentsequal segments
&lt;ABM =&lt;LAC if &lt;ANC = &lt;ALB, medians (2011 Kyiv City MO 8.4.1)

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7/14/2020
The medians AL,BMAL, BM, and CNCN are drawn in the triangle ABCABC. Prove that ∠ANC=∠ALB\angle ANC = \angle ALB if and only if ∠ABM=∠LAC\angle ABM =\angle LAC.
(Veklich Bogdan)
geometryequal anglesMedians
4 incircles in convex ABCD concurrent if ABCD is rhombus (2013 Kyiv City MO 8.5)

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8/4/2020
Let ABCDABCD be a convex quadrilateral. Prove that the circles inscribed in the triangles ABCABC, BCDBCD, CDACDA and DABDAB have a common point if and only if ABCDABCD is a rhombus.
geometryrhombusconcurrentcirclesconcurrency
perimeter of triangles inequality, isosceles (2012 Kyiv City MO 7.4)

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8/3/2020
Given an isosceles triangle ABCABC with a vertex at the point BB. Based on ACAC, an arbitrary point DD is selected, different from the vertices AA and CC . On the line ACAC select the point EE outside the segment ACAC, for which AE=CDAE = CD. Prove that the perimeter ΔBDE\Delta BDE is larger than the perimeter ΔABC\Delta ABC.
geometryperimeterisoscelesgeometric inequality
equal angles wanted, intersecting circles (2012 Kyiv City MO 8.3)

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8/1/2020
On the circle γ\gamma the points AA and BB are selected. The circle ω\omega touches the segment ABAB at the point KK and intersects the circle γ\gamma at the points MM and NN. The points lie on the circle γ\gamma in the following order: A,  M,  N,  BA, \, \, M, \, \, N, \, \, B. Prove that ∠AMK=∠KNB\angle AMK = \angle KNB.
(Yuri Biletsky)
geometry
BP=CP when AD=AB+CD, angle bisectors related in ABCD (2014 Kyiv City MO 7.4)

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8/5/2020
In the quadrilateral ABCDABCD the condition AD=AB+CDAD = AB + CD is fulfilled. The bisectors of the angles BADBAD and ADCADC intersect at the point PP , as shown in Fig. Prove that BP=CPBP = CP. https://cdn.artofproblemsolving.com/attachments/3/1/67268635aaef9c6dc3363b00453b327cbc01f3.png
(Maria Rozhkova)
geometryangle bisectorequal segments
triangle inequality with integer AC in ABC, ACD (2014 Kyiv City MO 7.4.1)

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8/5/2020
The sides of triangles ABCABC and ACDACD satisfy the following conditions: AB=AD=3AB = AD = 3 cm, BC=7BC = 7 cm, DC=11DC = 11 cm. What values can the side length ACAC take if it is an integer number of centimeters, is the average in ΔACD\Delta ACD and the largest in ΔABC\Delta ABC?
triangle inequalitygeometrygeometric inequality
perpendicular wanted, starting with equilateral (2014 Kyiv City MO 8.5)

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8/5/2020
Given an equilateral ΔABC\Delta ABC, in which A1,B1,C1{{A} _ {1}}, {{B} _ {1}}, {{C} _ {1}} are the midpoint of the sides BC,  AC,  AB BC, \, \, AC, \, \, AB respectively. The line ll passes through the vertex AA, we denote by P,QP, Q the projection of the points B,CB, C on the line ll, respectively (the line l l and the point Q,  A,  PQ, \, \, A, \, \, P are located as shown in fig.). Denote by TT the intersection point of the lines B1P{{B} _ {1}} P and C1Q{{C} _ {1}} Q. Prove that the line A1T{{A} _ {1}} T is perpendicular to the line ll. https://cdn.artofproblemsolving.com/attachments/4/b/61f2f4ec9e6b290dfcd47e9351110bebd3bd43.png (Serdyuk Nazar)
geometryperpendicularEquilateral
MF: BK wanted, AK = AF, median related (2014 Kyiv City MO 8.5.1)

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8/5/2020
On the side ABAB of the triangle ABCABC mark the point KK. The segment CKCK intersects the median AMAM at the point FF. It is known that AK=AFAK = AF. Find the ratio MF:BKMF: BK.
geometryratiomedianequal segments
square in a right trapezoid construction (2015 Kyiv City MO 9.3)

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9/2/2020
It is known that a square can be inscribed in a given right trapezoid so that each of its vertices lies on the corresponding side of the trapezoid (none of the vertices of the square coincides with the vertex of the trapezoid). Construct this inscribed square with a compass and a ruler.
(Maria Rozhkova)
geometrytrapezoidconstructionsquare
&lt;ACB=60 iffAE + BD = AB, angle bisectors (2016 Kyiv City MO 8.5)

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9/5/2020
In the triangle ABCABC the angle bisectors ADAD and BEBE are drawn. Prove that ∠ACB=60∘\angle ACB = 60 {} ^ \circ if and only if AE+BD=ABAE + BD = AB.
(Hilko Danilo)
geometryangle bisectorangles
2 circles internally tangent to third one (2014 Kyiv City MO 9.3)

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8/6/2020
Two circles c1,  c2{{c} _ {1}}, \, \, {{c} _ {2}} pass through the center OO of the circle cc and touch it internally in points AA and BB, respectively. Prove that the line ABAB passes though a common point of circles c1,  c2{{c} _ {1}}, \, \, {{c} _ {2}} .
geometrycirclestangent circles
&lt;B= 30^o wanted, 2AC=AB,&lt; A = 2&lt;B, CL = ML (2019 Kyiv City MO 8.3)

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9/16/2020
In the triangle ABCABC it is known that 2AC=AB2AC=AB and ∠A=2∠B\angle A = 2\angle B. In this triangle draw the angle bisector ALAL, and mark point MM, the midpoint of the side ABAB. It turned out that CL=MLCL = ML. Prove that ∠B=30o\angle B= 30^o.
(Hilko Danilo)
geometryequal anglesanglesequal segments
AC = AA_1 if &lt;ABC = 45^o, AA_1= _|_ CC_1 (2016 Kyiv City MO 9.5)

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9/5/2020
On the sides BCBC and ABAB of the triangle ABCABC the points A1{{A} _ {1}} and C1{{C} _ {1}} are selected accordingly so that the segments AA1A {{A} _ {1}} and CC1C {{C} _ {1}} are equal and perpendicular. Prove that if ∠ABC=45∘\angle ABC = 45 {} ^ \circ, then AC=AA1AC = A {{A} _ {1}} .
(Gogolev Andrew)
geometryequal segmentsperpendicularity