VI GEOMETRICAL OLYMPIAD IN HONOUR OF I.F.SHARYGIN THE CORRESPONDENCE ROUND. SOLUTIONS

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1 VI GEOMETRIL OLYMPID IN HONOUR OF I.F.SHRYGIN THE ORRESPONDENE ROUND. SOLUTIONS 1. (.Frenkin) (8) Does there exist a triangle, whose side is equal to some its altitude, another side is equal to some its bisectrix, and the third side is equal to some its median? Solution. No, because the greatest side of a triangle is longer than any its median, bisector or altitude. Indeed, a segment joining a vertex of a triangle with an arbitrary point of the opposite side is shorter than one of two remaining sides. Thus each median or bisector is shorter than one of sides and so is shorter than the greatest side. This is correct also for the altitudes. 2. (D.Shvetsov) (8) isectors 1 and 1 of a right triangle ( = 90 ) meet at a point I. Let O be the circumcenter of the triangle 1 1. Prove that OI. Solution. Let 2, 2, 2 be the projections of 1, 1, I to (fig.2). Since 1 is a bisector, we have 2 =. On the other hand, 2 touches the incircle, thus the segment 2 2 = 2 2 is equal to the tangent to this circle from. Similarly 2 2 is equal to the same tangent, i.e. 2 is the midpoint of 2 2. y Phales theorem, 2 I meets segment 1 1 in its midpoint, which coincides with the circumcenter of triangle 1 1 because this triangle is right-angled. 1 O I Fig (F.Nilov) (8) Points,, lie on sides,, of a triangle. For a point X one has X = + and X = +. Prove that the quadrilateral X is cyclic. Solution. Let Y be the common point of circles и distinct from. Then since Y = π and Y = π, we obtain that Y = π, i.e. Y lies also on circle. Now note that Y = Y + Y = + = 2π = + = X (fig.3). Similarly Y = X, i.e. X and Y coincide. 1

2 X Fig (D.Shvetsov) (8) The diagonals of a cyclic quadrilateral D meet in a point N. The circumcircles of the triangles N and ND intersect the sidelines and D for the second time in points 1, 1, 1, D 1. Prove that the quadrilateral D 1 is inscribed into a circle centered at N. Solution. Since pentagon 1 N 1 D is cyclic, we obtain that 1 N = 1 N, because respective angles D and are equal. Similarly N 1 = ND 1. lso N 1 = D = D = DD 1 N (fig.4). Thus ND 1 = N 1, q.e.d. 1 1 N D 1 1 D Fig (D.Shvetsov) (8 9) point E lies on altitude D of triangle, and E = 90. Points O 1 and O 2 are the circumcenters of triangles E and E; points F, L are the 2

3 midpoints of segments and O 1 O 2. Prove that points L, E, F are collinear. Solution. Note that the medial perpendiculars to segments E and E are the medial lines of triangle E, thus they pass through F. So we must prove that F E is the median of triangle F O 1 O 2. ut O 1 O 2 because these two segments are perpendicular to D. Let the line passing through E and parallel to meet F O 1 and F O 2 in points X and Y (fig.5). Since F EX and F EY are parallelograms, then XE = F = F = EY. Thus F E is the median of triangles F XY and F O 1 O 2. L O 1 O 2 X E Y F Fig. 5 D 6. (D.Shvetsov) (8 9) Points M and N lie on side of a regular triangle (M is between and N), and MN = 30. The circumcircles of triangles M and N meet at a point K. Prove that line K passes through the circumcenter of triangle MN. Solution. Since M + N = MN and =, the reflection of in M coincides with the reflection of in N. Mark this point by L. Now LM = M = M, i.e. L lies on circle M. Similarly L lies on circle N and thus coincdes with K (fig.6). So KN = N = 30 M = 90 NM. ut the theorem on inscribed angle implies that we have the same angle between line N and the line connecting with the circumcenter of triangle MN. M N K 3

4 Fig (D.Shvetsov) (8 9) The line passing through vertex of triangle and perpendicular to its median M intersects the altitudes dropped from and (or their extensions) in points K and N. Points O 1 and O 2 are the circumcenters of triangles K and N respectively. Prove that O 1 M = O 2 M. Solution. onsider parallelogram D (fig.7). Since K = DK = D, points,, K, D lie on a same circle and O 1 M D. Similarly O 2 M D. lso since triangles D and D are equal, the distances from their circumcenters to point M also are equal. N K M O 2 O 1 D Fig (D.Shvetsov) (8 10) Let H be an altitude of a given triangle. Points I b and I c are the incenters of triangles H and H respectively; touches the incircle of triangle at a point L. Find LI b I c. Solution. We will prove that triangle LI b I c is right-angled and isosceles. Let L b, L c be the projections of I b, I c to, and r b, r c be the inradii of triangles H, H (fig.8). Since these triangles are right-angled, we have r b = (H+H )/2, r c = (H+H )/2 and r b r c = (H H)/2 ( )/2 = (H H)/2 (L L)/2 = LH. Thus I b L b = LI c = r b, I c L c = LI b = r c, i.e. triangles LI b L b and I c L c L are equal, LI b = LI c and I b LI c = 90. So LI b I c = 45. 4

5 I b I c L b L Fig. 8 H L c 9. (.Frenkin) (8 10) point inside a triangle is called "good" if three cevians passing through it are equal. Suppose the total number of good points is odd for an isosceles triangle ( = ). Find all possible values of this number. Solution. Since the reflection of any good point in the altitude from also is a good point and the total number of good points is odd, there exists a good point lying on this altitude. The cevian through this point from is not shorter than the altitude from. Hence the altitude from is not shorter than the altitude from and. lso can t be longer than the altitude from because in that case there exist two good points on this altitude. Suppose now that some good point doesn t lie on this altitude. Let,, be respective cevians, and 1, 1 be the altitudes. Then 1 = 1 and exactly one of points, lies between the foot of respective altitude and vertex. ut this implies that respective cevians are shorter than. So they are shorter than 1 and we obtain a contradiction. Thus there exists exactly one good point. 10. (I.ogdanov) (8 11) Let three lines forming a triangle be given. Using a two-sided ruler and drawing at most eight lines, construct a point D on the side such that D/D = /. Solution. onstruct lines a, b, c, parallel to,, and lying at the distance from them equal to the width of the ruler. Lines a, b,, form a rhombus, and its diagonal is the bisector of angle. Let E be the common point of this bisector with c, and F be the common point of diagonals of trapezoid formed by lines c,, and (fig.10). Then EF meets in the sought point D. 5

6 b a D F E Fig (.Frenkin) (8 11) convex n-gon is split into three convex polygons. One of them has n sides, the second one has more than n sides, the third one has less than n sides. Find all possible values of n. nswer. n = 4 или n = 5. Solution. It is clear that n > 3. Suppose that n > 5. Then one of three parts of n-gon has at least n + 1 sides, the second parts has at least n sides, the third part has at least three sides. If three pairs of the sides of these parts join inside the given polygon then at most three pairs can form its sides. If two pairs of sides join inside the polygon then at most four pairs can form its sides. In all cases the total number of sides of parts is not greater than n + 9. If n > 5 this isn t possible. The examples for n = 4, 5 are given on fig. 11. Fig (.linkov, Y.linkov, M.Sandrikova) (9) Let be the greatest leg of a right triangle, and H be the altitude to its hypothenuse. The circle of radius H centered at H intersects in point M. Let a point be the reflection of with respect to the point H. The perpendicular to erected at meets the circle in a point K. Prove that: a) M ; 6

7 b) K is tangent to the circle. Solution. a) Let N be an altitude of isosceles triangle HM. Then N = NM. Since H = H and NH, thus the line passing through and parallel to HN meets in point M (Phales theorem). Second solution. Since MH = MH = = = a, points, H, and M are concyclic. Thus M = HM = 180 2a and M = a q.e.d. b) From the right-angled triangle we have: H 2 = H H. Since H = H and KH = H then KH 2 = H H, i.e. triangles HK and KH are similar. This yields the assertion of the problem (S.erlov) (9) Given a convex quadrilateral D such that =. point K lies on the diagonal D, and K + K = +. Prove that K D = K D. Solution. Let L be a point on D such that L =. Since triangles L and D are similar we have L D = 2 = 2. Thus triangles L and D are also similar, i.e. L = and L coincides with K. The sought equality clearly follows from these two similarities. 14. (S.erlov) (9 10) Given a convex quadrilateral D and a point M on its side D such that M and M are parallel to and D respectively. Prove that S D 3S M. Solution. Since M = M = MD we have S M /S M = /M and S M /S MD = M/D. ut triangles M and MD are similar, so these two ratios are equal and SM 2 = S M S MD. y auchi inequality S M (S M + S MD )/2 which is equivalent to the assertion of the problem. 15. (D.Prokopenko,.linkov) (9 11) Suppose 1, 1 and 1 are the altitudes of an acute-angled triangle, 1 meets 1 1 in a point K. The circumcircles of triangles 1 K 1 and 1 K 1 intersect the lines and for the second time at points N and L respectively. Prove that a) the sum of diameters of these two circles is equal to ; b) 1 N/ L/ 1 = 1. Solution. a) Triangles 1 1, 1 1 and 1 1 are similar to triangle with coefficients cos, cos, cos respectively. Thus K 1 1 = K 1 1 = 90, and by sinuses theorem the diameters of circumcircles of triangles K 1 and 1 K 1 are equal to 1 K/ cos and 1 K/ cos respectively. So their sum is 1 1 / cos =. b) n equality proved in p.a) can be written as Dividing by we obtain sought relation. 1 N sin + 1L sin =. 16. (F.Nilov) (9 11) circle touches the sides of an angle with vertex at points and. line passing through intersects this circle in points D and E. chord X is parallel to DE. Prove that X passes through the midpoint of segment DE. Solution. Note that D = EX because the respective arcs lie between parallel chords. Furthermore since D = E, triangles D and E are similar and so 7

8 D/E = D/. Similarly D/E = D/, i.e. D E = D E = DE/2 (the last equality follows from Ptolomeus theorem). Now let X meet DE at point M (fig. 16). Then triangles D and ME are similar, thus D E = EM. From this and previous equalities we obtain EM = ED/2. X D M E Fig (S.Tokarev) (9 11) onstruct a triangle, if the lengths of the bisectrix and of the altitude from one vertex, and of the median from another vertex are given. First solution. Let l = L, h = H be the bisector and the altitude from vertex, m = M be the median from vertex and φ be the angle of right-angled triangle with hypothenuse l opposite to cathetus with length h. Let p be the line passing through and parallel to, and be the reflection of point in p. Suppose that triangle is constructed. Then L = φ or L = 180 φ, and in both cases M = 2φ. In fact, if L = φ, then M = = 2(180 L) = 2φ (Fig.17.1). The second case is similar. M L Fig

9 Since M = 2φ we obtain the following construction. onstruct two parallel lines with distance h between them. Let be a point lying on one of these lines and p be the second line. Now construct point and point M equidistant from two lines and such that M = m. onstruct angle φ and two arcs with endpoints and M equal to 360 4φ. If 1 and 2 are the common points of these arcs with line p, and i (i = 1, 2) is the reflection of i in M, then each of triangles 1 1 and 2 2 is sought. Indeed, the altitudes of these triangles from 1, 2 are equal to h, and segment M = m is their common median. Furthermore if L 1, L 2 are the feet of respective bisectors then our construction yields that one of angles 1 L 1 and 2 L 2 is equal to φ, and the second one is equal to 180 φ. Thus by definition of φ we obtain 1 L 1 = 2 L 2 = l. Note. It is evident that if l < h or m < h/2, then the solution doesn t exist. If l = h and m h/2, then the sought triangle is unique and isosceles (if m = h/2 it degenerates to a segment). When l > h and m = h/2, we obtain two equal triangles symmetric wrt line. If l > h and m = l/2 then one of two triangles is degenerated. In all other cases the problem has two solutions. Second solution. Having an altitude and a bisector from vertex, we can construct this vertex and line. onsider now the following map of this line to itself. For an arbitrary point X find a point Y such that its distances from X and are equal to the given median from and to the half of the given altitude (fig. 17.2). Now find a common point X for and for the reflection of Y in the bisector. Obviously this map is projective and is its fixed point. So we obtain the well-known problem of constructing the fixed point of a projective map. Y Fig = X 18. (D.Prokopenko) (9 11) point lies on a chord of a circle ω. Segments and are diameters of circles ω 1 and ω 2 centered at O 1 and O 2 respectively. These circles intersect ω for the second time in points D and E respectively. The rays O 1 D and O 2 E meet in a point F, and the rays D and E meet in a point G. Prove that line F G passes through the midpoint of segment. 9

10 Solution. Since D = E = 90, points D and E lie on the circle with diameter G. lso F DG = DO 1 = D = GED. Thus F D (and similarly F E) touches this circle (fig. 18). So GF is the symmedian of triangle GED. Since triangle GDE is similar to triangle G, this line is the median of the last triangle. F G D E O 1 O 2 Fig (V.Yasinsky, Ukraine) (9 11) quadrilateral D is inscribed into a circle with center O. Points P and Q are opposite to and D respectively. Two tangents drawn to that circle at these points meet the line in points E and F ( is between E and, is between and F ). Line EO meets and in points X and Y respectively, and line F O meets D and D in points U and V. Prove that XV = Y U. Solution. It is sufficient to prove that XO = OY. Indeed, we then similarly have UO = OV and so XUY V is a parallelogram. Let EO meet the circle in points P and Q (Fig. 19). The sought equality is equivalent to (P X; OY ) = (QY ; OX). Projecting line EO to the circle from point we obtain an equivalent equality (P ; ) = (Q; ). It is correct because P Q, and the tangent in concur. 10

11 Y Q P X E Fig (F.Ivlev) (10) The incircle of an acute-angled triangle touches,, at points 1, 1, 1 respectively. Points 2, 2 are the midpoints of segments 1 1, 1 1 respectively. Let P be a common point of the incircle and the line O, where O is the circumcenter of. Let also and be the second common points of P 2 and P 2 with the incircle. Prove that a common point of and lies on the altitude of the triangle dropped from the vertex. Solution. It is sufficient to prove that P =. Indeed, from this we obtain that line is the reflection of P in the bisector of angle. Similarly line is the reflection of P in the bisector of angle, and so the common point of these two lines lies on the reflection of line P in the bisector of angle, i.e. on the altitude. Let Q be the common point of line P with the incircle and S be the midpoint of arc 1 1 (fig. 20). onsider the composition f of the projections of incircle to itself from and 2. We have f( 1 ) = 1, f( 1 ) = 1, f(q) = and f(s) = S. Thus ( 1 Q; S 1 ) = ( 1 ; S 1 ), i.e. is the reflection of Q in 2, which proves the sought equality. 11

12 Q S P Fig (.kopjan) (10 11) given convex quadrilateral D is such that D+ D > + D. Prove that S D + S D > S + S D. Solution. If D then D = D and D =. Thus the given equality is equivalent to the fact that rays and D intersect, i.e. the distance from to line is less than the distance from D, and the distance from to line D is less than the distance from. So S D > S and S D > S D. 22. (.Zaslavsky) (10 11) circle centered at a point F and a parabola with focus F have two common points. Prove that there exist four points,,, D on the circle such that lines,, D and D touch the parabola. Solution. Take an arbitrary point lying on the circle and outside the parabola. Line F and the line passing through and parallel to the axis of parabola intersect the circle in points symmetric wrt the axis. Thus the tangents from to the parabola also intersect the circle in symmetric points and D. Similarly the second tangents from and D intersect the circle in point symmetric to. Thus,,, D are the sought points. 12

13 23. (N.eluhov, ulgaria) (10 11) cyclic hexagon DEF is such that F = 2 F, D E = 2DE, and EF D = 2F DE. Prove that lines D, E and F concur. Solution. Let P be the common point of F and D and G be the second common point of P with the circumcircle of the hexagon. Then 2 = (; F ) = (DF ; G) = (DF ; E), and so points G and F coincide. 24. (.kopjan) (10 11) Given a line l in the space and a point not lying on l. For an arbitrary line l passing through, XY (Y is on l ) is the common perpendicular to the lines l and l. Find the locus of points Y. Solution. Let the plane passing through and perpendicular to l intersect l in point. Let be the projection of Y to this plane. Hence XY, thus Y and by three perpendiculars theorem. So lies on the circle with diameter, and Y lies on the cylinder constructed over this circle. It is clear that all points of the cylinder are on the sought locus. 25. (N.eluhov, ulgaria) (11) It is known for two different regular icosahedrons that some six of their vertices are vertices of a regular octahedron. Find the ratio of the edges of these icosahedrons. Solution. Note that no icosahedron can contain four vertices of octahedron. Indeed, between four vertices of octahedron there exist two opposite vertices that together with each of remaining vertices form an isosceles right-angled triangle. ut no three vertices of icosahedron form such triangle. Thus one of given icosahedrons contains three vertices lying on one face of octahedron, and the other icosahedron contains three vertices lying on the opposite face. Now note that there exist only three different distances between the vertices of icosahedron: one is equal to the edge, the second is equal to the diagonal of a regular pentagon with the side equal to the edge, and the third is the distance between two opposite vertices. If three vertices form a regular triangle then the distance between them is one of two first types. Since two icosahedrons aren t equal, then some face of the octahedron is a face for one of them and a triangle formed by diagonals for the other. Thus the ratio of edges is the ratio of the diagonal and the side of a regular pentagon, i.e