XIII GEOMETRICAL OLYMPIAD IN HONOUR OF I.F.SHARYGIN The correspondence round. Solutions

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1 XIII GEOMETRIL OLYMPID IN HONOUR OF I.F.SHRYGIN The correspondence round. Solutions 1. (.Zaslavsky) (8) Mark on a cellular paper four nodes forming a convex quadrilateral with the sidelengths equal to four different primes. Solution. Take for example a quadrilateral with vertices ( 3, 0), (0, 4), (12, 1), D(12, 8). Its sidelengths are = 5, = 13, D = 7, D = (L.Shteingarts) (8) circle cuts off four right-angled triangles from rectangle D. Let 0, 0, 0 and D 0 be the midpoints of the correspondent hypotenuses. Prove that 0 0 = 0 D 0. Solution. Let the circle meet,, D, D at points K 1, K 2, L 1, L 2, M 1, M 2, N 1, N 2. Then K 1 K 2 M 2 M 1 is an isosceles trapezoid, i.e. K 1 DM 1 = K 2 M 2, or K 1 + M 2 = K 2 + DM 1. Hence the projections of segments 0 0 and 0 D 0 to, equal to (K 1 + M 2 )/2 and (K 2 + DM 1 )/2 respectively, are congruent. Similarly their projections to are congruent, therefore the lengths of these segments are equal. 3. (M.Plotnikov) (8) Let I be the incenter of triangle ; H, H the orthocenters of triangles I and I respectively; K the touching point of the incircle with the side. Prove that H, H and K are collinear. Solution. Since H and H are perpendicular to I, the quadrilateral H H is a trapezoid and its diagonals divide each other as H : H. Since the projections M, N of H, H to and respectively coincide with the projections of I to these lines, we obtain that M = K and N = K. lso since H M = H N = 90 /2, the right-angled triangles H M and H N are similar. Therefore H : H = K : K, and the diagonals of the trapezoid meet at K (fig. 3). 1

2 N H I M K Fig. 3 H 4. (.Zaslavsky) (8) triangle is given. Let be the vertex of an isosceles triangle with = 120 constructed on the other side of than, and be the vertex of an equilateral triangle constructed on the same side of as. Let K be the midpoint of. Find the angles of triangle K. nswer. 90, 30, 60. Solution. Let be a vertex of parallelogram. Then = =, = and = because the angle between and is equal to = 60. Therefore the triangles and are congruent, and the angle between their corresponding sidelines and is equal to 60 (fig. 4). Thus the triangle is regular, and since K is the midpoint of, we obtain that K K and K = 30. K Fig. 4 2

3 This reasoning can be also formulated in the following way. onsider the rotations around by 120 and around by 60. Their composition maps to, hence it is the reflection about K. Since it maps to, we obtain the indicated answer. 5. (.Frenkin) segment is fixed on the plane. onsider all acute-angled triangles with side. Find the locus of a) (8) the vertices of their greatest angles; b) (8 9) their incenters. nswer. a) The set of points lying inside or on the boundary of the intersection of two discs centered at and with radii, but outside the disc with diameter. b) The set of points lying inside the square KL, but outside the intersection of two discs centered at K and L with radii K. Solution. a) If the vertex of the greatest angle does not coincide with or then is the greatest side of triangle, i.e. and. On the other hand, since angle is acute, we obtain that lies outside the circle with diameter. b) Let I be the incenter of. Since angles and are acute, we have I < 45 and I < 45, i.e. I lies inside the square KL. On the other hand, since angle is acute, we obtain that I < 135 and I lies outside the intersection of the discs centered at K, L with radii K. 6. (N.Moskvitin) (8 9) Let D be a convex quadrilateral with = D = D; E and F the midpoints of and D respectively; O the common point of the diagonals. Prove that EF passes through the touching points of the incircle of triangle OD with O and OD. Solution. Let X, Y, Z be the touching points of the incircle with O, OD, D respectively. Then DY = DZ and therefore Y = Z = X. Furthermore OX = OY. pplying the Menelaus theorem to the triangle O and the line XY, we obtain that this line passes through E. Similarly it passes through F (fig. 6). E X O Y F Fig. 6 Z D 3

4 7. (.Frenkin) (8 9) The circumcenter of a triangle lies on its incircle. Prove that the ratio of its greatest and smallest sides is less than two. First solution. Since the circumcenter belongs to the given triangle, this triangle is not obtuse-angled. If it is right-angled then the circumcenter O is the midpoint of its hypothenuse and coincides with the touching point of the incircle. Therefore the triangle is isosceles and right-angled and the assertion of the problem is valid. Suppose that the triangle is acute-angled and O lies on one of three arcs between the touching points. Let this arc is faced to the vertex. onstruct the perpendiculars from O to and. The foot of each of them (the midpoint of the corresponding side) lies between and the touching point of the incircle ω with the side. Therefore, > and >. Now we have to prove that the ratio of each of sides, to is less than 2. For example let D be the midpoint of. Let us prove that D <. Let K and L be the touching points of ω with and. Then K = L, and we have to prove that DK < L. ut the perpendicular from D to passes through the point O on ω, hence DK is not greater than its radius. On the other hand L is greater than the radius, because the perpendicular from to does not intersect ω (the angle between and the tangent is acute). Q.e.d. Second solution. Use the Euler formula: OI 2 = R 2 2Rr, where I is the incenter, R, r are the radii of the circumcircle and the incircle. Since OI = r we obtain that r/r = 2 1. Each side of the triangle is a chord of the circumcircle tangent to the incircle. The greatest of these chords is equal to 2R, and the shortest one touching the incircle at the point opposite to O is 2 R 2 4r 2 > R. 8. (Ye.akayev) (8 9) Let D be the base of trapezoid D. It is known that the circumcenter of triangle lies on D. Prove that the circumcenter of triangle D lies on. Solution. Let the perpendicular bisector to meet D and at points K and L respectively. Then by the assumption LK = = D. Hence KL = D which yields the required assertion (fig. 8). K L Fig. 8 D 9. (.Zaslavsky) (8 9) Let 0 be the midpoint of hypotenuse of triangle ; 1, 1 the bisectors of this triangle; I its incenter. Prove that the lines 0 I and 1 1 meet on the altitude from. Solution. Use the following property of an arbitrary triangle. 4

5 Lemma. The line 0 I meets the altitude H at the point lying at the distance r from. In fact, let, be the touching points of side with the incircle and the excircle respectively, and 2 the point of the incircle opposite to. Point is the homothety center of the incircle and the excircle, and 2 and are the corresponding points of these circles, therefore, 2, are collinear. Furthermore 0 = 0, i.e. 0 I is the medial line of triangle 2, and 0 I 2. Hence the lines 2, 2 I, 0 I and H are the sidelines of a parallelogram, and we obtain the assertion of the lemma (fig. 9.1). 2 I 0 Fig. 9.1 H Return to the problem. Denote the common point of 0 I and H by H (fig. 9.2). Since H = r, the distances from H to, and are d a = r cos H = r cos = r /, d b = r / and d c = H r respectively. Since ( + +)r = H = 2S, we obtain that d c = d a +d b. It is clear that the distances from 1, 1 to, and also have the similar property. y the Thales theorem all such points are collinear. H 0 1 H I 1 Fig

6 10. (I.I.ogdanov) (8 10) Points K and L on the sides and of parallelogram D are such that KD = LD. Prove that the circumcenter of triangle KL is equidistant from and. Solution. The triangles KD and LD are similar by two angles, therefore K : L = D : D. Hence, when K moves along with constant velocity, L also moves along uniformly, and therefore the circumcenter of KL moves along some line. If K, L are the projections of D to and respectively, the circumcenter of KL coincides with the center of the parallelogram, and when K and L coincide with and respectively, the circumcenter lies on the perpendicular bisector to. Thus this perpendicular bisector is the locus of circumcenters. 11. (.Tolesnikov) (8 11) finite number of points is marked on the plane. Each three of them are not collinear. circle is circumscribed around each triangle with marked vertices. Is it possible that all centers of these circles are also marked? nswer. No. Solution. onsider the circle having the minimal radius. Let it be the circumcircle of triangle, and O be its center. If is not a regular triangle, then some of its angles, for example, is less than 60. ut in this case 60 < O < 120, i.e. sin O > sin, and by the sinus theorem the circumradius of O is less than the radius of circle, which contradicts to the definition of this circle. If is regular then the centers,, of circles O, O, O are also marked. ut for example the triangle O is regular, and its circumradius is less than the radius of circle. 12. (D.Shvetsov) (9 10) Let 1, 1 be the altitudes of triangle, 0 the common point of the altitude from and the circumcircle of ; and Q the common point of the circumcircles of and 1 1 0, distinct from 0. Prove that Q is the symmedian of. Solution. Since,, 1, 1 are concyclic we obtain that the lines, 1 1 and 0 Q concur at the radical center N of circles 1 1, and Let Q meet and 1 1 at points P and M respectively (fig. 12). Projecting the circumcircle of triangle from Q to, and projecting this line from to 1 1 we obtain te equality of cross-ratios ( 1 1 MN) = (P N) = ( 0 ) = : 0. Since 0 0 is the reflection of the orthocenter H of triangle about, the second fraction is equal to H/H = 1 / 1. Now applying the Menelaus theorem to the triangle 1 1 and the line N we obtain that 1 1 MN = 1 N/ 1 N, i.e. 1 M = 1 M. Therefore M is the median of triangle 1 1 and the symmedian of. 6

7 1 M 1 P N 0 Q Fig (.Zaslavsky) (9 11) Two circles pass through points and. third circle touches both these circles and meets at points and D. Prove that the tangents to this circle at these points are parallel to the common tangents of two given circles. Solution. Let the third circle touch two given circles at points X, Y, and their common tangent touch them at U, V (points X and U lie on the same circle). Since X is the homothety center of touching circles, the line XU meets the third circle at point P such that the tangent at this point is parallel to UV. Similarly Y V passes through P. lso X, Y, U, V are collinear, therefore P X P U = P Y P V. Hence P lies on and thus coincides with one of points, D (fig. 13). The proof for the second point is similar. 7

8 P Y X V Fig. 13 U 14. (N.Moskvitin) (9 11) Let points and lie on the circle with diameter D and center O on the same side of D. The circumcircles of triangles O and DO meet at points F and E respectively. Prove that R 2 = F DE, where R is the radius of the given circle. Solution. Since F O is cyclic and O = O, we have (fig.14) F O = sin OF sin O = sin F sin O = sin sin D. F E O D Fig. 14 8

9 Similarly, DE/OD = sin D/ sin D. Since D is cyclic, the product of these ratios is equal to (K.leksiev) (9 11) Let be an acute-angled triangle with incircle ω and incenter I. Let ω touch, and at points D, E, F respectively. The circles ω 1 and ω 2 centered at J 1 and J 2 respectively are inscribed into DIF and DIE. Let J 1 J 2 intersect at point M. Prove that D is perpendicular to IM. Solution. Since DJ 1, DJ 2 are the bisectors of triangles DI, DI respectively, we have J 1 /J 1 I = D/ID, IJ 2 /J 2 = I/. y the Menelaus theorem we obtain that the quadruple,,, M is harmonic, i.e. M lies on F E (fig.15). Since and D are the poles of lines EF and wrt the incircle we obtain that M is the pole of D, therefore D IM. F I E J 1 J 2 D Fig. 15 M 16. (P.Ryabov) (9 11) The tangents to the circumcircle of triangle at and meet at point D. The circle passing through the projections of D to,,, meet for the second time at point. Points, are defined similarly. Prove that,, concur. Solution. The pedal circle of point D coincides with the pedal circle of isogonally conjugated point D which is the vertex of parallelogram D. Hence is the projection of D to, i.e. the reflection of the foot of the altitude from about the midpoint of. Similarly, are the reflections of the feet of the altitudes from and about the midpoints of the corresponding sides. Therefore, and concur at the point isotomically conjugated to the orthocenter of the triangle. 17. (.Trigub) (9 11) Using a compass and a ruler, construct a point K inside an acute-angled triangle so that K = 2 K and K = 2 K. Solution. Let the circle centered at K and passing through meet and at points P and Q respectively, and let T be the midpoint of arc of the circumcircle. Then KP = KP = 2 KP, therefore KP = KP and P = P K = K. Similarly Q = QK = K. Since P = Q, T = T and P T = QT, the triangles T P and T Q are congruent i.e. T P = T Q and T lies on the circle P Q. 9

10 Hence the center K of this circle lies on the perpendicular bisector to T. Furthermore by the assumption K = 3 /2, i.e. K lies on the corresponding arc (fig.17). T P K Q Fig. 17 Now let us prove that the constructed point K is in fact the required one. Denote again the common points of the sidelines with the circle centered at K and passing through by P and Q. Since this circle passes through T, we obtain that P = Q. If P > P K = K then P K > P K, KP = KP > 2 K, K > 2 K and K < 3 /2 which contradicts to the construction of K. Similarly if P < P K we have K > 3 / (.Trigub) (9 11) Let L be the common point of the symmedians of triangle, and H be its altitude. It is known that LH = Prove that LH = Solution. Let 1, 1 be the altitudes of the triangle. Then the symmedians L, L are the medians of triangles 1 H, 1 H, i.e. they pass through the midpoints M, N of segments H 1, H 1 respectively. ut MNH = 1 1 H = 180 2, therefore LH = if and only if HLMN is cyclic. Similarly this is equivalent to the condition LH = (D.Prokopenko) (10 11) Let cevians, and of triangle concur at point P. The circumcircle of triangle P meets and at points M and N respectively, and the circumcircles of triangles P and P meet and for the second time respectively at points K and L. The line c passes through the midpoints of segments MN and KL. The lines a and b are defined similarly. Prove that a, b and c concur. Solution. y the assumption M = N and K = P = L. Hence KL MN and c passes through. Since MN and are antiparallel, this line 10

11 is the symmedian of triangle and so it divides into two angles with the ratio of sinuses equal to :. The similar relations for two remaining angles and the eva theorem yield the required assertion. 20. (V.Luchkin, M.Fadin) (10 11) Given a right-angled triangle and two perpendicular lines x and y passing through the vertex of its right angle. For an arbitrary point X on x define y and y as the reflections of y about X and X respectively. Let Y be the common point of y b and y c. Find the locus of Y (when y b and y c do not coincide). Solution. onsider the point X isogonally conjugated to X and its reflections U, V, W about,, respectively. Perpendicularity of x and y implies that U and V lie on y. Furthermore X, X are the perpendicular bisectors to UW, V W respectively. Therefore W lies on y b, y c, i.e. it coincides with Y (fig.20). Thus Y lies on the reflection of the isogonal image of x about. The required locus is this line without the points such that y b and y c coincide, i.e. the common point of this line with and the reflection of about. W = Y V X X Fig. 20 U 21. (N.eluhov) (10 11) convex hexagon is circumscribed about a circle of radius 1. onsider the three segments joining the midpoints of its opposite sides. Find the greatest real number r such that the length of at least one segment is at least r. Solution. Let be the hexagon in question, circumscribed about a circle ω with center I, and let M i be the midpoint of i i+1 (indices run modulo 6, so that, say, 7 1 ). If approaches an equilateral triangle and 4, 5, and 6 all approach the midpoint of 1 3 then the lengths of M 1 M 4, M 2 M 5, and M 3 M 6 all approach 3. We will show that r = 3 is indeed the answer to the problem. First we verify that I lies inside M 1 M 2... M 6. Suppose for example that it lies inside the triangle M 1 1 M 6. ut then ω is contained inside and cannot touch all sides of , a contradiction. 11

12 Let (D) denote the angle such that rotation by it counterclockwise about makes codirectional with D. Since all M i lie outside ω, we have IM i 1. Therefore, if 120 (IM i IM i+3 ) 240 for some i then M i M i+3 3 and we are done. Suppose now that this does not happen for any i. Let j be such that (IM j IM j+3 ) 120 and (IM j+3 IM j ) 240. Then there is some k, j k j+2, such that (IM k IM k+3 ) 120 and (IM k+1 IM k+4 ) 240. Without loss of generality, take k = 4. Then 120 IM 1 IM and consequently M 1 M 2 3. onsider the convex quadrilateral M 1 M 2 M 4 M 5. If angle M 1 is right or obtuse then M 2 M 5 > M 1 M 2 3 and we are done. If angle M 2 is right or obtuse then M 1 M 4 > M 1 M 2 3 and we are done. It remains to consider the case when angles M 1 and M 2 are both acute. In this case however 90 < (M 1 M 2 M 4 M 5 ) < 270. Since M 3 M 6 = M 1 M 2 + M 4 M 5 (because M 3 M 6 = M 3 M 4 + M 4 M 5 + M 5 M 6 and M 1 M 2 + M 3 M 4 + M 5 M 6 = 1( ) = 0), we have M 3 M 6 > M 1 M 2 3, and the proof is complete. 22. (M. Panov) (10 11) Let P be an arbitrary point on the diagonal of cyclic quadrilateral D, and P K, P L, P M, P N, P O be the perpendiculars from P to,, D, D, D respectively. Prove that the distance from P to KN is equal to the distance from O to ML. Solution. When P moves uniformly along, the lines KN and M L are translated uniformly and the point O moves uniformly as well. Thus d(p, KN) d(o, ML) is a linear function of the position of P. When P =, this function equals 0 by the Simson theorem, and when P is the common point of and D, it equals 0 because KLMN is circumscribed about a circle centered at P = O ( NKP = D = D = P KL because KP N and KP L are cyclic). 23. (I.Frolov) (10 11) Let a line m touch the incircle of triangle. The lines passing through the incenter I and perpendicular to I, I, I meet m at points,, respectively. Prove that, and concur. Solution. The polar transformation wrt the incircle maps,,, m to their touching points 1, 1, 1 M with the incircle. Since I is the polar of the infinite point of perpendicular line I, its common point with m is the pole of the line passing through M and parallel to I. Since I 1 1, the line is the polar of the projection of M to 1 1. Similarly the lines and are the polars of projections of M to 1 1 and 1 1 respectively. y the Simson theorem these projections are collinear, hence their polars concur. 24. (I.I.ogdanov) (11) Two tetrahedrons are given. Each two faces of the same tetrahedron are not similar, but each face of the first tetrahedron is similar to some face of the second one. Does this yield that these tetrahedrons are similar? nswer. No. Solution. Let t be some number close to 1. Then there exist two tetrahedrons such that their bases are regular triangles with side equal to 1, the lateral edges of the first tetrahedron are equal to t, t 2, t 3, and the lateral edges of the second one are equal to 1/t, 12

13 1/t 2, 1/t 3. It is clear that the assumption is valid for these tetrahedrons but they are not similar. 13

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