IX GEOMETRICAL OLYMPIAD IN HONOUR OF I.F.SHARYGIN.

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1 IX GEOMETRIL OLYMPID IN HONOUR OF I.F.SHRYGIN. THE ORRESPONDENE ROUND. SOLUTIONS. 1. (N.Moskvitin) Let be an isosceles triangle with =. Point E lies on side, and ED is the perpendicular from E to. It is known that E = DE. Find D. nswer. 45. Solution. y the external angle theorem ED = 90 + = (fig.1). Therefore, ED = (180 ED)/2 = 45. E D Fig.1 2. (L.Steingarts) Let be an isosceles triangle ( = ) with = 20. The bisectors of angles and meet the opposite sides in points 1 and 1 respectively. Prove that triangle 1 O 1 (where O is the circumcenter of ) is regular. Solution. On sides and take points and such that = O = O =. It is clear that, i.e. = = 80. lso O = O = O = 10. Thus O = 20 and O = 60, i.e triangle O is regular. Then = and = = (fig.2). Therefore coincides with 1. Similarly coincides with 1, q.e.d. O Fig.2 3. (D.Shvetsov) Let be a right-angled triangle ( = 90 ). The excircle inscribed into the angle touches the extensions of the sides, at points 1, 2 respectively; points 1, 1

2 2 are defined similarly. Prove that the perpendiculars from,, to 1 2, 1 1, 1 2 respectively, concur. Solution. Let I be the incenter of, and D be the fourth vertex of rectangle D. Since I 1 2, I 1 2, the perpendiculars from to 1 and from to 1 meet in the incenter J of triangle D. Then it is sufficient to prove that DI 1 1. Let X, Y, Z be the projections of I to,, D respectively. Then 1 = X 2 = ZD and 1 = Y = IZ, thus triangles 1 1 and IZD are equal, i.e. IDZ = 1 1 (fig.3), that proves the required assertion. Z D I 1 1 Fig.3 4. (F.Ivlev) Let be a nonisosceles triangle. Point O is its circumcenter, and point K is the center of the circumcircle w of triangle O. The altitude of from meets w at a point P. The line P K intersects the circumcircle of at points E and F. Prove that one of the segments EP and F P is equal to the segment P. Solution. Points O and K lie on the bisector of segment, thus OK P and OP K = P OK = OP. Therefore the reflection of in OP lies on P K. lso O = O, i.e. lies on the circumcircle of (fig.4). Thus coincides with one of points E, F. O P K Fig.4 5. (.Frenkin) Four segments drawn from a given point inside a convex quadrilateral to its vertices, split the quadrilateral into four equal triangles. an we assert that this quadrilateral is a rhombus? 2

3 nswer. Yes. Solution. Let D and O be the given quadrilateral and point. In equal triangles the angles opposite to equal sides are equal. Since O = O, angles O and O opposite to O are equal. Similarly DO = DO, thus D = D. Two remaining angles of the quadrilateral are similarly equal, therefore D is a parallelogram. There exist two adjacent angles with vertex O such that their sum is not less than π, suppose that these angles are O and O. The second angle is equal to some angle of triangle O. This can be only O, because its sum with any of two remaining angles of O is less than π. The sides of equal triangles O and O opposite to these angles are equal. Then = and D is a rhombus. 6. (D.Shvetsov) Diagonals and D of a trapezoid D meet at point P. The circumcircles of triangles P and DP intersect the line D for the second time at points X and Y respectively. Let M be the midpoint of segment XY. Prove that M = M. Solution. y condition, X = P = P D = Y D (fig.6). Thus XY is an isosceles trapezoid, which proves the required assertion. P X Fig.6 Y D 7. (D.Shvetsov) Let D be a bisector of triangle. Points I a, I c are the incenters of triangles D, D respectively. The line I a I c meets in point Q. Prove that DQ = 90. Solution. Lines I a and I c meet in the incenter I of. y the bisectrix theorem I a /I a I = D/ID, I c /I c I = D/ID. y the Menelaos theorem Q/Q = D/D = /. Therefore Q is the external bisectrix of angle, q.e.d. 8. (M.Plotnikov) Let X be an arbitrary point inside the circumcircle of a triangle. The lines X and X meet the circumcircle for the second time at points K and L respectively. The line LK intersects and at points E and F respectively. Find the locus of points X such that the circumcircles of triangles F K and EL touch. nswer. The arc of the circle passing through, and the circumcenter O of. Solution. Let the circles touche. Then the angles between their common tangent and lines and are equal to angles LE and KF respectively. Since these two angles are equal to 3

4 angles X and X, their sum is equal to angle and X = 2 = O. Similarly we obtain that for any point of the arc the correspondent circles touche. 9. (M.Plotnikov) Let T 1 and T 2 be the points of tangency of the excircles of a triangle with its sides and respectively. It is known that the reflection of the incenter of across the midpoint of lies on the circumcircle of triangle T 1 T 2. Find. nswer. 90. Solution. Let D be the fourth vertex of parallelogram D, J be the incenter of D, S 1, S 2 be the points of tangency of the incircle of with D and D. Then S 1 T 1, S 2 T 2 and T 1 JT 2 = S 1 JS 2 = π. lso DS 1 = DS 2, i.e. lines S 1 T 1, S 2 T 2 and DJ concur. Therefore J coincides with the common point of lines S 1 T 1 and S 2 T 2, i.e. = 90 (fig.9). T 2 J T 1 Fig.9 S 1 S (D.Shvetsov) The incircle of triangle touches the side at point ; the incircle of triangle touches the sides and at points 1, 1 ; the incircle of triangle touches the sides and at points 2, 2. Prove that the lines 1 1, 2 2, and concur. Solution. Since =, the incircles of triangles and touche at the same point. Therefore 1 = 2. lso 1 = 1, 2 = 2, and if we find the angles of quadrilateral , we obtain that it is cyclic. Thus 1 1, 2 2 and concur in the radical center of three circles: the circumcircle of and the incircles of triangles, (fig.10). D Fig.10 4

5 11. (P.Kozhevnikov) a) Let D be a convex quadrilateral and r 1 r 2 r 3 r 4 be the radii of the incircles of triangles, D, D, D. an the inequality r 4 > 2r 3 hold? b)the diagonals of a convex quadrilateral D meet in point E. Let r 1 r 2 r 3 r 4 be the radii of the incircles of triangles E, E, DE, DE. an the inequality r 2 > 2r 1 hold? nswer. a) No. b) No. Solution. a) Suppose that r 4 = r(). It is sufficient to prove that r()/2 < max{r(d), r(d The midpoint K of lies inside one of triangles D, D, for example inside D. Then triangle KL, where L is the midpoint of, lies inside triangle D, therefore r()/2 = r(kl) < r(d). b) Let r = r 1 be the inradius of triangle E. The diameters of the incircles of triangles E, DE, are less than the altitudes of these triangles coinciding with altitudes h a, h b of E. Thus it is sufficient to prove that one of these altitudes is less than 4r. Suppose that E E. Then the semiperimeter p < E + E 2E and h b = 2S/E = 2pr/E < 4r. omment. Note that the answer to both questions will be positive if we replace 2 to any smaller number. 12. (.Frenkin) (8 11) On each side of triangle, two distinct points are marked. It is known that these points are the feet of the altitudes and the bisectors. a) Using only a ruler determine which points are the feet of the altitudes and which points are the feet of the bisectors. b) Solve p.a) drawing only three lines. Solution. Preliminary hints. Since all points are distinct the triangle isn t isosceles. For each side, the foot of the altitude lies between the foot of the bisector and the smaller of two remaining sides. Thus it is sufficient to define the smallest and the greatest of the sides. We will denote the feet of the bisector and the altitude from vertex X as L X and H X respectively. Lemma. If > then lines L L and H H meet the extension of side beyond. Proof. Let L D be the perpendicular from L to, and H be the altitude. y the bisector theorem L D : H = : ( + ). Similarly if L E is the perpendicular from L to, then L E : H = : ( + ). Furthermore >, L D > L E, thus L L meets the extension of beyond. Points H, H lie on the semicircle with diameter. Since H < H, the distance from H to is less than the distance from H to. The lemma is proved. Simple solution of p.a). Joining the given points with the opposite vertices we obtain two families of concurrent lines. Take two points of the same family on two sides and draw the line through them. y the lemma this line meets the extension of the third side beyond the vertex lying on the smaller of two sides. Therefore we can define the smaller of any two sides. Solution of p.b). Take for each vertex the nearest marked points on two adjacent sides and join these points. We will prove that these lines meet the prolongation of the greatest side beyond the vertex of the medial angle and the extensions of two remaining sides beyond the vertex of the greatest angle. From this we can define the greatest and the smallest side. 5

6 Let us prove the above assertion. Suppose that > >. The marked points nearest to the vertex of the smallest angle are the feet of the bisectors, and the points nearest to the vertex of the greatest angle are the feet of the altitudes. y the lemma, the lines joining these points meet the extension of beyond and the extension of beyond. The marked points nearest to are H and L. y the lemma, line L L meets the extension of beyond in some point P. Ray H L passes inside triangle H P and thus intersects segment P, q.e.d. 13. (F.Ivlev) Let 1 and 1 be the tangency points of the incircle of triangle with and respectively, and be the tangency points of the excircle inscribed into the angle with the extensions of and respectively. Prove that the orthocenter H of triangle lies on 1 1 if and only if the lines 1 and are orthogonal. Solution. Suppose that 1. Then by Thales theorem the altitude from divides segment 1 1 in ratio 1 : = p c : p a. The altitude from passes through the same point. The inverse assertion is obtained similarly. 14. (D.Shvetsov) Let M, N be the midpoints of diagonals, D of right-angled trapezoid D ( = D = 90 ). The circumcircles of triangles N, DM meet line in points Q, R. Prove that the distances from Q, R to the midpoint of MN are equal. Solution. Let X, Y be the projections of N and M to. Than we have to prove that RY = XQ. Since NQX = N = D, triangles XQN and D are similar (fig.14). Thus XQ = NX/D. ut NX = D sin D/2 = D D/2, therefore XQ = D/2 = RY. D X N Q Fig a) (V.Rastorguev) Triangles and are inscribed into triangle so that 1 1, 1 1, 1 1, 2 2, 2 2, 2 2. Prove that these triangles are equal. b) (P.Kozhevnikov) Points 1, 1, 1, 2, 2, 2 lie inside triangle so that 1 is on segment 1, 1 is on segment 1, 1 is on segment 1, 2 is on segment 2, 2 is on segment 2, 2 is on segment 2 and angles 1, 1, 1, 2, 2, 2 are equal. Prove that triangles and are equal. Solution. a) Inscribe triangle into triangle in such a way that 2 2, 2 2, 2 2. It is clear that the corresponding sidelines of triangles 6

7 and are symmetric wrt the circumcenter of This symmetry maps to Therefore these triangles are equal and their circumcenters coincide. b) onsider the chords,,,,, of the circumcircle of lying on the lines 1 1, 1 1, 1 1, 2 2, 2 2, 2 2. y condition, arcs,,,,, are equal. Let their size be φ. The rotation around the circumcenter to φ maps,, to,, respectively, thus it maps to (fig.15) Fig.15 omment. In a special case when triangle degenerates to a point, also degenerates to a point, and the distances from these two points to the circumcenter are equal. These points are the rockard points of the triangle. 16. (F.Ivlev) The incircle of triangle touches,, at points,, respectively. The perpendicular from the incenter I to the median from vertex meets the line in point K. Prove that K. Solution. The polar transformation wrt the incircle maps the perpendicular from I to the median into the infinite point of this median, the image of line is point, and the image of the line passing through and parallel to is the common point P of and I. Thus we have to prove that P lies on the median. Since I = I, P I =, P I =, we have P : P = :. Since =, we have sin P : sin P = :, i.e. P bisects (fig.16). 7

8 P I Fig (.Zaslavsky) n acute angle between the diagonals of a cyclic quadrilateral is equal to ϕ. Prove that an acute angle between the diagonals of any other quadrilateral having the same sidelengths is smaller than ϕ. Solution. Let the diagonals of quadrilateral D meet in point P. Put P = a, P = b, P = c, P D = d and express the sidelengths of D through a, b, c, d and cos ϕ. Then D 2 2 = 2 cos ϕ(ab + bc + cd + da) = 2 D cos ϕ. y Phtolomeos theorem D D + D, and the equality holds only for a cyclic quadrilateral. 18. (.Ivanov) Let D be a bisector of triangle. Points M and N are the projections of and to D. The circle with diameter MN intersects in points X and Y. Prove that X = Y. Solution. Let,, X, Y be the reflections of,, X, Y in MN. Then the diagonals of isosceles trapezoid meet at point L, which is the reflection of in the circle with diameter MN. The diagonals of isosceles trapezoid XX Y Y inscribed into this circle also meet at L. The lateral sidelines of this trapezoid meet on the polar of L, passing through and parallel to the bases of the trapezoid. y symmetry is the common point of the sidelines, which implies the assertion of the problem (fig.18). Y Y X L X 8

9 Fig (D.Prokopenko) a) The incircle of a triangle touches and at points 0 and 0 respectively. The bisectors of angles and meet the perpendicular bisector to the bisector L in points Q and P respectively. Prove that the lines P 0, Q 0, and concur. b) Let L be the bisector of a triangle. Points O 1 and O 2 are the circumcenters of triangles L and L respectively. Points 1 and 1 are the projections of and to the bisectors of angles and respectively. Prove that the lines O 1 1, O 1 1, and concur. c) Prove that two points obtained in pp. a) and b) coincide. Solution. a) It is clear that P Q 0 0. lso P lies on the circumcircle of L. Thus P L = /2 and P L = 90 /2 = 0 0, where 0 is the touching point of the incircle with. Therefore the corresponding sidelines of triangles P QL and are parallel i.e., these triangles are homothetic (fig.19a). The homothety center S lies on line L 0. Thus lines P 0 and Q 0 meet in S, i.e. on line. P 0 Q 0 L 0 Fig.19а b) First prove that points 0, 0, 1 and 1 are collinear. In fact, since the reflection of in the bisector of angle lies on, point 1 lies on the medial line. lso we have 1 = /2, and therefore 1 = /2 = 0. This property defines the common point of and 0 0. Thus lines O 1 O 2 and 1 1 are parallel. Now quadrilateral 1 I 0 is cyclic, therefore 1 0 = 90 /2 = O 1 L and 0 1 LO 1. Similarly 0 1 LO 2 (fig.19b). Thus triangles O 1 O 2 L and are homothetic. 9

10 O 1 O L 0 Fig.19b c) oth homotheties of pp. a) and b) transform 0 to L, and line 0 0 to the medial perpendicular to L. Therefore their centers coincide. 20. (V.Yassinsky) Let 1 be an arbitrary point on the side of triangle. Points 1 and 1 on the rays and are such that 1 1 = 1 1 =. The lines 1 and 1 meet in point 2. Prove that all the lines 1 2 have a common point. Solution. y condition, quadrilaterals 1 1 and 1 1 are cyclic. Thus 1 1 = 1, 1 1 = 1, and therefore 2 = π, i.e. 2 lies on the circle passing through, and the reflection of wrt. lso 1 = 2, thus 1 passes through 2 (fig.20) Fig (D.Yassinsky) Let be a point inside a circle ω. One of two lines drawn through intersects ω at points and, the second one intersects it at points D and E (D lies between and E). The line passing through D and parallel to meets ω for the second time at point F, and 10

11 the line F meets ω at point T. Let M be the common point of the lines ET and, and N be the reflection of across M. Prove that the circumcircle of triangle DEN passes through the midpoint of segment. Solution. Firstly, project line to the circle from point D, and then project the circle to from point T. s a result we obtain that the image of is M, the image of infinite point is, and points and are fixed. From the equality of cross-ratios we obtain that M/M = (/) 2. Hence M = /( + ). Now let K be the midpoint of. Then N K = 2M( + )/2 = = D E, i.e. points D, E, K, N are concyclic. 22. (.Zaslavsky) The common perpendiculars to the opposite sidelines of a nonplanar quadrilateral are mutually orthogonal. Prove that they intersect. Solution. Let K, L, M, N be the feet of common perpendiculars lying on the sides,, D, D of quadrilateral D. The projection to the plane parallel to KM and LN transforms these lines to perpendicular lines K M and L N. y three perpendiculars theorem the projections of and D are perpendicular to K M, and the projections of and D are perpendicular to L N. Therefore the projection of D is a rectangle D, and K = D M, L = N. Thus K/K = DM/M, L/L = N/ND and by Menelaos theorem K, L, M, N are complanar. 23. (.Frenkin) Two convex polytopes and do not intersect. The polytope has exactly 2012 planes of symmetry. What is the maximal number of symmetry planes of the union of and, if has a) 2012, b) 2013 symmetry planes? c) What is the answer to the question of p.b), if the symmetry planes are replaced by the symmetry axes? nswer. a) b) c) 1. Solution. a) Estimation. The symmetry transposes polyhedrons and or fixes each of them. In the first case it transposes the centroids of polyhedrons, thus the symmetry plane is the perpendicular bisector of the segment between the centroids. In the second case this plane is a symmetry plane of both polyhedrons and. Thus we have at most =2013 planes. Example. Let be regular 2012-gonal pyramid. Take a point outside on its axis and construct a plane P passing through this point and perpendicular to the axis. Let the reflection of in P. Then all conditions are valid, and P and 2012 symmetry planes of are the symmetry planes of the union. b) Estimation. Since and have a distinct number of symmetry planes, they aren t equal and can t be transposed by a symmetry. Thus each symmetry is a symmetry of polyhedron, which has only 2012 symmetry planes. Example. Let be a regular 2012-gonal pyramid. Take a point outside on its axis, a plane passing through this point and perpendicular to the axis, and construct the reflection of the pyramid s base in this plane. Let be a prism with this reflection as the base, disjoint from. It is clear that has 2013 symmetry planes: one of them is parallel to the bases of the prism and equidistant from them and 2012 remaining planes coincide with the symmetry planes of. c) Estimation. Since and have a distinct number of symmetry axes they can t be transposed. Thus the sought symmetry fixes the centroid of each polyhedron. These centroids don t coincide because the polyhedrons are convex. Therefore the symmetry axis coincides with the line joining 11

12 two centroids. Example. Let be a regular 2011-gonal prism with horizontal bases. Then has one vertical and 2011 horizontal symmetry axes. Now let be a regular 2012-gonal prism with the same axis, disjoint from. Then has 2013 symmetry axes and the union of and has a vertical symmetry axis. 12