V GEOMETRICAL OLYMPIAD IN HONOUR OF I.F.SHARYGIN

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1 V GEOMETRIL OLYMPID IN HONOUR OF I.F.SHRYGIN Final round. First day. 8 form. Solutions. 1. (.linkov, Y.linkov) Minor base of trapezoid D is equal to side, and diagonal is equal to base D. The line passing through and parallel to intersects line D in point M. Prove that M is the bisector of angle. First solution. We have M = D = D = M (first and third equality follow from parallelism of M and, and D; second equality follows from = D). Thus, M = =, and M = M = M (fig.8.1). M Fig.8.1 Second solution. Let point P lie on the extension of side (beyond point ), and point K lie on the extension of diagonal (beyond point ). Then MK = D = D = M, i.e M is the bisector of angle K. Since bisects angle D and M, then M is the bisector of angle P. Thus M is the common point of two external bisectors of triangle, therefore M is the bisector of angle. 2. (.linkov) cyclic quadrilateral is divided into four quadrilaterals by two lines passing through its inner point. Three of these quadrilaterals are cyclic with equal circumradii. Prove that the fourth part also is cyclic quadrilateral and its circumradius is the same. Solution. Let the parts adjacent to vertices,, of cyclic quadrilateral D be cyclic quadrilaterals. Since angles and are opposite to equal angles in point of division L we have = = 90. So two dividing lines are perpendicular. Thus angle is also right and D is a rectangle. So the fourth quadrilateral is cyclic. Now the angles corresponding to arcs L, L, L are equal, and since the radii of these circles also are equal, we have L = L = L. So L is the center of the rectangle and the fourth circle has the same radius. 3. (.kopjan, K.Savenkov) Let H a and H b be the altitudes of triangle. Points P and Q are the projections of H a to and. Prove that line P Q bisects segment H a H b. Solution. Let H c be the third altitude of. Then H a H c = H b H c = because quadrilaterals H c H b and H c H a are cyclic. So the reflection of H a in 1 D

2 lies on H b H c. Similarly the reflection of H a in also lies on this line. Thus P and Q lie on the medial line of triangle H a H b H c (fig.8.3). Q H a H b H c P Fig (N.eluhov) Given is such that = 57, = 61 and = 62. Which segment is longer: the angle bisector through or the median through? First solution. Let K be the midpoint of arc in the circumcircle of. Let also the circumcenter of the triangle be O, and L and M be the angle bisector and the median. Define L K = N and let H be an altitude in K. Since <, lies inside arc K, therefore N lies inside segment L and L > N > H. Moreover H > KM as altitudes from a smaller and a greater angle in K. Finally, KM = MO + OK = MO + O > M, and the problem is solved: the angle bisector is longer (fig.8.4). 2

3 M N H L O K Fig.8.4 Second solution. Since >, we have M > 30. onstruct the altitude H and the perpendicular MK from M to. We have L > H = 2MK > M, because sin MK = MK > 1 M 2. Third solution. (K.Ivanov, Moscow). onsider regular triangle. Since ray lies inside angle, we have that the bisector of angle is longer that the altitude of regular triangle. In the other hand let M, N be the midpoints of and respectively. Since ray lies inside angle, we have MN > M. ut M > 90 because >. Thus N > M and the bisector of angle is longer that the median from. 3

4 V GEOMETRIL OLYMPID IN HONOUR OF I.F.SHRYGIN Final round. Second day. 8 form. Solutions. 5. (V.Protasov) Given triangle. Point M is the projection of vertex to bisector of angle. K is the touching point of the incircle with side. Find angle MK if = α Solution. Let I be the incenter of. Then quadrilateral M IK is cyclic because MI = KI = 90 (fig.8.5). Thus MK = MI = I + I = \+\ = 2 90 α 2. I M Fig (S.Markelov) an four equal polygons be placed on the plane in such a way that any two of them don't have common interior points, but have a common boundary segment? Solution. Yes, see fig.8.6. K Fig (D.Prokopenko) Let s be the circumcircle of triangle, L and W be common points of angle's bisector with side and s respectively, O be the circumcenter of triangle L. Restore triangle, if circle s and points W and O are given. 4

5 Solution. Let O be the circumcenter of. Then lines O O and O W are perpendicular to sides and, so the directions of these sides are known. lso OL = 2 L = 2 LW, thus OW = 90 (fig.8.7). Therefore is the common point of s and the circle with diameter OW. O O L W Fig (N.eluhov) triangle is given, in which the segment touches the incircle and the corresponding excircle in points M and N. If = 2 MN, show that = 2MN. Solution. We may assume that >, and therefore the points, N, M, lie on the line in this order. We will use the following well-known Lemma. Let K be the midpoint of, and I and J be the incenter and the excenter opposite to. Then N IK and M JK. Now the lemma shows that the original condition is equivalent to IKJ = 180 α/2. We will show first that if = 2MN then this is true. In this case, since the midpoints of and MN coincide, we have that M and N are midpoints of K and K, and therefore, IM and NJ are perpendicular bisectors of K and K. Thus triangles IK and JK are isosceles, and JK = 90 β/2, IK = γ/2, yielding the claim (fig.8.8). M J I K N Fig.8.8 Now, consider the circle (IJ). Given α, we see that IJ is determined as a diameter, and as an arc constituting angle 90 + α/2. When the chord runs along the circle, its midpoint K runs along a smaller circle. In the same time the locus of the points K 5

6 such that IKJ = 180 α/2, consists of two arcs of circles with endpoints I and J. Obviously, these loci intersect in four points, symmetric to each other with respect to IJ and its perpendicular bisector, thus corresponding to four equal quadrilaterals IJ. So this quadrilateral is completely determined by the condition IKJ = 180 α/2. ut the one obtained when = 2MN satisfies this condition, hence the claim. 6

7 V GEOMETRIL OLYMPID IN HONOUR OF I.F.SHRYGIN Final round. First day. 9 form. Solutions. 1. (.linkov, Y.linkov) The midpoint of triangle's side and the base of the altitude to this side are symmetric wrt the touching point of this side with the incircle. Prove that this side equals one third of triangle's perimeter First solution. Let a, b be the lengths of two sides, and the altitude divide the third side into segments with lengths x, y (if the base of the altitude lies out of the side then one of these lengths is negative). y the Pythagorean theorem x 2 y 2 = a 2 b 2. ut the touching point divides the side into segments with lengths p a and p b. So the condition of the problem is equivalent to x y = 2(a b). Dividing the first equality by the second one we obtain that x + y = (a + b)/2 = 2p/3. Second solution. Let c be the side in question, then r/r c = (p c)/p. Let K and P be the touching points of this side with the incircle and the excircle, I and Q be the centers of these circles. It is known that the midpoint of altitude H lies on line IP. Using similarity of two pairs of triangles we obtain that r = h/3, r c = h (fig.9.1). From the first equality we obtain the assertion of the problem. I P K Q Fig (O.Musin) Given a convex quadrilateral D. Let R a, R b, R c and R d be the circumradii of triangles D,, D, D. Prove that inequality R a < R b < R c < R d is equivalent to 180 D < < D. 7

8 Solution. Let the angles of the quadrilateral satisfy the given inequality. Then sin > sin D and so R b < R c. Since angle D is obtuse, this implies that point lies out of the circle D, thus D < D. s these angles are both acute, we have sin D < sin D and R c < R d. Moreover < D < 90, so R a < R b. onversely, from R b < R c it follows that angle lies between angles D and 180 D. If angle D is acute, we have D < D, and since R a < R d then D > 180 D. ut in this case we obtain by repeating previous argument that R b < R a < R d < R c. 3. (I.ogdanov) Quadrilateral D is circumscribed, rays and D intersect in point E, rays and D intersect in point F. The incircle of the triangle formed by lines, D and the bisector of angle, touches in point K, and the incircle of the triangle formed by lines D, and the bisector of angle, touches in point L. Prove that lines KL, and EF concur. Solution. Let the incircle of D touch sides and in points U and V. Then we have (E; KU) = EK K : EU U \E ctg 2 = ctg \ 4 : ctg \E 2 ctg \ 2 = (F ; LV ). This means that lines KL, EF, UV concur. Similarly lines, EF, UV concur (fig.9.3). U V K L E D Fig (N.eluhov) Given regular 17-gon Prove that two triangles formed by lines 1 4, 2 10, and 2 3, are equal. Solution. Firstly note that , , So we have to prove that given triangles are central symmetric. F 8

9 Let,,, D, E, F be the midpoints of 1 2, 3 4, 4 13, 6 14, 10 14, 15 2 respectively. Lines, DE, F as medial lines of three triangles are parallel to Lines D, E, F as axes of three isosceles trapezoids concur at the center of 17-gon. y dual Pappus theorem, D, EF concur at some point P (fig.9.4). ut these lines are the medial lines of three strips formed by parallel sidelines of given triangles. Therefore these triangles are symmetric wrt P. P F D 6 13 E Fig

10 V GEOMETRIL OLYMPID IN HONOUR OF I.F.SHRYGIN Final round. Second day. 9 form. Solutions. 5. (.Frenkin) Let n points lie on the circle. Exactly half of triangles formed by these points are acute-angled. Find all possible n. nswer. n = 4 or n = 5. Solution. It is evident that n > 3. onsider any quadrilateral formed by marked points. If the center of the circle lies inside this quadrilateral but not on its diagonal (call such quadrilateral "good"), then exactly two of four triangles formed by the vertices of the quadrilateral are acute-angled. In other cases less than two triangles are acute-angled. Therefore the condition of the problem is true only when all quadrilaterals are good. If n = 4 or n = 5 this is possible (consider for example the vertices of a regular pentagon). Now let n > 5. onsider one of marked points and the diameter. If point also is marked then the quadrilateral formed by, and any two of remaining points isn't good. Otherwise there exist three marked points lying on the same side from. The quadrilateral formed by these points and isn't good. 6. (.kopjan) Given triangle such that = 2. Let M be the midpoint of, and N be the base of the bisector from. Prove that M + N = 90. Solution. Let be the reflection of in N. Then = and by condition M/ = /. Thus triangles M and are similar and M = = 90 N. Furthermore using the formula for a median we obtain that M 2 =, so /M = M/. Therefore triangles M and M are also similar and M = M. Finally M = 180 M M = M q.e.d. (fig.9.6). M Fig (M.Volchkevich) Given two intersecting circles with centers O 1, O 2. onstruct the circle touching one of them externally and the second one internally such that the distance from its center to O 1 O 2 is maximal. Solution. Let O, r be the center and the radius of some circle touching the two given; r 1, r 2 be the radii of the given circles. Then OO 1 = r 1 r, OO 2 = r 2 + r, or OO 1 = r 1 + r, 10 N

11 OO 2 = r 2 r, and in both cases OO 1 + OO 2 = r 1 + r 2. Therefore we must find the point satisfying this condition with maximal distance from line O 1 O 2. It is known that the isosceles triangle has the minimal perimeter among all triangles with given base and altitude. Therefore the isosceles triangle has also the maximal altitude among all triangles with given side and the sum of two other sides. From this we obtain that the center of the required circle lies on equal distances (r 1 + r 2 )/2 from points O 1 and O 2, and its radius is equal to r 1 r 2 /2. 8. (.Pohoata,.Zaslavsky) Given cyclic quadrilateral D. Four circles each touching its diagonals and the circumcircle internally are equal. Is D a square? nswer. Yes. First solution. Let D = P, and let the incircles of the circular triangles P, P, DP, DP touch the circumcircle of D in K, L, M, N. onsider the segment. When a variable point X moves along the arc from to, the radius of the circle inscribed in the segment and touching the arc in X changes as follows: it increases until X becomes the midpoint of the arc, and then decreases. Therefore, each value of radius is reached in exactly two, symmetrically situated positions of X. Therefore K = L. nalogously N = M. So NK = LM. nalogously KL = MN. Now NL = NK+ KL = 180 i.e. NL is a diameter. nalogously KM is also a diameter. Now symmetry with respect to O sends the pair of circles touching the circumcircle in M and N in the analogous pair touching it in K and L. So the same symmetry sends the common external tangent of the first pair in that of the second namely it sends in. Therefore is a diameter and similarly, D is a diameter. So D is a rectangle. Its diagonals divide the circumcircle into four sectors with equal radii of incircles. Therefore these sectors are also equal and D is square. Second solution. Use the Thebault theorem: let point M lie on side of triangle and two circles touch ray M, line and internally the circumcircle of. Then two centers of these circles and the incenter of are collinear. pplying the Thebault theorem to triangles, D, D, D and the common point of diagonals we obtain that the inradii of these four triangles are equal. alculating the areas of triangles as product of semiperimeter by inradius and finding the area of quadrilateral in two ways we obtain that = D, so D is an isosceles trapezoid. Suppose that D, are its bases and D >. Then S D /S = D/ > (D + D + )/( + + ), and the inradii of these triangles can't be equal. Thus D is a rectangle. s in the first solution D must be a square. 11

12 V GEOMETRIL OLYMPID IN HONOUR OF I.F.SHRYGIN Final round. First day. 10 form. Solutions. 1. (D.Shvetsov) Let a, b, c be the lengths of some triangle's sides; p, r be the semiperimeter and the inradius of triangle. Prove an inequality ab(p c) + p ca(p b) + p bc(p a) p Solution. y auchi inequality the left part isn't less than abc (p a)(p b)(p c) p p 3 3 6r. = 3 3 4r 2 R. s R 2r we obtain the demanded inequality. 2. (F.Nilov) Given quadrilateral D. Its sidelines and D intersect in point K. Its diagonals intersect in point L. It is known that line KL pass through the centroid of D. Prove that D is trapezoid. Solution. Suppose that lines D and intersect in point M. Let X, Y be the common points of these lines with line KL. Then (D; MX) = (; MY ) = 1. Therefore relations X/XD and Y/Y are both greater or are both less than 1, and segment XY doesn't intersect the segment between the midpoints of D and. s this last segment contains the centroid of D, the condition of problem is true only when D. 3. (.Zaslavsky,.kopjan) The cirumradius and the inradius of triangle are equal to R and r; O, I are the centers of respective circles. External bisector of angle intersect in point P. Point Q is the projection of P to line OI. Find distance OQ. Solution. Let,, be the excenters of. Then I is the orthocenter of triangle,,, are the bases of its altitudes and so the circumcircle of is the Euler circle of. Thus the circumradius of is 2R, and its circumcenter O is the reflection of I in O. Furthermore points,,, lie on the circle. Line is the common chord of this circle and the circumcircle of, and the external bisector of is the common chord of this circle and the circumcircle of. So P is the radical center of three circles, and line P Q is the radical axis of circles and (fig.10.3). Therefore OQ 2 R 2 = (OQ + OO ) 2 4R 2. s OO = OI = R 2 2Rr, we have OQ = R(R + r)/ R 2 2Rr. 12

13 O O I P Q Fig (.Pohoata) Three parallel lines d a, d b, d c pass through the vertex of triangle. The reflections of d a, d b, d c in,, respectively form triangle XY Z. Find the locus of incenters of such triangles. First solution. When d a, d b, d c rotate around the vertices the symmetric lines rotate with the same velocity around the reflections of the vertices in opposite sidelines. Thus, firstly, the angles of XY Z don't depend on d a, d b, d c, so all these triangles are similar, and secondly, points X, Y, Z move with equal angle velocity along three circles. Therefore the incenter also moves along some circles and it is sufficient to find three points of this circle. Take d a, d b coinciding with line. Let, be the reflections of, in opposite sidelines. Then Z is the common point of lines and, Y and X are the common points of these lines with the line parallel to and lying twice as far from. Note that and circumcenter O of lie on equal distances from and, so the bisector of angle XZY coincides with line O. lso it is easy to see that the bisectors of angles ZXY and ZY X are perpendicular to and respectively. onsider the projections of O and of the incenter of XY Z to line. The projection 13

14 of O is the midpoint of. lso it is the projection of the common point of and d c, because these two lines form equal angles with. Thus the projection of X and the incenter of XY Z is symmetric to the midpoint of wrt (fig.10.4). Therefore the distance from the incenter to O is twice as large as the circumradius of. When d a, d b, d c are parallel to other sidelines of, we obtain the same result. So the demanded locus is the circle with center O and radius twice as large as the circumradius of. O Y Z X Fig.10.4 Second solution. s in the previous solution we obtain that when the direction of the lines d changes with a constant angular speed, so do the directions of XY, Y Z, ZX. Therefore, the vertex X of triangle XY Z traces a circle with chord, and the angle bisector of Y XZ rotates around the midpoint W a of the arc with constant angular speed, too. So do the angle bisectors of Y and Z around the midpoints W b, W c of the corresponding arcs and. Therefore, their intersection I traces in the same time the circumcircles of triangles IW a W b, IW b W c and IW c W a. So, these three circumcircles do in fact coincide, and we are left to describe the circumcircle of triangle W a W b W c. We will show that all the points W a, W b, W c are of distance 2R from O. Indeed, take W a. Let H b, H c be the altitudes in triangle, O a be the circumcenter of triangle H b H c, O be the reflection of O in, and M a be the midpoint of. The figures O, H b H c O a and W a are similar, and the figures H b 1 and H c 1 are also similar, therefore they are similar to O O a W a, and M a O a is a mid-segment in triangle O OW a. Since M a O a is a diameter of the Euler circle, and thus equals R, the claim follows. 14

15 V GEOMETRIL OLYMPID IN HONOUR OF I.F.SHRYGIN Final round. Second day. 10 form. Solutions. 5. (D.Prokopenko) Rhombus KLN is inscribed into triangle in such way that point L lies on side, point N lies on side, point K lies on side. O 1, O 2 and O are the circumcenters of triangles L, L and respectively. Let P be the common point of circles NL and KL, distinct from L. Prove that points O 1, O 2, O and P are concyclic. Solution. It is evident that L is the base of the bisector of angle, and lines LN, LK are parallel to sides,. Thus O 1 L = 2 L = = NL, so point O 1 lies on the circumcircle of triangle NL and coincides with the midpoint of arc NL. Thus, O 1 P L = P L+ O 1 P = + \+\ = π+\. Similarly O P l = π+\. Therefore 2 O 1 P O 2 = π. ut angle O 1 OO 2 is also equal to π, because lines OO 1, OO 2 are medial perpendiculars to and (fig.10.5). O 1 N O 2 K O P L Fig (.Zaslavsky) Let M, I be the centroid and the incenter of triangle, 1 and 1 be the touching points of the incircle with sides and, G be the common point of lines 1 and 1. Prove that angle GI is right if and only if GM. First solution. Let 1 be the touching point of incircle with side, 2 be the second common point of incircle with 1. Then G lies on segment 1. s there exists central projection transforming the incircle to some circle and G to the center of this circle, then the cross-ratio (G; 1 2 ) is the same for any triangle and regular triangle. So this cross-ratio is equal to 3. Therefore we have the chain of equivalent assertions: - GI = 90 ; - G is the midpoint of 1 2 ; - 1 = 3 2 ; 15

16 - 1 = 3G 1 ; - GM. Second solution. Let 1 = x, 1 = y, 1 = z. y Menelaus' theorem, where m = xy x+y. Now, ut, by Stewart's theorem, y + x x.g 1 G.z y = 1 G 1 G = k = xy z(x + y) = m z, IG = 90 I 2 r 2 = G 2 G 2 1 z 2 = = 1( 2 1 (1 + k) k2 2 (1 + k) ) = 2 2 1( 1 k 1 + k ) = 2 1( z m z + m ). 2 1 = Then, these two equations yield x x + y (z + y)2 + y x + y (z + x)2 xy = z(z + 4m). z 2 = z(z + 4m)( z m ) z(z + m) = (z + 4m)(z m) z + m as needed. 2zm = 4m 2 z = 2m k = 1 2, 7. (.Glazyrin) Given points O, 1, 2... n on the plane. For any two of these points the square of distance between them is natural number. Prove that there exist two vectors x and y, such that for any point i O i = k x+l y, where k and l are some integer numbers. Solution. y condition we obtain that for all i, j the product ( O i, O j ) is a half of an integer number. Thus for any integer m 1,..., m n the square of vector m 1 O m n O n is a natural number. onsider all points which are the ends of such vectors. Let X be the nearest to O of these points, Y be the nearest to O of considered points not lying on line OX. Divide the plane into parallelograms formed by vectors x = OX and y = OY. y definition of points X, Y all marked points are vertices of parallelograms, therefore x, y are demanded vectors. 8. (.Frenkin) an the regular octahedron be inscribed into regular dodecahedron in such way that all vertices of octahedron be the vertices of dodecahedron? nswer. No. Solution. If an octahedron is inscribed into a dodecahedron then their circumspheres coincide. Therefore two opposite vertices of the octahedron are opposite vertices of the dodecahedron, and all other vertices of the octahedrons are equidistant from these two vertices. ut the dodecahedron has no vertices equidistant from two opposite vertices. 16