VIII Geometrical Olympiad in honour of I.F.Sharygin

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1 VIII Geometrical Olympiad in honour of I.F.Sharygin Final round. First day. 8th form. Solutions 1. (.linkov) Let M be the midpoint of the base of an acute-angled isosceles triangle. Let N be the reection of M in. The line parallel to and passing through N meets at point K. etermine the value of K. nswer. 90. Solution. Let L be the common point of NK and (see Fig. 8.1). y means of the symmetry in we obtain M = M = N and M = N. Next, since LN, we have NL = LM, hence the triangle NL is isosceles, and LN = N = M. Thus, the segments M and LN are parallel and equal, hence the quadrilateral LNM is a parallelogram, and L MN L. Finally, by the symmetry in M we get K = L = 90. K L N I M Figure 8.1 Figure (.Karlyuchenko) In a triangle the bisectors and are drawn. fter that, the whole picture except the points,, and is erased. Restore the triangle using a compass and a ruler. First solution. Let I be the incenter of triangle. Then I = 180 ( I + I) = So, denoting by O the center of the circumcircle ω of triangle I, we get O = 180. Hence one can successively reconstruct points O and I (the latter is the meeting point of the smaller arc of ω with the bisector of, see Fig. 8.2). Finally, the points and can be reconstructed and the meeting points of I, and I, respectively. Second solution. Since is the bisector of, the point is equidistant from the lines and. Hence the circle with center tangent to is also tangent to. nalogously, line is tangent to the circle with center touching (see Fig. 8.2). So, to reconstruct the points and, it is sucient to draw a common outer tangent to these two circles (sharing dierent sides of with ) and to nd its common points with and. 3. (L.Steingarts) paper square was bent by a line in such way that one vertex came to a side not containing this vertex. Three circles are inscribed into three obtained triangles (see Figure). Prove that one of their radii is equal to the sum of the two remaining ones. Solution. ssume that square is bent by line XY ; denote the resulting points as in Fig Recall that in any right triangle, the incenter, the points of tangency of the 1

2 incircle with the legs, and the vertex of the right angle form a square; hence the inradius is equal to the segment of the tangent line from that vertex. Hence the indiameters of triangles UX, UP, and P V Y are d 1 = U + X XU, d 2 = U + P UP, and d 3 = P V + V Y P Y respectively. enote a = and notice that UX = X and V Y = Y ; therefore we obtain d 1 +d 3 d 2 = U+(a X) X+P V +Y P Y (a U) (a P Y Y )+(a P V ) = = 2(U + Y X). X K U P Y V Figure Figure Let K be the projection of Y onto. The points and U are symmetrical with respect to XY, hence XY U, and U = KY X. Moreover, KY = = a. onsequently, the right triangles U and Y KX are congruent, hence U = KX = X K = X Y. This means exactly that d 1 + d 3 2 d = 0. Remark. In the rst part of the solution, one may also argue as follows. The right triangles XY, V Y P, and U P are similar; hence the ratios of their inradii are the same as the ratios of their respective legs. Hence it suces to prove the equality X + V Y = U, or, equivalently, X + K = a U. The last relation follows from the relation U = KX which is proved in the second part of the Solution. 4. (.kopyan,.shvetsov) Let be an isosceles triangle with = 120. Points P and Q are chosen on the prolongations of segments and beyond point so that the rays Q and P intersect and are perpendicular to each other. Prove that P Q = 2 P Q. P Q P X Q = Figure 8.4 = Solution. Let us choose points X and Q on ray so that X = P and Q = P + Q. Then triangle P X is isosceles, and one of its angles is equal 60 ; hence 2

3 it is equilateral, so P X = P and P XQ = 120. Then triangles P Q and P XQ are congruent by SS, hence P Q = P Q and P Q = P Q. nalogously, choosing point P on ray so that P = P + Q, we get QP = QP and QP = QP (see Fig. 8.4). Now let us prolongate segments P and Q beyond the points P and Q by the length Q = P = P Q. Then we have = P +P = P +Q+P Q = Q +Q =. Now, triangles QP and P Q are isosceles, so P Q+ Q P = 1 2 ( QP + P Q ) = 1 2 ( P Q+ QP ) = 30. Hence the angle formed by the lines Q and P is equal to 180 ( P Q + Q P + + ) = 90. ut, if = <, then this angle should be less than 90, and if >, then it should be greater than 90, So we obtain =, =, and P Q = P Q = 2 P Q, QE. 3

4 VIII Geometrical Olympiad in honour of I.F.Sharygin Final round. Second day. 8th form. Solutions 5. (.kopyan) o there exist a convex quadrilateral and a point P inside it such that the sum of distances from P to the vertices of the quadrilateral is greater than its perimeter? nswer. Yes. Solution. onsider a quadrilateral such that = = = x, = = y < x/4, and a point P on the diagonal such that P = y (see Fig. 8.5). Then we have P + P = = x and P = P > P = x y, hence P + P + P + P > 3x 2y > 2x + 2y = W P H 1 Figure 8.5 Figure (.Tumanyan) Let ω be the circumcircle of triangle. point 1 is chosen on the prolongation of side beyond point so that 1 =. The angle bisector of meets ω again at point W. Prove that the orthocenter of triangle W 1 lies on ω. Solution. Let H be the second point of intersection of line 1 with ω. Since W is an angle bisector in the isosceles triangle 1, we have 1 H W. If the points and W share a common side of H, then W H = H = 90 W = 90 W, which implies W H 1 (see Fig. 8.6). If they share dierent sides, then W H = 180 H = 90 + W, which again follows W H 1. Finally, if these points coincide, then triangle W 1 is right-angled, and H = W is its orthocenter. Thus, in any case point H lies on two altitudes of triangle W 1 ; hence H is its orthocenter. 7. (.Shvetsov) The altitudes 1 and 1 of an acute-angled triangle meet at point H. Point Q is the reection of the midpoint of in line 1 ; point P is the midpoint of segment 1 1. Prove that QP H = 90. First solution. Let K be the midpoint of. Since KQ, the line KQ bisects the altitude 1. So, the diagonals of quadrilateral K 1 Q bisect each other and are perpendicular to each other, thus this quadrilateral is a rhombus. Moreover, from the symmetry we have HQ = HK. nalogously, let R be the reection of K in 1 ; then K 1 R is a rhombus, and HQ = HR (see Fig ). So the segments 1 Q, K, K, and 1 R are parallel and equal, hence Q 1 R 1 is a parallelogram, and P is a midpoint of RQ. onsequently, HP is a median, and thus an altitude, in the isosceles triangle HQR, QE. 4

5 1 R P P Q 1 Q H L K K Figure Figure H 1 Second solution. Let L be the midpoint of 1. Then P L is a midline in the triangle 1 1, so P LH = 1, and hence P LQ = 90 P LH = 1 H. On the other hand, points 1 and 1 lie on the circle with diameter ; therefore the triangles 1 1 H and H are similar; hence the two angles P H 1 and KH formed by their respective sides and medians are equal. Thus, QH = KH = P H 1, therefore P HQ = 1 H (see Fig ). So, P HQ = 1 H = P LQ, which implies that the points P, Q, L, and H are concyclic, and QP H = 180 QLH = (.Zaslavsky) square is divided into several (greater than one) convex polygons with mutually dierent numbers of sides. Prove that one of these polygons is a triangle. Solution. Suppose that a square is cut into n polygons. Then each of these polygons has at most one side lying on each side of the square; next, it shares at most one side with any of the other polygons. So the total number of its edges is at most 4 + (n 1) = n + 3. Thus, the number of sides of any polygon lies in the interval [3, n + 3]. If none of them is a triangle, then the numbers of sides should be equal to 4, 5,..., n + 3. So, there exists an n + 3-gon, and it should share a segment with every side of the square. Therefore, any other polygon can share the segments with at most two sides of the square, and the number of its sides is at most 2 + (n 1) = n + 1. Hence there is no (n + 2)-gon; a contradiction. 5

6 VIII Geometrical Olympiad in honour of I.F.Sharygin Final round. First day. 9th form. Solutions 1. (L.Steingarts) The altitudes 1 and 1 of an acute-angled triangle meet at point O. Let 1 2 and 1 2 be the altitudes of triangles O 1 and O 1 respectively. Prove that 2 2 is parallel to. Solution. y 1 = 90 = 1, the right triangles O 1 and O 1 are similar (see Fig. 9.1). So, their altitudes 1 2 and 1 2 divide the sides O and O in the same ratio. This exactly means that O Figure 9.1 Figure P 2. (.Shvetsov,.Zaslavsky) Three parallel lines passing through the vertices,, and of triangle meet its circumcircle again at points 1, 1, and 1 respectively. Points 2, 2, and 2 are the reections of points 1, 1, and 1 in,, and respectively. Prove that the lines 2, 2, 2 are concurrent. Solution. Let a, b, and c be the lines drawn through the points 1, 1 and 1 and parallel to,, and respectively. We claim that these lines are concurrent, and their concurrency point lies on the circumcircle of. Let c intersect the circumcircle at 1 and P (if c is tangent to the circumcircle, then P = 1 ). Then from 1 P and 1 1 we obtain P = 1 = 1 (here, by XY we denote the measure of the arc passing from X toy clockwise). This means exactly that 1 P, hence a passes through P. nalogously, b also passes through P (see Fig. 9.2). Next, the points 1 and P are symmetrical in the perpendicular bisector of, while the points 1 and 2 are symmetrical in ; this implies that the points P and 2 are symmetrical about the midpoint 0 of the segment. nalogously, the points 2 and 2 are symmetrical to P about the midpoints 0 and 0 of the other two sides of. Thus, 2 2 = = and analogously 2 2 =, 2 2 =. Therefore the triangles and are centrally symmetric to each other, and the lines connecting their respective vertices are concurrent at the symmetry center. 3. (V.Protasov) In triangle, the bisector L was drawn. The incircles of triangles L and L touch at points M and N respectively. Points M and N are marked on the picture, and then the whole picture except the points, L, M, and N is erased. Restore the triangle using a compass and a ruler. 6

7 First solution. Let K be the tangency point of the incircle of with the side (clearly, point K lies on the segment MN). Notice that MK = K M = + 2 L + L 2 = L + L 2 = LN. Next, let I a, I b, and I be the centers of the incircles ω a, ω b, and ω of triangles L, L, and respectively. y the angle bisector property, we get L IL = I a I a I = M MK, L MN so IL = M. Now we are ready to restore the triangle. It is easy to reconstruct successively points X, I (as the meeting point of the perpendicular to MN at K and a circle with center L), then I a and I b (as the meeting points of the bisectors of LI and LI with the perpendiculars to segment MN at its endpoints), then the circles ω a and ω b and, nally, the points (the intersection of the tangents to ω a drawn from and L) and (as the intersection of MN with a tangent to ω b from ). I a I I b M L K Figure 9.3 Second solution. First, let us prove the relation 1/M + 1/ML = 1/LN + 1/N. enote by x =, y = L, z = L the side lengths of triangle L, by p, S, and r its semiperimeter, area, and inradius respectively, and by h the distance from to 1 line. Then we have M + 1 ML = 1 p y + 1 p x = z. Next, by the (p x)(p y) S 2 rp zh/2 Äàëåå, (p x)(p y) = = p(p z) p(p z), which implies 1 M + 1 2(p z) = = ML rh 2. Now notice that in the triangle L, the angle at is the same, hence h tan( L/2) the value of 1/LN + 1/N is the same. Thus, knowing the lengths of segments M, M L, and LN, we may reconstruct the length of N and hence the point. Further, from the relations L = M LM and L = N LN we get the dierence = M LM N + LN = p of and, while the relation / = L/L = q provides their ratio. Now it is easy to nd the side lengths = triangle. p q 1 N and = pq, and to reconstruct the q 1 4. (.Frenkin) etermine all integer n > 3 for which a regular n-gon can be divided into equal triangles by several (possibly intersecting) diagonals. nswer. ll even n. 7

8 First solution. If n = 2k, then one may draw k main diagonals cutting the polygon into n congruent triangles. ssume now that such cutting exists for some odd n. onsider the obtained triangles sharing a side with the initial n-gon P. ll their angles opposite to these sides are equal; denote their value by α, and denote two other angles of an obtained triangle by β and γ. Two cases are possible. ase 1. ssume that the angles β and γ are dierent, say, β < γ. We call a side of n-gon β-side or γ-side according to the angle of the triangle at its left endpoint (if observing from the center of P ). hoose any β-side b, and consider the other side of angle β in the triangle adjacent to b. This side belongs to some diagonal of P, and the other endpoint of this diagonal also forms angle β with some side c of P (see Fig ). This angle β cannot be cut into some smaller angles, otherwise the angle adjacent to c is smaller than β; but it should be equal either to β or to γ > β, which is impossible. Thus, this angle β belongs to a triangle with c as its side, and c is a γ-side; let us put b and c into correspondence. onversely, considering angle β adjacent to any γ-side c we analogously nd a β-side b corresponding to it. Thus all the sides are split into pairs, and their total number is even. contradiction. c β γ γ β b Figure Figure ase 2. ssume now that β = γ, and consider a triangle containing the bottom side of P ; its angle at is α. The angle vertical to it is also α and belongs to some dierent triangle, whose side opposite to is equal and parallel to. On the other side of it we nd another triangle, and so on. Thus we obtain some chain of triangles (see Fig ); consider the last triangle U V W in this chain. If its orientation is dierent from that of, then its top side (which is parallel to ) is also a side of P, which is impossible for odd n (a regular n-gon has no parallel sides). Otherwise, the angle at the top vertex W is α, and W is a vertex of P. Then the sides UV and V W are (simultaneously) either sides of P or parts of its diagonals. In the rst subcase we get α = β = γ = 60, so the angle of n-gon is 60, which is impossible since n > 3. In the second subcase, the angle of P contains (as minimum) the angle equal to α and two angles equal β (adjacent to the sides sharing the vertex W ); thus α + 2β < 180. ut α + 2β = 180 as the sum of three angles of a triangle; a contradiction. Second solution. Here we present a dierent proof that the cutting is impossible for all odd n > 3. Notice that no two diagonals are perpendicular to each other; hence for any internal meeting point of two drawn diagonals, there should exist the third diagonal passing through that point: otherwise these diagonals form two dierent angles which sum up to 180 ; but they should be equal to two angles of some triangle. 8

9 Next, we claim that at least two diagonals should pass through each vertex of P. ssume rst that through some vertex, no diagonals are drawn. Then the triangle containing this vertex is a triangle formed by three consecutive vertices of P, and one of its angles is the angle of P. Then it is easy to see that every side of P belongs to some triangle which also contains one more side of P. Hence the sides are paired up, which is absurd. ssume now that exactly one diagonal passes through some vertex i. This diagonal splits the angle at into two dierent angles β < γ. oth these angles are adjacent to the sides of n-gon, therefore in any obtained triangle such angles are adjacent to a side equal to the side of P. Hence, the sum of all angles of triangles adjacent to the sides of P is n(β + γ), which equals to the sum of all angles of P. Therefore, through each vertex passes exactly one diagonal, and the vertices are paired up, which is absurd again. The claim is proved. Finally, assume that P is dissected into k triangles; the sum of all their angles is 180 kπ. The angles of P contribute 180 (n 2) to this sum, hence the sum of the angles at the internal points is 180 (k n + 2). Each such point contributes 360, so the number of internal points is (k n + 2)/2. On the other hand, each such point belongs to at least six triangles, while each vertex of P belongs to at least three of them. Hence the total number of triangles is at least (3(k n + 2) + 3n)/3 = k + 2 > k. contradiction. 9

10 VIII Geometrical Olympiad in honour of I.F.Sharygin Final round. Second day. 9th form. Solutions 5. (M.Kungozhin) Let be an isosceles right-angled triangle. Point is chosen on the prolongation of the hypothenuse beyond point so that = 2. Points M and N on side satisfy the relation M = N. Point K is chosen on the prolongation of beyond point so that N = K. etermine the angle between lines N K and M. nswer. 45. Solution. Let L be the projection of M onto. Notice that ML N = L K = = 1 ; 2 hence we also have L L + = K K + = 1. Thus, the right triangles ML and NK 2 are similar, and M L = N K (see Fig. 9.5). Therefore the angle between the lines NK and M is the same as the angle between K and L, which is equal to 45. M N L M K N K Figure 9.5 Figure (M.Rozhkova) Let be an isosceles triangle with = a and = = b. Segment is the base of an isosceles triangle with = = a such that points and share the opposite sides of. Let M and N be the bisectors in triangles and respectively. etermine the circumradius of triangle M N. nswer. ab a + b. Solution. hoose a point K on segment so that MK. Then M = M = MK, so MK = K. Moreover, by symmetry we get K = M. Next, by the bisector property we have K K = M M = a b = N. Hence, KN. Now we N may analogously obtain KN = K. Thus, K is the circumcenter of triangle MN, so its circumradius equals K = M = ab/(a + b) (see Fig. 9.6). 7. (.elov) convex pentagon P is divided by all its diagonals into ten triangles and one smaller pentagon P. Let N be the sum of areas of ve triangles adjacent to the sides of P decreased by the area of P. The same operations are performed with the pentagon P ; let N be the similar dierence calculated for this pentagon. Prove that N > N. Solution. Let be the initial pentagon, be the pentagon formed by its diagonals, and be the pentagon formed by the diagonals 10

11 of (see Fig. 9.7). We will enumerate all the vertices cyclically, thus, for instance, i+5 = i. For convenience, we will denote the area of polygon P by [P ]. Notice that N = i [ i i+1 i+2 ] [ ], since in the right-hand part the pentagon is counted with multiplicity 1, the triangles of the form i i+1 i+3 with multiplicity 1, and the triangles of the form i i+1 i+3 with zero multiplicity. Thus the desired inequality is equivalent to [ i i+1 i+3 ] > [ i i+1 i+2 ]. i i We will prove that [ i i+1 i+3 ] > [ i+2 i+3 i+4 ]; adding up ve such inequalities we will get the desired inequality. learly, it is enough to deal with the case i = 1. Let us glue a triangle to each of the triangles and ; we get two triangles and with a common base 1 3. Finally, the distance from 5 to the base is smaller than the distance from 2 ; hence, [ ] > [ ], QE P K O H L Q 1 5 T Figure 9.7 Figure (M.Plotnikov) Let H be an altitude of an acute-angled triangle. Points K and L are the projections of H onto sides and. The circumcircle of meets line KL at points P and Q, and meets line H at points and T. Prove that H is the incenter of triangle P QT. Solution. Let O be the center of the circumcircle Ω of triangle. From right triangles H and H we get K = H 2 = L, or K/L = /. Therefore, triangles LK and are similar, and KL =. Now, since O = π/2, we get O KL, which means that O is the perpendicular bisector to the chord P Q, so P = Q. This means that T is the bisector of P T Q (see Fig. 9.8). Thus, the incenter I of triangle P QT lies on T. Moreover, it is well-known that I = P. Thus, to prove that I = H it suces to show that H = P. Let be the meeting point of the lines O and KL, and let r be the radius of Ω. y the Pythagoras theorem, we have Q 2 r 2 = Q 2 OQ 2 = ( 2 + Q 2 ) (O 2 + Q 2 ) = 2 ( r) 2, which implies Q 2 = 2r. On the other hand, notice that H is a diameter of the circumcircle of KL since KH = LH = 90. Hence the similarity ratio of 11

12 triangles KL and equals H/(2r). The segments and H are the respective altitudes of these triangles, hence /H = H/(2r), or H 2 = 2r = Q 2, QE. Remark. The proof of the relation Q = H may be shortened by means of the inversion with center and radius Q. Under this inversion, the line P Q and the circle Ω interchange, hence the points and K also interchange, and Q 2 = K = H 2. 12

13 VIII Geometrical Olympiad in honour of I.F.Sharygin Final round. First day. 10th form. Solutions 1. (.Shapovalov) etermine all integer n such that a surface of an n n n grid cube can be pasted in one layer by paper 1 2 rectangles so that each rectangle has exactly ve neighbors (by a line segment). nswer. ll even n. Solution. onsider any even n. ivide each face into 2 2 squares, and paste each such square with two rectangles in such a way that the long sides of the rectangles in one square are adjacent to the short ones in a neighboring square. Let us show that such pasting is possible. It is easy to see that one may cover four side faces of the cube, leaving top and bottom faces uncovered. Next, one may paste one bordering row on the top face (the arrangement of the rectangles around the corner looks as on Fig. 10.1, or symmetrically to it). This row determines the arrangement of rectangles on the top face uniquely, and it is easy to see that all four bordering rows of squares will satisfy the conditions. The covering of the bottom face is analogous. Figure10.1 Now assume that a pasting is possible for some odd n. The total number of rectangles is 6n 2 /2 = 3n 2 ; if each of them has ve neighbors, then the total number of pairs of neighboring rectangles is 3n 2 5/2; but this number is not integer, which is absurd. 2. (.Zaslavsky,.Frenkin) We say that a point inside a triangle is good if the lengths of the cevians passing through this point are inversely proportional to the respective side lengths. Find all the triangles for which the number of good points is maximal. nswer. ll acute-angled triangles. Solution. Let 1, 1, and 1 be the altitudes of a triangle, and H be its orthocenter. onsider any good point P ; let P, P, P be the cevians passing through P. Then we have P / 1 = P / 1 = P / 1 ; hence the right triangles 1 P, 1 P, and 1 P are similar, so 1 P = 1 P = 1 P. There are two ways how these angles may be oriented. (Recall that an oriented angle (l, m) is the angle at which one needs to rotate l clockwise to obtain a line parallel to m.) ase 1. Suppose that ( 1, P ) = ( 1, P ) = ( 1, P ) (in particular, the triangle is acute-angled; if, for instance, π/2, then the angles 1 P and 1 P are acute, and their orientation s are opposite). The rst equality yields that the points P, H., are concyclic; analogously, P lies on the circumcircles of triangles H and H. ut these three circles have exactly one common point H; hence P = H. 13

14 X P H H P X Figure Figure ase 2. Now suppose that two of the oriented angles are equal, while the third (say, ( 1, P )) is opposite to them. Then, as in ase 1, the point P lies on the circumcircle Ω of triangle H (recall that (H, H) = (, ), so Ω is symmetrical to the circumcircle Ω of with respect to the line ). Let X be the second meeting point of Ω with the line H (then the points X and are also symmetrical in ; on Figs and two possible congurations are shown). Then (P X, X) = (P, H) = (P, X); if these angles are nonzero, then this relation shows that the triangle P X is isosceles, P = P X. ut then the point P lies on the perpendicular bisector of segment X, which is impossible. Thus, (P, H) = 0, and P = H. onsequently, a point inside the triangle is good only if it is the orthocenter (and, obviously, the orthocenter of an acute-angled triangle is good). So, in an acute-angled triangle there exists exactly one good point, and there are no good point i other triangles. Remark. In ase 2, one may apply a shorter (but less elementary) argument. The locus of points P satisfying the relation ( 1, P ) = ( 1, P ) is an equilateral hyperbola circumscribed about triangle. Two such hyperbolas may have at most four common points, and these points are,,, and H. 3. (.Karlyuchenko) Let M and I be the centroid and the incenter of a scalene triangle, and let r be its inradius. Prove that MI = r/3 if and only if MI is perpendicular to one of the sides of the triangle. First solution. Let 1 and 2 be respectively the tangency points of side with the incircle ω and excircle ω of triangle. enote by the midpoint of. It is well known that 1 = 2. Next, consider a homothety with center mapping ω to omega; under this homothety, point 2 maps to a point 3 on ω opposite to 1 (since the tangents in 1 and 3 to ω are parallel; see Fig ). Then I is a midline of triangle 1 2 3, hence I 2. Thus, under a homothety with center M and coecient 2, the point I maps to a point N lying on 2 (analogously, N lies on the segments connecting other vertices with the corresponding point of tangency of other excircles; N is called the Nagel point of triangle ). Therefore, N is obtained from M by a homothety with center I and coecient 3. Now we turn to the problem. Suppose that MI = r/3. Then point N lies on ω. Without loss of generality we may assume that the tangent at N to ω intersects sides and ; then ω and share dierent sides of this tangent, and hence N = 3. Since I 3, we obtain MI (see Fig ). 14

15 3 N M I N = 3 M I 1 2 Figure Figure onversely, if IM, then N lies on the line I 2 ; moreover, it also lies on the line 2. Since the triangle is scalene, these lines are distinct, hence N = 2, and thus r = IN = 3IM. Second solution. Suppose that IM. y the Pythagoras theorem, M 2 M 2 = ( M 2 ) ( M 2 ) = (p a) 2 (p b) 2 = c(b a) (here, again, 1 is the tangency point of with the incircle). Using the standard formula for the median length, we get M 2 = 1 9 (2b2 + 2c 2 a 2 ) and M 2 = 1 9 (2a2 + 2c 2 b 2 ), whence c(b a) = 1(b a)(a+b), or a+b = 3c, that is, p = 2c. It is easy to show that the converse is also true: 3 namely, if p = 2c, then IM. Finally, from c(im + r)/2 = S M = S /3 = pr/3 we obtain IM + r = 4r/3, or IM = r/3. onversely, assume that MI = r/3. Notice that I 2 + I 2 + I 2 = ( IM + M) 2 + ( IM + M) 2 +( IM + M) 2 = M 2 +M 2 +M 2 +2 IM ( M+ M + M)+3MI 2 = M 2 + M 2 + M 2 + 3MI 2. So, if MI = r/3, then I 2 + I 2 + I 2 = M 2 + M 2 + M r2 = 1 3 (a2 + b 2 + c 2 + r 2 ). Next, by the Pythagoras theorem I 2 = r 2 + (p a) 2. Finally, using the relation r 2 = S 2 /p 2 = (p a)(p b)(p c)/p, we obtain or a 2 + b 2 + c 2 3 a 2 + b 2 + c 2 + r 2 3 = (p a) 2 + (p b) 2 + (p c) 2 + 3r 2, (p a) 2 (p b) 2 (p c) 2 = 8r2 3 8(p a)(p b)(p c) =, 3p which rewrites as (p 2a)(p 2b)(p 2c) = 0. It was mentioned above that if some expression in the brackets vanishes then IM is perpendicular to the corresponding side. 4. (.Frenkin) onsider a square. Find the locus of midpoints of the hypothenuses of rightangled triangles with the vertices lying on three dierent sides of the square and not coinciding with its vertices. nswer. ll the points of a curvilinear octagon bounded by the arcs of eight parabolas with foci at the vertices of the square and directrices containing a side (non-adjacent to the focus); the midpoints of the sides of the square should be excluded (see Fig ). 15

16 Y O Z Figure Figure Solution. Notice rst that the midpoint of the hypothenuse lies inside a square. If the endpoints of a hypothenuse lie on opposite sides of the square, then its midpoint lies on the midline of the square. Now assume that the endpoints X and Y of the hypothenuse of a triangle XY Z lie on sides and of a square respectively, while the vertex Z lies on the side (see Fig ). enote by O the midpoint of XY. The points and Z belong to the circle with diameter XY ; hence O = OX = OY = OZ, and the distance from O to is less that the distances to the other vertices of the square, but is not less than the distance from O to the line. The locus of points equidistant from and is the parabola with focus and directrix (the vertex of this parabola is the midpoint of ). So, the point O lies between this parabola and in the quarter of the square closest to. Point O may lie on the parabola, but it cannot lie on the midline of the square (otherwise Y = ). nalogously, one may consider other arrangements of points; taking the union of the obtained sets, we obtain the curvilinear octagon P bounded by the arcs of eight parabolas. The vertices of P are the midpoints of the sides of the square (they do not belong to the locus) and the points of intersection of parabolas with the diagonals of the square. Since the midlines of the square also lie in P, we obtain that the total locus lies in P. It remains to show that each point O in P (distinct from the midpoints of the sides) belongs to the locus. If O lies on the midline parallel to, and it is not farther from than from, then one may take its projections onto and V as the midpoints X and Y of a hypothenuse, and nd Z as a meeting point of with the circle with diameter XY. If O lies in the quarter closest to, between the parabola with focus and directrix and the corresponding midline, then one chooses X and Y as the second intersection points of sides and with the circle with center O and radius O, while Z may be chosen as any point of intersection of the same circle with side (such a point exists since the distance from O to is less than O, but O > O). X 16

17 VIII Geometrical Olympiad in honour of I.F.Sharygin Final round. Second day. 10th form. Solutions 5. (F.Nilov) quadrilateral with perpendicular diagonals is inscribed into a circle ω. Two arcs α and β with diameters and lie outside ω. onsider two crescents formed by the circle ω and the arcs α and β (see Figure). Prove that the maximal radii of the circles inscribed into these crescents are equal. X Y K O L Figure Figure Solution. Let X and Y be respectively the midpoints of the arc α and the arc of the circle ω. enote by O the center of ω. Then the crescent with vertices and is situated between the concentric circles centered at O with radii OY and OX (see Fig ). Hence the diameter of any circle inscribed into this crescent is at most XY ; on the other hand, the circle with diameter XY lies inside the crescent. Thus the maximal diameter of a circle inside this crescent equals XY. Since, the sum of arcs and of the circle ω equals 180. Let K and L be the midpoints of segments and respectively; then OK = 90 OL = OL, hence the right triangles OK and OL are equal by hypothenuse and acute angle. So OX = OK + KX = OK + K = ( + )/2, and therefore XY = ( + )/2 r, where r is the radius of ω. nalogously we obtain that the maximal radius of a circle inscribed into the second crescent also equals ( +)/2 r. 6. (V.Yassinsky) onsider a tetrahedron. point X is chosen outside the tetrahedron so that segment X intersects face in its interior point. Let,, and be the projections of onto the planes X, X, and X respectively. Prove that K X L M Figure 10.6 Figure 10.7 Solution. Since (X), we get = 90 ; analogously, = 90 (see Fig. 10.6). Hence the points and lie on the sphere with diameter, and the distance between them is does not exceed the diameter: K. nalogously, we get and. dding up these inequalities we get the desired one. 17

18 7. (F.Ivlev) onsider a triangle. The tangent line to its circumcircle at point meets line at point. The tangent lines to the circumcircle of triangle at points and meet at point K. Prove that line K bisects segment. Solution. It is well known that in any triangle XY Z, the symmedian XX (that is, the line symmetrical to the median from the vertex X with respect to the angle bisector of angle X) passes through the common point of the tangents to the circumcircle of XY Z at Y and Z. Thus, the line K is a symmedian in triangle. Next, the triangles and are similar. So, denoting by L and M respectively their medians from, we get K = L = M, which implies that M lies on K. 8. (.Shvetsov) point M lies on the side of square. Let X, Y, and Z be the incenters of triangles M, M, and M respectively. Let H x, H y, and H z be the orthocenters of triangles X, Y, and Z. Prove that H x, H y, and H z are collinear. M X H y P Q H z Y H x Z Figure 10.8 Solution. learly, the points X and Y lie on the diagonals and respectively. Hence the lines and contain some altitudes of triangles X and Y respectively. Let us choose points P and Q on the segments M and M respectively so that P = Q =. Then X is an angle bisector (and hence an altitude) of an isosceles triangle P. Thus, the orthocenter H x is the common point of the lines P and. nalogously, H y is the common point of the lines Q and. Finally, from similar arguments we get that Z P and Z Q, so H z is the common point of the lines Q and P (see Fig. 10.8). Now let us apply the esargues' theorem to the triangles P and Q. Since the lines, P, and Q which connect the corresponding points of these triangles are concurrent at M, we get that the common points of the lines containing the corresponding sides of these triangles are collinear. ut these points are exactly H x, H y, and H z. 18