International Mathematics TOURNAMENT OF THE TOWNS

Transcription

1 International Mathematics TOURNMENT OF THE TOWNS Junior -Level Paper 1 Fall n angle is said to be rational if its measure in degrees is a rational number triangle is said to be rational if all its angles are rational Prove that there exist at least three different points inside an acute rational triangle such that when each is connected to the three vertices of the original triangle, we obtain three rational triangles 2 The incircle of triangle BC touches the sides BC, C and B at D, E and F respectivel If D = BE = CF, does it follow that BC is equilateral? 3 What is the maximum number of knights that can be place on an 8 8 chessboard such that each attacks at most seven other knights? 4 On a blackboard are written four numbers The are the values, in some order, of x +, x, x and x where x and are positive numbers Prove that x and are uniquel determined 5 K is a point on the side BC of triangle BC The incircle of triangle BK touches BC at M The incircle of triangle CK touches BC at N Prove that BM CN > KM KN 6 Two persons share a block of cheese as follows The take turns cutting an existing block of cheese into two, until there are five blocks Then the take turns choosing one block at a time The person who makes the first cut also makes the first choice, and gets an extra block Each wants to get as much cheese as possible What is the optimal strateg for each, and how much is each guaranteed to get, regardless of the counter measures of the other? 7 We have man copies of each of two rectangles If a rectangle similar to the first can be made b putting together copies of the second, prove that a rectangle similar to the second can be made b putting together copies of the first, with no overlapping in both instances Note: The problems are worth 4, 5, 6, 6, 7, 8 and 8 points respectivel 1 Courtes of nd Liu

2 Solution to Junior -Level Fall We remark that if a triangle has two rational angles, then the third angle must also be rational, so that the triangle is rational Let P be a point inside an acute rational triangle BC, which is then divided into triangles P BC, P C and P B In each of the following cases, it is sufficient to prove that P BC is a rational triangle, since we can prove that P C and P B are also rational in an analogous manner Suppose P is the incentre of triangle BC Since P B bisects the BC and CP bisects BC, both of which are rational, both P BC and P CB are rational Hence triangle P BC is rational Suppose P is the circumcentre of triangle BC Since BC is acute, P is indeed inside it Now CP B = 2 CB is rational, so that P CB = 1 2 (180 CP B) is also rational Hence triangle P BC is rational Suppose P is the orthocentre of triangle BC Since BC is acute, P is indeed inside it Now P BC = 90 BC and P CB = 90 BC are both rational Hence triangle P BC is rational If BC is not equilateral, then its incentre, circumcentre and orthocentre are disinct points Thus we have the required three points If BC is equilateral, there exist infinitel man points P on the perpendicular bisector of BC such that P CB is rational n three such points will meet the requirement of the problem P B C 2 ssume that BE = CF but B C In triangles BE and CF, BE = CF, E = F and BE = CF Since B C, BE and CF are not congruent triangles Hence BE CF but we do have BE + CF = 180 Hence either BE or CF is obtuse, which means that either E > B or F > C Since E = F, either E > C or F > B This is a contradiction It follows that B = C, and we can prove in a similar wa that D = CF implies BC = B, so that BC is indeed equilateral if D = BE = CF F B D C E

3 3 Let us start with a knight on each square of the 8 8 chessboard If we remove the 4 knights in the central 2 2 subboard, we are left with 60 knights each of which attacks at most 7 others We now show that 60 is indeed the maximum gain, we start with a knight on each of the 64 squares Note that a knight can attack 8 other knights onl if it occupies one of the squares in the central 4 4 subboard We put these 16 knights on a black list In the following diagram, the number on each square shows the maximum number of knights on the black list that can attack that square Note that all the numbers are 4 or less Thus the removal of a knight can take at most 4 other knights off the black list Even if the removed knight itself is on the black list, we can take at most 5 knights off Hence removing at most 3 knights will not clear the black list Note that (x + ) + (x ) = 2x while (x)( x) = x2, and that onl x can be non-positive We consider three cases Case 1 ll four numbers are positive Let a, b, c and d denote x +, x, x and x in some order Choose a pair of them and check if the square of their sum is four times the product of the other two numbers The pair can be chosen in six was There are three subcases Subcase 1a This is satisfied b two disjoint pairs We ma assume that we have (a + b) 2 = 4cd and (c + d) 2 = 4ab dding these two equations ields (a b) 2 + (c d) 2 = 0 so that a = b and c = d Substituting back into (a + b) 2 = 4cd, we have a = ±c Since all four numbers are positive, we must have a = b = c = d This is a contradiction since x + x Subcase 1b This is satisfied b two intersecting pairs We ma assume that we have (a + b) 2 = 4cd and (a + c) 2 = 4bd with b c Then we have b(a + b) 2 = 4bcd = c(a + c) 2, or equivalentl (b c)( + 2a(b + c) + (b 2 + bc + c 2 )) = 0 This is a contradiction since b c 0 while + 2a(b + c) ( b 2 + bc + c 2 ) > 0 Subcase 1c This is satisfied b onl one pair We ma assume that (a + b) 2 = 4cd Then we know that the larger one of a and b is x + and the smaller one x We can determine x and uniquel Case 2 One of the numbers is 0 We know that x = so that x = 1 must also be among the four numbers The other two are x + = 2x and x = x 2 Since their product is 2x 3, we can determine x = uniquel

4 Case 3 One of the numbers is negative We know that x < and x < 1 Check how man numbers in S = {x +, x, x } lie strictl between 0 and 1 There are three subcases Subcase 3a There is exactl one such number We know that this number is x, and we can determine x and uniquel from x and x Subcase 3b There are exactl two such numbers We cannot have x+ < 1 Otherwise, we must have x < 1 and < 1 so that x < 1, but then all three numbers in S lie strictl between 0 and 1 Hence x + > 1 is the largest number in S, and we can determine x and uniquel from x and x + Subcase 3c There are exactl three such numbers From x + < 1, we have x < 1 and < 1 so that x < x + and x < x Hence the smallest number in S is x, and we can determine x and uniquel from x and x 5 First Solution: Note that BM = B+BK K, CN = C+CK K, KM = K+BK B and KN = K+CK C Hence BM CN > KM KN is equivalent to C KB+B KC > K(KB+KC) = K BC, or C KB + B KC > K Let L be the point on C such that KL is parallel to B Then BC BC triangles BC and LKC are similar Hence L = C KB and KL = B KC B the Triangle BC BC Inequalit, L + KL > K, which is the desired result L Second Solution: B M K N C Construct the tangent to the incircle of triangle CK parallel to B and closer to C than to, cutting BC at D and C at E Let O and P be the respective incentres of triangles BK and CK Note that OK is perpendicular to P K since the bisect BK and KC respectivel Hence triangles MKO and NP K are similar, so that OM = KN Since B is KM P N parallel to DE, BD + BDE = 180 The are bisected respectivel b OB and P D, which are thus perpendicular to each other Hence triangle MOB is similar to triangle NDP, so that BM Multiplication ields KM KN = BM DN < BM CN = P N OM DN B M O P N K D E C

5 6 Let the total amount of cheese be 1, the first plaer be lexei and the second plaer be Boris lexei can be assured of getting at least 3 if he cuts 1 into 3 and 2 We consider two cases Case 1 Boris cuts 2 into x and 2 x, where 0 x 1 lexei cuts 3 into x and 3 x Now the four pieces are of sizes x = x 2 x < 3 x 5 No matter how Boris makes his second cut, the second smallest piece is at most x, and the second largest piece is at most 2 x since 2( 2 x) 3 x Hence Boris can get at most x + ( 2 x) = Case 2 Boris cuts 3 into x and 3 3 x, where 0 x If 0 x 1, lexei cuts 2 into x and 2 x, and this is the same as in Case 1 Hence we ma assume that 1 < x 3 lexei cuts 3 x into 2 x and 1 Now the four pieces are of 5 10 sizes 2 x < 1 < x < 2 There are four subcases Subcase 2a Boris cuts 2 into and 2, where 0 1 We have either 2 x < 1 < x 2, in which case Boris gets ( 2 x) + x = 2, or x 1 2 x, in which case Boris still gets + ( 2 ) = Subcase 2b Boris cuts x If 1 remains the third largest piece, lexei gets at least = 3 If it becomes the second 5 largest piece, Boris gets at most = 2 Subcase 2c Boris cuts 1 into and 1 1, where Since 2 x, the second smallest piece is at most 2 x Hence Boris gets at most 5 5 ( 2 x) + x = Subcase 2d Boris cuts 2 x 5 lexei gets at least = 3 We now show that Boris can be assured of getting 2 We consider three cases 5 Case 1 lexei cuts 1 into 3 x and x, where 0 x Boris cuts 3 x into 2 + x and 1 2x If lexei cuts 1 2x, Boris gets at least 2 + x 2 If lexei cuts one 2 + x, Boris cuts the other 2 + x in the same wa and gets at least 2 + x 2 5 Case 2 lexei cuts 1 into 3 + x and 2 x, where 0 x 1 Boris cuts 2 x into 1 and 1 x If lexei cuts either of these two pieces, Boris cuts 3 x into 5 halves and gets at least 3 + x + 1( 1 x) = 2 If lexei cuts 3 + x into and 3 + x where x, Boris cuts the latter into 1 + x and 2 If 1 x x, 10 2 Boris gets +( 2 ) = 2 If 1 x 1 1 +x 2, Boris still gets ( 1 x)+( 1 +x) = 2 Case 3 lexei cuts 1 into 4 + x nd 1 x, where 0 x 1 Boris cuts 4 + x into 3 + x and 1, and this is the same as Case 2 7 Suppose we have an a 1 rectangle and a b 1 b 2 rectangle B n rectangle P QRS that can be constructed from copies of has dimensions (u 1 a 1 + u 2 ) (v 1 a 1 + v 2 ) for some non-negative integers u 1, u 2, v 1 and v 2 If P QRS is similar to B, then b 1 b 2 = u 1a 1 + u 2 v 1 a 1 + v 2 We first consider the case where a 1 is rational, so that it is equal to m 1 m 2 for some positive integers m 1 and m 2 Then b 1 b2 = u 1m 1 +u 2 m 2 v 1 m 1 +v 2 m 2 = n 1 n 2 for some positive integers n 1 and n 2, so that it is also rational Using n 1 n 2 copies of B, we can construct a square of side s = n 2 b 1 + n 1 b 2 Using m 1 m 2 copies of this square, we can construct an sm 1 sm 2 rectangle which is similar to

6 We now consider the case where a 1 is irrational We claim that in constructing the rectangle P QRS with copies of, all the copies must be in the same orientation Let P T UV be the largest subrectangle of P QRS that can be constructed with copies of all in the same orientation Suppose U is in the interior of P QRS, as illustrated in the diagram below P T Q V U S If the line T U can be extended without cutting in interior of a cop of, then the space immediatel below UV must be filled with copies of in the same orientation as those above, as otherwise it contradicts the irrationalit of a 1 However, now it contradicts the maximalit of P T UV Hence T U cannot be so extended, but this implies that V U can, and we have a contradiction as well It follows that U must lie on QR or RS We ma assume b smmetr that it lies on QR, so that T coincides with Q However, the space immediatel below UV must be filled with copies of in the same orientation as those above This contradicts the maximalit of P T UV unless U coincides with R and V with S Thus our claim is justified Suppose this construction uses k 1 k 2 copies of in k 1 rows and k 2 columns for some positive integers k 1 and k 2 Then k 1a 1 k 2 = b 1 b2 so that k 2b 1 k 1 b 2 = a 1 Hence we can construct a rectangle similar to using k 1 k 2 copies of B in k 2 rows and k 1 columns R