SMO(J) Mock Paper 1 Duration: Three hours hours May 31, 2012
1 1. The sequenc sequencee {x } satisfies x1 = , x 2 n
+1
n
=x + n
x2 . Prove that x2001 < 1001. n2 n
2. Let n1 , n2, · · · , n1998 be positive integers such that 2 2 = n1998 n21 + n22 + · · · + n1997 .
Show that at least two of the numbers are even. 3. Consider a convex convex polygon having n vertices, n ≥ 4. We arbitrarily decompose the polygon into triangles having all the vertices among the vertices of the polygon, such such that that no two two of the triangle triangless have have interio interiorr points points in com common mon.. We paint paint in black the triangles that have two sides that are also sides of the polygon, in red if only one side of the triangle is also a side of the polygon and in white those triangles that have no sides that are sides of the polygon. Prove that there are two more black triangles that white ones. 4. Let AL and BK be angle bisectors bisectors in the non-isosc non-isosceles eles triangle ABC (L lies on the side BC , K lies on the side AC ). ). The perpendicular bisector of BK intersects the line AL at point M . Point N lies on the line BK such that LN is parallel to M K . Prove that LN = N A. 5. On a given 2012 × 2012 chessboard, every unit square is filled with one of the letters S,M,O,J. The resulting board is called harmonic if every 2 × 2 subsquare contains all four different letters. How many harmonic boards are there?
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Solutions
1 1. (CWMO 2001) 2001) The sequence {x } satisfies x1 = , x 2 x2001 < 1001. n
Solution.
First we have x1 = x
x2 = x + 2 . Prove Prove that that n n
+1
n
n
n 1 9 < 1. If x < , then and x2 = 2 16 2 n
x2 n n2 n+1 1 < . =x + 2 < + 2 × n 2 n 4 2 n
+1
n
n
Hence by induction we must have x < n
n
2
. So x2001 < 1000 .5 < 1001.
2. (JBMO 1997) 1997) Let Let n1 , n2 , · · · , n1998 be positive integers such that 2 2 = n1998 n21 + n22 + · · · + n1997 .
Show that at least two of the numbers are even. Firstly, if exactly one of n1 , n2 , · · · , n1998 is even, then n1998 must not be even since it is a sum of 1997 odd numbers. If exactly one of x1 , x2 , · · · x1997 is even, then x1998 must be even since it is a sum of one even number and 1996 odd numbers. Hence there cannot be exactly one even number. Solution.
If all numbers are odd, we observe that the remainder when an odd square number is divided by 8 is always 1. As such, 2 2 n21 + ... + n1997 ≡ 1997 ≡ 5 ≡ n1998
(mod 8)
which which is a contradictio contradiction. n. Hence there must be at least two of the numbers numbers which are even. 3. (JBMO 2004) 2004) Consider a convex convex polygon polygon having having n vertices, n ≥ 4. We arbitrarily decompose decompose the polygon polygon into triangles triangles having all the vertices vertices among among the vertices of the polygon, such that no two of the triangles have interior points in common. We paint in black the triangles that have two sides that are also sides of the polygon, in red if only one side of the triangle is also a side of the polygon and in white those triangles that have no sides that are sides of the polygon. Prove that there are two more black triangles that white ones. Let b be the number of black triangles, r be the number of red triangles, w be the number of white triangles. The sum of interior angles of all the triangles is equal to the sum of interior angles of the polygon. Hence, we have
Solution.
b + r + w = n − 2.
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Moreover, since each black triangles share 2 common sides with the polygon while each red triangles share 1 common side with the polygon, we have 2b + r = n. Subtracting the two equations we obtain b = w + 2 and hence there are 2 more black triangles than white triangles. 4. (JBMO (JBMO 2010) Let AL and BK be angle bisectors in the non-isosceles triangle ). The perpendicular bisector ABC (L lies on the side BC , K lies on the side AC ). of BK intersects the line AL at point M . Point Point N lies on the line BK such that LN is parallel to M K . Prove that LN = N A. First, we construct the circumcircle of AKB and suppose it intersects with line AL at point M . We note that M is the midpoint of the minor arc BK . Hence, M lies above the perpendicular bisector of BK and hence M coincides with poin p ointt M .
Solution.
This shows that ABMK is a cyclic cyclic quadrilate quadrilateral. ral. From there we have have ∠LAB = cyclic quadrilateral quadrilateral too. Finally Finally,, since ∠BK M = ∠BN L and hence ABNL is a cyclic ∠ABN = ∠LBN , we have LN = N A. 5. (CGMO 2008) 2008) On a given given 2012 × 2012 chessboard chessboard,, every every unit square is filled with one of the letters S,M,O,J. The resulting board is called harmonic if every 2 × 2 subsquare contains all four different letters. How many harmonic boards are there? Solution.
We consider the problem in the following two cases:
(a) If the first row row only only contai contains ns two two letter letters. s. Withou Withoutt loss loss of genera generalit lity y, let us assume that S, M are selected selected on the first row. In this case, the first row must must be in the form of SMSM · · · SM or MSMS · · · M S . There are 2 possibilities for the next row, which is either OJOJ · · · OJ or JOJO · · · J O. Similarly Similarly,, there will be 2 possibilities possibilities for each row on the chessboard. chessboard. The total number number 4 of possibilities in this case is × 22012 = 6 × 22012 . 2
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(b) If the first row contains contains three or more letters, we note that the entire chesschessboard board is determin determined ed just by the first first row row of the chess chess board. board. In this case, case, we observe that the first column of the chess board only contains 2 letters. Hence, this case is similar to the previous case, just that the chessboard is being rotated clockwise by 90 . We still need to deduct the number of cases where both the first row and column only has 2 letters. The total number of 4 cases is × 22012 − 4 × 3 × 2 = 6 × 22012 − 24. 2 ◦
Hence, the total number of cases is 12 × 22012 − 24.
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