2018 H3 Solution

Suggested Solution Remarks 1i             1 1 1F0 1 11 1 1 1 1 1 11 ...2 2 3 1 11 1 n n r n r rr rr nn n                   As n , 1 01n  and so  F 0 1n  . 1ii (a)                 1 1 1 2 2F 11 11 1 1 1 1 11 ...2 2 3 1 2 2 2 2 2 2 ...1 1 1 2 1 1 11 1212 11 123 11 n n r x r r rx rx nn x x x nx nx n nx n nx                                    1ii (b) Since  1lim 0 1n n  and   2, if 02lim 1 0, if 0n x nx x   ,   1, if 0lim F 3, if 0 nn xx x   . There’s a need to distinguish the special case when 0x . 2i Let u a x .         0 0 0 0 f d f d f d f d a a a a a x x u u u u x x        2ii Since f is symmetrical about 1 2xa ,    ff x a x for 0 xa . -------------------- (*) Since f is continuous and x is also continuous , the function given by  fxx is also continuous on  0,a .                             00 0 00 00 00 f d f d by i f d by * f d f d 2 f d f d f d f d 2 aa a aa aa aa x x x a x a x x a x x x a x x x x x x x x a x x ax x x x x            

Alternative Since f is symmetrical about 1 2xa ,    11ff 22x a x a    for 11 22a x a   . Let 1 2v x a . Note that        g : f f g 22 aav v v v v v         , which shows that g is an odd function on ,22 aa .                 2 0 2 22 22 2 2 0 f d f d 22 f d f d2 2 2 0 f d since g is odd22 fd2 a a a aa aa a a a aax x x v v v a a av v v v v aa vv a xx                    2iii Since sin x and 2 1 1 cos x are continuous on  0, π , so is their product   2 sinh: 1 cos xx x  . We can also show that h is symmetrical about π 2x :        22 sin π sinh πh 1 cos π 1 cos x xxx xx       Applying the result in (ii):         ππ π1 22 000 2 11 sin π sin πd d tan cos221 cos 1 cos π π π π πtan 1 tan 12 2 4 4 4 x x x x x xxx                  3 A triangle with sides of lengths p, q and r exists iff p q r , q r p and r p q . 3i WLOG, let 0c b a   . Note that 1 1 1 c b ac b a c b a       . It suffices to show that 1 1 1 a b c a b c   . Indeed, 1 1 1 1 1 1 a b a b a b c a b c c c c           .

3ii WLOG, let 0c b a   . Note that c b a c b a     . It suffices to show that a b c . Indeed, we have   2 2a b a b ab a b c       which implies that a b c . 3iii By (ii), if a triangle with sides of lengths  a b c a ,  b c a b and  c a b c exists, then so does a triangle with side of lengths  a b c a ,  b c a b and  c a b c . By symmetry, it suffices to show that      a b c a b c a b c a b c        .              2 2 2 2 2 2 22 2 0 of sides of lengths , , exists, an d a b c a b c a b c a b c ab ac a bc ab b ac bc c c ab a b c a b c a b c b a a b c c a b c b a                                    4i (a) Bijection: Place 7 identical balls (counters to select T-shirts) into 4 distinct boxes (T-shirts of different colours). 7 4 1 10 4 1 3Number of ways C C 120    4i (b) Bijection: With 1 ball in each of the four boxes, place 3 more identical balls (counters to select T -shirts) into the 4 distinct boxes (T -shirts of different colours). 3 4 1 6 4 1 3Number of ways C C 20    4ii (a) 7Number of ways 4 16384 4ii (b) 6Number of ways 4 3 2916   4ii (c) Let iA be the set of arrangements without using T-shirts of colour i.       7 1 2 3 4 7 ,, 7 4 7 4 7 4 7 1 2 3 Number of ways 4 4 4 C 3 C 2 C 1 8400 i i j i j k i i j i j i k j k A A A A A A A A A A                         

5i (a) If a tessellation exists, then the pq rectangle is formed completely by ab rectangles. Hence, the area of the pq rectangle, pq, is a multiple of the area of the ab rectangle, ab. Hence, ab is a factor of pq. 5i (b) If a tessellation exists, then the leftmost column of p squares must be formed by columns and/or rows of ab rectangles, i.e. 1a and/or 1b . Hence , p a b for some 0,  . Similarly, the bottommost row of q squares is also formed by columns and/or rows of ab rectangles. Hence, q a b for some 0,  . 5i (c) If a tessellation using ab rectangles exists, then there is a tessellation using 1a rectangles. Each of these 1a rectangles has 1 shaded square. Similar to the argument in (i)(a), the number of 1a rectangles is pq a , and hence, the number of shaded squares is pq a . 5ii (a) Place the rs rectangle in the bottom left corner of the pq rectangle. This rs rectangle will have t shaded squares, namely  1,1 ,  2, 2 , …,  ,tt . The remainder of the pq rectangle can be tessellated with 1a rectangles that contain exactly 1 shaded square each. [The s columns above the rs rectangle fitted with 1a rectangles, and the remaining  p q s rectangle to be fitted with 1 a rectangles.] Using the argument in (i)(a), the number of such 1a rectangles is pq rs a  . Hence, the total number of shaded squares is given by pq rs ta   . 5ii (b) Using the results in (i)(c) and (ii)(a), pq rs pq taa   . This yields rs ta  and we claim that 0t . Suppose instead that 1t , and since tr or ts , then we must have 1s a  or 1r a  , which are both impossible since ,r s a . Hence, 0rs ta  which gives 0r  or 0s . Consequently,  0 modpa or  0 modqa , i.e. a is a factor of either p or q.

6a Assumption: We will assume that friendship is symmetric relation (i.e. if A is a friend of B, then B is a friend of A). In a group of 2n students, if a student has 0 friends, there cannot be a student with 1n friends; and similarly, if a student has 1n friends, there cannot be a student with 0 friends. Hence, the number of friends each of the n students can possibly have must be from an  1n - element set (  1, 2,..., 1n or  0,1,..., 2n ). By the pigeonhole principle, there must be at least 2 students with the same number of friends. 6b Let iA be the interval 1,ii nn   , for 1,2,...,in . These iA ’s form the n pigeonholes, while the fractional parts form the n pigeons. Case 1: There is a fractional part, say px px (with 1 pn ) in 1A . 110 pxpx px x n p pn       Take a px and bp . Case 2: There is no fractional part in 1A . By the pigeonhole principle, there must exist two fractional parts, say px px and qx qx (with 1n p q   ) in some kA . 1,kkpx px nn   and 1,kkqx qx nn           1 1 p q x px qx px px qx qx n px qxx pq p q n                                   As 1 p q n   , we may take a px qx       and b p q .

7i (1): 2 dd 1dd yyyx xx   , 0x and let d d yt x . 2 1yt xt Differentiating w.r.t. x on both sides:   2 2 2 2 d dd 2d d d ddd 2d d d d20 d ddd2 or 0 rej. 0 dd d y ttt y t xtx x x ytty xt t tx x x ty xt x ytty xt xx x            2 2 d2 d 12dd ln 2ln , where e or e CC yyx x xyxy x y C y AAx y Ax            Hence, d2 d yyA x  and by (1), 2 2 1 1 42 2 4 A A Ax x Ay Ax         . Therefore, 2 4yx . 7ii “  ” 2 d 2:4 d yS y x xy   Equation of tangent:  00 0 2y y x x y   , where  00,xy lies on S This satisfies (1):  00 00 2 0 2 0 0 2 0 2 0 d 22 d 422 4d 11d yy y x xx y y xx y y xyx x y                    “  ” Suppose a straight line l