DHS EJC RVHS 2025 Prelim (Answers)

©RIVER VALLEY HIGH SCHOOL 9820/01/2025 Solutions 1 Solution [11] Inequalities (a) By AM-GM inequality, 3 3 2 3 3 2 3 36 3(6) 18 a b c a b c abc + +    = = = (b) By Cauchy Schwarz inequality   ( )   ( ) 2 2 2 2 2 2 2 2 2 2 2 49 ( 3 ) (3 2 ) (2 ) 2 33 3 2 2 49 4 2 6 2 33 3 2 2 4 9 2 3 since 2 3 03 3 2 2 2 a b c b c c a a b a b cb c c a a b a b c a b c a b cb c c a a b a b c a b c a b cb c c a a b  + + + + + + +  + + + + +  + + + +  + + + + + +++ +  + + + + + Hence 2 2 249 93 3 2 2 a b c b c c a a b+ + + + + Equality holds for AM-GM inequality when 23a b c== . Substituting into 36abc= yields 2(2 ) 363a a a  = 3 27 3 a a = = 3, 6, 2a b c = = = Check against original expression 2 2 2 2 2 24 9 4(3) (6) 9(2) 3 3 3 93 3 2 2 6 3(2) 3(2) 2(3) 2(3) 6 a b c b c c a a b+ + = + + = + + =+ + + + + +

(c) Consider the substitution 1 1 1,,a b cp q r= = = Then 1 1 1 1 36abc p q r pqr=   = = ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 3 3 3 3 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 22 1 1 1 9 3 36 3 2 4 2 1 1 1 369 3 36 3 2 4 2 49 3 3 2 2 1 1 1 49 1 1 13 3 2 2 11 14 9 1 1 1 13 3 2 p r q q r q p pqrp r q q r q p qr pr pq p r p q q q r q r p qr pr pq p p p p r p q q q r q r p p p p pq r pr pr pr qr q r qr p q r r r qr q r q p r +++ + + = + +  + + + = + ++ + + =  +  + + + + = + + ++ ( ) ( ) 2 2 2 2 112 49 3 3 2 2 9 pq a b c b c c a a b + = + ++ + +  by the result of the earlier part.

2 Solutions [11] Differential Equations, Conics (a) Differentiating the substitution with respect to x , dd dd yu xx ux=+ Substituting dd dd yu xx ux=+ and y ux= into the differential equation, we have ( ) 2 2 2 2 2 3 2 2 2 d2 d 2 d2 d 12 d 2 2 d 12 21d d 12 u x uux x x x u uuxu x u u u u ux x u uuux x u += − =− − −+= − + = − Hence, ( ) 2 2 2 2 1 2 1 dd 21 1 4 1 dd21 ln ln21 u ux xuu u uxu u x u Axu − = + −= + =+   2 2 22 , where 21 2 2 u Bx B u yy B Bxxx y By B A x == + =+ =+  (b) Case 1: 0B . ( ) 22 22 2 2 2 22 2 11 24 16 1 0 4 1 8 11 4 y By Bx x y B x B B y B B =+ + − =  −− +=       The general solution represents an ellipse centre at 10, 4B   with horizontal radius 1 8B and vertical radius 1 4B with 2x y for all ,xy   . Case 2: B = 0 The particular solution is the line 0y= .

3 Solutions [12] PIE, Pigeonhole Principle, Distribution Qns (a)(i) Total number of grids 162 65536== (a)(ii) Label the squares as shown below Let iA be the number of grids with a 3-by-3 shaded square that’s top-left square is labelled i. 72iA = , 4 1 2 1 3 2 4 3 4 2A A A A A A A A= = = =  , 2 1 4 2 3 2A A A A==  , 2i j kA A A =  and 1 2 3 4 1A A A A =  . By PIE, ( ) ( ) ( ) 1 2 3 4 4 11 1 2 3 4 1 4 7 4 42 . 2 2 2 24 2 4 1 447 4 ij ik i i i i i j ik A A A A A A A A A A A A A A        = =− +− = − + =  +   −      (b) We require more than 12 from the following construct: Now we show that every grid with 13 shaded unit square will satisfy the required condition. Consider the partitioning for the grid by the four 2-by-2 squares as shown below: 1 2 3 4

We have 13 shaded squares (pigeons) and four 2 -by-2 squares (pigeonholes). By pigeonhole principle, there will be one 2 -by-2 square with at least 13 44  = shaded squares. This would be we have at least one shaded 2-by-2 square if we have 13 shaded square in the grid. Therefore, at minimum, we need 13 shaded squares. (c)(i) Method 1: This is equivalent to 1 2 3 4 4c c c c+ + + = , where 0ic +  . Hence, 4 4 1 3541 +− = − . Method 2: By listing cases,     4 3 1 column with 4 2 columns with 1,1 with 2 44 1 column with 3, 1 with 1 2 columns with 2 4 colums 2 with 1 2 4 3 1 35 C PC + ++ +=  (c)(ii) As the number of white squares is 16− the number of shaded squares. Counting this is equivalent to counting if the grids have at most 4 white squares. By symmetry this is the same as counting the grids that have at most 4 shaded squares. Method 1: So, we require, 4 4 1 3 4 1 2 4 1 1 4 1 0 4 1 4 1 4 1 4 1 4 1 4 1 70. + − + − + − + − + −         + + + +         − − − − −          =

Method 2: This is equivalent to 1 2 3 4 5 4c c c c c+ + + + = , where 0ic +  . We treat 5c as the number of shaded squares not in the grid if there are less than 4 shaded squares in the grid. 4 5 1 70.51 +− = − Method 3: By listing cases: ( ) ( ) ( ) ( ) ( ) 22 4 4 4 4 3 2 2 4 4 3 1 4 4 4 4 1 70 C P C P C +  + + + + + + + + + + =

4 Solutions [12] Probability (a) In constructing a subset of S, we have 2 choices for each of the n integers from 1 to n: include or exclude. By multiplication principle, there are 2n subsets of S. (b) (i) Then, ( ) ( ) ( ) ( ) 1 2 1 2 12 2 1 1 (by independence) 11 2 3 4 P i A A P i A A P i A P i A   = −   = −   =−  = Alternatively, Consider the partition: ( ) ( ) ( ) ( )1 2 1 2 1 2 1 2A A A A A A A A     Then i has equal probability to be in one of these four sets. Therefore, ( )12 3 4P i A A  = . (b) (ii) ( ) ( )1 2 1 2 1 3 4 n i n P A A P i A A =  = =   =  (b) (iii) Let the k elements of 1A be 1,..., kaa . ( ) ( )1 2 1 2 1 | 1 2 k i i k P A A A k P a A =  = = =  =  Alternatively,

( ) ( ) ( ) 1 2 1 1 2 1 1 and | 2 4 2 1 2 nk n n k P A A A k P A A A k P A k n k n k −  = =  = = = =     =     =   (b) (iv) ( ) ( ) ( )1 2 1 1 2 1 0 0 0 | 11 (there are subsets with elements; part (bii))22 1 2 n k n nk k n nk k P A A P A k P A A A k nn kkk n k = = + =  = = =  = =    =        =     From (bi), we have 0 13 24 nn nk k n k + =  =  . (c) (i) WLOG, let the k elements of their intersection be 1, …, k. Then, we must have 12i A A for 1 ik and 12j A A for 1k j n+   . By (bi), ( ) ( )1 2 1 2 13 and 44P i A A P i A A  =   = . Moreover, there are n k   ways to pick the k elements. So ( )12 13 44 k n k nP A A k k −      = =         (c) (ii) By (ci), 12 1~ ( , ) 4A A B n . So ( )12 1 44 nE A A n =  = .

Q5 Solutions [14] Polynomial, Numbers, Induction (a)(i) ( ) ( )ff x − is a polynomial. When x = , ( ) ( )f f 0−= . By factor theorem, x − is a factor of ( ) ( )ff x − . i.e. ( ) ( ) ( ) ( )0f f fx x x − = − for some integer polynomial ( )0f x . When x = , ( ) ( ) ( ) ( )0f f f 0    − = −  (modulo − ). Alternatively By remainder theorem, ( ) ( ) ( ) ( )f = g +fx x x − , where g is a polynomial with integer coefficients. Sub x = : ( ) ( )ff  (modulo − ). (a)(ii) Suppose there exists such a polynomial. ( ) ( )f 3 f 1 0− (modulo 2) from (a)(i). But ( ) ( )f 3 f 1 5 2 3 1− = − =  (modulo 2) a contradiction. (a)(iii) ( ) ( ) ( ) ( ) ( ) f 2025 f 1 0 modulo 2024 f 2025 0 modulo 2024 −  Since f(2025) is a multiple of 2024, it cannot be strictly between 0 2024 0= and 1 2024 2024= . (b)(i) 24 (modulo 7) 22 2 1 7 0+ + =  (modulo 7) And 24 4 1 21 0+ + =  (modulo 7) So, 2 and 4 are incongruent roots modulo 7 of 2 1xx++ . (b)(ii) Consider ( ) ( )hhx − . From (a)(i) we have that ( ) ( ) ( ) ( )0h h hx x x − = − for some integer polynomial ( )0h x with degree 1n− . We note that 1nx − coefficient of ( )0h x must be 1 when we compare the nx coefficient in the equation ( ) ( ) ( ) ( )0h h hx x x − = − . ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 0 0 h h h h 0 h modulo h h modulo x x x x x x p x x x p    − = − −  − − Alternatively ( ) ( ) ( ) ( )0h = h +hx x x − by Remainder Theorem, where h is a degree 1n− polynomial with integer coefficients. h is monic by