Imo Prelim Answer (2001-2002)

1.For a number to be divisible by 11, the rule is the sum of the digits in the odd positions minus the sum of the digits in the even posiions should be a multiple of 11. So we need (2n+7+9)-(n+2)=n+14 divisible by 11. Hence the least n is 8. (Briefly, the rule is because 10^k=(11-1)^k=11m+(-1)^k by the binomial theorem and so N=a_m10^m+...+a₁10+a₀=11M+a_m(-1)^m+...+a₁(-1)+a₀=11M+(a₀+₂+..)-(a₁+a₃+...)

2.Suppose 1 armchair cost $a, 1 bookcase cost $b and 1 cabnet cost$c. Then 8a+11b+2c=875 and 3a+2b+5c=343. Taking the first equation and adding it to 3 times the second equation, we get 17a+17b+17c=1904. So a+b+c=112.

3. T₁+T₂+T₃+...+T₂₀₀₂

  =F(1²)-F(1)+...+F(2002²)-F(2002)

  =F(1²+...+F(2002²)-F(1)-,,,-F(2002)

  =200(1+4+9+6+5+6+9+4+1+0)+1+4-200(1+2+...+9+0)-1-2

  =2

4.2²⁰⁰²(2²⁰⁰³-1)=4×16⁵⁰⁰(8×16⁵⁰⁰-1). Now M=16⁵⁰⁰-1 is divisible by 16-1=15. Thus 2²⁰⁰²(2²⁰⁰³-1)=4(M+1)(8M+7)=32M²+60M+28 has last two digits equal 28 because 32M² and 60M are divisible by 4×5×5=100.

5.The equations can be put in the form

  z²=x²+y²-2xy cos30^o and y²=x²+z²-2xz cos45^o,

which remind us the cosine law. So we consider triangle with AB=z, BC=x, CA=y, ¡çB=45^o and ¡çC=30^o. By sine law, y:z=sin¡çB : sin¡çC =2^1/2:1

6.The equation can be rewritten as (x³-b)²=100, which implies x=(b¡Ó10)^1/3. Thus 2=(b+10)^1/3-(b-10)^1/3. Cubing both sides, we get

  8=b+10-3×[(b+10)²(b-10)]^1/3+3×[(b+10)(b-10)²]^1/3-(b-10)

  8=20-3×(b²-100)^1/3)[(b+10)^1/3-(b-10)^1/3]=20-6×(b²-100)^1/3.

Then (b²-100)^1/3=2, which implies b = ¡Ô108=6¡Ô3.

7.Upon drawing an accurate figure and using a protractor, ¡çGFC seems to be 20^o. To confirm this, let H, I be the feet of perpendiculars from F and A to BC respectively. Now ¡çBCA=180^o - ¡çBAC - ¡çABC=80^o and BI = BA cos 60^o=½BA. Applying sine law to triangle ABC, we get (BA/sin 80^o)=(BC/sin40^o. So BA=(BC sin 80^o/sin40^o)=2BC cos 40^o and BI=BCcos40^o. In triangle ABC, ¡çBFC=70^o=¡çBCF. So BC=BF. Then BI = BF cos 40^o = BH yielding H=I. Then AF is perpendicular to BC. So G=H=I and ¡çGFC=¡çHFC=20^o.

8.Since the figure is same after 90^o, 180^o or 270^o rotation about the centre of ABCD, the region must be a square. Let this square has side length s. Now AF=(AB²+BF²)^1/2=29/21. The area of AFCH is AH×AB=1/21 and is also AF×s=29/21s. So s=1/29 and s²=1/841.

9.Let w=½x, then w²+y²=1. So w=cos£c and y=sin£c for some £c. Then x²+2xy+4y²+x+2y=4+4cos£csin£c+2cos£c+2sin£c. Now let u=cos£c+sin£c=¡Ô2[sin(£c+45^o)], then|u|¡Ø¡Ô2, u²-1=2cos£csin£c. Thus x²+2xy+4y²+x+2y=4+2(u²-1)+2u=2(u²+u+1)¡Ø2(2+2^1/2+1)=6+2¡Ô2 with equality if u=¡Ô2, £c=45^o, x=¡Ô2, and y =½¡Ô2. So the mzximum is 6+2¡Ô2

10.Let the length, width and height of cuboid be L, W, H respectively. Then x=(L-2)(W-2)(H-2), y=2[(L-2)(W-2)+(W-2)(H-2)+(H-2)(L-2)] and z=4[(L-2)+(W-2)+(H-2)]. If we let a=L-2, b=W-2, c=H-2, then 1994=x-y+z-8=abc-2(ab+bc+ca)+4(a+b+c)-8=(a-2)(b-2)(c-2). Since no face is a square and 1994=997×2×1, it follows a=999, b=4, c=3 and LWH=1001×6×5=30030.

11.Attempting to draw a figure, it leads to the suspicion that AB=AC. This is the case because the circumradius of triangle BMA being (BM/2sin¡çBAM) would equal the circumradius of triangle CAN so that both circumcentres are symmetic with respect to the perpendicular bisector of BC, which must then contain A. Since AM, AN bisect¡çBAN, ¡çCAM respectively, we get AN/AB=3/4=AM/AC.If AC=x, then AM=¾x=AN. By cosine law, 9=2(¾x)²-2(¾x)²cos¡çMAN and 16=(¾x)²+x²-2(¾x)x cos ¡çNAC. Since ¡çMAN=¡çNAC, solving the euqtions, we get AC=x=8.

12.Let a_n be the number of n-digit positive integers with the two properties. First, notice that a₁=0 and a₂=1. For n¡Ù3, of the a_n integers, there are a_n-1 of them begin with 2, there are a_n-1 of them begin with 12 and there are 2^n-2 of them begin with 11. So a_n=a_n-1+a_n-2+2^n-2 for n¡Ù3. Then a₃, a₄=8, a₅=19, a₆=43, a₇=94, a₈=201, a₉=423, a₁₀=880.

13.Let y=2001+n². Without loss of generality, assume n¡Ù0. Since x>y, we have x=2001+(n+1)². Now, d divides x, y implies d divides x-y=2n+1. Then d divides 2y-n(2n+1)=4002-n and 2(4002-n)+(2n+1)=8005. Hence the maximum d can be is 8005. Then n=4002+8005k and x=2001+(4003+8005k)². Since x<5×10⁷, k=0 and x=2001+4003²=16026010.

14.Let {x}=x-[x]. Note 0¡Ø{x}<1. If x>1, then the equation becomes x={x}/(x-1). This leads to {x}=x(x-1)>x-1¡Ùx-[x]={x}, a contradiction. Also, if x<-1, then the equation becomes -x-2={x}/(1-x). If x¡Ø-3, then{x}=(-x-2)(1-x)¡Ù1×4>{x}, a contradiction. If -3<x<-2, then {x}=x+3 and the equation becomes -x-2=(x+3)/(1-x), which has the solution x=-¡Ô5. So the largest possible |x| is ¡Ô5.

15.Let [XY...Z] denote the area of polygon XY...Z. Since [SAP]/[SAB]=AP/AB=k/(k+1) and [SAB]/[DAB]=SA/DA=1/(k+1), we get [SAP]/[DAB]=k/(k+1)². Similarly, [PBQ]/[ABC]=[QCR]/[BCD]=[RDS]/[CDA]=k/(k+1)²2. Now [PQRS]=[ABCD]-([SAP]+[PBQ]+[QCR]+[RDS])=

So (k²+1)/(k+1)²=0.52. Solving for k and noting k<1, we get k=2/3.

16.There are C^₃30 ways of choosing integers a, x, d such that 1¡Øa<x<d¡Ø30. The equation a+d=x+x is possible if and only if a and d are both even or both odd, which account for C^₂15+C^₂15 cases. In the remaining C^₃30-2C^₂15=3850 cases, we have a+d=x+x' with x is not equal to x' and a<x, x'<d. So he answer is 3850/2=1925.

17.Suppose . Let x and y be the integers with digits a₁...a_n and a_n+1...a_n+7 respectively. Then 10ⁿ/p=x+[y/(10⁷-1)]. So 10ⁿ(10⁷=1)=p[x(10⁷-1)+y]. Then p divides 10 or 10⁷-1==3²×239×4649. Since 2, 3, 5 are not such prime, so if p is not equal to 4649, then p can be the prime 239. As 1/239=0.004184100418410041841..., the only other choice of p is 239.

18.Consider triangle ABC with a=BC=4/3, b=AC=2^1/2, c=AB=10^1/2/3. Let D be the midpoint of BC, E on BC such that AE is perpendicular to BC and F on AC such that FD is perpendicular to BC. Consider folding along MN is perpendicular with M on BC and N on AB or AC. If N is on AB, then the maximum overlapped area occurs when N=A. If N is on CF, then the maximum overlapped area occurs when N=F. So for the problem, we may assume N is on FA. By cosine law, cos C=(a²+b²-c²)/2ab=1/¡Ô2 and cos B=(a²+c²-b²)/2ac=1/¡Ô10. Then ¡çC=45^o and sin B = 3/¡Ô10.

Let C' be on line BC such that M is the midpoint of CC' and G be the intersection of AB and NC'. In triangle BC'G, ¡çBGC'=¡çABC-¡çNC'C=¡çABC-45^o. So sin ¡çBGC'= sin B cosC - sinC cosB=1/¡Ô5.

, M is on segment DE and N is on segment FA.) In that case[MBGN]=4/15.