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Key Stage 3 - Team Contest
1. Let ( ) 9 9 3 x x f x = + . Calculate f( ) ( )
20171 + f 20172 + +... f( ) ( )
20172015 + f 20162017 . 【Submitted by Bulgaria】 【Solution】 Notice that, 1 1 9 9 ( ) (1 ) 9 3 9 3 9 9 9 9 3 9 9 3 9 3 9 9 3 3 9 1 9 3 1 9 3 9 3 a a a a a a a a a a a a a a a f a f a − − − − − − + − = + + + × = + + × + × = + + × + = + = + + So,( ) ( ) ( ) (
1 2016 1 1 1)
1 2017 2017 2017 2017 f + f = f + f − = ,( ) ( ) ( ) (
2 2015 2 1 2)
1 2017 2017 2017 2017 f + f = f + f − = , ....,( ) ( ) ( ) (
1008 1009 1008 1 1008)
1 2017 2017 2017 2017 f + f = f + f − = and( ) ( )
1 2( ) ( )
2015 2016 =1008. 2017 2017 2017 2017 f + f + +⋯ f + f Answer: 1008 2. In △ABC, point M is between A and B suchthat AM : MB = 1 : 2. Points N and P are between C and M such that CN : NM = 3 : 2 , CP : PM = 1 : 5. Segments AN and BC intersect at point Q. Segments PQ and AC intersect at point L. Find the ratio CL : LA. 【Submitted by Bulgaria_SMG】 A B C N L M P Q
【Solution 1】
From the given, we know that CP : PN : NM = 5 : 13 : 12. We apply Menelaus theorem to △MBC and line AN, △ABQ and line CM, and to △ANC and the line PQ.
We have BQ CN AM 1
CQ×MN × AB = , so 2 BQ
CQ = ; 1
AM BC QN
BM ×QC × AN = , and taking into account the previous equality, we have 2
3 QN
AN = . Finally, from 1
AQ NP CL
NQ×CP× LA=
and the previous equality, we obtain 2 13 CL
LA= .
【Marking Scheme】
Apply Menelaus theorem correctly, 10 marks Observe that BQ 2 CQ = , 10 marks Observe that 2 3 QN AN = , 10 marks Observe that 2 13 CL LA = , 10 marks A B C N L M P Q A B C N L M P Q A B C N L M P Q
【Solution 2】
From the given, we know that CP : PN : NM = 5 : 13 : 12. We can solve this by applying mass point geometry or “Law of the lever.” Consider the system ABC and line AQ, CM. If the force component on point M is 3a, then the force component on point A is 2a and the force component on point B is a. Hence, the force component on point C is 2a and the force component on point N is 5a. So, the force component on point Q is 3a. We get
3 2 AN NQ = and 2 1 BQ QC = .
Next, consider the system AQC and line QL, CN. If the force component on point N is 5a, then the force component on point C is 13a and the force component on point P is 18a. Hence, the force component on point A is 2a and the force component on point Q is 3a. So, the force component on point L is 15a. We
get 2 13 CL LA = . Answer: 2 13 CL LA= 【Marking Scheme】
Apply Mass Point Geometry or “Law of the lever” correctly, 10 marks
Find the force components on each point of the system ABC and line AQ, CM, 10 marks
Find the force components on each point of the system AQC and line QL, CN, 10 marks Observe that 2 13 CL LA = , 10 marks
3. Find the largest integer p such that 142017 +22017 is divisible by 2p. 【Submitted by Bulgaria_FPMG】
【Solution】
(
)
2017 2017 2017 2017
14 +2 =2 7 +1 and 72017 ≡ ×7
( )
72 1008 ≡7(mod8), therefore,(
)
2017 2017 2020
14 +2 ≡0 mod 2 , but 72017 + ≡ ×1 7
( )
72 1008 ≡ + ≡7 1 8 mod 2(
4)
.Answer: 2020. A B C N M Q 2a a 3a 2 1 2 3 2a 5a 3a 2 3 2 1 A B C N L P Q 13 5 13a 2 3 2 1 5a 18a 2a 3a 15a
4. In pentagon ABCDE, points M, P, N and Q are midpoints of AB, BC, CD and DE respectively. While points K and L are midpoints of QP and MN, respectively, as shown in the figure below. If KL=25cm, find the length of EA, in cm.
【Submitted by Bulgaria】
【Solution】
Connect BE and CE. Let T be the midpoint of BE. Connect QN, NP, PT and TQ.
In triangle ABE, 1 2 TM = AE and TM//AE. In triangle ECD, 1 2 QN = CE and QN//CE. In triangle BEC, 1 2 TP= CE and TP//CE. So, TPNQ is parallelogram. Connect NT and TM. In triangle NTM , 1 2 KL= TM and KL//TM
since K is the midpoint of QP and L is the midpoint of diagonal NT. Hence, 1
4
KL= AE, i.e. EA=4KL=100cm.
Answer: 100 cm 【Marking Scheme】
Plot point T, 10 marks
Show that TPNQ is parallelogram, 10 marks
Observe that 1 2 KL= TM , 10 marks Observe that 1 4 KL= AE, 5 marks
Get the correct answer, 5 marks.
A B C N L D K E M P Q A B C N L D T K E M P Q
5. Let x and y be positive integers, where 0< < <x y 2018. How many ordered pairs (x, y) are there such that x2 +20182 = y2 +2017 ?2 【Submitted by
Bulgaria_SMG】
【Solution】
The given equation is transformed to y2 −x2 =20182 −20172, so that
(
y+x)(
y− =x)
4035= × ×3 5 269.Because y+ x and y− x are positive integers, it follows
{
5 269 3 y x y x + = × − = or{
3 2695 y x y x + = × − = or{
15269 y x y x + = − = or{
13 5 269 y x y x + = × × − =Solve them and get 671 674 x y = = or 401 406 x y = = or 127 142 x y = = or 2017 2018 x y = =
But the last pair (2017, 2018) does not satisfy the conditions, thus, there are only 3 pairs of positive integers (x, y).
Answer: 3 pairs 6. Points A, B, C, D and E are on the circumference.
Chord AC is a diameter of the circle, as shown in the figure below. If ∠ABE= ∠EBD= ∠DBC,
16 cm
BE= and BD=12 3 cm, find the area of pentagon ABCDE. 【Submitted by Indonesia】
【Solution】
Since ∠ABE= ∠EBD= ∠DBC and AC is divided into three equal lengths, so, AE=ED=DC. ABCD is a cyclic quadrilateral, so ∠ + ∠ =A D 180° and∠ + ∠ =E B 180°. E D B B′ B′′ C =E′ A=D′ A E D C B
Look at triangle △BED, since AE= ED, so if we rotate △BED clockwise, with center of rotation a point E, then point D placing point A and form a new triangle
BED′
△ as shown in the figure below.
And since BE =B E′ , and ∠EBD′= ∠EB D′ ′ so △BEB′ is an isosceles triangle. 30
ABE EBD DBC
∠ = ∠ = ∠ = °,
and altitude of △BEB′ is equal to 1 8 cm
2BE= , BB′ =16 3 cm The area of 16 3 8 2 64 3 cm 2 BEB′= × = △
If we do the same action for △BED counter clockwise with central of rotation at point D, so point E placing point C and form a new triangle △BDB′′ as shown in the figure below.
And since BD= B D'' , and ∠DB E′′ ′= ∠DBE′= °30 , △BDB′′ is an isosceles triangle. The altitude of △BDB′′is equal to 1 6 3 cm
2BD= , so BB′′ =36 cm. The area of 36 6 3 2 108 3 cm 2 BDB′′= × = △ .
The area of △BED=
1 16 12 3 sin 30 2 48 3 2 2 BD BE× × ° = × × = cm2.
Hence, the area of pentagon ABCDE = Area△BEB′+ Area△BDB′′– area △BED = 64 3 108 3+ −48 3=124 3 cm2
Answer: 64 3 108 3+ −48 3 =124 3 cm2 【Marking Scheme】
Found the area of △BEB′, 10 marks Found the area of △BDB′′, 10 marks Found the area of △BED, 10 marks Correct answer, 10 marks
D
B C =E′ B′′ B
E
7. A 10 10× chessboard is dissected into thirty-three 1 3× or 3 1× rectangles and one unit square. In how many different positions can this unit square be, if the chessboard may not be reflected or rotated? 【Submitted by Central Jury】
【Solution】
We first label the 100 squares A, B or C diagonal by diagonal in two ways, as shown in the diagram below. In each labelling, there are 34 As, 33 Bs and 33 Cs. Since each rectangle covers one square with label A, one with B and one C, the 1 1× square must occupy a square with label A in both labellings. There are 16 possible positions for it. Each position is attainable. Just add three 1 3× rectangles in the same row, and fill the remaining three groups of three rows with 3 1× rectangles.
A B C A B C A B C A A B C A B C A B C A B C A B C A B C A B C A B C A B C A B C C A B C A B C A B C B C A B C A B C A B A B C A B C A B C A A B C A B C A B C A B C A B C A B C A B C A B C A B C A B C C A B C A B C A B C B C A B C A B C A B A B C A B C A B C A A B C A B C A B C A B C A B C A B C A B C A B C A B C A B C C A B C A B C A B C B C A B C A B C A B A B C A B C A B C A A B C A B C A B C A Answer: 16 8. Prove the inequality:
2 100 99 101 98 102 ... 1 199 4 π × + × + + × < 【Submitted . by Bulgaria_SMG】 【Solution】
Consider a quarter circle with radius 1.
Now, inscribe a row-like figure consisting of 99 rectangles, each of them having a base length 1
100. The area of the first rectangle is:
2 1 2 99 101 1 100 S =OB×AB=OB −OB = × ,
the area of the second rectangle is:
2 2 2 1 2 98 102 1 ( ) 100 100 100 S = − = ×
and so on. The area of the last one is 99 1 1 (99 )2 1 1992 100 100 100
S = − = × .
So, the total area of the figure is less than 1
4 of the circle area, i.e. O
A
B
2 2 2 99 101 98 102 1 199 100 100 100 4 π × + × + + × < ⋯ . 【Marking Scheme】
Construct a quarter of a circle and inscribe the rectangles, 15 marks. Find the rules of the areas of the rectangles, 15 marks.
Conclude the inequality, 10 marks.
9. A computer randomly chooses three different points on the given grid (all points have the same chance of being chosen). Let p
q be the probability to form a triangle with these
points (this fraction is written in its irreducible form). Find the sum of p and q. 【Submitted by Central Jury】
【Solution】
There are C321 =1330 ways of choosing 3 points out of 21. They will form a triangle if not all three are on the same line. We can see that there are C35× =6 60 ways to choose three points on one of the 6 lines (black) having exactly 5 points of the grid,
4
3 4 16
C × = ways to choose three points on one of the 4 lines (blue) having exactly 4 points and C33× =14 14 ways of choosing three points on one of the 14 lines (red) having exactly 3 points.
Thus, the number of ways in which a triangle can be formed is 1330−60 16 14 1240− − = . The probability is then 124
133 p
q = , and so q+ =p 257. Answer: 257
10. Jane has 12 marbles, where in one is fake. She are not certain if the fake marble is heavier or lighter than the real marble. What is the minimum number of weightings needed to find the fake marble and determine whether the fake marble is heavier or lighter than the real marble? Explain your answer.
【Solution】
Jane would need to divide the 12 marbles into three groups (A A A A1, 2, 3, 4), (B B B B1, 2, 3, 4), and (C C C C1, 2, 3, 4).
She begins by balancing A A A A1, 2, 3, 4 and B B B B1, 2, 3, 4. If both balance, then she would know that one of C C C C1, 2, 3, 4 is fake.
Now, she chooses 3 marbles from C C C C1, 2, 3, 4 (assume that she has chosen C C C1, 2, 3) and balance it against any 3 from of the known genuine marbles (assume that she choseA A A1, 2, 3). If they balance, so the fake one is C4. Otherwise, if they don't
balance and that C C C1, 2, 3 is heavier toA A A1, 2, 3, she can conclude that the fake one is heavier and she chooses 2 from C C C1, 2, 3 and do the weighing (assume she chose
1 and 2
C C , if they balance, the fake marble is C3, otherwise, the fake one is the heavier marble. (it’s the same way scenario if one out of the set is lighter).
Now, suppose that she balances A A A A1, 2, 3, 4 and B B B B1, 2, 3, 4 and it didnt’t balance, so C C C C1, 2, 3, 4 are all genuine. Suppose that A A A A1, 2, 3, 4 was heavier than
1, 2, 3, 4
B B B B . For her second balance, she replaces 3 marbles from A A A A1, 2, 3, 4
(suppose she has chosenA A A1, 2, 3) with 3 marbles from C C C C1, 2, 3, 4 (suppose she has chosenC C C1, 2, 3), and in addition, swap A4 to one from B B B B1, 2, 3, 4 (suppose she choose B4). So, for the second balance we do weighing C C C B1, 2, 3, 4 and B B B A1, 2, 3, 4. If it balances, she knows that one of the one from 3 balls A A A1, 2, 3 is fake and
heavier, then he only need to know a third weighing to know the fake one.
If C C C B1, 2, 3, 4 is lighter, one of the 2 balls swapped (A4orB4) is fake, so on the third weighing , weigh one marble from C C C C1, 2, 3, 4 withA4, if it balance, we know that
4
B is fake and lighter, if not, thenA4 is fake and heavier. If B B B A1, 2, 3, 4 is still lighter, she would know that one of B B B1, 2, 3 is lighter, then she only needs the third weighing to know the fake marble.
In any case, minimum three balances are required.
Answer: 3 weighings 【Marking Scheme】
Reason why we cannot do with 2 weightings, 10 marks
Give a solution with 3 weightings , up to 30 marks (partial credit to be given depending on progress)
Show a solution with 4 weightings, which of course is not the right solution, 10 marks
