注意：
允許學生個人、非營利性的圖書館或公立學校合理使用
本基金會網站所提供之各項試題及其解答。可直接下載
而不須申請。
重版、系統地複製或大量重製這些資料的任何部分，必
須獲得財團法人臺北市九章數學教育基金會的授權許
可。
申請此項授權請電郵
ccmp@seed.net.tw
Notice:
Individual students, nonprofit libraries, or schools are
permitted to make fair use of the papers and its
solutions. Republication, systematic copying, or
multiple reproduction of any part of this material is
permitted only under license from the Chiuchang
Mathematics Foundation.
Requests for such permission should be made by
Key Stage 3 - Individual Contest
Section A.In this section, there are 12 questions. Fill in the correct answer in the space provided at the end of each question. Each correct answer is worth 5 points. 1. If both a and b are positive integers greater than 1, find the smallest possible sum
of a and b, such that a a =b.【Submitted by Bulgaria】 【Solution】
2 3 4
a a =b⇒a a =b ⇒a =b . Then b is a perfect cube. When b=23 =8, we get
( )
43 3
2 16
a = ⇒а = . So, a+ = + =b 16 8 24.
Answer: 24
2. Find the largest positive integer d, in which there exists at least one integer n such that d divides both n2+1 and (n+1)2 +1.【Submitted by Sri Lanka】
【Solution】
Let d n| 2 +1 and d| (n+1)2 +1, or d n| 2+2n+2.
Then, d| (n2 +2n+ −2) (n2 +1), or d| 2n+1⇒d| 4n2 +4n+1, So, d| 4(n2 +2n+ −2) (4n2 +4n+1), or d| 4n+7.
Then, d| (4n+ −7) 2(2n+1), or d| 5, so d can only be 1 or 5. The largest d is 5. Answer: 5
3. A regular hexagon is inscribed in a circle with radius 12 cm. The hexagon is divided into 6 congruent equilateral triangles and each of them has a small circle inscribed in it. Another small circle is then drawn touching all the inscribed circles. What is the area of the shaded region? Take π =3.14.【Submitted by
【Solution】
The radius of smaller circle is 1 12 3 2 3 3 2
r = × = cm.
The area is Sgc − ×7 Ssc =122π − ×7
( )
2 3 2π =(
144 84−)
π =188.4 cm2.Answer: 188.4 cm2
4. Find the smallest positive integer n so that 2n is a perfect square and 7n is a perfect 7th power. 【Submitted by Thailand】
【Solution】
Let 2n=a2 and 7n=b7, where a and b are positive integers.
a is divisible by 2 →a2 =2n is divisible by 22 and n is divisible by 2. The same way n is divisible by 7.
Since we want to find the smallest positive integer n , we can let n=2 7p q where p and q are positive integers.
Then, 2n =2p+17q =a2, 7n=2 7p q+1 =b7.
If 2p+17q is a perfect square, then p+1 and q must be divisible by 2. Let p+ =1 2p1, q =2q1, where p1 and q1 are positive integers.
If 2 7p q+1 is a perfect 7th power, then p and q+1 must be divisible by 7 and r must be perfect 7th power. Thus, p =2p1− =1 7s and q+ =1 2q1+ =1 7t.
By inspection, we see that the smallest integer p for which 1 2p1−1 is divisible by 7 is p1=4, p= × − =2 4 1 7.
So, the smallest integer q in which 1 2q1+1 is divisible by 7 is q1 =3, 2 3 6
q = × = .
Therefore, the smallest positive integer n is n=2 77 6.
Answer: 2 7 7 6
5. Find all possible integer values of x, in which
2 21 21 2 2 2 x 42 x x − − − = + . 【Submitted by Bulgaria】 【Solution】 2 2 21 21 21 2 2 2 1 x 43 2 1 x 43 x x x − − − + = + ⇔ − − = + , i. е. 2 21 3 x 43 x − = + .
Since x is an integer, its left side accepts only whole values. Therefore, x divides 21, i.е. 1; 3; 7; 21
x= ± ± ± ± . When x= −7 and x=21 the number x+43 is a perfect square. It is easy to check that −7 is a solution, but 21 is not a solution. So, x= −7.
Answer: −7
6. If the equation
(
a2 +3b2)
x2 −(
4a+6b x)
+ =7 0has a root of 2017, where a and【Solution】
If a= =b 0, we get an equation, which has no solutions. So, at least one of the numbers a, b is different from zero and then the equation is quadratic. Discriminant
(
)
2(
2 2)
(
2 2)
(
)
24 6 28 3 12 4 4 12 2 0
D= a+ b − a + b = − a − ab+ b = − a− b ≤ for any a and b. On the other side, given equation has a real root when D≥0. It is possible only when D=0, from where 2017 is the only solution of equation and a =2b. Therefore,
2 2 2 4 6 2( 3 14 1 201 ) 7 14 a b b b b a b = + = = + , and 1 2017 b = and 2 2017 a = . Answer: 3 2017
7. If x>1, find all possible values of x that satisfies the following equation:
2017 2018 2017 2018 2018 2017 2018 2017 x x x x − − − = − − − .【Submitted by Bulgaria】 【Solution】 Substitute 2017 2018 x y = − and 2018 2017 x z= − . Then y z 1 1 z y − = − , i.e. (y−z yz)( − =1) 0. Therefore, (1) y= z, i.е. 2017 2018 2018 2017 x− = x− , and x =4035. (2) yz=1, i.е. 2017 2018 1 2018 2017 x− × x− = , and x2 −4035x =0. Since x>1, x=4035. Answer: 4035.
8. In the figure below, ABC is an isosceles triangle with base AB. Its orthocentre H divides its altitude CD into two segments CH and HD, where CH =7cm and
9
HD= cm. Find the perimeter of triangle ABC, in cm.【Submitted by
Bulgaria_SMG】
A B
C
H
【Solution】
If CD intersects the circumcircle of △ABC at point N, then △ADH≅△ADN.
So DN =DH =9. For chords АВ and CN,
2 ( ) 16 9 AD DB× =CD DN× ⇒ AD = × 12 AD ⇒ = 2 2 2 2 2 12 16 20 AC = AD +CD = + ⇒ AC= . Hence, the perimeter of △ABC is
20+20+24=64cm.
Answer: 64 cm.
9. In quadrilateral ABCD, ∠ABD= ∠ACD= °90 and a point P is on AD so that APB CPD
∠ = ∠ , as shown in the figure below. If AP=24cm and DP=19cm, find the value of PB PC× .【Submitted by Thailand】
【Solution 1】
Let M and N be on AD so that BM ⊥ AD and CN ⊥ AD. Let O be the midpoint of
AD. Connect OB and OC.
Since OB and OC are medians of the hypotenuse of the right triangles ABD and ACD, respectively, 1
2
OB=OC =OA=OD= AD. Now let AM = y and DN =x, then
A B C N H D A B C P D A B C O P N D M
from OB2 −OM2 =BM2 =PB2 −PM2, we have 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ( ) ( ) 2 2 2 2 2 2 24 5 576 5 OB OA AM PB PA AM OB OA OA AM AM PB PA PA AM AM OA AM PB PA PA AM PA PB PA AM OA AM PB y PB y − − = − − − + × × − = − + × × − × × = − + × × − = × × − × × − = = −
Similarly, we also have PC2 =5x+361 from OC2 −ON2 =PC2 −PN2.
Since ∠MPB= ∠NPC and ∠BMP= ° = ∠90 CNP, triangles BMP and CNP are
similar. Hence, 24 120 5 19 95 5 PB PM AP AM y y PC PN CP CN x x − − − = = = = − − − .
Observe that 120 5− y =PB2 −456 and 95 5− x=456−PC2, so
2 2 456 456 PB PB PC PC − = − . Hence, PB(456−PC2)=PC PB( 2 −456), i.e. (PB+PC PB PC) × =456(PB+PC). Thus, PB PC× =456. 【Solution 2】
Since ∠ABD= ∠ACD= °90 , we know that points A, B, C and D are concyclic. Let point C′ be symmetric point of C along AD, as shown in the figure below. Then, we have ∠C PD' = ∠CPD= ∠APB. Hence, points B, P and C’ are collinear. Using the Intersecting Chord Theorem, we get
' 24 19 456 BP PC× = BP PC× = AP PD× = × = . Answer: 456 A C P D C′ B
10. A girl tosses a fair coin 100 times and a boy tosses a fair coin 101 times. The boy wins if he has more heads than the girl has, otherwise, he loses. Find the probability that the boy win this game. 【Submitted by Central Jury】 【Solution】
Let us pause the game just before the boy makes his last toss. At this point, both players has made 100 tosses or a total of 200 tosses in all. If the girl got more heads than the boy did, the last toss wouldn’t help him. If he got more heads than the girl did, the last toss is unnecessary. By symmetry, these two scenarios are equally likely. The last toss only matters if both have the same number of heads at this point. On the last toss, the boy wins if he gets a head, and loses if he gets a tail. Since the coin is fair, either is equally likely. Overall, there is a 50% probability for the boy to win the game.
Answer: 50%
11. A 6 6× chessboard is formed by using 36 unit squares. How many different combinations of 4 unit squares can be selected from the chessboard so that no two unit squares are in the same row or column? 【Submitted by Thailand】 【Solution】
After one of the 36 blocks is chosen, 25 of the remaining blocks do not share its row or column. After the second block is chosen, 16 of the remaining blocks do not share a row or column with either of the first two, 9 of the remaining blocks do not share a row or column with either of the first two. Because the four blocks can be chosen in any order, the number of different combinations is 36 25 16 9 5400
4!
× × × = .
12. If abcd is a 4-digit number, where each different letter represents a different digit such that a<b, c <b and c<d. How many such 4-digit numbers are there?【Submitted by Bulgaria_FPMG】
【Solution】
Observe that a≠0 since a is the leading digit of a four-digit number.
If b =2, then a=1, c=0 and hence 3≤ ≤d 9. There are 7 such numbers. If b =3, then a =1 or 2.
As a =1, then c=0 or 2 . When c =0, then d =2 or 4≤ ≤d 9. There are 7 such numbers. When c=2, then 4≤ ≤d 9. There are 6 such numbers.
As a =2, then c=0 or 1 . When c=0, then d =1 or 4≤ ≤d 9. There are 7 such numbers. When c=1, then 4≤ ≤d 9. There are 6 such numbers.
Hence there are totally 2 (7× + =6) 26 such numbers.
If b =4, then there are totally 3 (7× + + =6 5) 54 numbers by similar argument. If b =5, there are totally 4 (7× + + + =6 5 4) 88 numbers by similar argument. If b =6, there are totally 5 (7× + + + + =6 5 4 3) 125 numbers by similar argument. If b =7, there are totally 6 (7× + + + + + =6 5 4 3 2) 162 numbers by similar argument. If b =8, there are totally 7 (7× + + + + + + =6 5 4 3 2 1) 196 numbers by similar argument. If b =9, there are totally 8 (7× + + + + + + =6 5 4 3 2 1) 224 numbers by similar argument. So, there is a total of 7+26 54 88 125 162 196+ + + + + +224=882 numbers.
Section B.
Answer the following 3 questions, each question is worth 20 points. Partial credits may be awarded. Show your detailed solution in the space provided. 1. Let α, β and γ be the three roots of the polynomial x3 − +5x 1. If p and q are
relatively prime positive integers such that
(
)(
) (
)(
) (
)(
)
3 3 3 6 1 6 1 6 1 6 1 6 1 6 1 p q α β γ β γ α γ β α − = + + + + + + + + .Find the sum of p and q.【Submitted by Bulgaria_FPMG】 【Solution】
Note that
(
x−α)(
x−β)(
x−γ)
= −x3 5x+1⇒α β γ+ + =0, αβ βγ γα+ + = −5 and α β γ× × = −1.Also, it follows that,
(
)
2 2 2 2(
)
2 2 2 0= α β γ+ + =α +β + +γ 2 αβ βγ αγ+ + ⇒α +β +γ =10. Now(
)(
) (
)(
) (
)(
)
(
)(
) (
)(
) (
)(
)
(
)(
)(
)
(
)
(
)
(
) (
)
3 3 3 2 2 2 6 1 6 1 6 1 6 1 6 1 6 1 5 1 6 1 5 1 6 1 5 1 6 1 6 1 6 1 6 1 30 3 216 36 6 1 297 395 p q α β γ β γ α γ β α α α β β γ γ α β γ α β γ α β γ αβγ αβ βγ αγ α β γ − = + + + + + + + + − + + − + + − + = + + + + + − + + − = + + + + + + + = − Hence,p+ =q 297+395=692. Answer: 692 【Marking Scheme】Full solution with correct answer, 20 marks. Correct answer only, 10 marks.
Finding 0 5 1 α β γ αβ αγ γα αβγ + + = + + = − = − , 5 marks or 3 3 3 2 2 2 5 1 5 1 5 1 10 α α β β γ γ α β γ = − = − = − + + = , 5 marks
2. A 21 21× table contains 21 copies of each of the numbers 1, 2, 3, . . . , 20 and 21. The sum of all the numbers above the main diagonal (diagonal from the top-left cell to the bottom-right cell) is equal to three times the sum of all the numbers below the main diagonal. Find the sum of all the numbers on the main diagonal of the table.【Submitted by Central Jury】
【Solution】
There are 21 numbers on the main diagonal, 21 10× =210 numbers above it and 210 numbers below it. The sum of the largest 210 numbers is
21 (12 13 14 15 16 17 18 19× + + + + + + + +20+21)= ×21 165, while the sum of the smallest 210 numbers is
21 (1 2× + + + + + + + + +3 4 5 6 7 8 9 10)= ×21 55.
Since the former is exactly three times as large as the latter, the largest 210 numbers should all be located above the main diagonal and the smallest 210 numbers are all located below the main diagonal. In other words, every number on the main diagonal should be 11. Hence, the sum of the numbers on the main diagonal is 11 21 231× = .
Answer: 231
【Marking Scheme】
Identified how many numbers are there above, below and on the main diagonal, 5 marks
Calculated the largest possible sum of 210 numbers, 5 marks Calculated the smallest possible sum of 210 numbers, 5 marks
Deduced that the numbers on the diagonal should all be 11 and arrived with the correct answer, or correct answer only (without any explanation), 5 marks.
3. In the figure below, ABCD is a square. Points E, Q and P are on sides AB, BC and
CD, respectively, such that PE⊥ AQ and △AQP is equilateral triangle. Point F
is inside △PQC such that △PFQ and △AEQ are congruent. If EF = 2cm, find the length of FC, in cm.【Submitted by Bulgaria】
A B C P Q F E D
【Solution】
From AD= AB and AP= AQ, it follows
△ADP≅△ABQ and ∠DAP= ∠QAB= °15 . △AQP is an equilateral triangle, so PE is a symmetrical to AQ and △AQE is isosceles ( AE= EQ). Therefore,
15
QAE AQE
∠ = ∠ = ° and ∠AEQ=150°.
Now, from △PFQ≅△AEQ it follows that F lies on the bisector of ∠PCQ, i.e. on diagonal AC.
15 60 15 90
FQE FQP PQE AQE
∠ = ∠ + ∠ + ∠ = ° + ° + ° = °
and from △PFQ≅△AEQ we have FQ=QE, i.e. △FQE is isosceles right-angle triangle. Therefore
2 2 EF
АЕ=FQ=EQ= = .
But △PCQ is isosceles right-angle triangle, so
45 15 30
FQC PQC PQF
∠ = ∠ − ∠ = ° − ° = °.
From ∠AFE=180° − ° −45
(
180° − ° − ° = °45 30)
30 and ∠BAC= ∠ACB= °45 , it follows that triangles FAE and QCF are similar with coefficient EF 2FQ = . Therefore, 2 2 EF АЕ FQ FС FC = = = , i.e. FC=1cm. Answer: 1 cm 【Marking Scheme】
Correctly establish relationships of different triangles/angles (without any mistakes), up to 5 marks.
Establish the relationship between EF and FQ, 5 marks
Establish the fact that FAE and FQC are similar triangles, 5 marks Finding the correct answer with explanation, or 5 marks
Correct answer only (without any explanation), 0 marks
A B C P Q F E D
