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Solution Key to Second Round of 4th IMAS
Junior Division
1. Which of the following expressions is equal to
(a1)(b 1) (b 1)(c 1) (c 1)(a1)? (A)ab bc ca3 (B) ab bc ca (C)ab bc ca2a2b2c3 (D)ab bc ca2a2b2c3 (E)ab bc ca 2a2b2c3 【Suggested Solution #1】 ( 1)( 1) ( 1)( 1) ( 1)( 1) ( 1) ( 1) ( 1) 3 a b b c c a ab b a bc c b ca a c ab bc ca 【Suggested Solution #2】
When a b c 1, it follows that the value of the original expression will be 0 and the final results for the expression in each option are:
(A) ab bc ca 3 1 1 1 3 0 (B) ab bc ca 1 1 1 3
(C) ab bc ca2a2b2c 3 1 1 1 2 2 2 3 12 (D) ab bc ca2a2b2c 3 1 1 1 2 2 2 3 6 (E) ab bc ca2a2b2c 3 1+1+1 2 2 2 3 0 Hence, only options (A) and (E) will yield a result of 0.
Now assuminga b c 1, so that the value of the original expression will still produce 0 and the results of the expression in each option are as follow:
(A) ab bc ca 3 1 1 1 3 0 (B) ab bc ca 1 1 1 3
(C) ab bc ca2a2b2c 3 1 1 1 2 2 2 3 0 (D) ab bc ca2a2b2c 3 1 1 1 2 2 2 3 6 (E) ab bc ca2a2b2c 3 1 1 1+2 2 2 3 12
Here, only options (A) and (C) have the same value as that of the original expression, which is 0.
Therefore, only option (A) will produce the same result as the original expression. Answer: (A)
2. In triangle ABC, D is the midpoint of BC. E is an arbitrary point on CA, and F is the midpoint of BE. If the area of triangle ABC is 120 cm2 and the area of the quadrilateral AFDC is 80 cm2, what is the area, in cm2, of triangle BDF? (A)10 cm2 (B)15 cm2 (C)17.5 cm2 (D)20 cm2 (E)25 cm2 F E D C B A
【Suggested Solution】
Connect FC. Let S represent the area. From the given information, we know that
ABF AFE S△ S△ , S△BCF S△FCE, which is 1 60 2 AFC ABC S△ S△ cm2. Therefore, 80 60 20 BDF FDC AFDC AFC S△ S△ S S△ cm2. Answer: (D)
3. Let m be a positive integer such that m3 can be expressed as a sum of m consecutive odd integers. For instance, 23 3 5, 33 7 9 11 and
3
4 13 15 17 19 . If 999 is one of the consecutive odd integers in the expression for m3, what is the value of m?
(A)30 (B)31 (C)32 (D)33 (E)34
【Suggested Solution】
When m3 was expressed as the sum of m consecutive odd integer and assuming the first term in the series as 2k1, then the last term in this series will be represented as 2k 1 2(m 1) 2k2m1, so the sum of the series is
3 (2 1 2 2 1) , 2 k k m m m it follows 2m3 4k2 ,m then 2 , 2 m m k this implies the first term becomes 2k 1 m2 m 1 and the last term becomes
2
2k2m 1 m m 1, hence m2 m 1 999m2 m 1. Using the completing of square again, then we have
2 1 2 3 2 1 2 5 1 ( ) 999 1 ( ) . 2 4 2 4 m m m m m m That is, 2 2 2 2 1 1 ( ) 998 1024 32 2 4 1 1 961 31 1000 ( ) 4 2 m m Thus, we have m32. Answer: (C)
4. The perimeter of an equilateral triangle is a cm while the perimeter of a square is
b cm. If the area of the square is half the area of the triangle, what is the value of
2 2 a b ? (A)3 3 8 (B) 3 3 4 (C) 3 3 2 (D) 3 3 3 (E) 6 3 【Suggested Solution】
From the given information, the side length of the given equilateral triangle is 3
a
, by
Pythagoras Theorem, the altitude of this equilateral triangle is 3 3 3 2 6
a
a
area of this equilateral triangle is 1 3 3 2 2 3 6 36
a
a a . Likewise, the side length of the given square is
4
b
, it follows the area of this square is
2 2 , 4 16 b b then we have 2 2 1 3 16 2 36 b a , this implies 2 2 3 3 2 a b . Answer: (C)
5. Mindy has two boxes, containing 0 and n pieces of candy respectively, where n is a positive integer. She adds 4, 3 and 2 pieces of candy to one of the boxes in that order, always adding to the box containing fewer pieces of candies. If the two boxes have the same number of pieces of candy, then she adds to either of them. In the end, there is 1 more piece of candy in one box than in the other. How many possible values of n are there?
(A)2 (B)3 (C)4 (D)5 (E)6
【Suggested Solution #1】
The usual approach to solve this problem is to generate an equation as follows: |||n 4 | 3| 2 | 1, apply the property of absolute value to the outer of the given absolute value equation, we have ||n 4 | 3| 2 1 or ||n 4 | 3| 2 1, it follows ||n 4 | 3| 3 or ||n 4 | 3| 1, apply the same property again, then
|n 4 | 3 3, 1, 1, 3, this implies |n 4 | 0, 2, 4, 6. Hence n4 will be equal to some of the values in the following: 6 , 4, 2, 0, 2, 4, 6.
Since n must be positive integers, thus n must be either 2, 4, 6, 8, 10 which is a total of 5 possible values.
【Suggested Solution #2】
Let us solve this problem using Working Backwards, we know that the last two boxes have a total of n 4 3 2 n 9 candies and the number of candies in one box is 1 more than the number of candies in the other box, then the total number of candies in the two boxes must be an odd number, it follows n must be an even number. The box that was empty at the start will then have at most 4 3 2 9 candies at the ending while the box containing n pieces of candies at the start will have at least n pieces of candies at last, then n 9 1, or n10. There are 5 even numbers that are less than 10: 2, 4, 6, 8, 10 and each of them meets the condition of the problem.
Hence, the values of n may be 2, 4, 6, 8 and 10. Therefore, there are 5 possible values of n.
Answer: (D)
6. Let a, b and c be real numbers such that x2 5x3 is one of the factors of the polynomial x3ax2 bxc. What is the numerical value of a b 2c?
【Suggested Solution #1】
From the given information, we know that x3 ax2 bxc can be factored as the product of x2 5x3 and a linear factor.
or x3on both sides of the above assumption identity, we have p1, now equating the coefficients of like powers of other terms, we obtain a q 5, b5q3,
3
c q.
Thus, a b 2c(q 5) (5q 3) 2( 3 )q 2. 【Suggested Solution #2】
We know that x3 ax2 bxc can be expressed as the product of x25x3 and a linear factor. Since both the coefficient of x3 and x2 in x3 ax2 bxc and
2
5 3
x x are equal to 1, it follows the coefficient of x in the linear factor must be equal to 1 as well, this implies the constant in the linear factor is
3 c , so that 3 2 2 3 2 2 3 2 ( 5 3)( ) 3 5 5 3 3 3 5 (5 ) (3 ) 3 3 c x ax bx c x x x c c x x x x x c c c x x x c then 5 , 3 c a 3 5 . 3 c b Thus, 5 2 5 (3 ) 2 2. 3 3 c c a b c c Answer: 2
7. How many triples of integers (x, y, z) are such that |xyz|6?
【Suggested Solution】
The given equation can be rewritten as | | | | | | 6x y z ; which can be interpreted as the product of three positive integers 6, the possible product is either 1 1 6 or 1 2 3 . Consider the three positive integers 1, 1 and 6: there are 3 possible
arrangements in the solution of | |x , | |y , | |z . When the three positive integers are
1, 2, 3: there are 6 possible arrangements in the solution of | |x , | |y , | |z . Thus,
there are a total of 9 possible arrangements. Lastly, the value of x, y, z may be assigned either positive or negative. Thus, the original equation has a total of
3
9 2 72 triples of integer solution.
Answer: 72
8. In triangle ABC, AB7cm, AC 8cm and BC9cm. A circle with centre A intersects AB at F and AC at E. The circles with centres B and C and radii BF and
CE, respectively, are tangent to each other at a point D on BC. What is the total
area, in cm2, of these three circles? (Taking 3.14)
A
F E
D C
【Suggested Solution】 Let AF AExcm,
then BF BD
7x
cm, CECD
8 x
cm, so that (7 x) (8 x)9cm, it follows x3cm.This implies the radius of three circles are 3 cm, 4 cm, 5 cm.
Therefore, the area of three circles is (32 42 5 )2 50 157cm2.
Answer: 157 cm2
9. There are 2 counters in the first row, and each subsequent row has one more counter than the preceding row. If there are 2015 counters altogether, how many rows of counters are there?
【Suggested Solution】
Assuming all counters are arranged in n rows, then there are n1 in the last row, we obtain the sum of all counters as (2 1) 2015
2
n n
, it follows n62 (n 65 is an extraneous root).,
【Note】We may use estimation method to solve for n62.
Answer: 62 rows
10. In a book fair, the organizers give a book to each participant. Each male
participant gives every other male participant a book, and each female participant gives every other female participant a book. If the total number of books received by the male participants is 31 more than the total number of books received by the female participants, how many participants are there altogether?
【Suggested Solution】
Assuming there are m male participants and n female participants attended the book fair. From the given information, we have mm m( 1) n n n( 1) 31,
rearranging the terms we get m2 n2 31, then (m n m )( n)31. From 0
m n we have m n 0, but 31 is a prime number, then m n 1, m n 31. Therefore, there are 31 participants.
Answer: 31 participants
11. D is a point on the side BC of triangle ABC such that BAD 76 . When the point C is reflected across AD to the point C, ABC D is a parallelogram. What
is the measure, in degrees, of ADC?
【Suggested Solution】
Connect CC. From the given information, △ ADC△ADC, then AC AC and
CAD C AD
. Since the line segments of angle bisector, perpendicular and median of an isosceles triangle coincide together with
ADCC, AD BC// , we conclude that BCCC.
From the corresponding parts of congruent triangles are equal we haveDCDC, then
DCC DC C
. But BC D is the complement of DC C and C BD is the C A
B
D
complement of DCC, hence BC D C BD .
Using Properties of Parallelogram, ADB C BD BC D BAD 76 . Thus, ADC180 76 104.
Answer: 104
12. Let k be a non-zero integer such that the equation
2 9 81 10 k x k x has two
distinct integer roots.What is the difference when the smaller root is subtracted from the larger one?
【Suggested Solution】
The original equation can be written as x2 10k x (9k2 81)0, since both roots are integers, so the discriminant, (10 )k 2 4(9k2 81) must be a perfect square integer. Let m2, where m is a positive integer. This implies m2 64k2324, if follows (m8 )(k m8 )k 324.
From the equation , we know that m is positive integer, then both m8k and 8
m k are positive integers, obviously these two integers are distinct due to parity checking. Since 324 34 22, so we have the following four cases:
Case 1. When m8k 162 and m8k 2, solving the equation , we get 82
m and k 10, then the roots of the given quadratic equation are 91
and 9, meet the condition of the problem, so the difference is 82. Case 2. When m8k 54 and m8k6, solving the equation , we get
30
m and k 3, then the solutions of the given quadratic equation are 30 and 0, but 0 is an extraneous root, which contradicts the given condition, reject the answer.
Case 3. When4 m8k 6 and m8k54, solving the equation , again, this is same with Case 2, an extraneous root appears, this is a
contradiction of the given, reject the answer.
Case 4. When m8k 2 and m8k162, solving the equation , it is the same as Case 1, the roots of the given quadratic equation are 9 and 91, meeting the condition of the problem, the difference is 82.
From the above four cases, we conclude the difference of two roots is 82.
Answer: 82
13. The integers 1, 2, 3, …, 20 are divided into two groups. The sum of all the numbers in one group is n, while the product of all the numbers in the other group is also n. What is the maximum value of n?
【Suggested Solution #1】
Let us show that when n192, it meets the condition of the problem. Let 2, 4 and 8 as the numbers in the second group while all the remaining numbers are in the first group, so the product of the three numbers in the second group is 4 6 8 192 and the sum of all the remaining numbers in the first group is
(1 2 20) (4 6 8) 192 , which meets the condition of the problem.
We know the prime factor of each n193, 194, 197, 199, 201, 202 and 203 is greater than 20 and they are 193, 97, 197, 199, 67, 101 and 29; respectively, which cannot be the product of the second group.
When n195, one of the prime factors of n is 13, the second group must contain the number 13, so the sum of all numbers after removing from the second group will be
(1 2 20) 195 13 2, then the second group must contain numbers 2 and 3,
but 2 13 195 , a contradiction!
When n196, two of the prime factors of n are 7, the second group must contain 7 and 14, but the sum of all the numbers in the first group is at most
(1 2 20) (7 14) 189 , again a contradiction!
When n198, one of the prime factors of n will be 11, then 11 must be included in the second group, but the sum of all numbers after removing from the second group will be (1 2 20) 198 11 1 , that is, 1 and 11 will be the two numbers in the second group, but 1 11 198 , a contradiction!
When n200, two of the prime factors of n are 5, so there are two numbers in second group that are multiple of 5, the sum of all the numbers in the first group is at most (1 2 20) (5 10) 195 , a contradiction!
When n204, sum of all the numbers in the second group cannot be more than
(1 2 20)2046, their product is obviously not equal to n, which is also a contradiction.
In summary, the maximum value of n is 192. 【Suggested Solution #2】
Since 4 6 8 192 1 2 3 5 7 9 10 11 20, then the largest possible value of n is at least 192.
There is another case: 1 2 3 4 8 192 5 6 7 9 10 11 20. If n192, then the number n must be expressed as the product of m positive integers such that the sum of these m distinct positive integers is at most 17.
From 1 2 3 4 5 6 21, we know that m6.
When m5, we have n 6 5 3 2 1 180, a contradiction! When m3, we have n 7 6 4 168, a contradiction! When m2, we have n 8 9 72, also a contradiction!
When m4, suppose one of the addends is in 17 is 1 with the largest possible product n 1 7 5 4 140, a contradiction!
The expressions are 8 4 3 2 , 7 5 3 2 and 6 5 4 2 with 192, 210 and 240 as their respective product. From those three products, only 210 and 240 are greater than 192.
But n 1 2 3 20210, then the product of these four positive integers cannot be equal to 210 or 240.
Now consider this case, when the sum of four positive integers is 16, then the largest possible product is n 6 5 3 2 180, which is also a contradiction!
In summary, the largest required number n is 192.
14. E is a point on the side BC and F is a point on the side CD of a square ABCD such that the perimeter of triangle CEF is equal to half the perimeter of ABCD. G is the point on AE such that FG is perpendicular to AE, and H is the point on FG such thatAH EF. Prove that AH is perpendicular to EF.
【Suggested Solution】
Refer to the diagram, extend CB to K such that BK DF
and connect AK. Connect AH to meet EF at L. Since
AB AD, ABK D 90 so that △ADF △ABK.
Hence, DFBK, AF AK, BAK DAF. It follows that
90
FAK FAB BAK FAB DAF DAB
. But the perimeter of △CEF is one-half the perimeter of square ABCD,
then EF 2AB(CECF)BEDFBEBK EK. But AE AE, AF AK, then △AFE△AKE.
Thus, 1 45
2
FAE EAK FAK
.
We know that FGAE, then △AGE is an isosceles right-angled triangle, so that
AGFG.
Hence, AH EF, FAK AGH FGE 90 and then △ AGH △FGE. So that FLH 180 FHL HFL180 AHG GAH 90 .
Therefore, AH EF.
【Marking Scheme】
Able to show thatEF EK, reward 5 marks. Able to show thatFAE 45 , reward 5 marks. Able to show that AG FG , reward 5 marks. Able to show thatAH EF, reward 5 marks.
15. Each side of an equilateral triangle is divided into 5 equal parts by 4 points, and these points are joined by lines parallel to the sides of the triangle, dividing into 25 small equilateral triangles. A tetriamond is a shape formed from 4 small equilateral triangles joined side to side. There are three tetriamonds as shown in the diagram below on the left.
A B C D E L F K G H
(a) Show that if 7 of the small triangles are painted, then it will be impossible to fit any tetriamond inside the large triangle without covering any part of the painted small triangles. (4 marks)
(b) Prove that if 6 of the small triangles are painted, then it is always possible to fit a tetriamond inside the large triangle without covering any part of the painted small triangles. (16 marks)
【Suggested Solution】
(a) Let us paint 7 of the small triangles black as shown in the figure below, then we cannot put any of the tetriamond in the big triangle by preventing covering any of black-shaded small triangles. When we flip and rotate the biggest triangle, and still is unsuccessful to put any of the tetriamond in the big triangle.
(b) Assuming the conclusion does not hold, then there exists a certain way to color the six of the small equilateral triangles such that it will be impossible to put the tetriamond inside the biggest triangle and will cover all the small equilateral triangle that were painted.
Refer to the figure at the lower left, in the two hexagons whose sides are bounded by red color, there will be at least two small equilateral triangles that must be colored, otherwise it will be easy to locate those tetriamonds that were inside them.
For those small equilateral triangle that were located inside the triangle whose sides were bounded by red, it is necessary to have at least a small equilateral triangle to be colored, for the concave hexagon whose sides are bounded red must also have at least one small equilateral triangle to be colored.
It means the all the 6 small equilateral triangles that will be colored must be inside the triangle whose sides were bounded by red, this implies the two small triangles located on the left side at the center cannot be colored, or else the number that will be colored will exceed 6 pieces.
yellow equilateral triangle that is not to be painted, so the little green equilateral triangle must have been painted, otherwise we can find four pieces of equilateral triangle, but there are 9 small green equilateral triangles, which is a contradiction! In summary, the selection of any 6 small equilateral triangles and color them, they must be able to place in a block of four equilateral triangles and will not cover any of those painted small equilateral triangles in the large equilateral triangle. QED.
【Marking Scheme】
(1) In the first sub-problem: if the method in painting black to the respective small triangles are correct, then reward 4 marks.
(2) For sub-problem:
Able to point out at least 2 small equilateral triangles in each of the two convex hexagons whose sides are bounded by red must have dig and removed, then reward 4 marks.
Able to point out that at least one of those small equilateral triangles that are inside those concave hexagons whose sides are bounded by red must dig and removed, then reward 4 marks.
Able to point out that at least one of those small equilateral triangles that are inside those triangles bounded by red color must be dig and removed, then rewards 4 marks.
Able to point out that all the small equilateral triangles painted green must be dig and removed, the reward 4 marks.