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China Mathematical
Competition
There are 8 short-answer questions and 3 word problems, which should be solved in 120 minutes with a full score of 120 marks, in the first round test of 2016 China Mathematical Competition (CMC). While the scope of these questions and problems does not exceed the teaching requirements and content stipulated in the “Mathematics Syllabus for Full-time Ordinary Senior Middle Schools” issued by the Ministry of Education in 2000, smarter methods are needed to deal with them, so that contestants’ mastery of basic knowledge and skills, as well as their abilities to comprehensively and flexibly apply these knowledge and skills in practice is examined. There are 4 word problems (including one in plane geometry), which should be solved in 150 minutes with a full score of 180 marks, in the second round test (Complementary Test) of CMC. Their scope is in line with the International Mathematical Olympiad, plus some content of the Mathematical Competition Syllabus.
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2 Mathematical Olympiad in China (2017–2018)
Part I Short-Answer Questions (Questions 1–8, eight marks each) 1 Given real number a satisfies a < 9 a^3 − 11 a < |a|, the range of a is .
Solution From a < |a| we have a < 0. Then the original inequalities become
1 >
9 a^3 − 11 a a
|a| a
or − 1 < 9 a^2 − 11 < 1. Therefore a^2 ∈
. Since a < 0 we have
a ∈
2 Suppose complex numbers z, w satisfy
|z| = 3 and (z + w)(z − w) = 7 + 4i,
where i is the imaginary unit, and z, w are conjugates of z, w, respec- tively. Then the modulus of (z + 2w)(z − 2 w) is.
Solution We have
7 + 4i = (z + w)(z − w) = |z|^2 − |w|^2 − (zw − zw).
Since |z|^2 and |w|^2 are real, and Re (zw − zw) = 0, then
|z|^2 − |w|^2 = 7 and zw − zw = −4i.
As |z| = 3, |w|^2 = 2. Then
(z + 2w)(z − 2 w) = |z|^2 − 4 |w|^2 − 2(zw − zw) = 9 − 8 + 8i = 1 + 8i.
Finally, we get the modulus of (z + 2w)(z − 2 w) is
12 + 8^2 =
3 None of three positive real numbers u, v, w being 1, if
loguvw + logvw = 5 and logv u + logw v = 3,
then the value of logwu is.
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4 Mathematical Olympiad in China (2017–2018)
Solution From ∠ABC = 90◦^ we know AC is the diameter of the bottom circle, whose center is denoted by O. Then P O⊥ plane ABC. Since AO = 1 2
AC = 1, we have P O =
AP 2 −AO^2 = 1.
Let H be the projection of M on the bottom surface. Then H is the midpoint of AO. If HK⊥BC is taken at point K in the bottom surface, then M K⊥BC is known from the three perpendicular theorem, so ∠M KH equals the dihedral angle M - BC-A. Since M H = AH =
and HK//AB, we have
HK
AB
HC
AC
, i.e.,
HK =^3 4
. Then tan∠M KH = M H HK
=^2
, which means the dihedral angle
M - BC-A is arctan^2 3
(^6) Given f (x) =
sin
kx 10
cos
kx 10
, where k is a positive integer. If for any real number a,
{f (x) | a < x < a + 1} = {f (x) | x ∈ R},
then the minimum of k is.
Solution We have
f (x) =
sin^2
kx 10
kx 10
− 2sin^2
kx 10
cos^2
kx 10
= 1 −
sin^2
kx 5
cos
2 kx 5
f (x) reaches the maximum if and only if x =^5 mπ k
(m ∈ Z). Since any open interval (a, a + 1) with length 1 contains at least one maximum point, so k > 5 π. On the other hand, when k > 5 π, any open interval (a, a + 1) will contains an entire period of f (x). Therefore, the minimum of k is [5π] + 1 = 16.
7 Let hyperbola C: x^2 −
y^2 3
= 1, with the left and right focal points being F 1 , and F 2 , respectively. Crossing F 2 draw a line that inter- sects with the right half of the hyperbola at points P and Q, mak- ing ∠F 1 P Q = 90◦. Then the inscribed circle radius of ΔF 1 P Q is .
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China Mathematical Competition 5
Solution By the property of a hyperbola, we know F 1 F 2 = 2 ×
1 + 3 = 4 and P F 1 − P F 2 = QF 1 − QF 2 = 2. Since ∠F 1 P Q = 90◦, then P F 12 + P F 22 = F 1 F 22. We have
P F 1 + P F 2 v =
2(P F 12 + P F 22 ) − (P F 1 − P F 2 )^2
2 × 42 − 22 = 2
Therefore, the inscribed circle radius of ΔF 1 P Q is
r =
(F 1 P + P Q − F 1 Q) =
(P F 1 + P F 2 ) −
(QF 1 − QF 2 )
8 Let a 1 , a 2 , a 3 , a 4 be four different integers among 1, 2 ,... , 100 , sat- isfying (a^21 + a^22 + a^23 )(a^22 + a^23 + a^24 ) = (a 1 a 2 + a 2 a 3 + a 3 a 4 )^2. Then the number of such ordered sequence (a 1 , a 2 , a 3 , a 4 ) is .
Solution By Cauchy inequality, we know (a^21 + a^22 + a^23 )(a^22 + a^23 + a^24 ) ≥ (a 1 a 2 + a 2 a 3 + a 3 a 4 )^2. The equality holds if and only if a^1 a 2
= a^2 a 3
= a^3 a 4
, so the original question is equivalent to: find the number of ordered (a 1 , a 2 , a 3 , a 4 ), such that {a 1 , a 2 , a 3 , a 4 } ⊆ { 1 , 2 , 3 ,... , 100 } and a 1 , a 2 , a 3 , a 4 is a geometric sequence. Now suppose the common ratio of the sequence is q = 1, where q is a rational number. Then we can write q = n m
, where m, n are positive integers satisfying m = n and (m, n) = 1.
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China Mathematical Competition 7
10 Given that f (x) is an odd function on R, f (1) = 1, and f
x x − 1
xf (x) for any x < 0, find the value of
f (1)f
f
f
f
Solution Define an = f
n
(n = 1, 2 , 3 ,.. .); then a 1 = f (1) = 1. For
x = −
k
(k ∈ N+), Note that
x x − 1 =
k −
k −^1
k + 1 ,^ and^ f^ (x) is an odd function.
We have
f
k + 1
k ·^ f
k
k ·^ f
k
and that means
ak+ ak
k
. Therefore,
an = a 1 ·
n∑− 1
k=
ak+ ak
n∑− 1
k=
k
(n − 1)!
Then ∑^50
i=
aia 101 −i =
∑^50
i=
(i − 1)! · (100 − i)!
∑^49
i=
i! · (99 − i)!
∑^49
i=
Ci 99 =
∑^49
i=
2 (C
i 99 + C
99 −i 99 )
×
× 299 =
11 As shown in Fig. 11.1, in the plane rectangular coordinate system xOy, point F on the positive semi-axis of the x-axis is the focus and O is the vertex of parabola C. Through P on C in the first quadrant and Q on the negative semi-axis of the x-axis, draw line PQ tangent
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8 Mathematical Olympiad in China (2017–2018)
Fig. 11.
Fig. 11.
to C, and |P Q| = 2. Circles C 1 and C 2 are tangent to line OP at P and are both tangent to the x-axis. Find the coordinates of F so that the sum of the areas of circles C 1 and C 2 takes the minimum value.
Solution Suppose the equation of parabola C is y^2 = 2px(p > 0), Q(−a, 0) (a > 0), and O 1 (x 1 , y 1 ), O 2 (x 2 , y 2 ) are the centers of C 1 , C 2 , respectively. Let the equation of line PQ be x = my − a (m > 0). Combining it with the parabola equation to eliminate x, we get
y^2 − 2 pmy + 2pa = 0.
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10 Mathematical Olympiad in China (2017–2018)
Let t = 1 − a^2. Then from 4t = 4 − 4 a^2 = 2pa > 0, we know t > 0. Therefore,
T =
(3t + 1)(t + 1) t
= 3t +
t
3 t · 1 t
The equality holds if and only if t =
. Then a =
1 − t = √( 1 −
, when πT reaches the minimum. Then by 1 we have
p 2
1 − a^2 a
t √ 1 − √^1 3
3 t √ 3 −
Therefore the coordinates of F is
√^1
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China Mathematical
Competition
While the scope of the test questions in the first round of the 2017 China Mathematical Competition does not exceed the teaching requirements and content specified in the “General High School Mathematics Curriculum Standards (Experiments)” promulgated by the Ministry of Education of China in 2003, the methods of proposing the questions have been improved. The emphasis is placed on testing the students’ basic knowledge and skills, and their abilities to integrate and flexibly use them. Each test paper includes eight fill-in-the-blank questions and three answer questions. The answer time is 80 minutes, and the full score is 120 points. The scope of the test questions in the second round (Complementary Test) is in line with the International Mathematical Olympiad, with this some expanded knowledge, plus a few contents of the Mathematical Compe- tition Syllabus. Each test paper consists of four answer questions, including a plane geometry one, and the answering time is 150 minutes. The full score is 180 points.
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China Mathematical Competition 13
Solution Clearly, we have A(3, 0) and F (0, 1). Let the coordinates of P be
(3 cos θ,
10 sin θ), θ ∈
0 , π 2
Then
SOAP F = SΔOAP + SΔOF P =
10 sin θ +
· 1 · 3 cos θ
10 sin θ + cos θ) =
sin(θ + ϕ),
in which ϕ = arctan
When θ = arctan
10, quadrilateral OAPF has the maximal area as 3
4 A three-digit positive integer is called a “stable number” if any two consecutive digits differ by at most 1. The number of stable numbers is.
Solution Let abc be a stable number. If b = 0, then a = 1, c ∈ { 0 , 1 }, and there are 2 stable numbers. If b = 1, then a ∈ { 1 , 2 }, c ∈ { 0 , 1 , 2 }, and there are 2 × 3 = 6 stable numbers. If 2 ≤ b ≤ 8, then a, c ∈ {b − 1 , b, b + 1}, and there are 7 × 3 × 3 = 63 stable numbers. If b = 9, then a, c ∈ { 8 , 9 }, and there are 2 × 2 = 4 stable numbers. Therefore, the number of stable numbers is 2 + 6 + 63 + 4 = 75.
5 In tetrahedron P - ABC, AB = 1, AP = 2. Plane α through AB cuts the tetrahedron into two solids with equal volumes. Then the cosine of the angle between line PC and plane αequals.
Solution Let K, M be the midpoints of AB, PC, respectively. Clearly, plane ABM is exactly plane α. By the Apollonius’s theorem,
AM^2 =^1 2
(AP^2 + AC^2 ) − 1
PC^2 =^1
(2^2 + 1^2 ) − 1
× 22 =^3
and hence
KM =
AM 2 − AK^2 =
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14 Mathematical Olympiad in China (2017–2018)
Furthermore, MK is the projection of line PC on α, MC = 1, KC =
Therefore
cos ∠KMC =
KM 2 + M C^2 − KC^2
2 KM · M C
4 + 1^ −^
√^4
i.e., the angle between line PC and plane α has cosine value
6 A plane point set is given by K = {(x, y) | x, y = − 1 , 0 , 1 }. Choose 3 points from K at random. Then the probability that there are two points among them with distance
5 is.
Solution There are 9 points in K, and the number of ways of choosing 3 points from K is C^39 = 84. Let the points be A 1 , A 2 ,... , A 8 and O, as shown in the figure below.
There are 8 pairs of points at distance
- Let us choose one pair among them. By symmetry, we may assume the chosen pair are A 1 and A 4. Since there are 7 ways to choose the other point, there are 7 × 8 = 56 unordered triples. On the other hand, each Ai(i = 1, 2 ,... , 8) is at distance
5 from points Ai+3, Ai+5 (indices modulo 8). This implies that there are 8 triples {Ai, Ai+3, Ai+5}(i = 1, 2 ,... , 8)
being counted twice. Hence the total number of unordered triples is 56 − 8 = 48, and the probability is^48 84
=^4
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16 Mathematical Olympiad in China (2017–2018)
Solution Clearly, a 1 , a 2 , b 1 are positive integers, and a 1 < a 2. Since 2017 > b 10 = 2^9 · b 1 = 512b 1 , b 1 ∈ { 1 , 2 , 3 }. Repeatedly use the recurrence formula of {an} to derive
a 10 = a 9 + a 8 = 2a 8 + a 7 = 3a 7 + 2a 6 = 5a 6 + 3a 5 = 8a 5 + 5a 4 = 13a 4 + 8a 3
= 21a 3 + 13a 2 = 34a 2 + 21a 1.
Thus
21 a 1 ≡ a 10 = b 10 = 512b 1 ≡ 2 b 1 (mod 34).
As 13 × 21 = 34 × 8 + 1, we have
a 1 ≡ 13 × 21 a 1 ≡ 13 × 2 b 1 = 26b 1 (mod 34). 1
On the other hand, notice that a 1 < a 2 , which implies 55a 1 < 34 a 2 + 21 a 1 = 512b 1 , or
a 1 <
b 1. 2
If b 1 = 1, 1 , 2 become a 1 ≡ 26(mod 34), a 1 <
; no integer solution exists. If b 1 = 2, 1 , 2 become a 1 ≡ 52(mod 34), a 1 <
55 ; we have a unique solution a 1 = 18, and a 1 + b 1 = 20. If b 1 = 3, 1 , 2 become a 1 ≡ 78(mod 34), a 1 < 1536 55
; we have a unique solution a 1 = 10, and a 1 + b 1 = 13. As a summary, a 1 + b 1 could be 13 or 20.
Part II Word problems (Questions 9–11, 56 marks in total for three questions) 9 (16 marks) Let k, m be real numbers. If |x^2 − kx − m| ≤ 1 holds for all x ∈ [a, b], prove b − a ≤ 2
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China Mathematical Competition 17
Solution Let f (x) = x^2 − kx − m, x ∈ [a, b]. Then f (x) ∈ [− 1 , 1], and in particular f (a) = a^2 − ka − m ≤ 1 , 1
f (b) = b^2 − kb − m ≤ 1 , 2
f
a + b 2
a + b 2
− k ·
a + b 2
− m ≥ − 1. 3
Take 1 + 2 − 2 × 3 to get (a − b)^2 2
= f (a) + f (b) − 2 f
a + b 2
Hence b − a ≤ 2
10 (20 marks) Let x 1 , x 2 , x 3 be nonnegative real numberss satisfying x 1 + x 2 + x 3 = 1. Find the maximum and minimum values of (x 1 + 3x 2 + 5x 3 )
x 1 +
x 2 3
x 3 5
Solution By Cauchy inequality, we have (x 1 + 3x 2 + 5x 3 )
x 1 +
x 2 3
x 3 5
x 1 ·
x 1 +
3 x 2 ·
x 2 3
5 x 3 ·
x 3 5
= (x 1 + x 2 + x 3 )^2 = 1, where the equality holds at x 1 = 1, x 2 = 0, x 3 = 0. Therefore, the minimum value is 1. On the other hand, (x 1 + 3x 2 + 5x 3 )
x 1 +
x 2 3
x 3 5
(x 1 + 3x 2 + 5x 3 )
5 x 1 +
5 x 2 3
[
(x 1 + 3x 2 + 5x 3 ) +
5 x 1 +
5 x 2 3
)] 2
6 x 1 +
x 2 + 6x 3
(6x 1 + 6x 2 + 6x 3 )^2 =^9 5
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China Mathematical Competition 19
Test Paper B, the First Round
Part I Short-Answer Questions (Questions 1–8, eight marks each)
1 Suppose a geometric sequence {an} satisfies a 2 =
2, a 3 = 3
Then
a 1 + a 2011 a 7 + a 2017 equals^.
Solution The common ratio of {an} is q = a^3 a 2
, and hence
a 1 + a 2011 a 7 + a 2017
a 1 + a 2011 q^6 (a 1 + a 2011 )
q^6
2 Let complex number z satisfy the equation z + 9 = 10¯z + 22i. Then |z| equals.
Solution Let z = a + bi, a, b ∈ R. Clearly,
(a + 9) + bi = 10a + (− 10 b + 22)i.
Comparing the real and the imaginary parts, we have
a + 9 = 10a, b = − 10 b + 22, which gives a = 1, b = 2, z = 1 + 2i. So |z| =
3 Let f (x) be defined on R, such that f (x) + x^2 is an odd function, and f (x) + 2x^ is an even function. Then f (1) equals.
Solution By definition of odd and even functions, we have
f (1) + 1 = −[f (−1) + (−1)^2 ] = −f (−1) − 1 ,
f (1) + 2 = f (−1) +
Adding the above equations to cancel f (−1), it follows 2f (1) + 3 = − 1 2
, or
f (1) = − 7 4
4 In ΔABC, sin A = 2 sin C, and the side lengths a, b, c form a geo- metric progression. Then cos A equals.
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20 Mathematical Olympiad in China (2017–2018)
Solution By the law of sines, a c
= sinA sinC
= 2. From b^2 = ac, we have a : b : c = 2 :
2 : 1. Then by the law of cosines,
cos A =
b^2 + c^2 − a^2 2 bc
2)^2 + 1^2 − 22
2 ×
2 × 1
5 Given regular tetrahedron ABCD, points E, F are on AB, AC, respectively, such that BE = 3, EF = 4, and EF is parallel to plane BCD. Then the area of ΔDEF is.
Solution Clearly, EF is parallel to BC. Since all the faces of ABCD are equilateral triangles, we have
AE = AF = EF = 4, AD = AB = AE + BE = 7.
By the law of cosines,
DE =
AD^2 + AE^2 − 2 AD · AE · cos 60◦
=
and DF =
37 likewise. Let DH be an altitude of isosceles ΔDEF, H being the foot on EF, as shown in the figure.
Then EH =
2 EF^ = 2, and
DH =
DE^2 − EH^2 =
We conclude SΔDEF =^1 2
· EF · DH = 2
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