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Mathematical Excalibur, Vol. 12, No. 2, May 07 – Aug. 07 Solution. Jeff CHEN (Virginia, USA).
cos A cos B cos C + + sin 2 A sin 2 B sin 2 C
= ≥
⎞ R ⎛ b2 a2 c2 b2 a2 c2 ⎜⎜ + + + + + − a −b − c⎟⎟ 2rs⎝ a b b c c a ⎠ R R by (*). ( a + b + c) = 2rs r
Commended solvers: CHEUNG Wang Chi (Singapore). Problem 274. Let n < 11 be a positive integer.
Let p1, p2, p3, p be prime
numbers such that p1 + p3n is prime. If
p1 + p2 = 3p, p2 + p3 = p ( p1 + p3 ) n 1
and p2> 9, then determine p1 p2 p3n . (Source: 1997 Hubei Math Contest) Solution. CHEUNG Wang Chi (Singapore), NG Eric Ngai Fung (STFA Leung Kau Kui College), YIM Wing Yin (HKU, Year 1) and Fai YUNG. Assume p1≥ 3. Then p1+p2 > 12 and 3p is even, which would imply p is even and at least 5, contradicting p is prime. So p1=2 and p2=3p−2. Modulo 3, the given equation p2 + p3 = p1n(p1+p3) leads to 0 ≡ 3p = p2+2 = 2n(2+p3)+2 = 2n+1 + 2 + (2n−1)p3 ≡ (−1)n+1+2+((−1)n−1)p3 (mod 3). The case n is even results in the contradiction 0 ≡ 1 (mod 3). So n is odd and we get 0 ≡ p3 (mod 3). So p3 = 3. Finally, the cases n = 1, 3, 5, 7, 9 lead to p1 + p3n = 5, 29, 245, 2189, 19685 respectively. Since 245, 19685 are divisible by 5 and 2189 is divisible by 11, n can only be 1 or 3 for p1+p3n to be prime. Now p2 = p1n(p1+p3) −p3 = 2n5−3 > 9 implies n = 3. Then the answer is p1 p2 p3n = 2 ⋅ 37 ⋅ 33 = 1998. Problem 275. There is a group of children coming from 11 countries (at least one child from each of the 11 countries). Their ages are from 7 to 13. Prove that there are 5 children in the group, for each of them, the number of children in the group with the same age is greater than the number of children in the group from the same country.
For i =7 to 13 and j = 1 to 11, let aij be the number of children of age i from country j in the group. Then 11
13
j =1
i =7
bi = ∑ aij ≥ 0 and c j = ∑ aij ≥ 1
are the number of children of age i in the group and the number of children from country j respectively. Note that 13
c j = ∑ aij = ∑ aij , where i =7
bi ≠ 0
∑
is
bi ≠0
used to denote summing i from 7 to 13 skipping those i for which bi=0. Now 11 ⎛1 1⎞ aij ⎜ − ⎟ ∑∑ ⎜c b ⎟ b ≠ 0 j =1 i ⎠ ⎝ j i
11
=∑
∑
bi ≠ 0
j =1
cj
11
13
j =1
i =7
aij
∑
11
−∑
bi ≠ 0
j =1
aij
Problem 8. A 2n×2n square is divided into 4n2 unit squares. What is the greatest possible number of diagonals of these unit squares one can draw so that no two of them have a common point (including the endpoints of the diagonals)?
From How to Solve It to Problem Solving in Geometry (II) (continued from page 2)
Idea: By examining the conditions given, we may see that the point C is not too important. C
bi
B'
≥ ∑1 − ∑1 = 4.
Since aij(1/cj − 1/bi) < aij/cj ≤ 1, there are at least five terms aij(1/cj − 1/bi) > 0. So there are at least five ordered pairs (i,j) such that aij > 0 (so we can take a child of age i from country j) and we have bi > cj.
Olympiad Corner (continued from page 1)
Problem 4. (Cont.) After that an obtused-angled triangle (or any of two right-angled triangles) is deleted and the procedure is repeated with the remained triangle. The player loses if he cannot do the next cutting. Determine, which player wins if both play in the best way. Problem 5. AA1, BB1 and CC1 are the altitudes of an acute triangle ABC. Prove that the feet of the perpendiculars from C1 onto the segments AC, BC, BB1 and AA1 lie on the same straight line. Problem 6. Given real numbers a, b, k (k>0). The circle with the center (a,b) has at least three common points with the parabola y = kx2; one of them is the origin (0,0) and two of the others lie on the line y=kx+b. Prove that b ≥ 2. Problem 7. Let x, y, z be real numbers greater than 1 such that xy 2 − y 2 + 4 xy + 4 x − 4 y = 4004, and xz 2 − z 2 + 6 xz + 9 x − 6 z = 1009. Determine all possible values of the expression xyz+3xy+2xz−yz+6x−3y−2z.
Q
A
P
B
P'
We will focus on how to represent the condition AB + BP = AQ + QB in the diagram. For that, we construct points P′ and B′ on AB and AQ extended respectively so that PB = P′B and QB′ = QB. Then
AB + BP = AQ + QB ⇒ AB + BP′ = AQ + QB′ ⇒ AP′ = AB′ ⇒AP′B′is equilateral (as∠B′AP′= 60o). Solution outline: (1) Let ∠ABQ = ∠QBP = θ. Since PB = P′B, we have ∠PP′B = θ. (2) Since AP bisects ∠QAB and ΔAB′P′ is equilateral, it follows that B′ is the reflected image of P′ about AP. So, PP′ = PB′ and ∠QB′P =∠AP′P = θ.
(3) Since QB = QB′ and ∠QBP = θ
=∠QB′P, by Example 2, P lies on either BB′ or the perpendicular bisector of BB′. If P does not lie on BB′, we will have PB = PB′ = PP′. This will imply ΔBPP′ is equilateral, θ = 60o and ∠QAB + ∠ABP = 60o + 2θ = 180o, which is absurd. So, P must lie on BB′. Therefore, B′ = C.
(4) Since QB=QB′=QC, ∠QCB =
∠QBC = θ. So ∠QAB + 2θ + θ = 180o ⇒ 60o + 3θ = 180o ⇒ θ = 40o. Therefore, ∠ABC = 80o, ∠ACB = 40o.