"OBRIGADO DEUS PELA VIDA,PELA MINHA FAMILIA,PELO TRABALHO,PELO PÃO DE CADA DIA,PROTEGENOS DO MAL"

Matematicas preuniversitarias,fisica preuniversitaria,algebra,geometria,trigonometria
mathematics,physics,geometry,Математика,College,Pre-College,vestibular universidades,olimpiadas de matematicas,Mathematical Olympiad,Algebra Problems,Geometry Problems,High School Geometry,Trigonometry Problems,Descriptive Geometry,Problems In Calculus Of One Variable,ECUACIONES DIFERENCIALES,problemas de fisica,Problems On Physics,Linear Algebra,Problems In Elementary Mathematics,Inequalities,Mathematics for high school students,EXAMENS DE ADMISION ALGEBRA.
   

https://picasion.com/
https://picasion.com/

BLOG DO ENG. ARMANDO CAVERO MIRANDA -BRASIL


terça-feira, 1 de outubro de 2013

19th Chinese Mathematical Olympiad 2004 Problems & Solutions

A1.  ABCD is a quadrilateral. E, F, G, H are points on AB, BC, CD, DA respectively such that (AE/EB)(BF/FC)(CG/GD)(DH/HA) = 1. The lines through A parallel to HE, through B parallel to EF, through C parallel to FG, and through D parallel to GH form a quadrilateral E1F1G1H1 (E1 is the intersection of the lines through A and B, F1 is the intersection of the lines through B and C, and so on). Find CF1/CG1 in terms of AH1/AE1.
A2.  a1/, a2, a3, ... is a sequence of integers with a1 positive and an = [(1 + 2/n)(an-1 - 1)] + 1. Find an in terms of a1 and n.
A3.  S is a set of n points. It contains a convex 7-gon and given any convex pentagon in S, there is a point of S inside it. Find the minimum possible value of n. (Note that 5 points are not considered to form a convex pentagon if any of them lies in the convex hull of the other 4, so in particular they cannot form a convex pentagon if three of them are collinear.)
B1.  a is a real. The sequence of reals x0, x1, ... , xn+1 satisfies x0 = xn+1 = 0 and (xi-1 + xi+1)/2 = xi + xi3 - a3 for i = 1, 2, ... , n. Show that the sequence is uniquely determined by n and a and that |xi| ≤ |a|.
B2.  a1 < a2 < ... < an is a sequence of positive integers with ∑ 1/ai ≤ 1. Prove that (∑ 1/(ai2 + x2) )2 ≤ 1/(2a12-2a1+2x2) for any real x.
B3.  Show that any sufficiently large integer can be expressed as a sum of distinct positive integers a1, a2, ... , a2004 such that each ai is divisible by its predecessor. 

Solutions

Problem A1
ABCD is a quadrilateral. E, F, G, H are points on AB, BC, CD, DA respectively such that (AE/EB)(BF/FC)(CG/GD)(DH/HA) = 1. The lines through A parallel to HE, through B parallel to EF, through C parallel to FG, and through D parallel to GH form a quadrilateral E1F1G1H1 (E1 is the intersection of the lines through A and B, F1 is the intersection of the lines through B and C, and so on). Find CF1/CG1 in terms of AH1/AE1.
Answer
CF1/CG1 = 1/(AH1/AE1)
Solution
Let the diagonals AC, BD meet at O. Suppose the lines EF and AC meet at P (we consider later what happens if they are parallel). Then the heights of A and C above EF are proportional to PA and PC, so area AEF/area CEF = PA/PC. But area AEF/area BEF = AE/BE, and area CEF/area BEF = CF/BF, so PA/PC = (AE/EB)(BF/CF). Similarly, if GH meets AC at Q, then QA/QC = area AGH/area BGH = (AH/DH)(GD/CG). But (AE/EB)(BF/CF) = (AH/DH)(GD/GC), so PA/PC = QA/QC and hence P and Q coincide.
Let E1F1 and G1H1 meet at R. Then area PAD/area PDH = AD/DH = area ARH1/area HRH1. Similarly, area PBA/area PBE = BA/BE = area ARE1/area ERE1. Hence (area PAD/area PDH)(area PBE/area PBA) = (area ARH1/area HRH1)(area ERE1/area ARE1). But area PAD/area PBA = DO/BO and area ARH1/area ARE1 = AH1/AE1, so (DO/BO)(area PBE/area PDH) = (AH1/AE1)(area ERE1/area HRH1). Now the bases PE and RE1 of PBE and ERE1 are parallel and the heights are the same, so area PBE/area ERE1 = PE/RE1. Similarly, the bases PH and RH1 of PDH and HRH1 are parallel and the heights are the same, so area PDH/area HRH1 = PH/RH1. Also triangles PHE and RH1E1 are similar (corresponding sides are parallel, so the angles are equal). Hence PE/RE1 = PH/RH1. Hence AH1/AE1 = DO/BO.
A similar argument gives CG1/CF1 = DO/BO. Hence AH1/AE1 = CG1/CF1.
It remains to consider the case where EF is parallel to AC. In that case AE/EB = CF/FB. Hence AH/DH = CG/GD, so GH is also parallel to AC. Hence E1F1 and H1G1 are parallel to AC. So AH1/E1H1 = area AH1G1/area E1H1G1 = area CH1G1/area F1H1G1 = CG1/F1G1. Hence AH1/AE1 = CG1/CF1.
Thanks to Polly T Wang

Problem A2
a1/, a2, a3, ... is a sequence of integers with a1 positive and an = [(1 + 2/n)(an-1 - 1)] + 1. Find an in terms of a1 and n.
Answer
If a1 = 3k, then an = 1 + [(n+2)2/4] + (k-1)(n+1)(n+2)/2
if a1 = 3k+1, then an = 1 + k(n+1)(n+2)/2
if a1 = 3k+2, then an = (n+1)(1 + k(n+2)/2)
Solution
It is convenient to put bn = an - 1, so bn = [bn-1(n+2)/n]. It is a trivial induction that bn = k(n+1)(n+2)/2 is a solution for k a positive integer. Note that in this case we do not need the [ ], so we can add this solution to others. It gives b1 = 3k and hence a1 = 3k+1.
We claim that bn = n is also a solution. For [(n-1)(n+2)/n] = [(n2+n-2)/n] = [n+1-2/n] = n (for n ≥ 2). We can add this to the previous solution to get bn = n + k(n+1)(n+2)/2 which gives b1 = 3k+1 and hence a1 = 3k+2.
Finally, we claim that bn = [(n+2)2/4] is also a solution. That implies b2m = (m+1)2 and b2m-1 = m(m+1). Then b2m+1 = [(m+1)2(2m+3)/(2m+1)] = [(m+1)2 + 2(m+1)2/(2m+1)] = [(m+1)2 + (m+1)(2m+1)/(2m+1) + (m+1)/(2m+1)] = [(m+1)(m+2) + (m+1)/(2m+1)] = (m+1)(m+2). Also b2m+2 = [(m+1)(m+2)(2m+4)/(2m+2)] = [(m+2)(m+2)] = (m+2)2, which proves the claim. It gives b1 = 2. Adding to the first solution we get the solution bn = [(n+2)2/4] + (k-1)(n+1)(n+2)/2 which gives b1 = 3k-1 and hence a1 = 3k.
The sequence is obviously determined by a1 and the recurrence rule, so we have found solutions for all positive integers a1


Problem A3
S is a set of n points. It contains a convex 7-gon and given any convex pentagon in S, there is a point of S inside it. Find the minimum possible value of n. (Note that 5 points are not considered to form a convex pentagon if any of them lies in the convex hull of the other 4, so in particular they cannot form a convex pentagon if three of them are collinear.)
Answer
11
Solution
Thanks to Polly T Wang
We show that there must be at least 4 points inside the convex 7-gon.
There must be at least one, because there must be a point inside the convex hull formed by 5 of the points. If there is just one, X, then take a line through the point which does not contain any other points in the set. At least 4 points must lie on one side of the line. With X they form a convex pentagon, which must have a point inside. Contradiction.
If there are just two, X and Y. Take the line through X and Y. It can pass through at most 2 of the 7 points, so at least 3 of the 7 points must lie on one side of the line. With X and Y they form a convex pentagon, which must have a point inside. Contradiction.
Finally, suppose there are just three, X, Y and Z. Let HZ be the open half-plane on the opposite side of XY to Z. Similarly, let HY be the open half-plane on the opposite side of ZX to Y, and HX the open half-plane on the opposite side of YZ to X. Then HX ∪ HY ∪ HZ is the whole plane outside XYZ. So all 7 vertices of the convex 7-gon belong to HX ∪ HY ∪ HZ. Hence one of HX, HY, HZ contains at least 3 vertices of the 7-gon. With the two points on the line they form convex 5-gon. Contradiction.
So we have shown that 11 points are necessary. The diagram shows that they are sufficient.

Problem B1
a is a real. The sequence of reals x0, x1, ... , xn+1 satisfies x0 = xn+1 = 0 and (xi-1 + xi+1)/2 = xi + xi3 - a3 for i = 1, 2, ... , n. Show that the sequence is uniquely determined by n and a and that |xi| ≤ |a|.
Solution
Suppose first a = 0. Suppose some xi > 0. Then take xj to be the largest xi. We have xj > 0 and hence j ≠ 0 or n+1. So (xj-1 + xj+1)/2 = xj + xj3 > xj. So xj-1 or xj+1 > xj. Contradiction. So all xi ≤ 0. Similarly, if some xi < 0, the take xj to be the smallest. But then (xj-1 + xj+1)/2 = xj + xj3 < xj. Contradiction. So all xi ≥ 0. Hence all xi = 0. It is easy to check that for a = 0, x0 = x1 = ... = xn+1 = 0 satisfies the conditions and so is the unique solution for any n.
Now suppose a > 0. Suppose some xi < 0. Then take xj to be the smallest xi. But then (xj-1 + xj+1)/2 = xj + xj3 - a3 < xj. Contradiction. So all xj ≥ 0. Suppose some xi = 0 for i ≠ 0 or n+1. Then (xi-1 + xi+1)/2 = xi + xi3 - a3 = -a3 < 0, so xi-1 or xi+1 < 0. Contradiction. Thus all the terms except the first and last are positive. Let xj be the largest term. Then we must have xj ≤ a, because otherwise (xj-1 + xj+1)/2 = xj + xj3 - a3 > xj. Contradiction. In fact we must have xj < a, because if xj = a, then (xj-1 + xj+1)/2 = xj + xj3 - a3 = xj and hence xj-1 = xj+1 = a. Repeating, we get eventually x0 or xn+1 = a. Contradiction.
Now suppose x1 = x. We ignore for the moment the requirement that xn+1 = 0. We show by induction on k that xk and xk-xk-1 are continuous, strictly increasing functions of x. That is obvious for k=1. Put k=2. We have x2 = 2x + 2x3 - 2a3, and x2-x1 = x + 2x3 - 2a3, which are both continuous, strictly increasing functions of x. Suppose it is true for k. Then xk+1 = 2xk + 2xk3 - 2a3 - xk-1 = xk + 2xk3 - 2a3 + (xk - xk-1) and xk+1 - xk = 2xk3 - 2a3 + (xk - xk-1). Since both xk and (xk - xk-1) are both continuous, strictly increasing functions of x, so are xk+1 and (xk+1 - xk).
In particular, if follows that xn+1 is a continuous, strictly increasing function of x. Put xn+1 = f(x). Now if we take x = a, then x2 = 2a > a, and by a trivial induction xk > a for all k > 1. So, in particular, f(a) > a. On the other hand, if we take x = 0, then x2 = -2a3 < 0, and by a trivial induction xk < 0 for all k > 1. So, in particular f(0) < 0. Thus there must be a unique x in the interval (0,a) such that f(x) = 0. Since x uniquely determines the entire sequence, we have established that the entire sequence is uniquely determined by n and a.
Now suppose a < 0. If xi is the sequence uniquely determined by n and -a, then it is clear that the sequence -xi is a solution for n and a. Conversely, if yi is a solution for n and a, then -yi is a solution for n and -a and hence must be the unique sequence xi. So the result is also true for a negative. 

Problem B2
a1 < a2 < ... < an is a sequence of positive integers with ∑ 1/ai ≤ 1. Prove that (∑ 1/(ai2 + x2) )2 ≤ 1/(2a12-2a1+2x2) for any real x.
Solution
Appying Cauchy-Schwartz, we have lhs = (∑ (√ai/(ai2+x2))(1/√ai) )2 ≤ (∑ ai/(ai2+x2)2 )(∑ 1/ai) ≤ ∑ ai/(ai2+x2)2.
Now (ai2+x2)2 > (ai2+x2)2 - ai2 > 0, so ∑ ai/(ai2+x2)2 < (1/2) ∑ (1/(ai2+x2-ai) - 1/(ai2+x2+ai) ).
Now ai+1 ≥ ai+1, so ai+12-ai+1+x2 ≥ ai2+ai+x2 and hence ∑ (1/(ai2+x2-ai) ≤ 1/(a12+x2-a1) + (∑ 1/(ai2+x2+ai) ) - 1/(an2+x2+an) ≤ 1/(a12+x2-a1) + (∑ 1/(ai2+x2+ai) ). Hence lhs < (1/2) 1/(a12+x2-a1), as required. 

Thanks to Li Yi

WEBSITE ORIGINAL SOURCE
http://www.kidsmathbooks.com/2011/01/19th-chinese-mathematical-olympiad-2004.html 

Nenhum comentário:

Postar um comentário