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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

1747.0. "1992 Olympiad" by RUSURE::EDP (Always mount a scratch monkey.) Tue May 04 1993 08:32

Article 43960 of sci.math:
From: [email protected] (W. Jose Castrellon G.)
Newsgroups: alt.uu.math.misc,k12.ed.math,sci.math
Subject: 1992 Olympiad Problems
Organization: The Ohio State University, Math.Dept.(student)
Lines: 34


Recently somebody asked in Benjie's IAMS for the problems of last year's
Olympiad. I got them today; here they are:

============================================================================

1. Find all integers   1 < a < b < c   such that  (a-1)(b-1)(c-1) is a 
   divisor of  abc-1.

2. Find all functions  f: R --> R  satisfying  f(x^2 + f(y)) = y + (f(x))^2

3. Consider  9  points in space, each pair joined by an edge, and all edges
   are colored either blue or red, or left uncolored. Find the smallest  n
   such that when  n  edges are colored, there is a subset of them containing
   a triangle of a single color.

4. Given a circle  C  on the plane together with a tangent  L  and a point
   M  on  L, find the locus of points  P  such that there exist two points
   Q  and  R  on  L  with  M the midpoint of QR  and  C  inscribed in the
   triangle  PQR .

5. Let  S  be a finite set of points in space. Let  Sx, Sy, Sz  be the
   projections of  S  on the  yz, xz, xy planes respectively. Prove that
   |S|^2  <  |Sx| |Sy| |Sz|  + 1

6. For each positive natural n,  S(n)  is the largest integer such that for
   all positive naturals  k < S(n) + 1 ,  n^2  can be written as the sum of
   k  non-zero squares.
   a. Prove that S(n) is not greater than   n^2 - 14 .
   b. Find  n  with   S(n) = n^2 - 14 .
   c. Show that there are infinitely many  n's  as in  b.

=============================================================================
T.RTitleUserPersonal
Name		DateLines
1747.1typing errors correctedHERON::BUCHANANThe was not found.Tue May 25 1993 08:2123
>3. Consider  9  points in space, each pair joined by an edge, and all edges
>   are colored either blue or red, or left uncolored. Find the smallest  n
>   such that when  n  edges are colored, there is a subset of them containing
>   a triangle of a single color.

	n = 33

Steps in proof:

	Suppose we've got 33 edges across 9 points.
	Then there's a clique of 6 points which form a complete graph.
	If we colour the edges of this subgraph blue/red, there must be a
monochromatic triangle.

	Alternatively, we know that we can colour a complete graph on 5 points
without admitting a monochromatic triangle.
	"Clone" 4 of the vertices, to yield 9 vertices. (One vertex is uncloned)
	"Clone" the corresponding edges.
	Between each vertex x and its clone x', draw an uncoloured edge (a
total of 4 uncoloured edges drawn.)
	Here's a graph with 32 coloured edges but no monochromatic triangle.

Andrew.
1747.22. f(x)=xTAV02::NITSANOne side will make you largerWed Jun 02 1993 16:4632
 > 2. Find all functions  f: R --> R  satisfying  f(x^2 + f(y)) = y + (f(x))^2

(1) f is a 1:1 function, because:
    f(a)=f(b) ==> f(5^2+f(a))=f(5^2+f(b)) ==> a+(f(5))^2=b+(f(5))^2 ==> a=b

(2) f covers all R, because for every t in R:
    t = t - (f(5))^2 + (f(5))^2 = f(5^2 + f(t-(f(5))^2))

(3) y+(f(x))^2 = f(x^2+f(y)) = f((-x)^2+f(y)) = y+(f(-x))^2 ==>
    ==> (f(x))^2 = (f(-x))^2 ==> f(x)=f(-x) or f(x)=-f(-x)
    If x/=0 (not equal 0) then x/=-x and by (1) we are left with f(x)=-f(-x)

(4) What is the z such that f(z)=0?
    z/=0 ==> f(-z)=-f(z)=0=f(z) ==> -z=z ==> z=0   Hence f(0)=0

(5) For every x: f(x^2) = f(x^2+f(0)) = 0+(f(x))^2 = (f(x))^2

(6) For every y: f(f(y)) = f(0^2+f(y)) = y+(f(0))^2 = y+0^2 = y

(7) For any x>=0:  f(x+y) =
     f(sqrt(x)^2+f(f(y))) = f(y)+(f(sqrt(x)))^2 = f(y)+f(sqrt(x)^2) = f(y)+f(x)

(8) Let x>=0. Let f(x)=x+a. We then get:
    x = f(f(x)) = f(x+a) = f(a)+f(x) = f(a)+x+a ==> 0 = f(a)+a ==> f(a) = -a
    If a>0 then: (f(sqrt(a)))^2  = f(sqrt(a)^2)  = f(a)  = -a < 0
    If a<0 then: (f(sqrt(-a)))^2 = f(sqrt(-a)^2) = f(-a) =  a < 0
    Both ways we get a contradiction, because something^2 cannot be < 0.
    Therefore a=0 and therefore f(x)=x.

    For x<0 also:  f(x) = -f(-x) = -(-x) = x

    Thus the only function satisfying the above is f(x)=x.
1747.3AUSSIE::GARSONnouveau pauvreWed Jun 02 1993 19:388
re .2
    
>(8) Let x>=0. Let f(x)=x+a. We then get:
    
    Since (4) shows that f(0) = 0, it follows immediately that a=0.
    
    However I don't see where you have really shown that f(x) is of the
    form x+a.
1747.4TAV02::NITSANOne side will make you largerThu Jun 03 1993 03:3513
re .3
    
> >(8) Let x>=0. Let f(x)=x+a. We then get:
>    
>     Since (4) shows that f(0) = 0, it follows immediately that a=0.
>    
>     However I don't see where you have really shown that f(x) is of the
>     form x+a.

    Space & font limitations. What (8) meant to say is: Suppose for some
    x we have f(x)=x+a, then for that specific x we get (by the way
    demonstrated in (8)) that a=0. Thus, for *any* x it follows that
    f(x)=x.