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Home » Mathematics » High School Pre-Olympiad (Ages 14+)
 
Proof contest
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1=2
Birch & Swinnerton Dyer
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#181
here is what I have got:
-1\leq f(1)\leq 1
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PostPosted: Sat Oct 25, 2008 2:15 pm  Back to top 
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Temperal
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#182
hint
1+a+b+c=\prod^{3}_{i=1}(1-r_i)

PostPosted: Sun Oct 26, 2008 2:25 pm  Back to top 
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ghjk
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#183
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Since nobody gives a complete solution, I will present my approach:
From the hypothesis, applying Vieta's formula we have:
x_1+x_2+x_3= -a
x_1.x_2+x_2.x_3+x_3.x_1=b
x_1.x_2.x_3= -c
From the inequality given above we have: -2=<-(x_1+x_2+x_3)+x_1.x_2+x_2.x_3+x_3.x_1-x_1.x_2.x_3=<0
Rewriting it in this form: 0=<(1-x_1)(1-x_2)(1-x_3)+1=<2 (1), we will examine 4 cases:
Case 1: 2 roots<0, 1 root>2, then we easily see that (1-x_1)(1-x_2)(1-x_3)+1<0(contradicts to (1))
Case 2: If 3 roots<0, we have: (1-x_1)(1-x_2)(1-x_3)+1>2(contradicts to (1))
Case 3: If 1 root <0, 2 roots>2, we have: (1-x_1)(1-x_2)(1-x_3)+1>2(contradicts to (1))
Case 4: If 3 roots>2, then we easily see that (1-x_1)(1-x_2)(1-x_3)+1<0(contradicts to (1))
From 4 cases above, we can conclude that at least one root must lie on the interval [0,2](Q.E.D)
@chess64: Nice problem! I will try generalizing this problem for n roots also. And don't worry if no one answers your problem since everybody is very busy at this time(because of contests or homework). Just try to post as many hard+nice problems as you can. I believe that most of them will be solved by MLinkers Wink

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PostPosted: Sun Oct 26, 2008 6:28 pm  Back to top 
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chess64
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#184
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f(x) = x^3 + ax^2 + bx + c = (x-r)(x-s)(x-t) = 0.

f(1) = 1 + a + b + c = (1-r)(1-s)(1-t). (Temperal)

r \not \in [0,2] \implies r<0 \,\vee\, r>2 \implies |1-r| > 1.

\therefore r,s,t \not\in [0,2] \implies |(1-r)(1-s)(1-t)| > 1

\implies |1+a+b+c| > 1 \implies (a+b+c < -2) \vee (a+b+c > 0).

\therefore 0\le a+b+c \le 2 \implies \textsc{ (w.l.o.g.) } r \in [0,2].

PostPosted: Mon Nov 03, 2008 8:35 pm  Back to top 
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chess64
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#185
6. Find all continuous functions f : \mathbb{R} \to \mathbb{R} such that f(x)-f(y) is rational iff x-y is rational.

PostPosted: Sat Apr 11, 2009 10:20 am  Back to top 
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mathwizarddude
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#186
Rzeszut wrote:
It's a nice solution, because it is elementar. I solved this problem by complex numbers.
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Let the point O be the zero point on the complex plane and let the three vertices a,b,c of the trangle satisfy |a| = |b| = |c| = 1. Let's notice that the orthocenter corresponds with the number h = a + b + c, because the line passing through h,a is perpendicular to the line passing through b,c (analogously for other two vertices). It can be proved by fact that \frac {h - a}{b - c} = \frac {b + c}{b - c} is purely imaginary, because \frac {b + c}{b - c} + \overline{\left(\frac {b + c}{b - c}\right)} = 0 after some computations. Now we clearly see that |OH| = |h| = |a + b + c| = 3\cdot \left|\frac {a + b + c}{3}\right| = 3\cdot |OG|.

The (vector) sum of the sides of any triangle is 0, but then why would the othocenter be a+b+c (which means it's always at the origin, which doesn't seem correct)?

PostPosted: Sun Jun 07, 2009 2:03 pm  Back to top 
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Altheman
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#187
a,b,c are not vectors representing the sides, they are the vectors OA, OB,OC
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PostPosted: Sun Jun 07, 2009 11:02 pm  Back to top 
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Altheman
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#188
chess64 wrote:
5 The equation x^3 + ax^2 + bx + c = 0 has three real roots for some a,b,c \in \mathbb{R}. Show that if - 2 \le a + b + c \le 0, at least one of these roots lies in the interval [0,2].

Please turn in something by next friday.


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Let y=x-1 so x=y+1, then the equation is f(y)=y^3+(3+a)y^2+(3+2a+b)y+a+b+c+1 which also has three real roots. Rephrased, we must prove if |f(0)|\le 1, then a root lies in [-1,1]. Suppose that all the roots are outside that interval, then |r|,|s|,|t|>1. But 1\le |f(0)|=|rst|=|r||s||t|>1 which is a contradiction.
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PostPosted: Sun Jun 07, 2009 11:09 pm  Back to top 
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Bugi
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#189
mathwizarddude wrote:
Rzeszut wrote:
It's a nice solution, because it is elementar. I solved this problem by complex numbers.
Click to reveal hidden content
Let the point O be the zero point on the complex plane and let the three vertices a,b,c of the trangle satisfy |a| = |b| = |c| = 1. Let's notice that the orthocenter corresponds with the number h = a + b + c, because the line passing through h,a is perpendicular to the line passing through b,c (analogously for other two vertices). It can be proved by fact that \frac {h - a}{b - c} = \frac {b + c}{b - c} is purely imaginary, because \frac {b + c}{b - c} + \overline{\left(\frac {b + c}{b - c}\right)} = 0 after some computations. Now we clearly see that |OH| = |h| = |a + b + c| = 3\cdot \left|\frac {a + b + c}{3}\right| = 3\cdot |OG|.

The (vector) sum of the sides of any triangle is 0, but then why would the othocenter be a + b + c (which means it's always at the origin, which doesn't seem correct)?


There is a relation h+2o=a+b+c, and a,b,c are not OA,OB,OC they are vertexes A,B,C in complex plane. As o=0 then h=a+b+c

PostPosted: Tue Jun 09, 2009 5:55 am  Back to top 
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tiger on prowl
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#190
This is my solution to the functions problem:
replace x with y+1
we observe that x-y is rational(ie 1)
thus f(y+1)-f(y) must always assume rational values for all real x.
define g(x)=f(x+1)-f(x)

from continuity of f,continuity of g follows

CLAIM:g is a constant function
PROOF:if g were not constant.
say it assumes two values p,q for some real a,b(a<b)

since g is continuous on [a,b],there will exist a c in the interval [a,b] such that for any r in between p,q:
g(c)=r {intermediate value theorem}

choosing r as a real contradicts the fact that g always assumes rational values.
hence,our claim is proved.

therefore, g is constant.(infact rational constant)

NOW,we observe that similar proof can be extended to all functions h where
h(x)=f(x+C)-f(x) where C any rational of our choice.

thus h(x) is constant for any C(though it might be different constants for each rational C)

this strikes us an idea....
consider this expression:

{f(x+C)-f(x)}/C is a constant from what ever we have proved where C is rational

call this p(C)={f(x+C)-f(x)}/C

consider the sequence p(1),p(1/2),P(1/3),................
we observe that this sequence tends to the slope of tangent at x

observe that this sequence can be defined at every real x.
thus slope of tangent at every real x(limit of sequence) is a constant

therefore,f(x) must be a linear

thus,f(x)=ax+b where a is a rational and b is real



please consider this solution and tell me the flaws in it!

PostPosted: Tue Sep 08, 2009 7:14 am  Back to top 
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ocha
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#191
tiger on prowl wrote:


call this p(C)={f(x+C)-f(x)}/C

consider the sequence p(1),p(1/2),P(1/3),................
we observe that this sequence tends to the slope of tangent at x



have you assumed that the function is differentiable?

I think it can be fixed:
We can show that, given fixed c\in \mathbb{R}, f(x + c) - f(x) is constant for all x\in \mathbb{R}

Let g(c) = f(x + c) - f(x) which is continuous because f is continuous

Then g(a) - g(b) = f(x + a) - f(x + b) = f(x' - b + a) - f(x') = g(a - b) (substitution x = x' - b)

Hence we have a Cauchy equation g(a) - g(b) = g(a - b) which has solution g(x) = kx, k\in\mathbb{R} but since g(c) \in \mathbb{Q} iff c\in \mathbb{Q} we must have k\in \mathbb{Q}

So f(x + c) - f(x) = g(c) = kc let x = 0 then f(c) = kc + f(0)

\boxed{f(x) = kx + b} where k\in \mathbb{Q}, b\in \mathbb{R}

PostPosted: Wed Sep 09, 2009 11:44 pm  Back to top 
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praveen719
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#192
in this thread please help me also.......... prove mahematiclly that only one normal can be drawn from an outside point of the parabola and three from an inside point.
give your ans at 14+ pre olympiad forum

PostPosted: Thu Oct 15, 2009 12:46 am  Back to top 
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HeyThere1
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#193
good problems
these are good questions email me at progassman@gmail.com

PostPosted: Thu Nov 05, 2009 6:37 pm  Back to top 
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