Week 6 Challenging Problems

Problems for the week

Deadline 19.07.2020

Problem 1.

Pooh bear has 2N+1 honey pots. No matter which one of them he sets aside, he can split the remaining 2N pots into two sets of the same total weight, each consisting of N pots. Must all 2N+1 pots weigh the same?

Problem 2.

Suppose P is a point in the interior of triangle ABC such that \angle PAB =\angle PBC=\angle PCA=30^{\circ}. Prove that ABC is equilateral.

Problem 3.

Suppose that f:[0,1]\to\mathbb{R} has a continuous second derivative with f''(x)>0 on (0,1), suppose that f(0)=0. Choose a\in (0,1) such that f'(a) <f(1). Show that there is a unique b\in (a,1) such that f'(a)=f(b)/b.

Problem 4.

A natural number is perfect if it is the sum of its proper divisors. Prove that two consecutive number cannot both be perfect.

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We will post the solutuions next week along with a new set of problem.

7 Comments

  • Problem 3 :-

    Considering the function g:[a,1] → |R such that
    g(x) = [f(x)/x – f'(a)]
    It would suffice to show that this function has a unique root.

    g(a)= [f(a)/a – f'(a)]
    Using Lagrange’s mean value theorem f(a)/a = f'(c) such that c belongs to (0,a), since f”(x)>0, the function f'(x) is an increasing function and since c<a , f'(c)<f'(a)
    And hence
    g(a)0

    hence the values of g(x) on its endpoints have negative and positive signs at a and 1 respectively
    And hence it would suffice to show that the function g(x) is increasing function

    g'(x) = [ xf'(x)–f(x)]/x²

    Now considering the numerator G(x)=xf'(x)–f(x)
    G'(x)= xf”(x) >0 for all x element of (a,1) and
    G(a) = a[f'(a)–f(a)/a] >0 [ since g(a)0

    And hence there is a unique root to the function g(x), let it be denoted by “b” and hence
    f(b)/b = f'(a)

    • Edit :-
      Problem 3 :-

      Considering the function g:[a,1] → |R such that
      g(x) = [f(x)/x – f'(a)]
      It would suffice to show that this function has a unique root.

      g(a)= [f(a)/a – f'(a)]
      Using Lagrange’s mean value theorem f(a)/a = f'(c) such that c belongs to (0,a), since f”(x)>0, the function f'(x) is an increasing function and since c<a , f'(c)<f'(a)
      And hence
      g(a) 0
      hence the values of g(x) on its endpoints have negative and positive signs at a and 1 respectively
      And hence it would suffice to show that the function g(x) is increasing function

      g'(x) = [ xf'(x)–f(x)]/x²

      Now considering the numerator G(x)=xf'(x)–f(x)
      G'(x)= xf”(x) >0 for all x element of (a,1) and
      G(a) = a[f'(a)–f(a)/a] >0 [ since g(a) < 0 ]

      And hence there is a unique root to the function g(x), let it be denoted by “b” and hence
      f(b)/b = f'(a)

  • Problem 2-
    Let 30+ x = A, 30+y =B and 30 +z=C
    As AP, BP,and CP can be considered as 3 concurrent cevians we can apply ceva’s theorem. Let AP when extended meet BC at D similarly BE and CF be other two cevians. From ceva’s theorem : (AF. BD. CE)/(BF.DC.AR)=1. Clearly this is equivalent to ([ACP].[BAP]. [CBP])/([BCP].[CAP].[PBA]) = 1 ; where [.] denotes the area.
    Now consider [ACP] /[BCP] =(1/2.AC.CPsin(ACP))/(1/2BC.CPsin(BCP)) = (AC. Sin(ACP) /(BC. Sin(BCP))
    Similarly in a cyclic manner multiply corresponding ratios of area and that is equal to 1 (from ceva)
    Side ratios will cancel each other out and we would get a trignometric form of ceva’s theorem.
    Now put the values of angles and we would get : sinx.siny.sinz=1/8 ( also x+y+z =90°) Now using AM GM and basic trignometric inequalities we would get x=y=z=30° each. That completes the proof.

  • Problem 3:

    f''(x) \gt 0 implies f'(x) is increasing on (0,1). Now by LMVT, there exists a c \in (0,1) such that f'(c) = \frac {f(1)-f(0)}{1 - 0}= f(1) (as f(0)=0). Since we have chosen a in such a way that f'(a)< f(1), this implies a < c as f'(x) is increasing.
    Again similarly by LMVT \exists  d \in (0,a) such that f'(d) =\frac {f(a)}{a}Again asd <a, f'(d)<f'(a).  Now define g(x) = \frac{f(x)}{x}on[a,1]. Sinceg(x)is continuous, it has the intermediate value property. As \frac {g(a)}{a} < f'(a) < \frac {g(1)}{1}, there exists a b \in (a,1)such that f'(a)=\frac {g(b)}{b}$ by IVT.

  • Where are the solutions posted?

  • Where are the solutions posted?

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