Week 6 Challenging Problems

Fractions > 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

Rooptak

July 11, 2020 at 2:04 pm Reply

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)
- Rooptak
  
  July 13, 2020 at 12:14 am Reply
  
  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)
Uditanshu

August 5, 2020 at 6:50 pm Reply

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.
Leo

August 7, 2020 at 9:46 am Reply

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} $Again as$ d <a, f'(d)<f'(a) $. Now define$ g(x) = \frac{f(x)}{x} $on$ [a,1] $. Since$ g(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.
Anish

December 23, 2020 at 1:01 pm Reply

Where are the solutions posted?
Anish

December 23, 2020 at 1:01 pm Reply

Where are the solutions posted?
- FractionsHub
  
  December 25, 2020 at 1:48 am Reply
  
  The solutions were discussed with our students. We will post it publicly very soon.