Problems for the week

Problem 1.

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

Problem 2.

Suppose is a point in the interior of triangle such that . Prove that is equilateral.

Problem 3.

Suppose that has a continuous second derivative with on , suppose that . Choose such that . Show that there is a unique such that .

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.

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

• Rooptak

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

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

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

Problem 3:

implies is increasing on . Now by LMVT, there exists a such that (as ). Since we have chosen in such a way that , this implies as is increasing.
Again similarly by LMVT f'(d) =\frac {f(a)}{a}d <a, f'(d)<f'(a) g(x) = \frac{f(x)}{x}[a,1]g(x) \frac {g(a)}{a} < f'(a) < \frac {g(1)}{1} b \in (a,1) f'(a)=\frac {g(b)}{b}\$ by IVT.

• Anish

Where are the solutions posted?

• Anish

Where are the solutions posted?

• The solutions were discussed with our students. We will post it publicly very soon.