Some more problems...

Try the followings:

  • Consider a prism with a triangular base. The total area of the three faces containing a particular vertex AA is KK. Show that the maximum possible volume of the prism is K354\displaystyle \sqrt{\frac{K^3}{54}} and find the height of this largest prism.

  • Consider an n2×n2n^2 \times n^2 grid divided into n2n^2 subgrids of size n×nn \times n. Find the number of ways in which you can select n2n^2 cells from this grid such that there is exactly one cell coming from each subgrid, one from ach row and one from each column.

  • If a1,,a7(1,13)a_1, \cdots, a_7 \in (1,13) are not necessarily distinct reals, show that we can choose three of them such that they are lengths of the sides of a triangle.

  • Show that there cannot exist a non-constant polynomial P(x)Z[x]P(x) \in \mathbb{Z}[x] such that P(n)P(n) is prime for all positive integers nn.

  • (Calculus) Let ff be a twice differentiable function on the open interval (1,1)(-1,1) such that f(0)=1f(0)=1. Suppose that ff also satisfies f(x)0f(x) \geq 0 , f(x)0f'(x) \leq 0 and f(x)f(x)f''(x) \leq f(x), for all x0x \geq 0. Show that f(0)2f'(0) \geq - \sqrt{2}.

Note by A Brilliant Member
7 years, 3 months ago

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There must be a value of x(0,1)x \in(0,1) satisfying f(x0)=f(1)110f'(x_{0}) = \dfrac{f(1) - 1}{1-0}

For this xx , (f(x0))2=(1f(1))2(f'(x_{0}))^2 = (1 - f(1))^2


f(x)f(x)f(x)f(x)f(x)f(x)f''(x) \leq f(x) \Rightarrow f'(x) f''(x) \geq f'(x) f(x)


d((f(x))2)d((f(x))2)d((f'(x))^2) \geq d((f(x))^2)

Integrate in limits 0 to x0x_{0} to get:

(f(x0))2(f(0))2(f(x0))21(f'(x_{0}))^2 - (f'(0))^2 \geq (f(x_{0}))^2 - 1

Hence, (f(0))21+(1f(1))2(f(x0))2(f'(0))^2 \leq 1 +(1-f(1))^2 - (f(x_{0}))^2

Clearly, (f(0))22(f'(0))^2 \leq 2 ,or f(0)2f'(0) \geq - \sqrt{2}

jatin yadav - 7 years, 3 months ago

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A Brilliant Member - 7 years, 3 months ago

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hey paramjeet i cant get it can u explain me in detail

Shantanu Raut - 7 years, 3 months ago

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What needs explanation? And it's Paramjit. :)

A Brilliant Member - 7 years, 3 months ago

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Problem 3 is essentially USAMO 2012 Problem 1 in disguise. Can you see why?

Sreejato Bhattacharya - 7 years, 3 months ago

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