3 consecutive summing to 13

 You permute each of the digits $0, 1,2, \dots, 9$ and write them in a single row.

A) Show that no matter what the permutation, some 3 adjacent elements of the row sum to at least $13$.

B) Can you find a permutation where no 3 adjacent elements sum to more than $13$?

Scroll down for solution

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A) Say the permutation is $x_0, x_1, \dots, x_9$.

Now one of $x_0$ or $x_9$ is $< 9$. We can assume $x_0 < 9$.

Let $S_i = x_i + x_{i+1} + x_{i+2}$.

Since $x_0 < 9$ we must have that $S_1 + S_4 + S_7 > 36$ and thus $\text{max} \{S_1, S_4, S_7\} > 12$.

B)  9 3 1 7 4 2 6 0 5 8

Peter Winkler's factorial problem

 A cute problem from Peter Winkler's collection of math puzzles.

$$S = \{n! | 1 \leq n \le 100, n \in N\}$$

Can we remove a single element from $S$ such that the product of the elements of the resulting set is a perfect square?

Scroll down for a solution.

Product of elements of $S$ is

$$P = 1!  2! \dots 99!  100!$$

Pair up terms $(2n-1)! (2n)! = ((2n-1)!)^2 2n$

Thus

$$P = (1!  3!  5! \dots 99!)^2 (2 . 4 . 6 \dots 100)$$

$$=  (1!  3!  5! \dots 99!)^2 . 2^{50} . 50!$$

Thus removing $50!$ from $S$ will give us the desired result.

Volume of an n dimensional region

 What is the volume of the region in $R^n$ defined as follows

$$V_n = \{(x_1,x_2, \dots, x_n) \in R^n | x_i \ge 0 \text{ and } \sum x_i \leq 1 \}$$

Scroll down for a clever proof (if you know the source, please comment)

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Transform the space with $y_j = \sum_{1 \le i \le j} x_i$

This is a linear transformation that gives a new region defined as

$$W_n = \{(y_1,y_2, \dots, y_n) \in R^n | 0 \le y_1 \le y_2 \le \dots \le y_n \le 1 \}$$

$W_n$ is just a subset of the hypercube $[0,1]^n$. The hypercube can be split into $n!$ regions of equal volume, each region corresponding to a sort order among the coordinates. $W_n$ is one of them.

Thus volume of $W_n$ is $\frac{1}{n!}$

Since the determinant of the linear transformation from $V_n$ to $W_n$ is $1$, the volume of $V_n$ is $\frac{1}{n!}$ too!

Exactly 80% success

A basketball player is practising free throws. Their current success rate (ratio of successful throws to total) is exactly 70% (or 0.7 in terms of ratio). After a few more throws that success rate is 90%. 

Show that at some point the success rate was exactly 80%.

Scroll down for a solution

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Look at (misses, hits) on the integer lattice and x-y coordinate plane. If a miss occurs, we increment x coordinate, else we increment y coordinate.

Initially we are on the line $3y = 7x$ (70% hit rate) and reach the line $y = 9x$ (90% hit rate), crossing the line $y = 4x$ (80% hit rate) at some point, with some hit (i.e by incrementing the y-coordinate).,

The only way to cross the line $y = 4x$ vertically is to actually land on it first (every x = N line intersects y = 4x at (N, 4N) which is part of the integer lattice. Thus we achieve exactly 80%,

Product of three consecutive positive integers

 Can the product of three positive consecutive integers be a perfect square?

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Assume the three integers are $n-1, n, n+1$ and that $(n-1)n(n+1)  = n(n^2-1)$ is a perfect square.

Since $n$ is relatively prime to both $n-1$ and $n+1$ (and hence their product $n^2-1$), we must have that $n^2-1$ is a perfect square too.

A problem from INMO

 In a triangle $ABC$ (sides $a,b,c$ opposite $A,B,C$), angle $A$ is twice $B$.

Show that $$a^2 = b(b+c)$$

Try not to use trigonometry if possible.

Scroll down for a solution

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Let AD be the angular bisector of A, D lying on BC (might help to draw a figure).

Then by angular bisector theorem

$$BD = \frac{ac}{b+c}, DC = \frac{ab}{b+c}$$

BAD is isosceles, with $AD = BD$. Also triangle $ADC$ is similar to triangle $BAC$.

$AD/AB = DC/AC$ gives the result.

There are non-trigonometric proofs of the angular bisector theorem. For eg, prove for right angled triangles and use affine transform etc.