An integral with $\frac{1}{\log x}$ [Solution]

The problem was:

Suppose $n$ is a positive integer (though the result below does not really need that).

Show that

$$ \int_{0}^{1} \frac{x^n - 1}{\log x} \text{d}x = \log(n+1)$$

Note that the $\log x$ is the $\log$ to base $e$.

Solution

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This can be solved by the neat trick of differentiating under the integral sign.


Let

 $$ f(z) = \int_{0}^{1} \frac{x^z - 1}{\log x} \text{d}x$$

Differentiating under the integral sign gives us

 $$ f'(z) = \int_{0}^{1} \frac{d \frac{x^z - 1}{\log x}}{dz} \text{d}x$$

and so

 $$ f'(z) = \int_{0}^{1} \frac{x^z \log x}{\log x} \text{d}x  = \int_{0}^{1} x^z \text{d}x  = \frac{1}{z+1}$$

Thus $f(z) = \log(1 + z)$, since $f(0) = 0$.

[Note: There was some handwaving and the right theorems need to be applied, and right bounds on $z$ need to be assumed etc. That is left to the reader]

Don't cross the streams [Solution]

The problem: http://ruffnsluff.blogspot.com/2017/02/dont-cross-streams.html

$N$ red and $N$ blue 2D points, no three collinear. Show that we can pair them off  (red with blue) such that the line segment don't cross.

Solution

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This has an elegant existential solution (don't know the source).

Of all the possible pairings, pick the pairing which minimizes the sum of the lengths of line segments. No two line segments of this will cross!

Suppose $A,B,C,D$ are points with $A,B$ red and $C,D$ blue such that $AC$ and $BD$ cross (say at $E$).

In the triangles $BEC$ and $AED$ we have that $BE + EC \gt BC$ and $AE + ED \gt AD$ and so $AC + BD \gt AD + BC$.

Uncrossing will reduce the sum of the lengths of the line segments.

There are also constructive solutions, which lead to $O(n^2\log n)$ time algorithms.

$r^{th}$ roots of sides of a triangle

Prove or disprove:

If $a, b, c$ are sides of a triangle, then so are their $r^{th}$ roots $a^{1/r}, b^{1/r}, c^{1/r}$ where $r$ is a real number $\gt 1$.

Handling a 5-1 trump split

This is a first round hand from yesterday's open KO at the Bothell memorial day sectional.

Partner opens a 15-17 NT and you end up in 6C. LHO leads a high diamond spot (likely a doubleton/singleton).

This is what you see:

IMPS None	North ♠ AQ82 ♥ T2 ♦ KJ92 ♣ AJ7
	South ♠ K3 ♥ AKJ97 ♦ AT ♣ KT54

W	N	E	S
	1NT	P	2D1
P	2H	P	3C
P	4C2	P	4NT
P	5H3	P	6C4
P	P	P

1  transfer
2: ?
3: 2 Keycards, no Q
4: 6NT is better perhaps.

Anyway, LHO leads a high diamond spot, and RHO follows low.

You win the DT, and play a club to the J which wins! Now you cash the CA, RHO shows out, throwing a diamond.

How will you play?




If LHO is exactly 3=3=2=5 then you can make.

Cash AK heart, ruff a heart. Unblock DA, cash three rounds of spades (throwing a heart).  You are left with Hx, CKT in hand while LHO has to follow suit all along and is now left with CQ9x.

Now play the DK and  throw the last heart. LHO has to ruff this and lead into your KT.

So did we win IMPS on this? (or at least not lose big to 6NT)

No, we were actually missing the CT, so 6C was down one. LHO was indeed 3=3=2=5, so if we had the CT we would have made.

The other table was in 3NT.

Perhaps there are additional chances you have catered to? If so, please comment.

Odd points even triangles [Solution]

The problem was

You are given a set $S$ of $2005$ points (no three collinear) in the 2D plane. For each point $P$ in $S$, you count the number of triangles (formed by points in $S$) within which $P$ lies ($P$ must be strictly inside the triangle).

Show that this number is even, irrespective of the point $P$.


A possible solution


Given a point $P$ form a graph $G$ of the triangles that contain it.

In this graph, two triangles are adjacent if they share a side.

The claim is that each triangle has degree exactly $2001$ in $G$.

To show that,  consider a triangle $T = \triangle{ABC}$ which contains $P$ and another point $P'$.

The claim is that exactly one of the triangles $P'AB$, $P'AC$, $P'BC$ contains $P$ (haven't tried proving this rigorously, hence this is only a possible solution)

There are $2001$ such points $P'$ and thus degree of $T$ is $2001$.

Since the degree of each vertex in $G$ is 2001, the number of vertices must be even ($\sum d_i = 2|E| \implies 2001|V| = 2|E| \implies |V|$ is even).

 

Set of subsets of naturals with inclusion order [Solution]

The goal was to find an uncountable set $S$ such that each member of $S$ was a subset of the naturals, and given any two elements $A, B \in S$, either $A \subset B$ or $B \subset A$.

Solution

Consider the rationals in $[0, 1]$ which are countable, say $\{q_1, q_2, \dots\}$

For each real $x \in [0, 1]$ let $S_x = \{i : q_i \le x\}$.

$S = \{S_x : x \in [0, 1]\}$ is a set with that property.

Set of subsets of naturals with inclusion total order

Can you find an uncountable set $S$ with the following properties?

- every member of $S$ is a subset of the natural numbers.
- for any $A \in S$, $B \in S$ either $A$ is a subset of $B$, or vice-versa.

Odd points even triangles

This is a problem from the British (or maybe Balkan, no idea) math olympiad.


You are given a set $S$ of $2005$ points (no three collinear) in the 2D plane. For each point $P$ in $S$, you count the number of triangles (formed by points in $S$) within which $P$ lies ($P$ must be strictly inside the triangle).

Show that this number is even, irrespective of the point $P$.


[I don't remember the solution to this one, so medium.]

Sequence of product of sums and reciprocal sums

Similar to the last two problems, but no triangles this time.

Suppose $t_1, t_2, \dots, t_n, \dots$ are a sequence of positive reals.

Let

$$S_n = \sum_{i=1}^{n} t_i = t_1 + t_2 + \dots + t_n$$

and

$$R_n = \sum_{i=1}^{n} \frac{1}{t_i} = \frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n}$$


Show that:

If $$S_k R_k \lt k^2 + 1$$ for infinitely many $k$, then all the $t_i$s must be equal.

More about sides of a triangle

This problem is inspired by an IMO (don't know which year) and was the inspiration for the previous problem.

Anyway. Here goes:

Suppose $t_1, t_2, \dots, t_n$ are $n \ge 3$ positive real numbers such that

$$ (t_1 + t_2 + \dots + t_n)\left(\frac{1}{t_1} + \frac{1}{t_2} + \dots + \frac{1}{t_n}\right) \lt n^2 + 1$$

Show that:

Problem 1) For any distinct $i, j, k$ the numbers $t_i, t_j, t_k$ form the sides of some triangle. [This is the original IMO problem]

Problem 2) If $n \ge 13$, show that the triangles are all acute.

Problem 3) If $n \le 12$, the triangles are not necessarily acute.

A property of sides of a triangle?

Prove or disprove:

Proposition: $a, b, c$ are sides of a triangle if

$$ (a+b+c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \lt 10$$

Note that we can easily show that (using AM $\ge$ GM for instance)

$$ (a+b+c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \ge 9$$

for any positive reals $a, b, c$.

So we can restate the above statement as

Proposition: $a, b, c$ are sides of a triangle if

$$\left\lfloor (a+b+c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \right\rfloor = 9$$

where $\lfloor x \rfloor$ is the integer part of $x$.


[Note: Initially I had if only if, but I have changed the statements to just be an implication]

An integral with $\frac{1}{\log x}$

Suppose $n$ is a positive integer (though the result below does not really need that).

Show that

$$ \int_{0}^{1} \frac{x^n - 1}{\log x} \text{d}x = \log(n+1)$$

Note that the $\log x$ is the $\log$ to base $e$.

Ant in a room

This is a classic puzzle.

There is an ant in one of the corners of a cubic room (say of side 10ft). It wants to get to the diagonally opposite corner.

What is the shortest distance it can travel to achieve that? Note that it can only walk on the wall/floor (it cannot fly etc).

Partscores are important too...

This is a hand from one the sectional swiss teams in the Seattle area.

You are south and end up in 2S (bidding below).

IMPS N/S	North ♠ J2 ♥ T432 ♦ T432 ♣ J32
	South ♠ KQ9843 ♥ AQ ♦ A5 ♣ 654

W	N	E	S
		1C	1S
1NT	P	P	2S
P	P	P

Opponents cash 3 clubs, and RHO then switches to a low trump at trick 4. What will you do?

This is probably a simple hand as a puzzle.


You should stick in the 9 in case LHO holds AT7x.

If LHO has AT7x, then to defeat this contract, LHO needs to duck and not cover with the T. Maybe LHO will talk themselves into covering by thinking of some trump promotion scenario.

At the table LHO covered, and now it was easy to win the J, take the heart finesse and draw trumps, making 2 (losing 3 clubs, 1 diamond and 1 spade) [LHO did have AT7x].

If LHO does not cover, you could hope trumps are 3-2, and overtake the 9 with the J to take the heart finesse. [Or you could win the 9 in hand and play spade to J, hoping LHO will win and return a heart].

At the other table, the defence to first 4 tricks was the same (3 clubs, spade switch by East), but declarer didn't put in the 9 (or 8) and went down. 5 IMPs.