Surprising but easy

If

$$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = 1$$

Show that

$$\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} = 0$$

Solution [Click here to expand/collapse]

Consider $$a + b + c = (a+b+c)\left(\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}\right)$$
This is the same as $$a + b + c + \frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b}$$
And thus $$\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} = 0$$

A problem from the Vietnam School System

This is a problem given in an exam for middle-schoolers going to high school in Vietnam. It is not easy though.

Given that $a \ge 1, b \ge 1, c \ge$ and $$ab + bc + ac = K$$ where $K$ is a constant $\ge 3$, find with proof, the maximum value of $$a^2 + b^2 + c^2$$

Find $\tan x$

If $$a \cos x + b \sin x = c$$

Find the possible values of $\tan x$ in terms of $a, b, c$.

No calculus IVT for quadratic.

Suppose $$f(x) = x^2 + bx + c$$ and that $f(0) \lt 0$ and $f(1) \gt 0$.

Since $f$ is continuous, by the intermediate value theorem there is some $c \in (0,1)$ such that $f(c) = 0$.

The question here is to prove that in an elementary way, without using any calculus concepts.

Finding the prisoners' names

This is yet another of those crazy warden and mathematical prisoner puzzles.

This time, there are $2n$ prisoners (each with a distinct name).

The warden has a room with $2n$ boxes, each box has the name of exactly one prisoner, and each prisoner's name appears in some box (picked randomly by the warden).

The game the warden plays is that each prisoner goes independently (one by one) in a room and points at some $n$ of the boxes. The goal being that each prisoner should choose the box which contains their own name. If any one of the prisoners does not then they all lose the game. The prisoners aren't allowed to communicate in any way (with each other) what boxes they picked.

If each prisoner picked $n$ boxes at random, then the probability that they win the game (each picks their own name) is $\dfrac{1}{2^n}$ which is pretty small.

They are allowed to choose a strategy before any of them enter the room.

Show that there is a constant $c \gt 0$ (independent of $n$) and a strategy such that the prisoners win the game with probability at least $c$.

Limit of iterated x + 1/x

Suppose $x_1 = 1$ and

$$x_{n+1} = x_n + \frac{1}{x_n}$$

Find

$$\lim_{n \to \infty} \frac{x_n^2 - 2n}{\log n}$$

Expected number of correct coats

This is a classic:


$N$ people attend a party and deposit their ($N$) coats with the coat keeper.

At the end of the party, everyone is drunk (even the coat keeper). The coat keeper hands out a random coat to anyone who comes to claim a coat. Since the owner of the coats are drunk too, no one notices.

What is the expected number of people that get their correct coat back?

A curious inequality

I first saw this in UW's challenge of the week (if I remember correctly), where they used to post a nice math puzzle every week and give the winners (drawn at random from the correct solutions) a gift certificate to Baskin Robbins.

[That has now been discontinued and I believe the pages also have been taken down.]

Anyway, here is the puzzle.

If $x,y \gt 0$ are real numbers, show that

$$x^y + y^x \gt 1$$

Fibonacci property

Fibonacci numbers are defined as $f_0 = f_1 = 1$ and $f_{n+1} = f_n + f_{n-1}$.

Show that a number $F$ is a fibonacci number if and only if one of $5F^2 \pm 4$ is a perfect square.

Defend 3H in a BAM event

This is a hand from the recent Bothell Memorial day sectional. This is a hand from the BAM (board-a-match) teams event.

BAM N/S	Dummy ♠ AT2 ♥ 654 ♦ KQ83 ♣ 987
		You ♠ Q32 ♥ 932 ♦ JT92 ♣ QJ2

W	N	E	S
			1H
1S	2H	2S	3H
P	P	P

Partner leads AK of club and club to your Q (declarer following). What do you do now?



If partner has DA or a heart trick, we need to shift to a spade now.

What if partner does not have the DA or a heart trick?  Since this is BAM, overtricks are important.

If you shift to a low spade and declarer has Jx of spades, you will get squeezed in the pointed suits for declarers 10th trick!

To cater to that, you must shift to the SQ.

Minimum value of sum of trigonometric functions

What is the minimum value of

$$(\sin x + \cos x + \tan x + \csc x + \sec x + \cot x)^2$$


($x$ is real and takes only those values where the function is well defined.)

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

---

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

---

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.

Average free subset

A problem which someone posted in Google's internal GooglePlus page (yeah...).


Show that there is a subset $S$ of $\{1, 2, \dots, 3^k - 1\}$ such that, $S$ has at least $2^k$ elements and any three distinct elements $x, y, z$ of $S$ do not satisfy $x+y = 2z$

(i.e. no number is the average of some other two numbers).

Don't cross the streams!

There are $N$ red and $N$ blue points in the 2D plane, no three of which are collinear.

Show that we can pair off each red with a blue so that none of the $N$ line segments intersect.

[Solution]

Careful!

It has been a while since I posted a bridge hand.


This is a hand from the Seattle Round Robin (2016).

You end up in 6S and LHO leads a trump. Don't remember the bidding (but both sides were bidding). Scoring is IMPS, KO.

	North ♠ J2 ♥ A43 ♦ Q543 ♣ KJT9
	South ♠ AKQT9854 ♥ 2 ♦ A2 ♣ 32

How will you play on a spade lead?





Did you draw trumps first? Then you will go down!

We basically need the club finesse, but the first club could get ducked! If you draw trumps and then take the club finesse which gets ducked, how will you get back to hand? If you ruff a heart you lose your only dummy entry to cash the club for a diamond discard. If you play a diamond, they can cash the DK.

It is easy to go wrong at the table. In fact a very good player indeed drew trumps first and went down when my teammate (John Krah) made the excellent play of ducking the first club (CQ was onside). At our table we were in 4S and this turned out to be a huge swing board in a semi-final match which was decided by a single IMP!

Length of the creeper

There is a tree with a trunk that is a perfect cylinder. The circumference of the base being 6ft. The height of the cylinder is 8ft. A creeper wraps around the trunk, rising uniformly from the ground to the top, and in doing so, makes exactly one turn (so the end point is on the same vertical line as the starting point).

What is the length of the creeper?