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).