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.