England Selectors’ Ostrich Impression and Other Stuff

Some thoughts on the (in)action of the England selectors this week, some mathematical teasers and a few pictures.

INTRODUCTION

A couple of days ago I wrote about England’s series win over India and presented some problems and solutions. This post is on similar lines, dealing with the actual behaviour of the England selectors and my thoughts thereon.

AN OPPORTUNITY SQUANDERED

England, with the series already in the bag, had a diamond-encrusted golden opportunity to experiment with options to fill gaping holes in their top order. Cook’s announcement of his impending retirement from international cricket should have acted as an extra spur. Instead of which we see very little in the way of forward planning or of experimentation of any sort. Even with the certain knowledge that a new opener will have to come in to replace Cook the selectors persevere with the proven failure Jennings.

Three individuals who can feel more aggrieved than most by this behaviour are Rory Burns (another 90 against Essex yesterday after the latter won the toss and chose to bowl first), Dan Lawrence and Liam Livingstone

In view of Cook’s impending retirement I would have recognized openers at 1,2 and 3 (not a bad approach in test cricket anyway), with a view to the two other openers than forming a partnership in future matches. This is why in the previous post I mentioned Tammy Beaumont, a recognized opener who has been scoring stacks of runs recently. Batting is at least as much about timing and placement as it is about brute power, and that is why I believe (unlike in the case of fast bowling) a woman could mix it with the men even at the highest level, similarly with slow bowling and possibly wicket-keeping (for my money the best user of the gauntlets in world cricket across the board at the moment is Sarah Taylor). A number of the all-time greats of test match batting have been of diminutive stature (Bradman, Gavaskar, Tendulkar, Sehwag, Hanif Mohammad and several of the finest Sri Lankans spring to mind instantly). I am well aware that this super-radical option will not happen, but the alternatives that that leaves with are:

  1. Two brand new openers, neither of whom have any experience of international cricket.
  2. One new opener and one opener who has shown already that they are not actually good enough (Jennings)
  3. Two openers who gave failed to prove themselves (presumably Jennings and a recalled Stoneman). 

Of those three options, none of which massively appeal, my choice would number 1, which might end up working out well, and then the question is who to choose to open alongside Burns (whose case for selection is undisputable in the circumstances). 

Having taken the “ostrich option” re their top order difficulties the only outcome from this game that could be acceptable is not merely a win to make it 4-1 for the series but a win by a massive margin. The timidity of the England selectors means that at least one and possibly two England openers will be starting their careers on overseas tours, with their first home test series being against those well known softies, the Aussies.

PROBLEMS AND SOLUTIONS

I will start as usual with answers and solutions to the previous problems (all from brilliant.org) before offering up some new problems.

WHAT IS THE AREA OF THE QUADRILATERAL

Screenshot 2018-09-03 at 5.08.56 PM

First the answer:

quad answer

The hackers solution is that there are only two really serious possibilites since the shape is a square, namely 67 (giving an area of 289 = sides 17 units long) and 102 (giving an area of 324 = sides 18 units long), and since the question gave one three tries just enter those values for the first two tries (if your first entry does not come up right). Here, courtesy of Jeremy Galvagni is an elegant genuine solution:

quadsol

THE .99 STORE

First the answer:

Screenshot 2018-09-05 at 3.09.42 PM

The figure in front of the .99 part of the price can vary, so all we need to know is how many .99s add up to answer ending in .89, and the answer is 11 (11 x 99 = 1,089, so 11 x 0.99 = 10.89), and the next number of items after 11 that would give us an answer ending in .89 is 111, the lowest price total for which would be $109.89. Thus Marie purchased 11 items.

NEW PROBLEMS

First an astronomy themed problem:

astroproblem

Now a question that has got almost three-quarters of those who tackled in on brilliant, but is not actually difficult:

Bullets

PHOTOGRAPHS

Swimming MoorhenMoorhen on branchTwo MoorhensMoorhensMagpie