ERJ Brainteaser November 2019

First, a special mention to everyone who answered tricky Q2. However, Q4 was an extra nice teaser, with just enough steps to lead quite a few readers astray. So, for working out the six small steps for ant-kind, it’s giant congratulations to Andrew Knox, Michele Girardi, Hans-Bernd Lüchtefeld and France Veillette, joint winners of this month’s Brainiac of the Month title. 

Question 4: Ant on the run

An ant is on the square marked with a black dot.
The ant moves across an edge from one square to
an adjacent square four times and then stops.
How many of the possible finishing squares are
black?

Answer:  The only way the ant can end up on a black square is if on one of its moves the colour of the square does not change. Looking at these possible sequences we see that there are exactly 6 black squares that the ant can end up on after four moves. Well done to: Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Michele Girardi, Scame Mastaf Spa, Suisio, Italy; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany; France Veillette, chef environnement, Usine de Joliette, Bridgestone Canada Inc., Canada.

As explained neatly by Michele Girardi, the ant can reach:
- Two adjacent blacks, through a path:  black, white vertical, white vertical
- Two vertical blacks on first row through a path: black, white vertical, white vertical
- Two vertical blacks in second row, through a path: black, white vertical, black vertical.

And shown pictorially by Andrew Knox:

Question 3: Got the time?

Complete the following series

166 hours 40 minutes; 16hrs 40 minutes: 1hr 40 minutes; _____

Answer: Seems like readers either got this one straight away, or not at all: correct answers came in mostly early in the week, and then petered out. The key was to simply switch the hours into minutes to get the sequence: 10,000, 1000, 100 and then the answer 10. Well done to: John Bowen, consultant, Bromsgrove, UK: Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Michele Girardi, Scame Mastaf Spa, Suisio, Italy; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; Yoganand Nannapaneni, Mascot Systems Private Ltd, Mumbai, India; Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan; Stephen Pask, details not supplied; David Mann, Polymer Business Development, France.

Question 2: Squaring the circle

Find the area of the shaded region where the rectangle and circle overlap.

Answer: A bit trickier this week and very well done to the following readers who worked out the answer at 148.9 cm²: Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Michele Girardi, Scame Mastaf Spa, Suisio, Italy; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan; Martin Yonnone, Colmec USA, USA;  Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; David Mann, polymer business development / language coach, Arras, France: Ramasubramanian P, manager, marketing – mixer and LTKMPL products, rubber processing machinery, Larsen & Toubro Ltd, Vedal Village,  Kanchipuram, Tamil Nadu, India – and to everyone else who had a go.

Here are just three examples of the neat working out:

Jose Padron
Assume Shaded area = ½ Circle area - ½ Circular segment area. To calculate the circular segment area; it is necessary to know the angle delimiting the circular segment.
Using trigonometric calculations
R = 10 cm
Segment height = 8 cm
Angle = 2 * Cos -1 superior angle
Cos = 8/10 = 0.8 
Cos-1 = 36.87°
Angle = 2*36.87 = 73.74°
Then, Area of circular segment = R^2/2* (?/180*angle-sin?angle)?
Area of circular segment = 10²/2 ((?/180*73.74)-sin(73.74)) 
Area of circular segment = 16.3501 cm² and 
Area of circle = ?*R^2 = 314.16 cm²
Thus:
Area shaded area = 314.16/2 – 16.3501/2 = 148.905 cm²

David Mann
Using the equation of a circle x2 + y2 = r2 the coordinates of the intercept between the circle and the rectangle are (8, 6). The angle between the horizontal and the intercepting radius is calculated from tan A = 6 / 8; A = 36.9 degrees. Then the segment is Pi r2 x 36.9 / 360. From this the area of the triangle needs to be subtracted, and this is given by ½ 10 x 8 sin 36.9.
So the final shaded area is 157.1 – (32.2 – 24) = 148.9 cm²

Andrew Knox
Draw a radius of the circle to the intersection of the rectangle with the circle, leaving the shaded area as the sum of an angular segment of the circle and a simple right-angled triangle. 
For the height h of the unknown side of the triangle: (h)2 = (10)2 - (8)2 so h = square root of 100 - 64, so h = 6. Area of the triangle is therefore 8xh/2 = 24 cm2.
For the area of the segment of the circle:
Internal angle of the triangle at the centre of the circle is Cos 8/10 so angle is approx. 36.87 degrees. Therefore the internal angle of the segment of the circle is 180 - 36.87 = 143.13 degrees.
Area of full circle is Pie x r2, so area of segment is Pie x r2 x 143.13/360 = 3.1415926 x 100 x 143.13/360 = 124,9 cm2.
So shaded area in total is 124.9 + 24 = 148.9 cm2.

Question 1: Keep it country

What comes next in the following list?

8 Argentina, 7 India, 6 Australia.

If we forget about Monday, this was a fairly straightforward teaser-week. The answer was 5 Brazil in this list of countries in ascending order of surface-area. Well done to: Amparo Botella, Ismael Quesada SA, Spain; Bharat B Sharma, Sr VP Product Development & Technical Service (Elastomers), Reliance Industries Ltd, Vadodara, Gujarat, India; Yoganand Nannapaneni, Mascot Systems Private Ltd, Mumbai, India; John Bowen, consultant, Bromsgrove, UK: France Veillette, chef environnement, Usine de Joliette, Bridgestone Canada Inc., Canada: Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands;  Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan

Marks for style go to Jose Padron for this neat table: