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.

# ERJ Brainteaser November 2019

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* (π/180*angle-sin〖angle)〗

Area of circular segment = 10²/2 ((π/180*73.74)-sin(73.74))

Area of circular segment = 16.3501 cm² and

Area of circle = π*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: