A Fistful of Brain Teasers

Ok, to avoid the need for copying the previous diagrams or trying to visualize the appropriate rectangles I have produced the drawing above.

The three rectangles E, F and G (blue, orange and green) should enable you to "count squares" along the lines outlined by sjb and HH and thus work out the areas of fields E, F and G.

This will lead to the total area of land now owned by Farmer Watts.

Well, it should do !!

26, 20, 18, 9, 9, 9, 9  total = 100

Picked this up after your last post, wondering what on earth rectangles were for. I didn’t use them, and simply went back to the original post - but now I’ve done I recognise what you were trying to show - I just did base extensions for the perpendicular for the heights of the external triangles. If desired I’ll scribble out my working when I have time. Had no idea at start how to do the internal triangle, so looked up and found the cosine rulr (not had any use of that since school! Rest is basic geometry/trig.

But what is intriguing is that the areas of all the triangles are the same, and all are  round numbers yet adding up to 100 - there must be a relationship there I’m not aware of.

It suggested to me that the four triangles in this basic shape will always be equal regardlee of the sizes of the squares - so as on this occasion I’d used a spreadsheet to do the calcs, before closing I substituted three random numbers for the areas of the squares - and yes, indeed, the four triangles come out with the same areas.

I’ve scribbled the calcs, and done a diagram to illustrate - I’ll upload both when I’ve got time for the hassle of uploading the image 

For the second, more subtle approach…….(remember, my first approach was "grenade + excel)

26 is the sum of 5sq + 1sq

Using graph paper, I drew each side of sq A, 5 units by 1 unit

That’s why it looks slightly “skew-whiff” in the picture !

20 is the sum of 4sq + 2sq

Each side of sq B is drawn as 4 units by 2 units

18 is the sum of 3sq + 3sq

Again, the sides of sq C are drawn as 3 units by 3 units

Triangle D is formed by the sides of squares A, B and C

The area of triangle D is measured as follows :-

Draw a red rectangle around the triangle D

The red rectangle has sides 5 units by 4 units = 20 sq units

The three triangles formed within the red rectangle, but outside the triangle D

Are right triangles having sides 5x1; 4x2 and 3x3

Giving a total area for these three triangles of 11 sq units

The triangle D therefore has an area of 20 – 11 = 9 sq units

Likewise, the three external triangles E, F and G each have areas of 9 sq units

Each one comprises a rectangle , coloured Blue, Orange and Green

Less three surrounding triangles plus in the case of E, a small rectangle of 2 units

The total acreage is therefore 26 + 20 + 18 + (4x9) = 100

Innocent Bystander posted:

It suggested to me that the four triangles in this basic shape will always be equal regardlee of the sizes of the squares - so as on this occasion I’d used a spreadsheet to do the calcs, before closing I substituted three random numbers for the areas of the squares - and yes, indeed, the four triangles come out with the same areas.

 

It's true that all 4 triangle will have the same area. In the case of Farmer Watts, this was 9 acres. This provides a very satisfying solution to the overall acreage of 100.

But putting any three square fields together in the form of the initial triangle, then adding the three external triangles leads to four new fields, each with the same area.

I haven't been able to find a neat set of initial fields, other than the 26, 20 and 18 that gives such a satisfyingly precise overall acreage as 100. One or two close ones, but not 100.

Innocent Bystander posted:

I’ve scribbled the calcs, and done a diagram to illustrate - I’ll upload both when I’ve got time for the hassle of uploading the image 

I load my scribbles into Powerpoint and save them as JPEG.

I then load the JPEG into Flickr and transfer from Flickr to Naim using the <> tool in the top-right part of the Forum Post toolbar.

It's a right old pain !

Don Atkinson posted:

Innocent Bystander posted:

I’ve scribbled the calcs, and done a diagram to illustrate - I’ll upload both when I’ve got time for the hassle of uploading the image 

I load my scribbles into Powerpoint and save them as JPEG.

I then load the JPEG into Flickr and transfer from Flickr to Naim using the <> tool in the top-right part of the Forum Post toolbar.

It's a right old pain !

I use Imgur - doesn’t need an account, but still a hassle!

My spread sheet approach.      (a bit messy-looking, but it works !)

Clearly there are formulae in the cells with the red number "9"s    (and many of the other cells)

eg the first one (Heron's law) the cell contains "= (E10*H5*H6*H7)^0.5"        

in other words    √(6.906898 x 1.807879 x 2.434762 x 2.664257)

Lost count of how many attempts to load image so far, but I’ve spent far too long on it. So, holping this is OK by forum rules, here’s a link that should pull up the associated diagram:

https://m.imgur.com/a/LHDxk

NOtation below relates to this image, not any others.

a= SQRT 18
b= SQRT 26
c= SQRT 20

Internal triangle:
Cosine rule I had to look up, for angles of non-right-angled triangle with all sides known
Cos B = ( c^2 + a^2 – b^2 ) / 2ac
= (20 + 18 – 26) / 2 x sqrt 18 x sqrt 20
Angle A = 52.12 deg
Angle B = 56.31 deg
Angle C = 71.57 deg
Internal triangle height h
Sin C = h/a, so h = a x Sin C
Area = 0.5 x b x h
=9

External triangle:
Angle D = 360-90-90 –B = 123.69 deg
Angle E = 180- D = 56.31 deg
Height of external triangle (k)
Sin E = k/c, so k = c x Sin E
Area external triangle = 0.5 x base x height
= 0.5 x a x (c x Sin E)
=9

Similarly for other external triangles.

This might help

Oh ! and neat solutions IB

And big thanks to sjb for the breakthrough moment  well done !

And well done to HH for tidying up sjb's breahthrough !

Thanks, Don for fixing my sketch - I don’t know why I couldn’t get it to open from Imgur - did you do it from my upload there, or copy it elsewhere?

I copied it from imgur and loaded it into Flickr.

I then copied it from Flickr into the forum post.

it seems that Richard’s “standard” instructions don’t work for me !

Well, after that struggle with Farmer Watts and his 100 acre field system, I thought something a tad more "gentle" on these old brains might be in order.......

This one's for those with a "language" background.

and for those with a more mathematical bent.......

33.

JRHardee posted:

33.

You have the same number as do I.

But for the benefit of others it will need to be explained.

Shall I ? or will you ?

3-2 = 1. 5-3 = 2. 9-5 = 4. 17-9 = 8. 

The series of differences, 1, 2, 4, 8, suggests 16 as the next member. 33-16 = 17.

As for the previous puzzle, either "A" for August or "D" for December is a reasonable answer. 

JRHardee posted:

3-2 = 1. 5-3 = 2. 9-5 = 4. 17-9 = 8. 

The series of differences, 1, 2, 4, 8, suggests 16 as the next member. 33-16 = 17.

Yep, that's the way I did it too.

Others have used "Double the previous number and subtract one" it works just as easily.

The number ones, including this one, seem to be a whole lot easier than the Language ones !

0.333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333

Etc

JRHardee posted:

As for the previous puzzle, either "A" for August or "D" for December is a reasonable answer. 

Yes, either would fit