A Fistful of Brain Teasers
Posted by: Don Atkinson on 13 November 2017
A Fistful of Brain Teasers
For those who are either non-British, or under the age of 65………. The UK used to have a brilliant system of currency referred to as “Pounds, Shillings and Pence”. Simplified to £ ״ s ״ d. No! Don’t ask me why the “Pence” symbol is a “d”, just learn it and remember it !
A £ comprised 20 Shillings and a Shilling comprised 12 Pence. Thus a £ comprised 240 Pence. I reckon that both Microsoft and Apple would have difficulty with these numbers in their spreadsheets, more so if we included Guineas, Crowns, Half-Crowns and Florins. However, I digress..............
The purpose of the explanation is to assist with the first two or three teasers that follow. So just to ensure a reasonable comprehension has been grasped…. ….. if each of three children has £3 − 7s − 9d, then collectively they have £10 − 3s − 3d Got the idea ? Good ! Just try 5 children, two each with £4 − 15s − 8d and three each with £3 − 3s − 4d. How much do they have between them ? (this isn’t the first brain teaser, just the basic introduction with some “homework”, the Teasers follow)
There are one or two ways of approaching the Hexagonal Tiles & Hoop Teaser. But fundamentally it is a case of discovering the ratio of the areas of two triangles.
The diagram that I posted a few days ago and repeated above, showed how a hoop could occupy a hexagon WITHOUT crossing any boundary line. The lightly shaded area of that hexagon represents the area within which the centre of such a hoop must lie.
The “guts” of the solution are that :-
The probability the Hoop DOES NOT lay across two or more tiles (ie that it lies wholly within a single hexagon is ….. Area of COD/Area of AOB. And is represented in mathematical shorthand as P(Ā).
Each hexagon can be considered as a group of six equilateral triangles. The ratio of the triangular areas will be the same as the ratio of the hexagonal areas.
More importantly, we don’t actually need to work out the area of these triangles. Each pair of triangles such as COD and AOB are “similar”. Therefore all we need is a CHARACTERISTIC LENGTH of each triangle, eg the length of a side. The SQUARE of this characteristic length is sufficient to represent the area of the associated triangle (and hexagon). Some solutions find the length of a side of the triangles, others use the height of the triangles. It doesn’t matter. Either will work !
Continued on next post......
In the diagram above, I have opted to use the lengths of CD and AB ie the sides of the two triangles, COD and AOB.
It is also worthwhile recalling that Tan 30° = √3/3
In the diagram above, the two red triangles have 30° angles at C and D
Tan C = Tan 30° = √3/3 = x/(d/2) = 2x/d
Hence 2x = d.√3/3
Base CD = L – d.√3/3
Base AB = L
P(Ā) = CD2/AB2
P(Ā) = (L – d.√3/3)2/(L)2
Now for a little reminder about algebra
A2/B2 can be re-written as (A/B)2
Therefore P(Ā) can be re-written
P(Ā) = ((L – d.√3/3)/L)2
This can be simplified
P(Ā) = (1 – d/L.√3)2
Now remember, we have been looking at the probability of a hoop NOT lying across two or more tiles, ie being wholly within one hexagon.
The probability of it either lying across more than one tile, or being wholly within one hexagon is a certainty ie 1
The probability of the hoop lying across two or more tiles is therefore 1 minus the probability of it being contained within a single tile.
Probability of hoop lying across two or more tiles = P(A)
P(A) = 1 – P(Ā)
P(A) = 1 - (1 – d/L.√3)2
You are facing a cube. The cube can be rolled left, right, forward (towards you) or backwards (away from you). There is a dot on the bottom of the cube.
- imagine that you roll the cube: forwards, left, left, forwards, right, backwards, right,. Where is the dot now ?
I hope so, if I can figure out how and when to "Log in".....could be a REAL brain teaser 