Perhaps I should add that the above ‘Teaser’ came from a recent Maths Challenge competition. It was passed to me by my 14 year old grandson.

As so often these days, it came with a choice of 5 answers

A) 108

B) 96

C) 90

D) 84

E) 72

I think the answer is 72, ie a third of the total area.

Difficult to explain without a labelled diagram but I’ll do my best:-

If the side of the large hexagon is A, and the side of the small hexagon is B.

B is one side of an equilateral triangle. The other sides (also = B) create a 30degree angle with A.

Looking at the triangle to the side of the equilateral triangle, BCos30=A/2, and given that Cos30=(sqrt3)/2, so A/B = (sqrt3)/1.

The area of a hexagon H versus its side S is given by the formula H=(3xsqrt3xSsquared)/2.

So the ratio of the area of the large hexagon divided by the small hexagon is determined by the square of their respective sides.

If A/B is (sqrt3)/1, then Asquared/Bsquared is 3/1.

Therefore the large hexagon is 3 times the area of the small hexagon.

Hope that makes some kind of sense

Looks good to me.

I used a visual approach, assuming the diagram is a piece of paper.

Fold the white equilateral “petals” inwards, fully covering the blue hexagon. Therefore the blue hexagon has the same area as six equilateral triangles.

Cut the wider triangles in half. Rotate one half of each through 180 degrees to make equilateral triangles. Hence the area of a wide triangle is the same as an equilateral triangle.

We now have 18 triangles all of the same area. The six blue ones are one third so the answer is 72.

Well done Steve, and Ravvie. I got the same answer 72.

I’ll post the bones of my solution below and in the following two or three posts. It’s in the form of diagrams and text and starts with a refresher of the problem.

Thanks for the reminder. My answer is 4 x 2178 = 8712

My solution is too long to type up before cooking dinner, I will do so later, wine consumption permitting!

… so, we’ll look forward to your solution sometime next week …

DCBA <= 9876/4 hence D = 1 or 2

D must be even as ABCD needs to divide by 4. Hence D = 2

2CB8 x 4 = 8000 + 400C + 40B + 32 fits, since first three terms are a multiple of 10.

2CB9 x 4 = 8000 + 400C +40B + 36 doesn’t work since D = 2 not 6

Hence A = 8

8000 + 100B + 10C + 2 = 4x2000 + 400C + 40B + 32, gives:

6B = 39C + 3

Using Douglas Adams’ Theorem:

42 = 6B = 39C + 3, gives B = 7 and C = 1

Mrs R had a go, thinking it a good idea to solve it in her head.

She got that D = 2 and A = 8 before deciding pen and paper were needed.

Not bad after 2 glasses of wine!

The ski hill (Silver Star) in Vernon BC opened to skiing last week. The grandchildren have already forgotten the sun and sand of Sandymouth Bay back in August and have christened their new ski boots.

The ratio of male (boys) to female (girls) pupils going on their next school ski trip is 5:3

Four male teachers and 9 female teachers are also going on the trip

The ratio of males to females going on the trip is 4:3 (including the teachers)

How many female pupils are going on the trip ?

As a guide, the above teaser can be solved quite easily by “Trial & Error”

On the other hand, if you can remember how to do simultaneous equations, you can avoid the “Trial & Error”.

I’ll post my solution to the number of school girls going on a skiing trip, later this weekend.

Meanwhile, here is an “upgrade” to a recent teaser, ie from four digits to five digits !

School Ski Trip:-

If B is the number of boys and G is the number of girls.

Equation1 is B/G = 5/3, so 3B = 5G.

Equation2 is (B+4)/(G+9) = 4/3, so 3B + 12 = 4G + 36.

Substituting 1 into 2 gives 5G + 12 = 4G + 36, so G = 24.

So 40 male pupils and 24 female pupils went on the trip (ratio 5:3).

Including teachers, there were 44 males and 33 females in total (ratio 4:3).

Well done Steve.

And very nicely set out and explained.