Brain Teasers are Back!

6%20circles%20and%20a%20triangle%20Solution%20jpeg

I have added a few triangles and highlighted a couple of lines, all of which I have used in my solution.

You might find them helpful ?

PS, I am off to Oman for a couple of weeks and shall be incommunicado. So plenty of time for somebody, or a team effort, to come up with a solution !

Starting with the square.

Let the radii of the circles be denoted by their uppercase colour letters.

Let x be the horizontal distance between the contact points of the Red circles and the centre of the Yellow, then

x = (R+Y)2 - R2 by Pythagoras

and the square requirement

2x + 2Y = 4R

Y2 + 2RY = 4R2 - 4RY + Y**2

Y = 2R/3

The areas of the other triangles are easily calculated in terms of x and y the lengths from the tangent on the hypotenuse to the sides and similar triangles on Don’s drawing.

2/3*(Ry +Rx) + R**2 = 1/3(x + R)(y + R)

R(x + y) + 2R**2 = xy

ditto Blue and similar triangles with the ratio of x to y the same and the proportions being B to R

Should have enough equations now.

Phil

I came up with x = 5/2 R and y = 3R from the base of the red triangle length of 4R.

B/R = 4/3 or B = 4/3R = 2Y

Phil

The area method has a flaw. Pythagoras works to solve x = 2R then the second post gives the answer. Can’t figure why the areas went wrong.

Phil

The flaw was area half rather than third. Stupid me!

Phil

Hi Phil,

I just found a local wi-fi spot to use. Looks like you have cracked it.

I won’t be back for a couple of weeks, but I’ll post my workings when I do.

Well done ,

2 Likes

Back and my solution follows below. (You might need to make a copy and mark up the dimensions as they appear in the text below)

Circle radii Yellow, Red and Blue are a, b and c respectively

In the diagram, I have drawn a few triangles and highlighted a couple of lines.

The square inside the main triangle has side 4b

The triangle within the square has a hypotenuse = b + c and the lengths of the other two sides are b and 2b – a.

From Pythag (b + a)^2 = b^2 + (2b – a)^2

This expands then contracts to give 3a = 2b

And finally b = 3a/2 (1)

At the top of the main triangle I have drawn three radii each = b

This splits the top triangle (which contains a red circle) into a b x b square and two kite shapes.

The bottom edge of the left hand kite is 3b. And based on symmetry, so is the long sloping top edge of that kite. The short edges of that kite are b

If the long sides of the right-pointing kite are “x” then using Pythag on the top triangle

(3b + x)^2 = (b = x)^2 + (4b)^2

This can be expanded, then simplified to give

4bx = 8b^2

And finally x = 2b

The vertical side of the top triangle is x + b which = 2b

The vertical side of the bottom triangle (which contains the blue circle) is 4b

The top triangle and bottom triangle are “similar” and therefore the ratio of their sides which are 3b/4b ie ¾ must equal the ratio of the radii of their inscribed circles which is b/c

Hence ¾ = b/c or c = 4b/3 (2)

So now I have c in terms of b and b in terms of a

Therefore c in terms of a must be c = 4b/3 = (4/3) (3/2)a = 2a

QED

1 Like

Just noticed a small typo…

The line that reads “The vertical side of the top triangle is x + b which = 2b”

should read “The vertical side of the top triangle is x + b which = 3b”

Apologies, Don

1 Like

A set of circles

It’s pretty obvious that the maximum number of different sized circles (excluding infinitely large circles that have become straight lines!) that can be made to ‘just touch’ one another is four. (each and every circle just touches the other three).

Three circles Red, Blue and Yellow have radii of 300, 200 and 100mm respectively. What are the radii of the two ‘fourth’ circles that can be made to just touch each of Red, Blue and Yellow circles ? (note the two ‘fourth’ circles will not touch each other, ‘cos that would break the golden rule!!) I have illustrated them in Green.

Oh! BTW, there is a standard way and an elegant way to solve this little problem. The normal way will do for most, but a bonus point for the “elegant” solution.

PS Fatcat is NOT permitted to participate in this specific teaser – he knows why !

Every Saturday I buy 4 lottery tickets, which equals 208 tickets per year.

Would I increase my chances of winning over a period of say 5 years if bought all 208 tickets on one Saturday per year.

No. The total number of draws is the same for the same pool of draws.

I think it will depend on how many tickets are in each draw. If we assume the number is always the same then the odds should be equal.

Also, best not to use the same numbers on all the tickets. If you use the same 4 combination of numbers, then buying them each week will be a better approach.

Lots of “if’s” in the Lottery teaser. eg…

208 tickets, each with the same combination of 6 numbers, all placed on the same day, would probably lead to winning most of that day’s Jackpot (assuming the numbers came up - 1 in 45 million), even if a few other contestants had picked that combination.

208 tickets, each with a different combination of 6 numbers, all placed on the same day, would improve the chances of holding a winning ticket to 208 in 45 million, but you might have to share any jackpot with one or two other lucky winners.

The distribution of selected numbers (evenly spread or tightly grouped) might affect the probability of winning if all 208 entries are made on the same day.

Are we only considering the Jackpot, or are we considering maximising our prospects of winning even small amounts.

… as I say, lots of “if’s” at the moment.

I was going to post an entry in the “What book are you reading now” thread. It’s a non-fiction book, first published about 50 years ago, a bit like Arabian Sands by Wilfred Thesiger. And it was Bruce W’s mention of Arabian Sands that triggered my thinking to post.

But then I thought…why not post a few abstracts and little hints about the book and let participants recall books they might have read, or can search and find x-references to, on the internet.

Who knows, it might provide a welcome relief from Brexit, Farage and Boris (we live in hope !) It might bring back forgotten memories for one or two, or even interest or inspiration for others. Then again, it might just fall flat on its face !!

So my next post will be a starter for “10”. Just guess the Title/Author or simply ask questions such as “does it take place in South America” (it doesn’t) but not “where does the action take place”.

PS Plenty of scope/time for others to select one of their favourite “mystery tomes” to tease our semi-dormant brains !

The “starter for ten”

.it is an ancient road………… running below a sheer rock face over 3,800 feet high. On this cliff,In 1835 and 1836 Rawlinson copied the greater part of the old Persian inscription, which was the easiest of access, scrambling three and four times a day up the (500 feet high) steep rock face to the foot of the inscription, balancing on a ladder upon a ledge, only eighteen inches wide, which ran along the foot of the inscription, and then standing upon the topmost rung of the ladder ‘with no other support’, he tells us, ‘than steadying the body against the rock with the left arm, while the left hand holds the notebook and the right hand is employed with the pencil’.

If you have ever read this book, you should recall this passage imediately. But 10 marks nevertheless !

Lesley Adkins, Empires of the Plain

Now that’s not a bad start, but it’s not the book that I am reading.

The “Rawlinson” is indeed the young Major who copied and deciphered the cuneiform above the sculpture of Darius and proved to the astonished Victorian public that people and places in the Old Testament really existed (and that documents and chronicles had survived from well before the writing of the Bible).

The text I wrote is an exact copy from the book I am reading. I first read the book about 50 years ago at about the time it was published. The cuneiform writing that Rawlinson deciphered, provided clues to the existence of various other civilizations, that had been “lost” with the passage of time. One or two of these “lost” civilizations is the subject of the book and the author is seeking their whereabouts. A bit like searching for Atlantis, only it’s not Atlantis !

When I first read the book and again now, I was taken by the risk that Rawlinson took in being able to copy the cuneiform inscription. The more difficult part of the inscription could only be got by spanning one ladder across a gap in the 18" ledge and balancing a second ladder on the first, all 500 feet above the almost sheer drop below !

I’ll provide the next clue later, but a good start, well done !

Not a book about which I have any hope of guessing, however more intriguing than copying the inscription is its original inscribing…!

Yes, that crossed my mind as well. The relief was created about 2,500 years ago, so quite some feat. I understand that after the sculpture was created, part of the rock wall below was removed, to make the sculpture less accessible and less likely to be vandalised.

I recall that some years ago, I was leading a small convoy of five Land Rovers from Ras al Khaimah through this next bit of territory, without letting the TOS know. I think they were more relieved than myself when they intercepted us between Masafi and Dibba. But the next passage from the book, which took place a few years earlier, reminded me not only of that encounter and the location, but the events surrounding it.

Now the Shihu were massing, the Musafi adjutant told us, in the hills above Dibba. The boundaries between Muscat, Sharjah and Fujaira all met at Dibba, and had never been defined to the satisfaction of all parties. The Shihu claimed Sharjah had occupied part of their territory, and were threatening to redress their grievance by force of arms. For two months the squadron of the Scouts had been sitting encamped on the plain south of Dibba, hoping that tempers would cool and negotiations take the place of threats. But tribal war could break out at any time.

9 points if this leads to the book title !