Square root of 50.25 = 7.08 and a bit m. Time for beer and music.
Well done Dozey and Mike.
Yes, I was a bit surprised at the answer. Didn’t seem intuitive !
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Perhaps a shade more challenging ?
(The “wing” area is coloured red, the “lens” area is coloured grey.)
Let Q denote the quarter circle and S the semicircles. Let Area denote the function that generates the area of a shape.
Now Area(4Q) = 4Area(2S) or Area(Q) = 2Area(S)
Area([Q-S]) = Area(S) = Area([A +[S-B]])
Area(A) = Area(B)
Nice explanation Phil, neat !
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The picture above incorporates five sizes of circle. From smallest to largest there are :-
Six yellow circles
Seven red circles
Three blue circles
One dotted circle that just sits inside the triangle
One large circle with a solid perimeter
Please accept the limitations of my Power-point drawing, the circles and triangle “just” touch each other tangentially. No “overlaps” no “slight” gaps.
If the diameter of the smallest circles (yellow) is 1.0 what are the diameters of each of the other circles ?
The crux seems to be to calculate the diameter of the red circles. Everything else follows. I suspect to do this you have to use the triangle, but I don’t remember much about the properties of triangles.
I’ve got as far as the blue being 2.5 times the red, by using the diameters of the circles.
The blue is two reds plus 2 yellows, or two reds plus 2.
I think the red diameter is R=2 and the blue diameter is B=6.
From the outer circle one gets B=2R+2 ( as mentioned).
Consider the right triangle from the centre of a blue circle to the tangent line. The radial line is B/2 and the angle 30 degrees so the hypotenuse is B. This makes the distance from the centre (centroid) to the vertex of the triangle B+B/2+R/2. This length is twice 3R/2 +2 from the small circles.
This gives the second equation relating R and B which yields the solution.
Sorry I know diagram would make it clearer but I’m on the move.
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Hi Mike, Dozey and Ian,
Good efforts. It does need a logical approach and a bit of intuitive insight.
I started by considering Radii rather than Diameters. It seemed easier to generate relationships based on Radii. But to avoid introducing “halves”, I simply re-designated the Radius of the yellow circles as unity (ie 1). To my mind, it’s easier to manage equations without “halves”.
And equilateral triangles are sometimes easier to work with than right angled triangles.
But you designated the diameter of the yellow circle as 1, not the radius!
I thought Ian’s answer was correct, even if he implicitly used the centroid theorem (1/3rd of way along each of the lines from midpoint to vertex).
Phil
You can get the same answer using similar triangles and avoiding the centroid theorem, but it was easier to write in words 
Hi Mike, Dozey, Ian and Phil,
First the apologies. When I said “good efforts”, I was recognising that between you, you had worked out three of the five circles Yellow = 1; Red = 2 and Blue = 6. (sorry Mike - you were pretty close) and I had kind of assumed the other two (dashed and brown) would follow very quickly with 10 and 14 respectively, to earn the traditional “well done”.
My other comments were simply to outline a couple of things I found useful, eg using Radii instead of Diameters with the yellow Radius set at unity to match its unity diameter. And also that I had used equilateral triangles (or even a rhombus) rather than the centroid rule. I have found people have forgotten some of these rules (even Pythagoras) and using equilateral triangles avoids having to use/prove some of these rules and makes the solution easier for some people to grasp - but it does need a diagram to make sense of it.
I’ll post a picture later.
But meanwhile, Well done !
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In time honoured fashion I failed to re-read the question and missed the request for the other two circles. So having cracked the hard bit I assumed I was done. Schoolboy error - literally
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Hi Ian,
Yes, you cracked the hard bit and the last two were pretty straightforward (trivial really) so I really couldn’t justify pushing any of you to the indignity of having to count the diameters (or radii) along the “Y” shaped “arms”. 
I thought I had better post my solution to the Yellow, Red, Blue etc Circles in a Triangle teaser. I know Ian cracked it, but perhaps a diagram or two and a few words would help others following this thread to grasp the solution ?
First a picture…
Label the circles Yellow, Red, Blue, Black Dash and Brown with corresponding radii a, b, c, d and e
Symmetry shows that :-
triangle LMN is equilateral and that
angles XOY = YOZ = ZOX = 120 deg
and that :-
triangle OPQ is equilateral because :-
angle POQ = ZOM = 60 deg and
QO = QP = b + c
Since triangle OPQ is equilateral it follows that :-
QO = QP = OP = b + c = 2b + 4a (1)
OH = d = 4a + 3b (2)
OG = e = b + 2c (3)
OY = e = 5b + 4a (4)
Four equations and four unknowns (we know that a = 1)
(3) and (4) give b + 2c = 5b + 4a
Which reduces to 4a + 4b = 2c
Or 2a + 2b = c (5)
Substitute c in (1) gives b + 2a + 2b = 2b + 4a
Which reduces to 2a = b (6)
And substituting in (2) gives d = 4a + 3(2a) = 10a (7)
We can now see that if :-
a = 1 (given)
b = 2 (6)
c = 6 ( 5)
d = 10 (7)
e = 14 (4) or (3)
PS I noticed a couple of typos in the diagrams eg XOX should read ZOX and YOX should read YOZ. I also noticed two "P"s in the second diagram. Fortunately these didn’t transfer over into the text. Hopefully, there aren’t too many more typos !
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Now, since a few of you obviously enjoyed the triangle/circle/geometry teaser above, I thought you might like this triangle/square/circle geometry one below… ?
By posting this teaser, I have made three posts in a row. I have a suspicion that this new forum will only allow three posts in a row before blocking that author ie “me”. So, unless somebody has a stab at a solution, or makes some sort of comment (not too indecent !) I might not be able to post the solution.
Over to you !
“Enjoy” may be streching it a bit…
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