# Find The Radius Of All The Circles

Find the diameters of all the circles shown, given the information provided in the diagram below. All diameters are of integer length. P.S: the signature in the bottom right is mine and I own this.

• It's not clear which apparent intersection points are intentional and which are not. I assume line segment DE is a diameter of the yellow circle, and EF of the green, so that E is the point where those circles' perimeters cross; but if you look, that's not actually how it's drawn. Similarly points D and especially C and B. Leaving the muddle in the middle, where all the circles come somewhat together, unclear as to where exactly their perimeters intersect. That precision is going to be important, by the look of it. – Rubio Jan 6 '17 at 12:06
• We will need a scale rather than a hint here :D – Techidiot Jan 6 '17 at 12:12
• Adding Few More Information {AB,BC,CD,DE,EF) are diameter of respective circles .i.e B is connecting point of two circle with diameter(AB & BC) C is connecting point of two circle with diameter(BC & CD) D is connecting point of two circle with diameter(CD & DE) – Dharmesh Jan 6 '17 at 12:17
• The informative comment would be good to include in the puzzle statement, along with a note that the circles are not necessarily to scale – humn Jan 6 '17 at 12:49
• I thought it helpful to note this observation: you use the word “diameter” consistently in the text of the question, yet the title and the image use “radius”. Also, the plural form is ‘radii’. – can-ned_food Mar 21 '17 at 10:39

I think we can assume from inspection, despite the diagram's vagueness, that

$\frac{DE}2 + \frac{BC}2 = CD$
$CD + \frac{BC}2 ≤ EF$
and of course,
$AB < BC < CD < DE < EF$
$AB + BC + CD + DE + EF = 35$

We can conclude

BC and DE are both even or both odd, as that's required for all integer solutions.

A solution for all the above is

AB=4; BC=5; CD=7; DE=9; EF=10

which is the closest fit for the actual pixel widths of the circles, though they are not well-drawn. Pixel widths are: 272, 368, 494, 594, 676 within a reasonable approximation of accuracy, on the full sized image. These give a pixel->unit conversion of around 74:1. That agrees with the numbers given above.

My original answer here said: 5,6,7,8,9. That's actually wrong, both by the relationships given earlier, and by pixel approximation. And yet that was an accepted answer. Meh!

• CD + BC/2 = 10 which is not less than or equal to EF in your solution. – LeppyR64 Jan 6 '17 at 12:36
• @LeppyR64 Huh. That's actually true isn't it; I missed that. I blame the diagram. There won't be a valid solution with integer entries that total 35 that exactly fits that diagram, is my guess. I'm half tempted now to start counting pixels. – Rubio Jan 6 '17 at 12:40
• @Rubio I already counted the pixels. See my answer. They don't add up. – Marius Jan 6 '17 at 12:45

I cheated a bit and used Firefox Developer toolbar to measure the diameters.
The results were these (I may have added a pixel here or there or subtracted one so I get multiples of 5)

AB = 140px
BC = 190px
CD = 245px
DE = 305px
EF = 345px

Addind all of them we end up with

1225px = 35, so this means 1 unit equals 35px.

Dividing every value by 35 we get the following diameters

AB = 140px = 4
BC = 190px = 5.43
CD = 245px = 7
DE = 305px = 8.71
EF = 345px = 9.86

Divide these by 2 and you get the radius.
conclusion

the image may be distorted so rounding the values somehow to integers we get the values

AB = 140px = 4
BC = 190px = 5
CD = 245px = 7
DE = 305px = 9
EF = 345px = 10

These look like the only reasonable integer values.

PS.

I will measure again in case I missed something.

The bottom of the largest (green) circle is at the same height as the top of the middle (orange) circle. This means that DE=(CD+EF)/2, or CD, DE, EF are an arithmetic progression. Similarly for the middle three circles - the right side of the yellow and the left side of the blue look like they have the same x coordinate. For the smallest three circles it is less clear, but I don't know if that is a slight inaccuracy of the drawing. Lets assume they are an arithmetic progression too.

This means that the whole sequence is an arithmetic progression, with an average of 35/5=7. Therefore the middle circle has diameter 7, and the sequence is 5,6,7,8,9 or 3,5,7,9,11, or 1,4,7,10,13. Since the smallest circle is about half the largest, I think it is 5,6,7,8,9.

It looks like

3 + 5 + 7 + 9 + 11 = 35

is the only possible solution;

the difference between the radiuses is approximately constant, and one of the requirements is that they're all integer. AB + BC is about as long as DE, so 5+6+7+8+9 is not possible.

On the other hand,

AB is longer than half of CD, so I don't think this has a proper solution.

• This is much better :) +1 Though, I am not sure this is anywhere close to be called as puzzle. – Techidiot Jan 6 '17 at 12:16
• I don't think this is it. Just from the picture AB looks to be a little more than half of CD. Or at least half of CD – Marius Jan 6 '17 at 12:17
• Yeah, that's what I realized later (see the edit). With the restriction of integer lengths, I don't think this problem has a solution at all. – Glorfindel Jan 6 '17 at 12:18

Possible Assumption(not sure if I am missing something)

AB = 2
BC = 6
CD = 8
DE = 9
EF = 10

Gives

$2+6+8+9+10=35$

Or another assumption -

$2+3+8+10+12 = 35$

Note - It's pure assumption and provided just by looking at the picture. :)