I would like to start a new column of mathematical puzzles. Scientific American has one; New Scientist has one; so I hope Science 2.0 will be happy to host one!

Preamble over, here's your "started for 10".

The diagram shows five circles, each with integer radius, all touching the base of the large triangle. The four smaller circles all touch their two neighbouring circles, with the large circle touching all four. The two sides of the triangle each touch two of the circles.

Let the radii of the circles be a, b and c, such that a > b > c.

Given that c has a length of 4 units, find the area of the large triangle. You may leave it as an exact solution.

Supplementary Question

Using the relationship between a, b and c, found in the first question, we can set the value of c to any square number and this will generate values for a and b.

However, if we set c=9 the triangle that encloses the circles flips over. The horizontal line used to construct the circles is now no longer the base of the whole triangle. We can redefine the triangle: the two equal sides are still the tangents of circles A and B, but the base is now the horizontal tangent to A so that the triangle still encloses the five circles.

What are the values of a and b for c=9 and why does the triangle flip over?

What happens for higher values of c, such as c=16, 25, 36 and so on?

Do the areas of the sequence of triangles generated by setting c as a sequence of squares tend towards a fixed vaue, or continue to increase indefinitely?

Full solution will be posted when a new question is published. There are, already, solutions below.

Update 7 March: have edited the above question for clarity and to include the extension problems discussed in the comments. Thanks to all participants.

11 March: a solution has now been published.

Circles Stuck in a Triangle

## Comments

Never mind "from". I would quite like to know how you get *to* that expression.From(a + b)^{2}= (a - b)^{2}+ [2(b*c)^{0.5}+ 2(a*c)^{0.5}]^{2}

derek_potter (not verified) | 03/03/13 | 18:26 PM

Well it wasn't entirely obvious you were talking about geometry :)

Okay, I guess you mean this:

Don't worry about me spoiling your "leave the geometry as an exercise for the reader" didactic: by far the most important step is the realization that you only need to plod through a few cases before the tangents fail to converge to the required triangle. Thanks for the neat explanation.

Umm, here's a thought. How would this puzzle go in [1 -1] space? :)

derek_potter (not verified) | 03/04/13 | 07:15 AM

Well it wasn't entirely obvious you were talking about geometryIf you hadn’t got there first, the formulae would have reminded me of

**Heron**— the 1st century Greek geometer who lived in Alexandria, not the one pictured below!

Another thought — this looks like a

**Diophantine**problem, named after an even later Greek.

—

Robert H. Olley
/ Quondam Physics Department
/ University of Reading
/ England
Robert H Olley | 03/04/13 | 12:47 PM

Well it looks to me as though c = 9, b = 18 and a = 72. As these are only the radii of the 3 circles then their diameters are double that, ie 18, 36 and 144 . As the area of a triangle is height x base divided by 2, the area of this triangle appears to be 324 X 216 divided by 2 which equals 34992 square units.

—

My 5 min film 'Hidden Dangers for ALS' entry in the AAN #2015Neurofilm Festival is listed no. 21 of 65 entries at https://www.youtube.com/playlist?list=PLYNSC2T2aPD7GH6pVWI7I6DjzwbbBKCWK
Helen Barratt | 03/03/13 | 13:03 PM

Well it looks to me as though c = 9, b = 18 and a = 72.Why?

derek_potter (not verified) | 03/03/13 | 17:50 PM

9 = (a*b)/((sqrt(a) + sqrt(b))^2)

this yields a solution of the circles being tangent to each other and the horizontal line if a=144, b=16 and c=9, however at these radius values the tangent lines only touching the A and B circles will not intersect at the top and the three lines will not form a triangle which encloses the circles.

Rick J (not verified) | 03/04/13 | 11:11 AM

Richard Mankiewicz | 03/04/13 | 21:35 PM

^{2}= (a - b)^{2}+ [2(b*c)^{0.5}+ 2(a*c)^{0.5}]^{2}it is suggested to use squares for c (as you did) and b in order to make a integer:

a = c*b > b > c

Eg:

36 > 9 > 4

144 > 16 > 9Now the area is straight forward.