Looped triangles – my solution

Up early on a Sunday for my long run (9 miles today, in case you wondered), I came across a tweet from @srcav:

As I had time to kill whilst my breakfast digested, I thought I’d have a go. Below is my solution. There has been some discussion as to whether this is right or not. I welcome any comments/corrections.

Initial thoughts – I looked at the site from which the puzzle was taken (http://fivetriangles.blogspot.co.uk/) and saw the following information:

“All Five Triangles problems share a characteristic: however opaque they may appear initially, none requires more than common mathematics skills learnt in classrooms throughout the world.”

I decided to solve the problem using angles and circle theorems. First, I worked out the angles I needed. I took the approach that there are 5 segments of 5cm in the diagram which run parallel to the line segments AC, CE, EB, BD, and AD. My belief (and this is where I may have gone wrong) is that this is effectively the tangent, where to string leaves the circumference of the circle. Because the tangent is always at 90 degrees to the radius, I thought we might be able to use this to find the sector of the circle the string does not touch, and use this work out the circumference touched by the string, and therefore the solution.

IMG_0002

Given that the interior angle of a pentagon is 108 degrees, the exterior must be 252 degrees. Because ABC is an isosceles triangle ( as AB = BC), angle BAC = 36 degrees. If we draw the line AD, we can see that angles CAD and DAE are also equal to 36 degrees. If we add in a point F so that segment AF bisects the reflex BAE, then FAE = 126 degrees, and FAD = 126 + 36 = 162 degrees. The same is true for FAC.

IMG_0003 (1)

I think a key part of the problem (and again, this may be a mistake in my thinking) is that AC is parallel to the segment of string running from the right hand side of the circle around point A to the right hand side of the angle around point C. The same is true for the other 4 5cm segments. I said above that FAC = 162 degrees. Drawing the tangent to the circle from point A gives a 90 degree angle, which leaves a remaining angle of 72 degrees. This means the difference between the angle from A to the tangent, and the angle created by drawing a perpendicular to FA, is 18 degrees. I therefore think that on each side of each circle, the string remains in contact for a further 18 degrees. This gives a total sector of 216 degrees.

(Interestingly enough, this means that the sector of the semi circle touched by the string is 108 degrees – I wonder if this is significant given the context of the problem?)

Taking this as a fraction of 360, and then multiplying by 5 (for there are 5 circles), gives an answer of 3. The circumference of each circle is 2 pi, so the total curved distance of the string is 6 pi. All that remains is to add the 25cm for each segment, giving a total of (25 + 6 pi)cm, or 43.85cm to 2 decimal places.

As I say, I may be wrong. Shoot away!

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