Isocseles I Saw

by Rich Holmes, posted in Geometry

Another Catriona Shearer puzzle.

13.
Isosceles I Saw

All 4 triangles are isosceles. What’s the angle?

Spoiler!

I have a feeling there is a simpler solution but this is all I’ve got…

The four isosceles triangles are: BCE, CEA, ADC, and BAC.

If the base of triangle BCE is b and its two other sides are s = 1, the base of triangle BAC is 1 and the two other sides are 1+b. They are similar, so b/1 = 1/(1+b), with solution b = (\sqrt{5}-1)/2. b/2 is the cosine of angle CBE; that angle then is 72^\circ.

As a consequence of the inscribed angle theorem, angles CBE and ADC are supplementary, so the latter is 108^\circ. I didn’t use the fact ADC is isosceles; in fact as long as D is on the circle between A and C, the angle remains the same.

One thought on “Isocseles I Saw”

  1. I just realized the whole thing could be part of a regular pentagram, so if there is an answer, it’d have to be that.

    Reply

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