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Recently I came across a question to show that an equilateral triangle can be inscribed in an epitrochoid (Calculus by Stewart, Chapter 10 Challenge Problems), along with a solution in this link:

http://www.stewartcalculus.com/data/CALCULUS%20Early%20Transcendentals/upfiles/WebChallengeProblems_5ET.pdf

(The link has both problem and answer.)

I've tried to do the same question and what I managed to come up with, via an argument on equidistances between points of angular separation $\frac{2\pi}{3}$ using polar coordinates, that every point of the epitrochoid is a vertex of an equilateral triangle with centroid lying on the circle centred at the origin with radius $b$. It's incredibly tedious, but it worked.

What I can't prove, however, is that the triangle must be inscribed within the epitrochoid, like the wording of the solution suggests. The height of the triangle is $4.5r$, where $r$ is as given in the question. The $y$-intercept distance, however, is $6r-2b$. It seems to me that since $0<b<r$, for $b > 0.75r$ the triangle's edges will intersect the epitrochoid as the triangle's vertices sweep out the boundary of the epitrochoid, so not every equilateral triangle must necessarily be inscribed in the epitrochoid.

Could anyone perhaps shed some light on this? Specifically, is my argument right, or is there some flaw in the logic I'm overlooking? Thanks!

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