Three circles and a triangle
Three circles with equal radii are centered at ( 1 , 0 ) , ( 0 , 0 ) , and ( 0 , 1 ) .
To 4 decimal places, what is the smallest radius of the circles such that we can pick a point within each circle--like A, B, C in the diagram--and connect them to form an equilateral triangle?
The answer is 0.1725.
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1 solution
You seem to be making some assumptions about similarity / symmetry. It's not obviously clear to me that they are justified.
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can u please post a solution to this problem?
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.
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This is a quite tidy solution which you should add as a separate thread. It's also quite easy to resolve using numerical methods.
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Numerical methods? Like you're talking about some monte carlo approach? Even still, isn't that a non-rigorous solution because you didn't prove that the answer is exactly 3 2 sin 1 5 ∘ ?
How is ON = 2 O A ?
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<CAB = 60 deg. , CA//MO , <CAW = <MOW = 30 deg.
ON is perpendicular to AC
Tr.ONA is 30-60-90 ---> ON/OA = sin30 = 1/2
I managed to come up with a solution with a smaller radius, 0.1645. My solution has B and C on the segment joining (1, 0) and (0, 1). In fact, B and C are the intersections of that segment and the the two circles. Here are details: A is (-R, -R), B is (R/sqrt(2), 1 - R/sqrt(2)) C is (1 - R/sqrt(2), R/sqrt(2))
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But A is not within the circle then.
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Thanks for checking this for me, Geoff. You are right, A should have been (-R/sqrt(2), -R/sqrt(2)), which gives me R = 0.1895.
All these solutions seem to hold some great wisdom that eludes me. I can see that the 75 degree angle features but ended up equating sides of the triangles which results in solving the quadratic
[2 * sqrt (6) + sqrt (2) - sqrt (14 + 8 * sqrt (3))]/6.
Easy with a calculator, unwieldy with pen and paper.