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Is it possible to draw a triangle from a circle so that it becomes its incircle ?

Many thanks,

Arthur

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    Thanks Mark, I will create a new question for it2011-09-25

3 Answers 3

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Draw tangent lines at three points on the circle, and see where they intersect.

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    Never thought about this be$f$ore, but the rule seems to be this. Pick the $f$irst point anywhere at all. Pick the second point anywhere _except_ antipodal to the $f$irst. Then the third point needs to lie in the open arc whose endpoints are antipodal to the $f$irst two points.2011-09-25
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It is. But it is not unique, i.e. infinitely many triangles can be drawn from a single circle. (To see this, draw many non-similar triangles, find their incircles, and then scale them so that the circles are all the same size. Then the triangles have the same incircle, though they're different).

I think the easiest would be to fit an equilateral triangle around it. From the center of the circle, mark off 120 degree arcs, and then place the tangent to each arc. These lines will form an equilateral triangle whose incircle is the desired circle.

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    @Michael: That wasn't an answer to the question - the rescaling was to show that the answer is not unique. If you notice, my equilateral triangle is exactly his answer: find three points, draw tangents. Mine just happen to be those of an equilateral triangle.2011-09-25
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Given a circle, and angles for your triangle of $\alpha, \beta, \gamma$ mark off radii with angles between them at the centre of the circle of $(180-\alpha)^\circ, (180-\beta)^\circ, (180-\gamma)^\circ$ [total $360^\circ$]. The tangents at the points where the radii meet the circle will make a triangle with the desired angles. The radii can be chosen to determine the orientation of the circle. Given a circle, it is therefore possible to draw a triangle which is both similar to, and similarly oriented to, any given triangle, and which has the given circle as its incircle.