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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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The edited version is very different from the original. –  Mark Bennet Sep 25 '11 at 19:20
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Thanks Mark, I will create a new question for it –  Arthur Mamou-Mani Sep 25 '11 at 19:22
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3 Answers

up vote 5 down vote accepted

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

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Thank you Robert, does this mean that a circle could be the incircle of several triangles ? –  Arthur Mamou-Mani Sep 25 '11 at 18:53
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Yes, it does. A three-parameter family of such triangles, in fact. –  Robert Israel Sep 25 '11 at 20:32
    
Never thought about this before, but the rule seems to be this. Pick the first point anywhere at all. Pick the second point anywhere except antipodal to the first. Then the third point needs to lie in the open arc whose endpoints are antipodal to the first two points. –  Michael Hardy Sep 25 '11 at 21:29
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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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Robert Israel's suggestion of drawing just one circle and then picking three points and drawing tangent lines seems simpler. There's no rescaling step. –  Michael Hardy Sep 25 '11 at 21:26
    
@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. –  mixedmath Sep 25 '11 at 23:22
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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.

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