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This puzzle is inspired by this great puzzle.

You are given a circle. You can draw two non-overlapping triangles of any size and shape inside that circle. What is the highest percentage of the circle area you can cover with the triangles?

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    $\begingroup$ Well, the largest 4-gon in a circle is a square.… $\endgroup$
    – msh210
    May 15 at 9:21
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    $\begingroup$ This problem was too easy. And I think that someone who found this should come on with more interesting mathematics problems. $\endgroup$ May 15 at 13:01
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    $\begingroup$ @VictorStafusa ok ok I will try harder next time for you :) I was thinking about 3 triangles, but thought it would be too hard and perhaps too similar to the previous puzzle. $\endgroup$ May 15 at 14:17

2 Answers 2

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The answer is intuitively clear:

The two triangles must form an inscribed square, i.e. the ratio is 2/pi.

Formal proof is also not all that difficult:

As triangles are convex two nonoverlapping triangles can be separated by a straight line. This line cuts the enclosing circle in two complementary sections. The problem reduces to inscribing a single triangle in a circle section. The area of such a triangle is bounded by half the area of any circumscribed rectangle, in particular, the minimal one that shares the cut with the circle section. This rectangle has the same height as the section and width at most the diameter of the circle. The heights of the two sections sum to another diameter. The sum of the triangle areas is therefore bounded by 1/2 x diameter2.

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Cut a square out of the circle. Divide it in the diagonal to make two equal triangles.

Area of circle:

$\pi r^{2}$

Diagonal of square:

$2r$

Side of square:

$2r \times \frac{\sqrt{2}}{2} = r\sqrt{2}$

Area of square:

$(r\sqrt{2})^{2} = 2r^{2}$

Part of square in the circle:

$\frac{\text{area of square}}{\text{area of circle}} = \frac{2r^2}{\pi r^{2}} = \frac{2}{\pi} \approx 0.6366198 \approx 63.66198\%$

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