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Reversing Malfatti Circles

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If you only want to construct one just draw the “outer” exterior tangents for each pair of circles. Since each circle is tangent to two sides there can only be one solution, otherwise the exterior tangents would not be congruent. Actually calculating the side lengths is tedious but not impossible.
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For each pair of circles, you can draw exterior common tangents. If they have same radius, you can construct a rectangle joining circle centers and using perpendicular radii to find the points of tangency that define the common tangents.

If they have different radius, ra > rb, then you can find the position of the point P where the common tangents intersect using similar triangles and the idea that ra/rb = AP/BP, and from that AP = ra/(ra - rb) * AB if you are thinking in vector terms.

Then you can find the common tangent intersection points by constructing a circle with diameter BP and where it intersects circle B, and circle with diameter AP and where it intersects circle A.

Then after you draw the six tangents affiliated with the three circles considered in pairs, you'll be able to see which three contain the sides of the circumscribed triangle.

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