B.5.RC:5.1 - Recovering an equilateral-triangle construction

A reader has two distinct points A and B in the Euclidean plane and needs an equilateral triangle on side AB. The description says to draw two circles, each centred at an endpoint and passing through the other endpoint, and use an intersection as the third vertex.

The reader recovers three operations: draw a circle with the given centre and radius; select a common point of the two circles; join two given points by a segment. The third vertex requires both circles. The final triangle requires that vertex together with A and B. The two circles share the segment length AB as radius.

For a small case, place A at (0,0) and B at (2,0). The circles have equations x²+y²=4 and (x−2)²+y²=4. Subtracting gives x=1, and substitution gives y²=3. Thus the two common points are (1,√3) and (1,−√3). Selecting C=(1,√3) supplies the vertex above AB. Joining A to C and B to C completes the construction.

The property follows from how C was obtained: AC and BC are radii of circles of radius AB, so AC=BC=AB. The coordinate calculation also supplies the intersection in the Euclidean-plane account used for this case. A description formulated under a different set of construction rules must obtain that intersection under those rules.

The recovered dependency is reusable for another positive side length. The value of the coordinates changes, while the two equal-radius circles and the common-point construction retain their roles. If A and B coincide, the initial requirement of a nondegenerate triangle fails; that case needs distinct endpoints before this construction can begin.

The Euclidean account supplies the circle and segment operations and their justification.