Let p be a point in the square ABCD, and the distances from point P to vertex ABC are 1, 2 and 3 respectively

Rotate △ BPC counterclockwise around point B to △ BEA, connect EP, so EP = 2, radical 2, EA = 3, AP = 1, ad ^ 2 + EP ^ 2 = AE ^ 2, so △ AEP is a right triangle, so ∠ ape = 90, so ∠ APB = 90 + 45 = 135, according to cosine theorem, AB ^ 2 = AP ^ 2 + BP ^ 2-2 * AP * BP * cos135 = 5 + 2 √ 2, so AB = √ (5 + 2 √ 2)

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