[1] https://www.youtube.com/watch?v=6Lm9EHhbJAY (Euclid's Big Problem - Numberphile)
For example, you might say: (1) If you have 4 different points P, Q, R, S so that the lines PQ and RS are not parallel, you may "acquire" their point of intersection; (2) If you have 4 different points P, Q, R, S and the circle with center P and radius PQ intersects the line RS then you may "acquire" the point(s) of intersection of the circle and the line; (3) If you have 4 different point P, Q, R, S and the circle with center P and radius PQ intersects the circle with center R and radius RS, you can "acquire" the intersection point(s) of the circles. ("Acquire" a point means roughly that your number system now contains the coordinates of the point.)
Notice that this removes the problem of people doing unintended tricks with or drawing horribly complicated diagrams with physical compasses or straightedges, and makes mathematically precise what is allowed.
The following is probably more than you want to know, even though I'll omit lots of the details. The idea is that circles are defined by quadratic equations, and lines by linear equations. At every step, you're solving (acquring the roots of) simultaneous linear or quadratic equations. Starting with the rational numbers, as you "acquire" new points, you "extend" the rationals to bigger fields "by extensions of degree 1 or 2". Since degrees multiply, at any point, you've "extended" the rationals by a total degree 2^n.
So the result is: Theorem. Any point you construct must (a) lie in an extension of the rationals of degree 2^n; (b) be algebraic, in the sense that its coordinates are roots of rational polynomials.
Now take something like trisecting a 60 degree angle. A third of 60 is 20, and to construct a 20 degree angle you need to construct t = cos 20 degrees. But from trig, t is a root of x^3 - 3 x - 1 (which is irreducible over the rationals), so t has degree 3 --- and 3 is not a power of 2. Hence, a 60 degree angle can't be trisected.
Squaring the circle means given a circle, construct (the side of) a square with the same area. Take the circle to have radius 1, so its area is pi. The square you need would have side pi^(1/2), but pi (and pi^(1/2)) isn't algebraic in the sense noted above. (Lindemann showed pi is transcendental.) So you can't square an arbitrary circle.
Duplicating the cube means given a cube (say with sides of length 1), construct (the side of) a cube whose volume is twice the volume of the original cube (so in this case, the new cube should have volume 2, and its sides should have length 2^(1/3). But 2^(1/3) is a root of the irreducible rational polynomial x^3 -2, so it has degree 3, and again, 3 is not a power of 2.
What's remarkable about this is that, once you prove the theorem (which isn't that hard, just a little fussy), you can dispose of those three old contruction problems so easily.
(Anyone who wants more details - I've left out a lot - can consult any book on Galois theory. Also, Nathan Jacobson's "Basic Algebra I" covers this [https://store.doverpublications.com/0486471896.html] --- it's a classic of abstract algebra which I like a lot, though a little old-fashioned.)
(a) implies (b) here, but I’m sure you know this.
From myself, I recommend Aluffi’s “Algebra. Chapter 0”, it’s more modern and I find its style much more matching my taste.