4 ms·
Monge's theorem
- maxk42 8y agoIs this the start of a new approach to public-key cryptography?
- throwawaymath 8y agoI strongly suspect that any cryptographically suitable problem derived from finding the intersection points of external tangent lines would reduce to the closest vector problem on lattices.
- dosy 8y agoOkay so circles to the line is a one-way function. But how to encrypt/decrypt a message?
- anonytrary 8y agoReminds me of flatland: There are three circular buildings. There's a line that all three people can be on such that each person thinks there are only two buildings and no one can agree on which two buildings are present. Edit: Nope, nevermind, two of the points both see the medium and small circles!
- rossdavidh 8y agoHowever, it does seem like they all see only two buildings, though. Interesting point!
- n4r9 8y agoThey all see at most two buildings. You can arrange the circles in a line so that each sees only one.
- ur-whale 8y agoThe wikipedia page has mostly geometric proofs. Are there algebraic proofs?
- contravariant 8y agoLet the circles be positioned at points p1, p2, p3 with radii r1, r2, r3 respectively. Note that we can get a family of circles with the same exterior tangents by simultaneously inter/extra-polating the position and radius of two other circles. Solving for a radius of 0 gives the following position for the intersection of those the tangents of circles 1 and 2. q3 = (r1 * p2 - r2 * p1) / (r1 - r2) (with similar expression for q1 and q2). Using homogenous coordinates we get the following list for q1, q2, q3: ( r2 p3 - r3 p2 ) ( r3 p1 - r1 p3 ) ( r1 p2 - r2 p1 ) ( r2 - r3 ), ( r3 - r1 ), ( r1 - r2 ) now note that r1 q1 + r2 q2 + r3 q3 = 0, showing that they're not linearly independent and hence collinear. A somewhat neater but more advanced proof, follows by making the positions pi also homogeneous and noting that (using Einstein summation notation): 0 = ε_ijk rj rk so we can determine the intersections to be qi = ε_ijk rj pk hence ri qi = ε_ijk ri rj pk = ε_jik rj ri pk = -ε_ijk ri rj pk = 0.
- fibo 8y agoI studied it at the University, it was introduced after Pappo's theorem and other theorem demonstrated using Descriptive Feometry most of all. On the other end it was a Projective Geometry course and at the end we used a lot the General Linear Group, in particular 4x4 matrices for bilinear forms, so it was like warming up with Geometry to arrive to the Algebra tools.
- pvg 8y agoHe's Pappus in English, for what it's worth.