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Decided by exhaustion — page 12

Questions with finitely many cases, settled by going through all of them — and what changes when a claim about every argument becomes a count.
Two hundred and three points from two. After two rounds: 203 points, 11 rational, 76 in ℚ(√3) but not rational, 116 needing a new square root (√2: 20, √5: 22, √7: 14, √11: 36, √13: 20, √35: 4); 10 lines and 16 circles in the second round. Computation

What two points build in two rounds

Start with two points and draw every line and circle they allow; mark every crossing; do it again. The first round gives six points, the second gives 203, and the third gives over 1.7 billion. Computed exactly, the 203 points show the tower of square roots growing in real time — eleven rational points, seventy-six that need √3, and the rest needing one of six new square roots.

 Fifteen logics from five axioms. K: rank 0, covered by D, KB, K4, K5; D: rank 1, covered by T, KDB, D4, D5; T: rank 2, covered by B, S4; KB: rank 1, covered by KDB, KB5; KDB: rank 2, covered by B; B: rank 3, covered by S5; K4: rank 1, covered by D4, K45; D4: rank 2, covered by S4, KD45; S4: rank 3, covered by S5; KB5: rank 3, covered by S5; S5: rank 4, covered by nothing; K5: rank 1, covered by D5, K45; D5: rank 2, covered by KD45; K45: rank 2, covered by KB5, KD45; KD45: rank 3, covered by S5. Logic

Five axioms and fifteen logics

Five familiar axioms of modal logic can be added to the basic system in thirty-two combinations. Only fifteen different logics come out, because the axioms are properties of arrows between worlds and some properties force others. Checking every frame with up to four worlds finds the fifteen, the order among them, and the small pictures that tell each from its neighbours.

 Five squares on one octagon. Octagon [[1.0756,0.5516],[0.6082,1.2964],[-0.466,1.2294],[-0.8758,0.3153],[-0.797,-0.3348],[-0.3706,-0.7683],[0.2718,-0.6558],[1.1693,-0.4453]]; squares with sides 1.4975, 1.4588, 1.4407, 1.3929, 1.3921. Topology

An odd number of squares on every polygon

Does every closed curve that does not cross itself pass through the four corners of some square? Toeplitz asked in 1911, and for polygons and smooth curves the answer is yes, because the squares can be counted and the count is odd. For rectangles of a fixed shape the count is even, which is why they needed a different proof — and for an arbitrary continuous curve the square is still not proved to exist.

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