Buridan’s Bridge

https://commons.wikimedia.org/wiki/File:Buridan%27s_bridge.jpg

Socrates wants to cross a river and comes to a bridge guarded by Plato. The two speak as follows:

Plato: ‘Socrates, if in the first proposition which you utter, you speak the truth, I will permit you to cross. But surely, if you speak falsely, I shall throw you into the water.’

Socrates: ‘You will throw me into the water.’

Jean Buridan posed this conundrum in his Sophismata in the 14th century. Like a similar paradox in Don Quixote, it seems to leave the guardian in an impossible position — whether Socrates speaks truly or falsely, it would seem, the promise cannot be fulfilled.

Some readers offered a wry solution: Wait until he’s crossed the bridge, and then throw him in.

| Oddities

Saving Time

https://commons.wikimedia.org/wiki/File:Recursive_maze.gif
Image: Wikimedia Commons

Above: A valid maze can be generated recursively by dividing an open chamber with walls and creating an opening at random within each wall, ensuring that a route can be found through the chamber. The secondary chambers themselves can then be divided with further walls, following the same principle, to any level of complexity.

Below: Valid mazes can even be generated fractally — here a solution becomes available in the third panel, but an unlucky solver might wander forever in the depths of self-similarity at the center of the image.

https://en.wikipedia.org/wiki/File:Wolfram_fractal_maze.svg
Image: Wikimedia Commons
| Science & Math

The Bride’s Chair

https://commons.wikimedia.org/wiki/File:Illustration_to_Euclid%27s_proof_of_the_Pythagorean_theorem2.svg

This is Euclid’s proof of the Pythagorean theorem — Schopenhauer called it a “brilliant piece of perversity” for its needless complexity:

  1. Erect a square on each leg of a right triangle. From the triangle’s right angle, A, draw a line parallel to BD and CE. This will intersect BC and DE perpendicularly at K and L.
  2. Draw segments CF and AD, forming triangles BCF and BDA.
  3. Because angles CAB and BAG are both right angles, C, A, and G are collinear.
  4. Because angles CBD and FBA are both right angles, angle ABD equals angle FBC, since each is the sum of a right angle and angle ABC.
  5. Since AB is equal to FB, BD is equal to BC, and angle ABD equals angle FBC, triangle ABD is congruent to triangle FBC.
  6. Since A-K-L is a straight line that’s parallel to BD, rectangle BDLK has twice the area of triangle ABD, because they share base BD and have the same altitude, BK, a line perpendicular to their common base and connecting parallel lines BD and AL.
  7. By similar reasoning, since C is collinear with A and G, and this line is parallel to FB, square BAGF must be twice the area of triangle FBC.
  8. Therefore, rectangle BDLK has the same area as square BAGF, AB2.
  9. By applying the same reasoning to the other side of the figure, it can be shown that rectangle CKLE has the same area as square ACIH, AC2.
  10. Adding these two results, we get AB2 + AC2 = BD × BK + KL × KC.
  11. Since BD = KL, BD × BK + KL × KC = BD(BK + KC) = BD × BC.
  12. Therefore, since CBDE is a square, AB2 + AC2 = BC2.

The diagram became known as the bride’s chair due to a confusion in translation between Greek and Arabic.

| Science & Math

Bid the Tree Unfix

https://commons.wikimedia.org/wiki/File:Buckland_Yew_1880.jpg

In 1880, an 800-year-old yew tree was threatening the west wall of the church of St Andrew at Buckland in Dover. The community called in landscape gardener William Barron, who solved the problem by boring tunnels under the trunk and then raising the tree’s entire 55-ton mass onto rollers by means of powerful screw jacks. Giant windlasses could then haul the tree 203 feet across the churchard to a safer location.

“The scale of this operation was probably never matched,” writes G.M.F. Drower in Garden of Invention, his 2003 history of gardening innovations. “[A]nd Barron, who had been rather more apprehensive than he let on, later admitted that all the other trees he had moved had been ‘chickens compared to the Buckland Yew.'”

| Technology

Note Taking

https://commons.wikimedia.org/wiki/File:Ut_Queant_Laxis_MT.png

Where did the familiar syllables of solfège (do, re, mi) come from? Eleventh-century music theorist Guido of Arezzo collected the first syllable of each line in the Latin hymn “Ut queant laxis,” the “Hymn to St. John the Baptist.” Because the hymn’s lines begin on successive scale degrees, each of these initial syllables is sung with its namesake note:

Ut queant laxīs
resonāre fibrīs
Mīra gestōrum
famulī tuōrum,
Solve pollūti
labiī reātum,
Sancte Iohannēs.

Ut was changed to do in the 17th century, and the seventh note, ti, was added later to complete the scale.

| Art · History · Language