Field

Geometry — page 4

Shapes, and the arguments that can be made by rearranging them.
Two curves with five crossings: x⁶ + (44/31)y³ − y and its mirror. Two curves in the positive quadrant, each the zero set of a trinomial, crossing at five marked points.

Five where four were promised

In one unknown, a polynomial with three terms has at most two positive roots, whatever its degree. The natural guess for two equations in two unknowns, each with three terms, was two times two: four. It is wrong. Bertrand Haas found two such equations in 2002 whose curves cross five times in the positive quadrant, all five crowded within a tenth of one corner, and five turned out to be the most there can ever be.

456 pairs at one distance among 144 grid points. A 12-by-12 grid of points with 456 segments joining every pair at distance √25, in 6 directions.

How many pairs can be one apart

Place n points in the plane and count the pairs exactly one unit apart. A square grid with its spacing as the unit gives two pairs per point; the same grid with the unit chosen as a length it realises in many directions — the square root of 5, of 25, of 65, of 325 — gives three, five, eight. Counting points on circles caps the total at a constant times n to the four-thirds. Between that cap and the grids the truth has not moved since 1984, and at every size anyone can draw, the two look alike.

The lowest mode of the regular pentagon. Level lines of the first Dirichlet eigenfunction of a regular 5-gon; λ times area 18.9191.

Three sides and four are proved

Among all shapes of a given area, the disc has the smallest lowest eigenvalue. Among triangles it is the equilateral one, among quadrilaterals the square — both proved by sliding chords to an axis. For five sides the regular pentagon wins every computation, every nudge raises its value by the square of the nudge, and there is still no proof.

Morley's construction on a sphere. A spherical triangle with angles 76.1, 97.3, 48.8 degrees, its angle trisectors as great-circle arcs, and the inner triangle they form, with sides 0.23902, 0.23881, 0.23902.

The trisectors on a sphere almost agree

Draw Morley's construction on a sphere and the triangle the trisectors cut out is not equilateral — but its sides agree to within a tenth of a percent even when the angles add to 222°. Every step of the rotation proof survives on the sphere except one, and that one is the angle sum.

The medial axis of a wobbly rectangle, whole and trimmed. A rectangle with sides wobbling by ±0.02: its medial axis (654 Voronoi vertices, Hausdorff distance 0.499 from the clean axis) and its λ-medial axis at λ = 0.2 (63 vertices, distance 0.078).

A skeleton that ignores small teeth

The medial axis of a shape grows a branch to every bump on its outline, however small, so a scanned shape has a skeleton made mostly of noise. Keep only the points of the axis whose nearest boundary points are spread across a circle of radius at least λ, and the branches to bumps smaller than about 2λ disappear while the real skeleton stays. That λ-medial axis moves only a little when the outline wobbles by much less than λ — and the price is λ itself, a size below which a feature is declared not to exist, which can even cut a shape's skeleton in two.

Malfatti's three circles against the three largest, one at a time. A triangle with angles 40, 70, 70 degrees: Malfatti's circles cover 69.1% of it, the greedy choice of incircle then largest remaining circles covers 73.7%.

Three circles that touch and are not the largest

In 1803 Gian Francesco Malfatti asked how to cut three round columns from a triangular block of marble with as little waste as possible, and answered with three circles each touching the other two and two sides. The circles exist in every triangle and are found by solving three equations. They are never the answer to his question: the plain greedy rule — the inscribed circle first, then the largest circle that still fits, twice — always does better, by 1.4 per cent on the equilateral triangle and by almost double on a thin one.

The butterfly: wings that cut a chord equally. A circle with chord PQ, its midpoint M, two chords through M, and the wings AD and BC meeting PQ at X and Y, each 0.476 from M.

Wings that cut a chord equally

Take a chord of a circle and its midpoint. Draw any two more chords through the midpoint, join their ends crosswise, and the two crossing lines — the butterfly's wings — cut the first chord at equal distances from the middle. The proof is the inscribed angle and the power of a point working together; move the point off the middle and what survives is a law about reciprocals; replace the circle by any conic and the theorem does not notice.

Two ways to cut a hexagon, one total of radii. A cyclic hexagon triangulated two ways with the incircles drawn; the inradii sum to 1.1622 both times.

A total hung in a temple

Put a polygon's corners on a circle, cut it into triangles, and add up the radii of the circles inscribed in the triangles. Cut it a different way and the total is the same — for all fourteen ways of cutting a hexagon, to every digit. The fact was painted on a wooden tablet in a Japanese temple around 1800, and the reason for it is a theorem about one triangle and the distances from its circumcentre to its sides.

The five lunes that can be squared. Five crescents with arc ratios 2:1, 3:1, 3:2, 5:1, 5:3 and their equal-area kites; θ = 45.000°, 34.265°, 26.812°, 23.439°, 16.794°.

The five crescents that can be squared

The circle cannot be squared, but some regions bounded by circular arcs can: Hippocrates of Chios found a crescent with exactly the area of a triangle around 440 BC. The trick is to make two multiples of π cancel, and it works for a crescent whose two arcs turn through angles in the ratio m to n exactly when an angle defined by a polynomial can be drawn with ruler and compass. That happens for five ratios and no others.

The point nearest in total to three corners, and the lines that find it. A triangle with equilateral triangles on its sides, the three lines from each corner to the opposite apex meeting at the Fermat point, where the sides subtend 120°; least total distance 7.7090.

The point nearest in total to three corners

Three towns want one depot with the least total road to all of them. The answer is not the centroid or any of the classical centres: it is the point from which every side of the triangle is seen at 120°, found by building equilateral triangles on the sides. Three straight lines through it have the same length, and that length is the least total. When an angle of the triangle reaches 120°, the answer jumps to that corner — and the same 120° rule shapes soap films and the shortest networks joining many points.

Seven towns joined with junctions and without. Seven random points: spanning tree 1.65407, Steiner tree 1.54997 with 2 junctions, ratio 0.93707.

What a junction can save

Join a set of towns by roads that may meet only at towns, and the shortest such network is easy to find. Allow junctions anywhere and the network gets shorter — but never, it is conjectured, by more than 13.4%, the saving the Fermat point makes on an equilateral triangle. Computed exactly on 1,304 random sets of up to seven towns, the average saving is about 3%. An adversarial search drives it down to the bound and no further, and a hexagon with its centre attains it exactly. The proof of the bound that was accepted for twenty years has a gap.

Every finite kaleidoscope on up to eight mirrors. 1: A₁; 2: A₂ B₂ H₂ G₂; 3: A₃ B₃ H₃; 4: D₄ A₄ B₄ H₄ F₄; 5: D₅ A₅ B₅; 6: E₆ D₆ A₆ B₆; 7: E₇ D₇ A₇ B₇; 8: E₈ D₈ A₈ B₈.

One determinant for every kaleidoscope

Three mirrors close up into a kaleidoscope when the angles of their triangle add to more than half a turn. In more dimensions there is no triangle to add up, but there is a matrix — the inner products of the mirrors' directions — and the mirrors close up exactly when it is positive definite. Trying every arrangement of up to nine mirrors against that one test finds Coxeter's list: three families in every dimension, a fork that runs out at eight mirrors because its determinant reaches zero at nine, and five exceptions. The same search finds the hyperbolic kaleidoscopes, and finds that they stop at five mirrors.

Two triangulations, the same sum of cos 3x, different sums of cos 2x. Cyclic heptagon, triangulations 0 and 29: Σcos3x 1.52896873 / 1.52896873; Σcos2x -0.789261 / -0.467086.

The sums a triangulation cannot change

The Japanese theorem says the inradii of any triangulation of a cyclic polygon add to the same total. Write the inradius as a sum of cosines of the angles, and the theorem becomes one case of a rule that fits in a line: a sum of f over every angle of every triangle ignores the triangulation exactly when f(x) + f(π − x) is constant. cos 3x qualifies, cos 2x does not, and the proof is the inscribed-angle theorem applied to a single flipped diagonal.

Billiards in an ellipse: every path keeps touching one confocal conic. Two billiard trajectories in the ellipse x² + y²/0.49 = 1, one tangent to a confocal ellipse and one to a confocal hyperbola.

The caustics only an ellipse keeps

A billiard ball in an elliptical table touches the same confocal ellipse or hyperbola with every segment of its path, for ever. Bend the table by a percent and a half and most of those curves survive, but the one turning a quarter-turn per bounce breaks into a chain of four islands whose width grows like the square root of the bend. Birkhoff asked in the 1920s whether the ellipse is the only table that keeps them all; it is still not known.

The tree of solutions of x² + y² + z² = 3xyz. Markov triples from (1, 2, 5) to depth 4; 231 Markov numbers below 10^15; first 1, 2, 5, 13, 29, 34, 89, 169, 194, 233, 433, 610, 985, 1325.

The numbers that come after the golden ratio

The golden ratio is the number worst approximated by fractions: no constant larger than √5 works in Hurwitz's theorem. Set it aside and √2 is worst, with √8. Set that aside and the next is a number whose continued fraction repeats 2, 2, 1, 1, with √221/5. Markov proved in 1879 that the list continues for ever below 3, one entry for each solution of x² + y² + z² = 3xyz.

Cells of points scattered at random. Poisson–Voronoi tessellation of 600 points on a torus; mean sides 6.000000000; side counts 3: 9, 4: 56, 5: 165, 6: 167, 7: 131, 8: 49, 9: 18.

The shape of a cell around a random point

Scatter points at random and give each the region nearer to it than to any other. The cells come in every shape from triangles to twelve-sided polygons, and almost nothing about them is fixed by a formula — except one thing: on a closed surface they have exactly six sides on average, every time, because three cells meet at every corner. Two hundred thousand cells measure the rest: 29.5 per cent hexagons, Lewis's law, and small cells surrounded by large ones.

A cube passes through a hole in a smaller cube. Cube passage: λ = 1.060660172 (3√2/4 = 1.060660172); outer hexagon, inner square.

A cube through a hole in a cube

Prince Rupert of the Rhine bet that a hole could be cut through a cube large enough for an equal cube to pass through it, and won. The trick is to look at the cube along a direction near its long diagonal, where its shadow is a hexagon wide enough to hold a square face with room to spare: a cube 6% larger can pass. All five Platonic solids turn out to have such a passage, found by searching over the directions from which a solid is seen — and in 2025 a solid was found that has none.

Where charges settle on a sphere. N=4: E=3.674234614, distinct minima from 8 starts: 1; N=6: E=9.985281374, distinct minima from 8 starts: 1; N=8: E=19.675287861, distinct minima from 8 starts: 1; N=12: E=49.165253058, distinct minima from 8 starts: 1; N=20: E=150.881568334, distinct minima from 8 starts: 1; five charges: bipyramid 6.474691, best square pyramid 6.483661.

Where charges settle on a sphere

Put equal electric charges on a sphere and let them push each other apart until nothing moves. Four settle at the corners of a tetrahedron, six of an octahedron, twelve of an icosahedron — but eight do not make a cube, and twenty do not make a dodecahedron. The cube loses to two squares turned half a corner apart, because charges on a sphere prefer triangles; and however many charges there are, exactly twelve more of them have five neighbours than have seven.

Right triangles of area five, six and seven. area 5: 3/2, 20/3, 41/6; area 6: 3, 4, 5; area 7: 24/5, 35/12, 337/60.

The areas a right triangle can have

A right triangle with sides 3, 4 and 5 has area 6; one with sides 3/2, 20/3 and 41/6 has area 5. No right triangle with rational sides has area 1, 2 or 3. Which whole numbers are areas is a thousand-year-old question, and its best answer — two counts of solutions to quadratic equations — rests on one of the great unproved conjectures.

The largest hexagon of diameter one. Regular hexagon of diameter 1: area 0.649519053; best hexagon: area 0.674981443, 6 pairs at distance 1; vertices (0.3438, -0.3253) (0.5000, 0.2115) (-0.0000, 0.6138) (-0.5000, 0.2115) (-0.3438, -0.3253) (-0.0000, -0.3862).

The largest polygon of diameter one

Among pentagons whose points are never more than one unit apart, the regular pentagon has the most area, and the same is true for every odd number of sides. For six sides it is false: a lopsided hexagon found in 1975 beats the regular one by four per cent. Symmetry wins half the cases and loses the other half, and the reason is parity.

Two valleys for sixteen charges. N = 16: 92.911655 (48 starts), 92.920354 (12 starts).

Valleys a search cannot count

Let equal charges on a sphere push each other apart from a random start, and they settle into an arrangement where nothing moves. Up to fifteen charges every start settles into the same one. Sixteen is the first number with two, a hundredth of a unit apart in energy; by fifty-two there are at least four, and for fifty-six charges only one start in ten finds the lowest. The other valleys sit a few parts in a million above it, and nothing in the arrangement a search returns says whether it is the bottom.

What junctions save in space. tetrahedron SMT/MST 0.81305; 12-point helix 0.79472.

A chain of tetrahedra saves more

In the plane, junctions never seem to shorten a network of roads by more than the equilateral triangle's 13.4%. In space the regular tetrahedron allows 18.7%, and the obvious guess — that it is the worst case, as the triangle is in the plane — is wrong. Points spaced along a chain of tetrahedra glued face to face do better with every point added: 20.5% at twelve, 21.4% at sixty, and about 21.6% in the limit. Random points come nowhere near, and a search that only goes downhill does not find the chain.

Numbers made of pyramids. below 1000000: 241 need five; first 17, 27, 33, 52, 73, 82, 83, 103, 107, 137, 153, 162; last 343867 = 341376+1771+680+20+20.

Five pyramids for every number

Stack balls in triangular pyramids and the piles hold 1, 4, 10, 20, 35, … balls. Frederick Pollock guessed in 1850 that every whole number is a sum of at most five such piles. Seventeen needs five — 10 + 4 + 1 + 1 + 1 — and so do 240 other numbers, the largest 343,867; past that, every number up to ten million needs four or fewer. Eight is proved to be enough. Five is not, and the reason is the same one that makes the three-dimensional cousins of the classical theorems hard.

A box with whole diagonals on every face. 44,117,240: face diagonals 267, 244, 125; space diagonal squared 73225.

A box with every face diagonal whole

Pythagoras gives a rectangle with whole sides and a whole diagonal — 3, 4, 5. A box with whole edges and whole diagonals on all three faces is far rarer: the smallest is 44 × 117 × 240, found in 1719, and there are only 113 essentially different ones with every edge below 300,000. Ask for the long diagonal through the box to be whole as well and the answer is unknown. No such perfect box has ever been found, and nobody has proved there is none.

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