Advanced Geometries

Solution.

1. Now view the scene through a pinhole camera at height $1$ above the ground, looking down the track. A ground point $(x,y,1)$ projects to the image point whose horizontal position stays proportional to $y$ but whose apparent size shrinks like $1/x$ as the point recedes. As $x\to \infty$, the images of both rails approach the same pixel on the horizon — the vanishing point — because the difference in their heights, proportional to $2/x$, goes to $0$.
2. Projective geometry records this by giving the two rails one shared "point at infinity" attached to their common direction. Sanity check: the two rails share a direction and now share exactly one added point, so they meet in exactly one point — consistent with the projective rule that any two lines meet once. Lines with a different direction (say $y=x$) get a different point at infinity, so they still only meet the rails at one finite point, never at the horizon.

Solution. Take the finite line $y=2x+3$ and homogenize it. Substitute $x\mapsto x/z$ and $y\mapsto y/z$ (which recovers affine coordinates when $z=1$) into $y=2x+3$:

Now find its point at infinity by setting $z=0$: the equation becomes $2x-y=0$, i.e. $y=2x$, giving the point $[1:2:0]$. Sanity check: the direction of the original line is $(1,2)$ (slope $2$), and $[1:2:0]$ encodes exactly that direction, so every line of slope $2$ passes through this same point at infinity — which is why all lines of a given slope are "parallel" only in the affine picture and concurrent in the projective one.□

Now apply the projective map $x\mapsto x^{\prime }=\frac{x}{x+1}$ (a fractional-linear map, the kind induced by a projection). The images are $A^{\prime }=0$, $B^{\prime }=\frac{1}{2}$, $C^{\prime }=\frac{2}{3}$, $D^{\prime }=\frac{4}{5}$. Recompute:

Sanity check: both computations give $\frac{3}{2}$, confirming the cross ratio is invariant even though every individual coordinate changed.□

Solution. Projective view. Homogenize: $L_1$ is $y=0$ and $L_2$ is $y=z$, i.e. $y-z=0$. Solve the two equations together: $y=0$ and $y=z$ force $z=0$, giving the point $[1:0:0]$. So projectively the lines do meet, at a point at infinity. Affine view. Now single out $z=0$ as the forbidden "line at infinity." The meeting point $[1:0:0]$ lives on it, so it is no longer a legal point. With it removed, $L_1$ and $L_2$ have no common point: they are parallel. Affine geometry is exactly this — projective geometry with one line demoted to "infinity," which is what first lets the word "parallel" mean anything. Euclidean view. Add a metric. Now we can say $L_1$ and $L_2$ are parallel and a perpendicular segment between them has length $1$; angles and distances become measurable on top of the affine structure.

Sanity check: each added layer only adds structure (a distinguished line, then a metric); the incidence facts from the layer below still hold.□

Check Axiom 1 (two points lie on a unique line) for the pair $\{2,7\}$: scanning the list, only $\{2,5,7\}$ contains both $2$ and $7$ — exactly one line, as required. Check Axiom 2 (two lines meet in a unique point) for $\{1,4,5\}$ and $\{3,4,7\}$: their only common element is $4$ — exactly one point. Check Axiom 3 (four points, no three collinear): take $1,2,4,7$; no line of the list contains three of them (e.g. $\{1,2,3\}$ misses $4,7$), so the non-degeneracy axiom holds.

Sanity check by counting: $7$ lines $\times$ $3$ points $=21$ point-on-line incidences, and $7$ points $\times$ $3$ lines $=21$ — the two counts agree, as they must. It satisfies every axiom yet is entirely finite, proving projective planes need not be the "real" one.□

For $q=2$ this gives $2^2+2+1=7$, recovering the Fano plane $\mathrm{PG}(2,2)$: $7$ points, $7$ lines of $3$ points each. Sanity check with $q=3$: $\mathrm{PG}(2,3)$ should have $3^2+3+1=13$ points and $13$ lines, each line holding $q+1=4$ points — the known order-$3$ projective plane.□

Solution.

1. Every projective line except $z=0$ itself, say $2x-y+3z=0$, restricts (set $z=1$) to the ordinary line $2x-y+3=0$. Two projective lines that used to meet on $z=0$ — for instance $y-z=0$ and $y-2z=0$, both passing through the infinity point $[1:0:0]$ — restrict to $y=1$ and $y=2$, which now never meet: they have become parallel. Sanity check: exactly the lines that shared a point on the deleted line at infinity are the ones that turned parallel, so "parallel" is precisely "meets on the removed line."

where $K$ is the Gaussian curvature and $\alpha ,\beta ,\gamma$ are the interior angles in radians. The right-hand side is exactly the angle excess. This one formula unifies all three geometries:

Euclidean ($K=0$): the integral is $0$, so $(\alpha +\beta +\gamma )-\pi =0$, i.e. angle sum $=\pi =180°$. Elliptic / spherical ($K=+1$ on the unit sphere): ${\iint }_{T}KdA=\text{Area}$, so $\text{Area}=(\alpha +\beta +\gamma )-\pi >0$, recovering Girard's excess law. Hyperbolic ($K=-1$): ${\iint }_{T}KdA=-\text{Area}$, so $-\text{Area}=(\alpha +\beta +\gamma )-\pi$, i.e. $\text{Area}=\pi -(\alpha +\beta +\gamma )>0$, the defect law.

Sanity check on the octant triangle from the spherical section: $K=1$ and $\text{Area}=\frac{\pi }{2}$, so the theorem predicts angle sum $=\pi +\frac{\pi }{2}=\frac{3\pi }{2}=270°$ — matching the three right angles found directly.□

Solution. Consider a hyperbolic triangle with angles $\alpha =\frac{\pi }{2}$, $\beta =\frac{\pi }{4}$, $\gamma =\frac{\pi }{6}$. First confirm it is genuinely hyperbolic: the angle sum is

so the sum is below $180°$, as every hyperbolic triangle must be. Its area is the defect:

Now push the angles toward $0$: an "ideal" triangle has all three vertices on the boundary at infinity, so $\alpha =\beta =\gamma =0$ and

Sanity check: shrinking the angles increased the area, and the ideal triangle's area $\pi$ is the supremum — no hyperbolic triangle can be larger, because the defect can never exceed $\pi$. The first triangle, with area $\frac{\pi }{12}$, is comfortably under this ceiling.□

Solution. Compute the distance between two points on the same vertical line, $P=(0,1)$ and $Q=(0,3)$. The geodesic is the vertical segment, along which $ds=dy$ and $x$ is constant, so

Sanity check the "stretching near the boundary" claim: the points $(0,0.1)$ and $(0,0.3)$ have the same Euclidean vertical gap scaled down by $10$, yet their hyperbolic distance is ${\int }_{0.1}^{0.3}\frac{dy}{y}=\ln 3\approx 1.099$ — identical. So a fixed hyperbolic distance corresponds to a shrinking Euclidean gap as you approach $y=0$; equivalently, the boundary is infinitely far away in the hyperbolic metric.□

Solution. Check the small-triangle limit explicitly using $\cosh u=1+\frac{u^2}{2}+O(u^4)$. For small $a,b$:

Dropping the fourth-order term $\frac{a^2{b}^2}{4}$ and matching with $\cosh c\approx 1+\frac{c^2}{2}$:

the Euclidean Pythagorean theorem. Sanity check with a concrete small case $a=b=0.1$: exactly, $\cosh c={\cosh }^2(0.1)=(1.0050042{)}^2=1.0100334$, so $c=\mathrm{arccosh}(1.0100334)=0.14154$. The Euclidean estimate is $\sqrt{0.1^2+0.1^2}=\sqrt{0.02}=0.14142$ — agreeing to three decimals, confirming the local flatness.□

By Girard's theorem the area on the unit sphere equals the excess, so $\text{Area}=\frac{\pi }{2}$. Sanity check against the whole sphere: the full sphere has area $4\pi$, and this triangle is one of the $8$ congruent "octant" triangles cut out by three mutually perpendicular great circles, so it should cover $\frac{1}{8}$ of the sphere. Indeed $\frac{1}{8}\cdot 4\pi =\frac{\pi }{2}$, matching exactly.□

The left side is $\cos \frac{\pi }{2}=0$ as well, so both sides equal $0$ and the identity holds — confirming $c=\frac{\pi }{2}$ is consistent. Sanity check: this is exactly the octant triangle from the excess example, and its three sides are three quarter-great-circles, each of length $\frac{\pi }{2}$, as found.□

By Girard's theorem the area is $\frac{\pi }{2}$ steradians on the unit sphere. Sanity check: three mutually perpendicular great circles cut the sphere into $8$ such triangles, and $8\times \frac{\pi }{2}=4\pi$, the total surface area of the unit sphere — so this triangle is exactly $\frac{1}{8}$ of the sphere. Notice all three angles were $90°$ even though the sides are "straight" (geodesic): this is impossible on a flat plane and is the signature of positive curvature.□

Concrete numbers: take two points both at latitude $45°$ N (so ${\theta }_1={\theta }_2=45°$) with longitudes differing by $\Delta \lambda =90°$. Then

so $c=\arccos \frac{1}{2}=60°$. On a sphere of Earth's radius $r\approx 6371$ km the distance is $r\cdot c=6371\times \frac{\pi }{3}\approx 6672$ km. Sanity check: $60°$ is less than the $90°$ longitude gap measured along the $45°$ parallel would suggest — the great-circle route "cuts the corner" over higher latitudes, which is exactly why it is shorter than following the parallel.□

Solution.

1. An individual triangle small enough to contain no antipodal pair, however, keeps its area: the octant triangle with three right angles has area $\frac{\pi }{2}$ on the sphere, and its image in the elliptic plane also has area $\frac{\pi }{2}$ (the antipodal map sends it to a different, disjoint octant, so the two are not glued to each other). Sanity check: Girard's excess law $\text{Area}=(\alpha +\beta +\gamma )-\pi =\frac{3\pi }{2}-\pi =\frac{\pi }{2}$ is a statement about angles, which the quotient preserves, so the same area must hold on both sides. The factor of two lives only in the global area, not in a single well-behaved triangle.