Decide whether each triangle below lies in the Euclidean plane, on a sphere (of radius 1), or in the hyperbolic plane

Question

A triangle has sides of length $a$, $b$, $c$ (given correct to $2$dp) and a right angle opposite $c$. Decide whether each triangle below lies in the Euclidean plane, on a sphere (of radius $1$), or in the hyperbolic plane

1. $a=b=1.00, c=1.27$
2. $a=0.50, b=1.20, c=1.34$
3. $a=1.0, b=2.40, c=2.60$
4. $a=b=3.00, c=0.20$

In each case, answer "Euclidean", "spherical" or "hyperbolic".

How to do this without calculator? I have an exam tomorrow please help.

Answer 1

The Pythagorean theorem says $a^2+b^2=c^2$ for the Euclidean case, the left hand side would be larger than the right one in spherical geomatry and smaller for the hyperbolic one. Thus you would have to calculate simply $\sqrt{a^2+b^2}$ and compare that to $c$.

That root would become $1.41$, $\ 1.3$, $\ 2.6$, $\ 4.23$ respectively.

You even could do all this without calculator. Because of $\sqrt{2}=1.41$ respectively $13^2=169$. The remainder is just a multiplication by $2$ respectively $3$.

--- rk

Answer 2

You may need a little bit of a calculator to square the numbers, but here is a rigorous proof that the proposition suggested in another answer works. To wit, in the spherical case $c^2\color{blue}{<}a^2+b^2$ while for the hyperbolic one $c^2\color{blue}{>}a^2+b^2$.

Consider the spherical case. The formula $\cos c=\cos a\cos b$ which we usually see applies only on a sphere whose radius matches the length unit for $a,b,c$. What if the sphere might have some other radius? If the triangle is spherical, we have to allow the possibility that $\cos(c/R)=\cos(a/R)\cos(b/R)$ for some other radius $R$ and we are not allowed to try to solve for $R$. The hyperbolic case gives a similar problem with $\cosh(c/R)=\cosh(a/R)\cosh(b/R)$. Even if a sphere is given to have unit radius (which I saw only later), we cannot use a calculator in this problem to check the cosine.

Pythagoras's Revenge: The Return of the Squares

Suppose we were to take the spherical law and use the double-angle formula to reintroduce the squared quantities from the familiar Pythagorean Theorem, to wit:

$[1-2\sin^2(c/2R)]=[1-2\sin^2(a/2R)][1-2\sin^2(b/2R)]$

and just multiply out the terms. The constant terms cancel out and we can ditch a factor of $2$ from the rest, thus

$\sin^2(c/2R)=\sin^2(a/2R)+\sin^2(b/2R)-2\sin^2(a/2R)\sin^2(b/2R)$

Split the term in half and apply the identity $\sin^2u+\cos^2 u=1$ to obtain

$\sin^2(c/2R)=[\sin^2(a/2R)-\sin^2(a/2R)\sin^2(b/2R)]+[\sin^2(b/2R)-\sin^2(a/2R)\sin^2(b/2R)]$

$\color{blue}{\sin^2(c/2R)=[\sin^2(a/2R)\cos^2(b/2R)]+[\sin^2(b/2R)\cos^2(a/2R)]}$

This is the half-argument form of the spherical law, which involves a sum of squares similar to the Euclidean Pythagorean Theorem.

So we have, since the cosines lie between $0$ and $1$:

$\sin^2(c/2R)<\sin^2(a/2R)+\sin^2(b/2R)$

Now consider this picture in which, for unit radius, the sine function is concave down while the hyperbolic sine arising from the hyperbolic case is concave up (created by the author).

Thus with the concave down relation for the sine, the last inequality above implies:

$\color{blue}{c^2<a^2+b^2}.$ (spherical right triangles)

I will let the reader handle the hyperbolic case, which is analogous. Watch your signs: in the hyperbolic world we have $\cosh =1+2\sinh^2 u$. Also watch your second derivative sign: the plot above shows the hyperbolic relation is concave up. You should find

$\color{blue}{c^2>a^2+b^2}.$ (hyperbolic right triangles)