Constructing Hyperbolic Triangles in the Upper Half Plane

From angle data to explicit geodesics in the upper half plane.

Hyperbolic triangles are completely determined by their angles. This is already different from Euclidean geometry, where fixing three angles leaves a whole family of similar triangles, one for every possible scale. In the hyperbolic plane there is no scaling ambiguity: fix the angles and the side lengths are fixed as well.

There is also a different condition on which angles are possible. The angles of a Euclidean triangle sum to $\pi$, while those of a hyperbolic triangle sum to less than $\pi$. In fact the deficit is exactly its area:

$\operatorname{area}(T)=\pi-(\alpha+\beta+\gamma)$

For every $\alpha,\beta,\gamma>0$ with $\alpha+\beta+\gamma<\pi$, there is a unique hyperbolic triangle with those angles, up to isometry. Our goal in this note is to construct that triangle explicitly. We will write down its three bounding geodesics in the upper half plane, each described by its endpoints in $\overline{\mathbb{R}}=\mathbb{R}\cup\{\infty\}$. From these coordinates we can draw the triangle directly.

Our first choice is a model. We will work in the upper half plane, where geodesics are vertical lines or semicircles centered on the real axis. The model is conformal, so the hyperbolic angles between geodesics are the Euclidean angles visible in the picture.

The Right-Angled Case

Right angles simplify the construction, so we begin with a triangle whose angles are

$\frac{\pi}{2},\qquad \beta,\qquad \gamma$

We only need one representative of its isometry class, so move the right-angled vertex to $i$, then rotate about $i$ until one side is the vertical geodesic

$\gamma_1\longmapsto\{0,\infty\}$

The other side through $i$ must cross the vertical at a right angle. Its tangent at $i$ is therefore horizontal, so its radius there is vertical. Since its center lies on the real axis, the center can only be $0$ and the radius can only be $1$:

$\gamma_2\longmapsto\{-1,1\}$

This fixes the first two sides. The third side must now supply both remaining angles.

Reading the Angles of a Circle

Write the third side as a semicircle with center $c$ and radius $\rho$. These are the two unknowns in the problem. Before solving for them, we first determine the forward relationship: given $c$ and $\rho$, what angles does this circle make with the two sides already placed?

Drag $c$ and $\rho$ to move through this two-dimensional space of circles and watch the two crossing angles respond.

Its crossing with the vertical occurs at height

$h=\sqrt{\rho^2-c^2}$

We want to express this angle in terms of $c$ and $\rho$, but the tangent and the vertical do not bound a triangle with those lengths. Rotate both through a quarter turn. The tangent becomes the circle’s radius, the vertical becomes a horizontal line, and their angle does not change because both lines were rotated by the same amount.

These two new directions are parallel to the sides meeting at the center of the Euclidean right triangle drawn below. Thus its angle at $c$ is the same $\theta_1$ we started with at the crossing. The triangle has hypotenuse $\rho$ and adjacent side $|c|$, so

$\cos\theta_1=\frac{|c|}{\rho}$

The crossing with the unit circle gives another Euclidean triangle: its three sides are the two radii $1$ and $\rho$, together with the distance $|c|$ between their centers. The ordinary law of cosines gives

$c^2=1+\rho^2-2\rho\cos\theta_2$

We now reverse this process, beginning with the two angles and solving for $c$ and $\rho$.

Solving for a Right Triangle

Now we run the two equations backwards. Asking for $\theta_1=\beta$ fixes the ratio of center to radius,

$|c|=\rho\cos\beta$

The two signs of $c$ give mirror-image pictures, so choose the center on the left:

$c=-\rho\cos\beta$

Asking for $\theta_2=\gamma$ and substituting this value of $c$ gives

$$\begin{aligned} \rho^2\cos^2\beta&=1+\rho^2-2\rho\cos\gamma\\[4pt] 0&=\sin^2\!\beta\,\rho^2-2\cos\gamma\,\rho+1 \end{aligned}$$

Thus the construction has come down to one quadratic:

$\rho=\frac{\cos\gamma\pm\sqrt{\cos^2\gamma-\sin^2\beta}}{\sin^2\beta}$

The expression under the square root is nonnegative exactly when

$\cos\gamma\geq\sin\beta=\cos\left(\frac\pi2-\beta\right)$

or equivalently

$\frac\pi2+\beta+\gamma\leq\pi$

Thus the quadratic reproduces the angle-sum condition. When the sum is less than $\pi$ there are two real roots; at equality they merge; when the sum is greater than $\pi$ there is no real circle satisfying the two angle conditions.

The two roots are the two circles shown in the figure. They are exchanged by the hyperbolic half-turn about $i$,

$H(z)=-\frac1z$

which preserves the first two sides and sends height $h$ on the vertical to $1/h$. Indeed, Vieta’s formula shows that the two crossing heights $h_\pm=\rho_\pm\sin\beta$ satisfy

$h_+h_-=1$

Thus one triangle lies above $i$ and the other below it. They are isometric, and our formula simply chooses the larger root.

At the limiting value $\frac\pi2+\beta+\gamma=\pi$, the two roots merge. Since their heights still satisfy $h_+h_-=1$, their common height must be $h=1$. The second vertex has collided with $i$, the third side passes through $i$ as well, and all three geodesics meet at one point. The triangle has collapsed to zero area.

The General Case

For a general triangle, keep the first side vertical and normalize the second side to be a circle of radius $1$. If its center is $a$, the same angle calculation as before shows that it meets the vertical at angle $\alpha$ when

$a=\cos\alpha$

The crossing has height $\sin\alpha$, so the second side is

$\gamma_2\longmapsto\{\cos\alpha-1,\cos\alpha+1\}$

At $\alpha=\pi/2$ the center returns to zero, recovering the previous picture.

The third side is again a circle of radius $\rho$ and center

$c=-\rho\cos\beta$

Only one distance in the previous calculation changes. The centers of the two circles are now $a$ and $c$, so their separation is $a-c$ rather than $|c|$. Requiring them to cross at angle $\gamma$ gives

$(a-c)^2=1+\rho^2-2\rho\cos\gamma$

Substitute $a=\cos\alpha$ and $c=-\rho\cos\beta$, expand, and use $1-\cos^2\alpha=\sin^2\alpha$:

$\sin^2\!\beta\,\rho^2-2\left(\cos\gamma+\cos\alpha\cos\beta\right)\rho+\sin^2\!\alpha=0$

This is the right-triangle quadratic with one extra term. To examine its discriminant, abbreviate

$B=\cos\gamma+\cos\alpha\cos\beta$

The discriminant is

$$\begin{aligned} B^2-\sin^2\alpha\sin^2\beta &=(B-\sin\alpha\sin\beta)(B+\sin\alpha\sin\beta)\\ &=\bigl(\cos\gamma+\cos(\alpha+\beta)\bigr) \bigl(\cos\gamma+\cos(\alpha-\beta)\bigr). \end{aligned}$$

For an admissible triple, $\alpha+\beta<\pi-\gamma$, so the first factor is positive, and the second is positive as well. At $\alpha+\beta+\gamma=\pi$, the first factor vanishes. Therefore the construction gives real circles throughout the hyperbolic range, and its two solutions merge exactly when the angle sum reaches $\pi$.

The two roots still describe the same triangle twice. Their vertical crossing heights satisfy

$h_+h_-=\sin^2\alpha$

and the half-turn about the first vertex $(0,\sin\alpha)$ sends $h$ to $\sin^2\alpha/h$ while preserving the first two geodesics. It exchanges the two roots, just as it did in the right-angled picture.

This also proves uniqueness. After the first two sides have been normalized, the angle conditions leave only these two isometric placements for the third. Equivalently, the second hyperbolic law of cosines reads

$\cosh \ell_\gamma=\frac{\cos\gamma+\cos\alpha\cos\beta}{\sin\alpha\sin\beta}$

where $\ell_\gamma$ is the side opposite $\gamma$. Thus the angles determine a side length on their own. Euclidean triangles have similarity; hyperbolic triangles do not.

Together these formulas give an explicit representative of the unique triangle associated to every point

$\{(\alpha,\beta,\gamma)>0:\alpha+\beta+\gamma<\pi\}$

in the space of hyperbolic triangles. Unlike the right-angled pentagons and hexagons considered in the notes that follow, there are no additional side-length parameters: for a triangle, the angles determine everything.