Download Geometry III/IV

Geometry III/IV
Exercises: Week 16, Feb 2013
Part A
Problem 1. Let ABC be a triangle. Let B1 ∈ AB and C1 ∈ AC be two points
such that AB1 C1 = ∠ABC. Show that ∠AC1 B1 > ∠ACB.
Solution. Consider the quadrilateral B1 BCC1 . If ∠AC1 B1 ≤ ∠ACB then the sum
of angles of B1 BCC1 is greater or equal to 2π. On the other hand, we can divide
B1 BCC1 by a diagonal into two triangles, each having a sum of angles less than π.
The contradiction proves ∠AC1 B1 > ∠ACB.
Problem 2. Show that there is no “rectangle” in hyperbolic geometry (i.e. no
quadrilateral has four right angles).
Solution. Suppose there is a rectangle ABCD. Then its sum of angles is 2π.
Decompose it into two triangles by a diagonal AC. The sum of angles of ABC is
smaller than π, the sum of angles of ACD is smaller than π but the sum of these
two sums of angles equal to the sum of angles of ABCD. Contradiction.
Part B
Problem 3. Given α, β, γ such that α+β+γ < π show that there exists a hyperbolic
triangle with angles α, β, γ.
Solution. Put a vertex A of angle α in the center of Poincaré disk model. Let AX
and AY be the ray forming the angle. Let T be a point on AX. Let T Z be a ray
emanating from T and such that ∠ZT A = β. Consider the point C(T ) = T Z ∩ AY .
When T is very close to A the hyperbolic line T Z is very close to a Euclidean line
through the same points, so the hyperbolic sum of angles of triangle AT C(T ) is very
close to π. As T runs away from A to Y , angle (T )T became smaller and smaller
(see Problem 1), then it tends to zero (when T Z is parallel to AY ) and disappears.
By the continuity we see that for some intermediate point T0 the sum of angles of
AT0 C(T0 ) coincides with α+β +γ which implies that ∠AC(T0 )T0 = γ and AT0 C(T0 )
is the required triangle.
Problem 4. Let l1 and l2 be two hyperbolic lines. Show that exactly one of the
following three possibilities hold:
• either l1 intersect l2 ,
• or l1 is parallel to l2 ,
• or there exists a unique line l orthogonal to both of l1 and l2 .
Solution. First, suppose that the lines neither have intersection nor are parallel.
Let XY and ZT be the ends of the lines so that XY ZT is a convex quadrilateral.
Then the diagonals XZ and Y T have an intersection point O. Put O into the center
of the Poincaré disk model. Then the triangles OXY and OZT are symmetric with
respect to O. So that the altitudes OP1, P1 ∈ XY and OP2 , P2 ∈ ZT lie on the
same line l. This line is orthogonal to both XY and ZT .
To show the uniqueness of this line, suppose there are two of them, l3 and l4 . If
l3 intersects l4 then the lines l1 , l3 , l4 form a triangle with two right angles, which is
impossible since the sum of angles of a triangle should be smaller than π. If l3 does
not intersect l4 then l1 , l2 , l3 , l4 bound a quadrilateral with four right angles, which
is also impossible (the sum of angles of a quadrilateral should be smaller than 2π).
This implies that at least one of the three possibilities hold.
We a left to show that no two of the three conditions are compatible. Clearly, that
two intersecting lines are not parallel. Suppose that there exist a line l orthogonal
to both of l1 and l2 , where l1 and l2 have a common point X either inside H2 or on
the absolute. Denote A = l ∩ l1 , B = l ∩ l2 . Then the sum of angles of the triangle
ABX is not smaller than π which is impossible.
Problem 5. Show that there exists a hyperbolic pentagon with five right angles.
Solution. Consider a Euclidean regular pentagon PEucl . Draw PEucl so that the
center of PEucl coincides with the center O of Poincaré disk model. Consider the
hyperbolic pentagon P spanned by the vertices of PEucl . When PEucl is very small
(but still centered at O) the angle of P are almost the same as the angles of PEucl .
When PEucl is inscribed into the absolute the angles of P are zero (as the adjacent
sides of P are tangent). Notice that the angles of PEucl are obtuse (more precisely,
they are equal to 3π/5). So, by continuity we see that there exist some intermediate
size of PEucl such the corresponding hyperbolic regular pentagon has right angles.
Problem 6. An ideal triangle is a hyperbolic triangle with all three vertices an the
absolute.
(a) Show that all ideal triangles are congruent.
(b) Show that three altitudes of an ideal triangle intersect in one point. (An
altitude is a straight line passing through a vertex and orthogonal to the
opposite side).
(c) Let O be a point of intersection of altitudes of an ideal triangle XY Z. Let
P be a foot of the altitude XP (P ∈ Y Z, XP ⊥ Y Z). Find the angles of
the triangle Y OP . Find the area of Y OP . Find d(O, P ).
(d) Show that an ideal triangle has an inscribed circle. Find its radius.
(e) Let c be a circle inscribed in a triangle ABC. Show that the radius of c does
not exceed √23 .
Solution. (a) There exists a hyperbolic isometry which takes any given three points
of absolute to any other three points of the absolute. So, it takes a triangle spanned
by the given three points to the triangle spanned by the other three points.
(b) Consider an ideal triangle T whose vertices are represented by vertices of a
regular Euclidean
triangle TEucl in the Poincaré disk model (say a triangle with
√
√
− 3−i − 3+i
vertices 1, 2 , 2 ). Then all three altitudes of the hyperbolic triangle T pass
through the center O of the disk. So, three altitudes have a common point O.
To see that the altitudes of any other ideal triangle have a common point, we map
the triangle to P by an isometry. The altitudes are mapped to the altitudes, so the
intersection point is mapped to the intersection point.
(c) The triangle Y OP has a right angle P (by definition of the altitude), a zero angle
X (as X is ideal vertex) and an angle O equal to π/3 (since there are 6 triangles
congruent to Y OP and tiling the neighbourhood of O).
The area of Y OP is a sixth part of the area of XY Z which is π, so S(Y OP ) = π/6.
(Or, one can find it by the formula S(Y OP ) = π−α−β−γ = π−1/2−1/3−0 = π/6).
To find d(O, P ) consider the triangle Y OP in the upper half-plane
model. With√
1+i 3
out loss of generality we may assume that Y = ∞, P = i, O = 2 . Then, by the
formula
|z − w|2
cosh d(z, w) = 1 +
2Im(z)Im(w)
we have
√
(1 − 23 )2 + ( 12 )2
1 + 3/4 + 1/4
2
√
√
=
=√ .
cosh d(O, P ) = 1 +
3
3
2 · 23 · 1
(d) Consider the ideal triangle in the Poincaré disk model, as in part (b). Consider
a circle centered at O. A small circle centered at O does not intersect any side of
XY Z. As the the radius of the circle grows, it first touchs a side and than get two
intersections with each of the sides of XY Z. It is clear from the symmetry, that
the circles touchs (is tangent to) all three sides of XY Z at the same moment. Then
this circle is the circle inscribed into XY Z. It is also clear (again by the symmetry)
that the radius of this circle equals to OP , i.e. by (c) it equals to √23 .
(e) We will show that any triangle may be enclosed into an ideal triangle. Than the
inscribed circle of an arbitrary triangle will be enclosed into ideal triangle, so it’s
radius does not exceed the radius of the circle inscribed into the ideal triangle.
Consider a triangle ABC in the upper half-plane model. Without loss of generality
we may assume that A and B lie on the imaginary axes, while C lie on the vertical
line x = 1 (to get this we may need to apply an isometry taking A to a point of
x = 0, the apply a rotation around A to put B on the line x = 0 and finally to scale
z → az so that C is mapped to the line x = 1).
Let as prove that ABC is contained in the ideal triangle with vertices 0, 1, ∞. To
prove it we only need to check that the lines AC and BC do not intersect the line
l connecting 0 and 1. Notice that all three points A, B, C lie on the same side with
respect to l, which is impossible if AC or BC intersects l. So, Neither AC nor BC
intersects l and ABC lie inside the ideal triangle with vertices 0, 1, ∞.