Download MTH 338 Penta-hebdomadal Quiz, Solutions

MTH 338 Penta-hebdomadal Quiz, Solutions
1. State the Euclidean Parallel Postulate or one of the statements equivalent to it.
Answer. Given a line ` and a point P not on `, there is a unique line through
P that is parallel to `.
2. State the Hyperbolic Parallel Postulate.
Answer. Given a line ` and a point P not on `, the angle of parallelism for
` and P is strictly less than 90◦ .
For the remaining problems, assume hyperbolic geometry.
Nonetheless a result may actually only require the neutral axioms.
3. Let `1 and `2 be parallel lines. Suppose Pi is a point on `i , for i = 1, 2,
and that P1 P2 is perpendicular to both `1 and `2 . Prove that if a transversal
crossing `1 and `2 contains the midpoint of P1 P2 , then the alternate interior
angles formed by that transversal are equal.
Answer. Let X be the midpoint of P1 P2 . Let the transversal through
X intersect `i in the point Qi . Using vertical angles at X, we see that
4Q1 P1 X ∼
= 4Q2 P2 X. The alternate interior angles are then corresponding
angles in the congruent triangles, hence congruent.
4. Prove that if one Saccheri quadrilateral S1 having summit angle measure
θ1 is completely contained in the interior of another Saccheri quadrilateral
S2 having summit angle measure θ2 , then θ2 < θ1 holds.
Answer. The defect of S1 is 180 − 2θ1 and the defect of S2 is 180 − 2θ2 .
Since defects are non-negative and additive, we conclude that the defect of
S1 is strictly less than the defect of S2 , so
180 − 2θ1 < 180 − 2θ2
from which follows θ2 < θ1 .
5. Let ` be a line and let P be a point not on `. Let Q be the point on `
such that P Q is perpendicular to `. Prove that if R is a point on ` distinct
from P , then P Q < P R.
Answer. Exactly one of P Q = P R, P Q > P R, and P Q < P R holds.
If P Q = P R, then 4P QR is isosceles and the angles at Q and R must
be congruent. That puts two right angles into the triangle, violating the
Saccheri–Legendre Theorem.
If P Q > P R, then by the Inverse of the Isosceles Triangle Theorem, the angle
at R is strictly larger than the angle at Q, violating the Saccheri–Legendre
Theorem (even more flagrantly than in the previous case).
The only viable possibility is P Q < P R, so that’s what holds.
6. Without using a compass, give a construction of an equilateral triangle,
and prove that your construction works.
Answer. Using the protractor, construct three rays from a point P that
make angles 120◦ with each other. On the rays, construct points A, B, and
C at distance 1 from P . Form the three isosceles triangles with 120◦ angles.
These triangles are congruent by SAS, so we conclude that AB = AC = BC.
Thus 4ABC is equilateral.
7. Prove that the distance between a pair of parallel lines is not constant.
Answer. Let ` and m be the lines. Arguing by contradiction, suppose the
lines are equidistant. Choose points A1 –A2 –A3 on `. For i = 1, 2, 3, let Bi
the point on m such that the line through Ai and Bi is perpendicular to m.
Since the lines are being assumed equidistant, we have Ai Bi = Aj Bj for any
i and j. We conclude that the quadrilaterals A1 A2 B2 B1 and A2 A3 B3 B2 are
Saccheri quadrilaterals with bases on m that share the side A2 B2 . In hyperbolic geometry, the summit angles of a Saccheri quadrilateral are acute, but
at A2 the summit angles from the two Saccheri quadrilaterals are supplementary. Thus we have reached a contradiction.