11_9

We began class by getting Lab 3 comments back and talking about common errors:

1) Constructions didn't correspond to proofs. Many people wrote proofs, but constructed the figure a different way in GeoGebra. 2) To bisect a segment, most people said "C bisects AB if A*C*B and AC is congruent to BC". People often constructed a perpendicular bisector in order to bisect the segment, which did not correspond with the written definition.

In Lab 3, one of the problems was to bisect an angle. Most people approached the problem in one of two ways.

__Method 1__: For angle CAB, construct segment CB. Construct a perpendicular bisector through the midpoint of CB. To construct the perpendicular bisector, first construct two circles, one centered at C and incident with B, and the other centered at B and incident with C. Construct a line through the two intersection points of the two circles. This line is a perpendicular bisector through the midpoint of CB. Call the midpoint of CB point D. Most people assumed this line would also be incident with point A, the vertex of the angle, which in fact would need to be proven. The resulting angles, angle CAD and angle DAB are then congruent, so angle CAB has been bisected.

__Method 2__: For angle CAB, the same method of constructing a perpendicular bisector to CB was used to determine the midpoint of CB, called D. Some people chose to construct a line through point A and point D. Angle CAD and angle DAB are congruent, so this method also bisects angle CAB.

After bisecting an angle, we naturally moved on to attempting to trisect an angle.We were able to determine that particular segments and angles were congruent, but were not able to prove that the angle would be trisected. We used the following construction to assess congruence relationships:

Pick point C and point B on angle CAB such that AC is congruent to AB. Construct segment CB. Chose point D and point E on CB such that C*D*E*B and CD is congruent to DE is congruent to EB. Since AC is congruent to AB, triangle CAB is an isoceles triangle, so angle ACB is congruent to angle ABC. Triangle CAD is congruent to triangle BAE by SAS (AC is congruent to AB, ACB is congruent to BAC, and CD is congruent to EB). By definition of congruent triangles, angle CAD is congruent to angle BAE. We now know that two out of the three parts of angle CAB are congruent. By definition of congruent triangles, AD is congruent to AE. Hense, triangle DAE is an isoceles triangle, so angle ADE is congruent to angle AED. We cannot make any conclusions about the third angle of triangle DAE, namely angle DAE. Hence, we do not know if angle DAE is congruent to angle CAD and angle BAE. Therefore, we do not know if the angle has been trisected.

It turns out that trisecting an angle is one of the famous "impossible" constructions using only a straightedge and a compass. The other two impossible constructions with these limitations are doubling the volume of a cube and constructing a square that has the same area as a given circle. However, it is possible to trisect a 90 degree angle in particular.

To trisect a 90 degree angle: Construct an equilateral triangle with its base on one ray of the angle using the same construction we have done previously for equilateral triangles. All of the interior angles in an equilateral triangle are 60 degrees, so the side of the triangle that is incident with the vertex of the 90 degree angle and on the interior of 90 degree angle forms a 30 degree angle with the ray of the 90 degree angle that is not incident with the equilateral triangle. Then use the same method of bisecting an angle, as described above, to bisect the 60 degree angle in the equilateral triangle whose vertex is incident with the vertex of the 90 degree angle. The result of bisecting the 60 degree angle is two 30 degree angles, so we have successfully trisected this particular angle.

If we change the tools available for construction, it is possible to trisect any angle if we have a marked straightedge. Trisect an angle with a marked straightedge using the following construction: For some angle XAB, construct the circle centered at A and incident with X and B. On the straightedge, mark the length of AX. Using the marked straightedge, construct FC, which is congruent to AX, such that F*A*B, C is incident with the circle, and F*C*X. The measure of angle CFA is one third the measure of angle XAB. We then spent time working in groups to prove that this is the case. First, determine which segments and angles are known to be congruent.AX, AB, and AC are all congruent because they are radii of the circle. FC is also congruent to these three segments by construction. Construct segment XB. Since AX is congruent to AB, triangle XAB is an isoceles triangle, so angle AXB is congruent to angle ABX. Since AX is congruent to AC, triangle XAC is an isoceles triangle, so angle AXC is congruent to angle ACX. Since AC is congruent to FC, triangle FCA is an isoceles triangle, so angle CFA is congruent to angle CAF. Label angle CFA and angle CAF as x. Knowing that the interior angles of triangle FCA add up to 180 degrees, angle FCA is represented by 180-2x. Since F*C*X, angle FCA and angle ACX are supplementary and must add up to 180 degrees. Hence, angle ACX is represented by 2x. Since angle ACX and angle AXC are congruent, angle AXC is also represented by 2x. Again using that the interior angles of triangle XAC add up to 180 degrees, angle CAX is represented by 180-4x. Since F*A*B, angle CAF, angle XAC, and angle XAB are supplementary and must add up to 180 degrees. Hence, angle XAB is represented by 3x. Since angle CFA being x implies that angle XAB is 3x, this construction does in fact trisect angle XAB.

Next we revisited the first problem about the hiker. A hiker is headed back to her tent after a long day's hike. What is the shortest path she can take to her tent? Label the hiker as point H, the tent as point T and A as any other point such that A, H, and T are noncollinear. Construct segments HT, HA, and AT to form triangle HAT. Call HT e. Call HA g. Call AT f. To determine that HT is the shortest path the hiker can take to the tent, we need to show that HT < HA + AT, which is the triangle inequality. Copy segment f such that one of its endpoints is A and H*A*T ' if T ' is the other endpoint of the segment. This new segment is of course congruent to segment f since segment f was copied. Segment HT ' is represented as g + f. First we need to show that angle HTT ' > angle HT ' T. Since AT is congruent to AT ', triangle T ' AT is isoceles and angle AT ' T is congruent to angle ATT '. By angle addition, angle HTT ' is the sum of angle ATT ' and angle HTA. Since ATT' is congruent to AT ' T, angle HTT ' > angle HT ' T. Call angle HT ' T angle alpha. Call angle HTT ' angle beta. Claim: If alpha < beta, then e < f + g. Lisa's Claim: The side opposite to the alrger angle is larger. Next we drew a new triangle, triangle ABC. BC is called a, AC is called b, and AB is called c. Claim: If angle ABC > angle BAC, then AC > BC, or equivalently, if angle beta > angle alpha, then b > a. To prove this claim, we utilized the Exterior Angle Theorem. For a triangle DCA, if D*A*B, then angle CAB is called an exterior angle. The Exterior Angle Theorem then says that an exterior angle is greater than either of the remote interior angles. This theorem applies to neutral geometry. In Euclidean geometry, the Exterior Angle Theorem is even stronger, stating that an exterior angle is equal to the sum of the remote interior angles. Proof: Assume b < a. By definition, there exists a point D such that B*D*C and CD is congruent to CA. Hence, triangle ACD is isoceles, so angle CDA is congruent to angle CAD. Since D is in the interior of angle CAB, this implies angle DAB < angle CAB < angle beta. Hence, angle DAB < angle beta. But angle ADC is an exterior angle to triangle ABD and is therefore (needs proof) larger than either remote interior angle. In particular, angle ADC > angle ABC = angle beta. This is a contradiction. Assuming the negation of the conclusion (b < a) leads to a contradiction, so the original statement is true. Therefore, if angle beta > angle alpha, then b > a.

New problems to think about: 1) A hiker is headed back to her tent after a long day's hike, but she needs to swing by the river on her way to refill her water bottles. She is on the same side of the river as her tent. What is the shortest path she can take to her tent? 2) A hiker is headed back to her tent after a long day's hike, but she needs to swing by the river on her way to refill her water bottles. She is really thirsty and wants to get to the river as fast as possible. What is the shortest path she can take to the river?

Upcoming axioms about area: A1) The area of any polygonal region is a unique positive number. A2) The area of any point or line segment is 0. A3) Congruent polygonal regions have the same area. A4) Area is additive (if we subdivide a figure into non-overlapping parts, or parts that share segments or points, then its area is the sum of the areas of its parts). A5) A square of length a units has area of a^2 square units.

Upcoming theorems about area: The area of a rectangle with sides a and b is ab. The area of a parallelogram is the product of the length of a base and the corresponding height. The area of a triangle is half the product of the length of the base and the corresponding height.

Corrections/Suggestions for class notes on 11/9/09 In addition to the corrections for Lab #3 as discussed in class, it was mentioned how important it is to ensure our definitions are specific, instead of a general statement. The example given was the definition of an angle bisector: An angle bisector of angle BAC is a ray AD such that D is in the interior of angle BAC and  angle BAD is congruent to angle DAC. In the explanation of trisecting an angle, the sentence “Triangle CAD is congruent to triangle BAE by SAS (AC is congruent to AB, angle ACB is congruent to angle BAC  , and CD is congruent to EB). I understand that the angles should be changed to either angle ACB is congruent is angle ABC or angle ACD is congruent to angle ABE.