#1
In this problem we are expected to construct a regular pentagon, starting with side AB of the pentagon. The Greeks used only compass and straight edge in such construction. Because of the complexity of only using a straight edge and compass to construct the vertex angle of the pentagon, they come out with a method the 'golden ratio' which is used even today. They noted that the golden ratio was equal to
which
the Greeks thought as the proportion, 
From this observations the Greeks arrived at the conclusion that the diagonal of two sides of pentagon is a golden ratio which they used in construction of a pentagon. The diagram below may give the inside what the Greeks had in mind. let us have point A as the starting point, construct a line through point A and construct a point on the line by using circle c3 and name it B as shown in figure 1. Using B as the center and AB as the radius of the circle, construct another circle c1.
At B construct a perpendicular line to AB which will meet circle c1 at point D as shown above. Cool, now at D construct another circle of the same radius as c1 and name it c2. It will be noted that circle c2 will meet the perpendicular line at Point C. Construct a segment BC which will be twice segment AB by construction. By employing pythagoras theorem, it will be noted that side AB and BC are the two short sides of the right triangle ABC. Which implies that if we take AB to be one unit and therefore BC two units. By pythagoras theorem, side AC will be root 5 (5^1/2). Finally, by constructing a line through point A and C to meet circle c3 at F and then construct segment FC. As per the golden ratio, it will be equal to a half FC i.e EC. Therefore we can use AB as the side of the polygon and EC as the diagonal connecting the two sides of the pentagon. To ply with what created the above image click here. With all this information, start at point A and use AB (figure 1) as one of the sides of the pentagon. By using circle c1 and and using point B as the the center EC as the radius of the circle, construct circle circle c4 as noted in figure 2. Check where circle c4 and circle c1 meet and name it C as shown in figure 2. Finally join point A, B and C to get the first two sides of the pentagon. Using the same size of the side of
the pentagon and the same size
of the diagonal, construct the same way by starting at point C
and keep on until the whole pentagon ABCDE (figure 2) is constructed.
To play with the file that constructed the image figure 2, click
here and here for the gsp script.
# 2
(a)
Given point A and B we can construct two points C and D to divide the line AB into 3 segments of equal length. Now to carry out this construction, start at A and construct point B, join the two points to form segment AB as shown in
Figure 1. From point B construct a line L1 which is not parallel to segment AB. Starting from point B construct any point a long the line and name it F. Using BF as the radius of the circle and F as the center, construct a circle to meet the line L1 at E, B and F as the center. Therefore with this points on the line, construct segment EF and FB. With all this, go back to point A and construct a line parallel to L1. Using BF as the radius of the circle and A as the center, construct a circle to meet the line at point G as shown above. and using the same radius and taking G as the center, construct a circle to meet the line at point K. Then construct segment AG and GK as shown in figure 1. Finally join point EG and FK to divide the line AB into three equal parts. To play with the gsp file click here and here for the gsp script.
(b)
Using the same approach like the one of constructing a line of three segment above, start at point A and construct a segment AB and from point B construct line L1 which is not parallel to AB and pick any point on line L1 and name it F. Using segment NF as the radius and F as the center of the circle, construct another segment by intersecting line L1 and the circle constructed and name the point E as illustrated in figure 1. Continue creating more point in the same way to construct n-1 segments on line L1 and then hide line L1. Next go to point A and construct a line L2 which is parallel to L1 and carry out the same construction of n-1 segments as on L1 and hide line L2 as shown in figure 1. Join the point on n-1 segment on line L1 to A and the last point on n-1 segment to B form a short of a parallelogram similar to figure 1 in part (a). The start joining from the last point on n-1 segment to the second point after point A. Continue joining this points which form a short of parallelogram within the figure until all points are joined. Note that the last two points on line L1 will both meet at the last point of line L2. You will notice that the lines that join points on L1 to L2 will cut the segment AB into n segment and finally hide L1 and L2.
For n = 7
As discussed in the general case of n, we can now discuss a specific case of n = 7. To carry out this division of a segment into 6 segment of equal length, we need to construct the segment of interest in this case AB as shown in figure 2. From point B construct line L3 which is not parallel to segment AB. Create a point on line L3 and name it C, from Point B to C construct a segment BC and use it as the radius and C as the center to construct a circle which intersect line L3 at B and D. Using point D as the center and BC as the radius, construct another circle to meet line L3 at point E. Using point D and E construct a segment. Proceed with the same construction until segment GH is constructed and hide line L3.
Next start at Point A and construct
line L4 which is parallel to line L3, then construct
a circle of length BC with its center at A to create point I.
Using I as the center and BC as the radius of a circle, construct
a circle to meet line L4 at Point J. Carry out this process
until segment TU is finally constructed. Construct a segments
HI, GJ, FK, ER,DT and finally CU. Construct the points of intersections
of the segments and segment AB (V, W, X, Y, Z and Q). It seems
from the construction that segment AB is divided into 7 equal
parts. To a see the file that created image of figure 2, click
here and here for the prove.
#3
The figure formed PQRS is a parallelogram. By midpoint theorem, it can be proved that in triangle ADC, if R is the midpoint of segment DC and S is the midpoint of segment DA, the SR is parallel to AC. Now if we consider triangle ABC, where P is the midpoint of AB and Q is the midpoint of BC then by the midpoint theorem, QP is parallel to CA hence parallel to RS.
Now taking a look at triangle DCB, if R is the midpoint of DC and Q is for CB then by the same argument of the midpoint theorem, RQ is parallel to SP. It should be noted that the shape of the quadrilateral ABCD, the inner quadrilateral is always a parallelogram. To see for yourself what I'm trying to say, click here.
As concerns the two areas, it seems that the area of the parallelogram is a half the area of the quadrilateral and here is the proof.
Given
1. A convex quadrilateral ABCD
2. S is the midpoint of AD
3. P is the midpoint of AB
4. Q is the midpoint of BC
5. R is the midpoint of DC, hence from gsp file area of SPQR/ABCD = 0.5
6. Let's mark
7. S( ASP) = spsin(DAB).
8. S(ADB) = (2s)(2p)*sin(DAB) = 2sp*sin(DAB)
9. S(ASP) = 1/2*S(ADB) i.e find the ratio of 7 and 8 hence with same reasoning we can deduce that: S(BPQ) = 1/2*S(BAC), S(CQR) =1/2*S(CBD), S(DRS) =1/2*S(DCA) , hence
10. S(ABCD) = S(ABC) + S(DBC)
11. S(ABCD) = S(ABD) + S(DBC)
12. 2*S(ABCD) = S(ABC) + S(ACD) + S(ABD) + S(DBC)
#4.
It appears that area plBK and Dipj have same area, that is the parallelograms that whose diagonal don't contains the point p. The other parallelograms keeps on changing their area (i.e as one area increases, the other decreases) as p moves along AC. To see he file that created the image, click here.
Note that since (a) angle DCA and angle CAB are equal(alternate angles), (b) angle CDA and CBA are equal (alternate angles) (c) angle DAC and ACB are also equal. Note also that side AD = BC, DC = AB ( property of a parallelogram) which means that area of triangle DAC is equal to triangle CAB hence congruent, then the following observation are made.
Triangle pLC and pCI are congruent by using their angles and sides . Also triangle AKp and JpA are congruent. Which means for triangle DCA to be equal to ACB, then the parallelogram pLBK should be equal to DIpJ otherwise triangle triangle DAC will not be equal to ACD.
#5
By constructing the points and using triangle MNL to find the center of the circle, I noticed that the points F,L,X,D,N,Z,M,E and Y appear to lie in the circle constructed To see the file that created the image below, click here.
