



Solutions for Session 1, Part C
See solutions for Problems: C1  C2  C3  C4  C5  C6 C7 C8

Problem C1  
Start by drawing a line segment. Then do the following:
a.  Fold the paper so that one of the endpoints of the line segment lies somewhere on the line segment. The crease created defines a line perpendicular to the original line segment. 


b.  Use the process above to construct a perpendicular line. Then use the same process to construct a line perpendicular to the new line, making sure that this second perpendicular is a different line from the original. Since this third line and the original are each perpendicular to the second line, they are parallel. 


c.  Fold the paper so that the endpoints of the line segment overlap. Draw a line segment along the crease, intersecting the original line segment. This new line segment is perpendicular to the original one and bisects it, because we used the same process that we used to construct the midpoint in the sample construction. 
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Problem C2  
Draw an angle on a piece of paper. Next, fold the paper so that the two sides of the angle overlap. The crease created defines a bisector of the angle.
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Problem C3  
For parts (a)(c), draw several triangles, at least one of which has an obtuse angle (to see that the definitions make sense in general). Then draw in the altitudes. Repeat with medians. Repeat with midlines.
a. altitudes:
b. medians:
c. midlines:
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Problem C4  
Draw five triangles on separate pieces of patty paper, and then do the following:
a.  Pick a side. Fold the paper so that the crease is perpendicular to the side [see Problem C1(a)] and so that it goes through the vertex opposite the side. You may have to extend the line segments of your triangle if the triangle has an angle larger than 90°. (See illustration for an example of what this looks like.) Connect the side with the vertex along the crease. The line segment drawn is the altitude corresponding to the side chosen. Now repeat with the other two sides. 


b.  Pick a side. Fold the paper so that the endpoints of the chosen side overlap. The midpoint of the side is the point where the side intersects the crease. Using a straightedge, connect the midpoint of the side with the vertex opposite it. Repeat with the other two sides. 
c.  Pick a side. Find the midpoint of the side by following the construction of question (b). Repeat this construction with the other two sides. Using a straightedge, connect the consecutive midpoints. 
d.  Pick a side. Construct a perpendicular bisector of the chosen side using the construction from Problem C1(c). Repeat with the other two sides. 
e.  Pick an angle. Fold the paper so that the two sides of this angle overlap. The crease defines a ray that bisects the chosen angle. Repeat with the other two angles. 
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Problem C5  
Draw a line segment, and then do the following:
a. 

b.  Construct a perpendicular bisector of the line segment. Choose any point on the perpendicular bisector and connect it with the endpoints of the original line segment. The resulting triangle is isosceles and has the original line segment as its base. 

c. 
Extend the line segment to form a line, being sure to mark the original endpoints of the line segment. Use this line to construct a perpendicular line through the endpoints of the original line segment [see Problem C1(a)]. 


Mark the point on the perpendicular line where the second endpoint falls on this line. (This defines one of the equal, perpendicular sides.) 



Perform the same process on the second perpendicular line to define the third side of the square. 





d. 
Construct the perpendicular bisector of the segment.
Flip your patty paper, and then mark that spot on the bisector.
Connect the marked spot with the two endpoints of the original line segment.

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Problem C6  
a.  Fold the paper in half to make a rectangle. Fold it in half again by bisecting the longer sides of the rectangle. The resulting square has onefourth the area of the original one. There are exactly four squares that fit exactly on top of each other, so they must have the same area. Since together they completely make up the original square, each must be onefourth of the original square. 
b.  Find the midpoints of all four sides. Connect the consecutive midpoints. The resulting square has onehalf the area of the original square. To see this, connect the diagonals of the new square. You will see four triangles inside the square and four triangles outside, all of which have the same area. Half the area of the original square is inside the new square. 
c. 
In order to obtain a square with exactly threefourths the area of the original square (sides = 1) we need to calculate the sides of the new square:
a • a = 3/4
a^{2} = 3/4
a =
So we are looking to construct a square whose sides are equal to .

Start with a piece of patty paper. Think of the bottom edge as the base of an equilateral triangle.



Fold the vertical midline. The third vertex of the equilateral triangle will be on this midline. 

Bring the lower right vertex up to the midline, so that the entire length of the bottom edge is copied from the lower left vertex to the midline. 


The distance from the bottom of the midline to this mark is /2. It is the height of an equilateral triangle with the side length equal to 1. This can also be easily calculated using the Pythagorean theorem for the right triangle with the sides 1 and 1/2. 


Next, fold down from this mark. This is the height of the square. 


Fold down the corner so that you can fold the right side in the same amount as the top. 


Fold over the right side. This square has area 3/4 of the original square. 



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Problem C7  
One way to do this is to use a straightedge to draw the two diagonals of the square. The centroid is the point of their intersection. Another is to draw the perpendicular bisectors of two consecutive sides of the square [the same construction as Problem C6(a)]. The intersection of these bisectors is the same centroid.
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Problem C8  
Noticing what appear as concurrencies in the folds may lead one to conjecture that concurrencies occur in general. Keep in mind any one construction that suggests this is a special case. Therefore, in order to convince ourselves that they do occur in general, we need to construct a formal proof.
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