MATH 5200/7200 Final Exam
Dr. McCrory, Tuesday, December 11, 2001, 3:30-6:30 pm
You may use our axioms or any theorems from class, homework, or exams. Clearly identify any axioms or theorems you use.
1. (15 points) A rhombus is a quadrilateral which has all four sides equal. Prove that the diagonals of a rhombus are perpendicular.
Proof:
Consider the rhombus ABCD. Since AB = AD, the
point A is the same distance from the points B and D. By the perpendicular
bisector theorem,
the point A is on the perpendicular bisector of the segment BD.
Similarly, since CB = CD, the point C is on the perpendicular
bisector of BD. Since
two points determine a line (the line axiom), the line AC is the
perpendicular bisector of the segment BD. So the segment AC is
perpendicular to the
segment BD.
Comment: Another approach is to use the four
isosceles triangles ABD, CBD, BAC, DAC, and the theorem that the
base angles of an isosceles
triangle are congruent, to prove that the four little triangles
formed by the diagonals are congruent. Thus the four angles at
the intersection of the
diagonals are congruent, and hence they are right angles.
2. (25 points) (a) What is the geometric mean of two positive real numbers x and y?
The geometric mean m of x and y is the square root of xy. In other words, m2 = xy.
(b) Given two line segments with lengths
x and y, explain how to construct a line segment whose length
is the geometric mean of x and y. (You may
use all the operations in the GSP Construct menu which follow
from the Greek construction rules.) Prove your construction works.
Construction
We are given two segments AB of length x and CD of length y.
(1) Construct a point E on the line L through
A and B so that BE = CD, and B is between A and E. (Construct
the line L and then the circle c1 with
center B and radius CD. Let E be the intersection point of c1
and L so that B is between A and E.)
(2) Construct a circle c2 with diameter AE. (Construct the midpoint F of AE, and then c2 is the circle with center F through the point A.)
(3) Construct the line M perpendicular to L through the point B.
(4) Let G be one of the two intersection points of M and c2.
If m = GB, then m is the geometric mean of x and y.
Proof that the construction works:
Since G lies on the circle with diameter AE,
we know (from homework 4, problem 2) that the triangle AGE is
a right triangle with hypotenuse AE
and right angle at G. Triangles AGE and ABG are similar (by angle-angle
similarity), since they are right triangles sharing the angle
A. Also,
triangles AGE and GBE are similar, since they are right triangles
sharing the angle E. Thus triangles ABG and GBE are similar. By
the definition of
similarity, AB/GB = GB/EB. Thus x/m = m/y, or m2 = xy.
3. (25 points) Prove that if a quadrilateral
is circumscribed about a circle (that is, the sides of the quadrilateral
are tangent to the circle) then the sum
of one pair of opposite sides is equal to the sum of the other
pair of opposite sides.
Proof:
Suppose the quadrilateral ABCD is circumscribed
about the circle c1. Let P, Q, R, S be the points of tangency
of the four sides AB, BC, CD, DA,
respectively.
Since AP is tangent to the circle c1 at P,
and AS is tangent to the circle c1 at S, AP = AS. We proved this
in homework 10, problem 1, but here's a
proof. Let O be the center of c1. Since AP is tangent to c1 at
P, AP is perpendicular to the radius OP. For the same reason,
AS is perpendicular to
OS. The two right triangles AOP and AOS have the same hypotenuse
OA, and the sides OP and OS are equal since they are radii of
the circle c1.
Thus, by the Pythagorean theorem, the sides AP and AS are equal.
The same argument shows that BP = BQ and CQ = CR and DR = DS.
Therefore,
AB + CD = (AP + BP) + (CR + DR) = (AS + BQ) + (CQ + DS) = (BQ + CQ) + (AS + DS) = BC + DA.
4. (25 points) (a) Prove that for a triangle ABC with sides a = BC, b = AC, and c = AB,
a = b cos C + c cos B.
There are two cases: (1) B and C are acute, or (2) B or C is obtuse.
Proof:
Case (1): Suppose that B and C are acute. If D is the foot of the the altitude from A, then D is between B and C.
Let a = BC, b = AC, c = AB, a' = BD, and a'' = CD. By the definition of cosine, cos B = a'/c and cos C = a''/b. Thus
b cos C + c cos B = b (a''/b) + c (a'/c) = a'' + a' = a.
Case (2): Suppose that B or C is obtuse. We
will assume that B is obtuse. (The case when C is obtuse is done
in the same way.) If D is the foot of
the altitude from A, then B is between D and C.
a = BC, b = AC, c = AB, a' = BD, a'' = CD. By the definition of cosine, cos B = -a'/c and cos C = a''/b. Thus
b cos C + c cos B = b (a''/b) + c (-a'/c) = a'' - a' = a.
Comment: Another
way to derive equation (a) is to use the law of cosines: c2 =
a2 + b2 -2ab cos C, and b2 = a2 + c2 -2ac cos B. Solve these two
equations for cos C and cos B, respectively, and check that equation
(a) holds by substituting these expressions for cos C and cos
B. (This proof
does not require two separate cases.)
(b) Use the law of sines to deduce from part (a) the sum formula
sin(B + C) = sin B cos C + sin C cos B,
provided that angles B and C have positive proper angle measure, and the sum of the proper angle measures of B and C is less than 180 degrees.
Proof: The hypotheses
on angles B and C imply that there is a triangle with angles B
and C. Let A be the third angle of this triangle. Let r = (sin
A)/a. The law of sines states that r = (sin B)/b and r = (sin
C)/c. Multiply both sides of the equation (a) by r :
ra = rb cos C + rc cos B
[(sin A)/a] a = [(sin B)/b] b cos C + [(sin C)/c] c cos B
sin A = sin B cos C + sin C cos B
By the triangle angle sum theorem, A + B + C = 180. Thus the angles A and B + C are supplementary, so sin A = sin (B + C). Therefore
sin(B + C) = sin B cos C + sin C cos B.
5. (10 points) The following questions are based on student project presentations.
(a) What are the first eight Fibonacci numbers?
1, 1, 2, 3, 5, 8, 13, 21
(b) How are the Fibonacci numbers related to the golden ratio?
The ratio of sucessive Fibonacci numbers approaches
the golden ratio. In other words, if fn is the nth Fibonacci number,
then the limit as n
approaches infinity of fn+1/fn is the golden ratio.
(c) Explain how to get each type of conic section by intersecting a plane and a cone.
Note: This question was for diagnostic purposes
only; I didn't take off points for incorrect answers. The students
who gave presentations about
conic sections made incorrect statements about these intersections.
Suppose that the cone C is obtained by rotating
the line L = {(x,y,z) | z = mx, y = 0}, m > 0, around the z
axis. Let theta be the angle between L and
the z axis, i.e. m = cot(theta).
Suppose that P is a plane which does not pass through the origin, and let phi be the angle between P and the z axis. If phi = 90 degrees, then the intersection of P and C is a circle.
(d) How do you make an ellipse by folding wax paper?
Draw a circle on the wax paper, and mark any
point P inside the circle. Mark lots of points on the circle,
and for each of these points Q fold the wax
paper so that Q is on top of P. The creases made by all these
folds form a collection of lines whose envelope is an ellipse.
(In other words all of these
crease lines are tangent to the ellipse.)
(e) Who invented the term "fractal"?
Benoit Mandelbrot
(f) Who were the two most important artists of the Cubism movement?
Pablo Picasso and Georges Braque