intersection point of N and P, and let H be the intersection point of O and P. Then AFHG is a golden rectangle.
MATH 5200/7200 Exam 1 Solutions
Dr. McCrory
Part A: IN CLASS EXAM
1. Explain how to construct a golden rectangle.
For a golden rectangle the ratio of the base to the height is the golden ratio tau = (1 + sqrt(5))/2.
Start with a segment AB. (Its length will be the height
of our golden rectangle.) Construct a line L through B perpendicular
to AB. Then construct a
circle c1 with center B which passes through the point A. Let
C be an intersection point of c1 and L. Construct a circle c2
with center C which passes
through the point B. Let D be the intersection of c2 and L (with
D not equal to B). Now BD = 2AB, so AD/AB = sqrt(5) by the Pythagorean
theorem
for the right triangle ABD.
Next extend the segment AB to a line M. Construct a circle
c3 with center B and radius AD. Let E be the intersection of c3
and M (with B between A
and E). Let F be the midpoint of AE. Now AF/AB = (1 + sqrt(5))/2.
Note: Jan White did a very neat construction using the figure
below. (It is a simplification of the preceeding construction.)
This is an ancient construction of the golden rectangle, and it's
the shortest one I know.
2. What are the Greek construction rules?
The Greek construction rules are about three types of objects
in the plane: points, lines, and circles. These rules determine
which objects can be
constructed starting from other objects. The three rules are "straight-edge,"
"compass," and "intersection."
1. Straight-edge: If two points A and B can be constructed, then the line through A and B can be constructed.
2. Compass: If two points A and B can be constructed, then the circle with center A which passes through B can be constructed.
3a. Line intersection: If two lines m and n can be constructed,
and they have a point in common, then the intersection point of
m and n can be
constructed.
3b. Circle intersection: If two circles c and d can be constructed,
and they have one or two points in common, then the points of
intersection of c and d
can be constructed.
3c. Line-circle intersection: If a line m and a circle c
can be constructed, and they have one or two points in common,
then the points of intersection of m
and c can be constructed.
3. (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, m is the positive real number such that m2 = xy, or x/m = m/y.
(b) Let ABC be a right triangle with right angle at the
vertex C, and let CD be the altitude on the hypotenuse AB. Prove
that the length of the altitude
CD is the geometric mean of the lengths of the two segments AD
and BD.
In triangle ABC, the angle C is a right angle, so angles A and B are complementary.
In triangle ACD, the angle u = CDA is a right angle, so
angles v = ACD and w = DAC are complementary. Angle A equals w,
so its complement B
equals v, and hence the triangles ABC and ACD are similar. It
follows that AC/BC = AD/CD.
In triangle CBD, the angle x = BDC is a right angle, so
angles y = CBD and z = DCB are complementary. Angle B equals y,
so its complement A
equals z, and hence the triangles ABC and CBD are similar. It
follows that AC/BC = CD/BD.
Thus AD/CD = CD/BD, so (CD)2 = (AD)(BD). Thus CD is the
geometric mean of AD and BD.
In triangle ABC, the angle C is a right angle, so angles A and B are complementary.
In triangle ACD, the angle u = CDA is a right angle, so
angles v = ACD and w = DAC are complementary. Angle A equals w,
so its complement B
equals v, and hence the triangles ABC and ACD are similar. It
follows that AC/BC = AD/CD.
In triangle CBD, the angle x = BDC is a right angle, so
angles y = CBD and z = DCB are complementary. Angle B equals y,
so its complement A
equals z, and hence the triangles ABC and CBD are similar. It
follows that AC/BC = CD/BD.
Thus AD/CD = CD/BD, so (CD)2 = (AD)(BD). Thus CD is the geometric mean of AD and BD.
The points A, B, C lie on the circumcircle of triangle ABC.
The point O is the center of this circle, so OA = OB = OC. Thus
the three triangles OBC,
OAC, OAB are isosceles.
Since triangle OBC is isosceles, its base angles x are equal.
Since triangle OAC is isosceles, its base angles y are equal.
Since triangle OAB is isosceles,
its base angles z are equal.
The sum of the angles of the triangle ABC is 180 degrees. Thus
(y + z) + (x + z) + (x + y) = 180,
2x + 2y + 2z = 180,
x + y + z = 90.
Thus
angle(CAB) = y + z = 90 - x.
The sum of the angles of the triangle OBC is 180, so
angle(COB) + 2x = 180,
angle(COB) = 180 - 2x.
Therefore
angle(CAB) = 1/2 angle(COB).
MATH 5200/7200 Exam 1 Solutions
Dr. McCrory
Part B: TAKE-HOME EXAM
1. Given a square S, construct a square T whose area is three times the area of S.
If the square S has side
length s then the area of S is s2. If the square T has side length
t then the area of T is t2. If the area of T is 3 times
the area of S then t2 = 3s2, so t is the square root of 3 times
s,
t = sqrt(3) s .
So to do this construction,
we have to construct a segment whose length is the square root
of 3 times the length of a side of the given
square S. This new segment will be a side of the square T.
First we construct the given
square S. Starting with two points A and B, we construct the segment
AB, the line L through A
perpendicular to AB, and the line M through B perpendicular to
AB. Let c1 be the circle with center A through the point B, and
let C
be an intersection of c1 and L. Let c2 be the circle with center
B through the point A, and let D be the intersection of c2 and
M, with D
on the same side of the line AB as C.
Next we construct a segment
with length sqrt(3) times AB. Construct the line AB, and the circle
c3 with center B through the point C.
The radius of this circle is BC, the diagonal of the square ABDC,
so the radius of c3 is sqrt(2). Let E be the intersection of c3
with the
line AB (with B between A and E). The legs BD and BE of the right
triangle BDE have lengths AB and sqrt(2)(AB). By the
Pythagorean theorem, the length of the hypotenuse is
DE = sqrt[(AB)2 + (sqrt(2)AB)2] = sqrt(3)(AB).
Now construct a square T= DEFG with side DE, using the same procedure as before. The area of T is then 3 times the area of S.
2. Given a circle C and
a point P outside C, construct the two lines through P which are
tangent to C. (Definition of tangent: Suppose
that C is a circle with center O, and A is an intersection point
of C and a line L. The line L is tangent to C at A if the radius
OA of C is
perpendicular to L.)
Suppose that L is a tangent
line to C through P, and A is the intersection of L and C. By
the definition of tangent, the angle OAP is a
right angle, so the triangle OAP is a right triangle with hypotenuse
OP. We know from homework 4 that all the right triangles with
hypotenuse OP have their right angles on the circle X with diameter
OP. So A must be an intersection point of X with the given circle
C.
Therefore we construct the
tangent lines as follows. Let Q be the midpoint of the segment
OP, and let X be the circle with center Q
through P. Let A and B be the intersection points of the circles
X and C, and let L = AP and M = BP. Then L and M are the lines
through P tangent to C.