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Math 303 Final Exam Solutions
Dec 14, 2009
- (10 pts) The figure below is a key element in a dissection proof of the Pythagorean theorem given by H. Perigal in 1873. Complete the proof by supplying the argument. In particular, if you want to use in your argument that the largest quadrilateral is a square, make sure you prove that this is so.
α
β α
β
β α
α
β
A
C B D E
F
G
H
I
b c
a
First note that ACDH and DEF I are squares by construction. To see that ABF G is also a square, note that
ABF = 180◦^ − (α + β) = 90◦ BF G = β + BF I = β − (90◦^ − β) = 90◦ F GA = α + β = 90◦ GAB = α + BAH = α − (90◦^ − α) = 90◦.
Now the following areas are equal:
ABF G + △ABC + △BEF = ACDH + DEF I + △IF G + △AHG
Since the four triangles mentioned are congruent by construction, they have the same areas and therefore cancel to give
ABF G = ACDH + DEF I =⇒ c^2 = a^2 + b^2.
- (5 pts each) (a) Prove that the straight line through the points (0, 0) and (1,
- passes through no point with integer coordinates, other than (0, 0), of the coordinate lattice.
Suppose that the line did pass through (x, y) such that x, y ∈ Z and (x, y) 6 = (0, 0):
(x,y)
Then the slope of this line is both √ 2 − 0 1 − 0
and y − 0 x − 0
x y
So
2 = x/y. But
2 is irrational and therefore cannot be written as a ratio of two integers.
(b) If x is the golden ratio (
5 − 1)/2, show that
x =
1 + x
1 + (^) 1+^1 x
1+x
= etc.
First, observe that 1 1 + x
√ 5 − 1 2
= x.
We can now use this to substitute x wherever we see 1/(1 + x): 1 1 + (^) 1+^1 x
1 + x
= x
and so on.
- (10 pts) Let S be a nonempty and finite collection of persons. By a person, we mean any man, woman, or child in S. By a club, we mean a nonempty subset of persons in S formed, perhaps, for some civic or other purpose. Let us assume the following axioms: P1. Any two distinct clubs have one and only one member in common. P2. Each person of S belongs to two and only two clubs. P3. There are exactly four clubs. Deduce the following theorem: T1. There are exactly six persons in S.
Now △OBC and △OAK are similar because they have the same angles. Hence OK OA
OB
OC
=⇒ OK =
(OA)(AB)
OC
a^2 c Draw another angle with vertex O and mark off L and M on it so that OL = b and OM = a. Connect K and L and repeat the procedure above to construct a parallel to KL through M. Let N be the point where this line intersects the line OK. By similar triangles ON OM
OK
OL
=⇒ ON =
(OM )(OK)
OL
a^3 bc
- (10 pts) The following computation of the volume of the frustum of a pyramid is from the Moscow papyrus from about 1850 B.C: “You are given a truncated pyramid of 6 for the vertical height by 4 on the base by 2 on the top. You are to square this 4, result 16. You are to double 4, result 8. You are to square 2, result 4. You are to add the 16, the 8, and the 4, result 28. You are to take one third of 6, result 2. You are to take 28 twice, result 56. See, it is right. You will find it right.” Generalize this procedure by turning it into a formula for the volume of a frustum of a pyramid like the one below. Show that the formula gives the correct result.
b
a
b
h
a
x
a
x
h
b
Complete the frustum to a pyramid as in the picture above and consider its cross section. By similar triangles a b
x + h x
h x
h x
a b
a − b b
=⇒ x = h
b a − b
Now the volume of the frustum is the difference between the volumes of the larger pyramid and the smaller pyramid:
V =
(x + h)a^2 3
xb^2 3
ha^2 + xa^2 − xb^2 3
ha^2 + x(a^2 − b^2 )
ha^2 + h
b a − b
(a^2 − b^2 )
h 3
a^2 +
b a − b
(a − b)(a + b)
h 3
(a^2 + b(a + b)) =
h 3
(a^2 + ab + b^2 ).
Observe that this formula fits the procedure described in the Moscow papyrus for the case a = 4, b = 2, h = 6.
- (20 pts) (a) Name the three mathematicians whose work resulted in the first solution of the general cubic equation.
Scipione del Ferro, Niccol´o Fontana (Tartaglia), and Girolamo Cardano.
(b) Recall that a depressed cubic equation is one of the form x^3 + mx = n where m and n are real numbers. Expand (t − u)^3 in a form that looks similar to this equation.
(t − u)^3 = t^3 − 3 t^2 u + 3tu^2 − u^3 = t^3 − u^3 − 3 tu(t − u) implies (t︸ −︷︷ u︸ x
)^3 + 3︸︷︷︸tu m
(t︸ −︷︷ u︸ x
) = t ︸^3 −︷︷ u^3 ︸ n (c) Solve the depressed cubic above by substituting x = t − u and taking advantage of the similarity between the resulting equation and the identity you obtained in part (b).
We will let 3 tu = m t^3 − u^3 = n. From the first equation, u = m 3 t. Substitute this into the second equation to get
t^3 −
( (^) m
3 t
= n.
Subtract n and multiply both sides by t^3 :
t^3 −
( (^) m 3 t
− n = 0 =⇒ t^6 − nt^3 −
m^3 27
This is a quadratic equation in t^3 and the quadratic formula gives
t^3 =
n ±
n^2 + 4 m 273 2
n 2
n^2 4
m^3 27
We will choose the positive square root and take the cube root. (In fact, it will be easy to verify in the end that choosing the negative sign would yield the same final result.)
t =
3
n 2
n^2 4
m^3 27
We can recover u from the second original equation:
u^3 = t^3 − n = −
n 2
n^2 4
m^3 27 so
u =
3
n 2
n^2 4
m^3 27
Finally
x = t − u =
3
n 2
n^2 4
m^3 27
3
n 2
n^2 4
m^3 27
- (10 pts) Extra credit problem. Recall that the Fibonacci sequence is defined by the following recursive formula: u 1 = 1 u 2 = 1 un = un− 1 + un− 2 for n ≥ 2. Prove that un can also be computed explicitly for all n ≥ 1 as
un =
)n −
)n) .
Since the recursive formula above determines the Fibonacci sequence uniquely, all we have to do is verify that the explicit formula gives the same initial values for u 1 and u 2 and satisfies the recursion.
u 1 =
= √^1
u 2 =
As for the recursion, first notice that ( 1 +
Hence
un =
)n −
)n)
)n− 2 −
)n− 2
)n− 2 −
)n− 2
)n− 1
)n− 2 −
)n− 1 −
)n− 2
= un− 1 + un− 2.
- (10 pts) Extra credit problem. Euclid’s 5th postulate says “if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.” Playfair’s Axiom says “through any point in the plane, there is at most one straight line parallel to a given straight line.” Prove that these statements are equivalent if Euclid’s first four postulates are assumed to hold. That is you need to show that Euclid’s 5th postulate implies Playfair’s axiom and vice versa.
This is currently left as an exercise for you. You can always ask me how to do it if you’re dying to know.
