Ch01
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Chapter 1 • Introduction to Proof Solutions for Selected Problems Exercise 1.1 4. c. Clearly if n 41 the expression becomes 412 41
41 or 41 43, which is composite. Actually, if n 40, we have 402 40 41 40 (40 1) 41 412. The expression gives a prime number for n 1, 2, 3, … , 39, but not for 40.
6. The expression to be tested is n2 9, n even.
For n 4, n2 – 9 7, which is not composite. The statement is not true.
7. The expression to be examined is n2 9, n odd.
For n 5, n2 9 16, which is divisible by 8. For n 7, n2 9 40, which is divisible by 8. For n 9, n2 9 72. For n 13, n2 9 160. For n 15, n2 9 216. All are divisible by 8.
8. The expression to be tested is n2 3.
For n 5, n2 3 22. For n 6, n2 3 33. For n 7, n2 3 46. For n 8, n2 3 61, which is prime. The expression is not a composite number for all values of n.
3. n3 4n n(n2 4)
n(n 2)(n 2). If n is even, then set n 2k, k 1, 2, 3, …. Now n3 4n 2k(2k 2)(2k 2) 8k(k 1)(k 1). Since k 1, k, k 1 are three consecutive integers for all values of k, one of them is divisible by 2 and one (possibly the same one) is divisible by 3. Then k(k 1)(k 1) is divisible by 6. Therefore n3 4n is divisible by 8 6 48.
4. Every odd integer n can be written as n 2p1,
k 0, 1, 2, …. Then n2 4p2 4p 1 4p (p 1) 1. Now p and p 1 are consecutive integers, so one of them is even. Then p(p 1) 2k. Therefore n2 8k 1, where k is an integer.
5. Since 71 ends in 7, 72 ends in 49, 73 ends in 43, 74
ends in 01, then 75 71 74 ends in 07, 76 72 74 ends in 49, 77 73 74 ends in 43, and 78 74 74 ends in 01. Then 79 ends in 07, 710 ends in 49, 711 ends in 43, and 712 ends in 01. This cycle continues. Hence 74k+1 ends in 07, 74k+2 ends in 49, and so on. Since 201 4 50 1, 7201 ends in 07. 6. If x and y are integers, then 2x is even and 4y is even.
Hence the left side of the equation is divisible by 2, which is impossible if the right side is 5. Then it is impossible for x and y both to be integers.
Exercise 1.2 2. Every odd integer n can be written as 2p 1, where
p 1, 2, 3, …. Now, p itself can be either even or odd. If p is even, set p 2k, k 0, 1, 2, 3, …. If p is odd, set p 2k 1, k 0, 1, 2, 3, …. Then every odd number can be written as n 4k 1 or n 4k 3, k 0, 1, 2, 3, …. If n 4k 1, then n 7 4k 8 4(k 2). If n 4k 3, then n 5 4k 8 4(k 2). Hence one of n 5, n 7 is always divisible by 4.
7. n5 5n3 4n n(n4 5n2 4)
n(n2 1)(n 2 4) (n 2)(n 1)n(n 1)(n 2). This is the product of 5 consecutive integers, so one of them is divisible by 5, and at least one is divisible by each of 4, 3, and 2, provided that n 3. Then the expression is divisible by 5 4 3 2 120, for n 3 and an integer.
Chapter 1: Introduction to Proof
1
8. Since p and q are odd primes, then p q is even, so 2
divides p q. pq Now, if q p, then lies between p and q. It is 2 larger than p and smaller than q. But every number between p and q is composite, pq since p and q are consecutive odd primes, so 2 has at least two divisors. Then p q has at least three divisors.
9. Every integer can be written as 3k, 3k 1, or 3k 2.
Hence there are two possibilities. The first is that there are three of a1, a2, a3, a4, a5, of the same form. That is, three of them are of the form 3k1, 3k2, 3k3, or they are 3k1 1, 3k2 1, 3k3 1, or they are 3k1 2, 3k2 2, 3k3 2. In each case their sum is divisible by 3. The second case is that there are not three of the same form, so there must be at least one of each form, say 3k1, 3k2 1, 3k3 2. But now this sum is 3(k1 k2 k3 1), which is divisible by 3. Hence there is always at least one subset of three numbers whose sum is divisible by 3.
a b x y
c 4. Let ∠ABC be 2x and ∠ACB be 2y.
Then 2x 2y 90 180 (angles in a triangle). Therefore x y 45. Now ∠BDC x y 180 (angles in a triangle). Therefore ∠BDC 135. ∠BDC 135. B xx
D
11. Every odd integer n can be written as n 2k 1,
k 2, if we wish integers greater than 5. Now n2 25 (2k 1) 2 25 4k2 4k 24 4(k)(k 1) 24. Now k and k 1 are consecutive integers, so one of them is even. Then 4k(k 1) is divisible by 8. Since 24 is also divisible by 8, then n2 25 is divisible by 8 for n 5.
y y
A
10. Consider n2 4 (n 2)(n 2).
If n 3, then each of n 2 and n 2 is greater than 1, and n2 4 is a composite number.
d
C
5. In ∆ PQR, ∠PQR ∠PRQ (isosceles triangle).
In ∆ QRS, ∠QRS ∠QSR (isosceles triangle). Then ∠SQR 2 ∠QSR 180 (angles in a triangle). Also ∠SQR ∠PQR 180. Then ∠PQR 2 ∠QSR. Then ∠PQS 3 ∠QSR. P
Exercise 1.3 Q
3. Let the angles in any quadrilateral have measures a,
b, c, d, and let the exterior angles at opposite vertices have measures x and y, as shown. Then a b c d 360 (angles in a quadrilateral). Also b x 180 and d y 180 (straight angles). Then b x d y 360. Therefore a x c y 0 by subtraction and a c x y. The sum of the exterior angles at opposite vertices is equal to the sum of the interior angles at the other two vertices.
2
Chapter 1: Introduction to Proof
S
R
6. If the polygon has n sides and n angles each of size x,
180(n 2) nx 180n nx 360 360 n . 180 x Note that (180 x) must divide evenly into 360 if the polygon is constructable. 7. Mark angles as shown, using the given information.
Since AB AD, ∠ADB b. Since AB BE, ∠BEA a. Now 2x a 180 1 so x 90 – a. 2 Also 2y b 180 1 so y 90 – b. 2 1 In ∆ ABD, b b a 90 – a 2 a4b 1 In ∆ ABE, a a b 90 – b 2 4a b Adding 5a 5b 360 a b 72. In ∆ ABC, a b c 180 Then c 180 72 108. ∠BCA is 108.
8. Mark angles as shown.
Since b 2x 180 b x 90 – . 2 a b In ∆ ABD, b 90 – t 180 (angle sum) 2 2 1 t (a b) 90 2 2t (a b) 180 In ∆ ABC, c (a b) 180 (angle sum) Subtracting 2t c 0 1 t C 2 1 or ∠APB ∠ACB. 2 A a a 2 2
180 180. 180 B
180.
c
b x x
C
t A
P
x
a x
9. a. For the polygon ABCDEFG the sum of the exterior
c
B b y y
C
D
angles is 360. But each exterior angle occurs twice. Hence ∠P ∠Q ∠R∠S∠T ∠U∠V720 7 180. Then ∠P ∠Q ∠R ∠S ∠T ∠U ∠V 540. P B
A
V
Q
G
C
E
R U
D F
E
S
T
Chapter 1: Introduction to Proof
3
b. In general, if there are n sides then there are n
small triangles surrounding the inner polygon. Let the sum of the angles at the tips of the star be S. Then S 720 n 180 S 180(n 4).
Exercise 1.4 6. Name the vertices as in the diagram, using properties
of the rectangle. 2 Then AC a b2 2 and BD a b2
8. Let ∆ ABC be any triangle and let its coordinates be
(2b, 2c), (0, 0), and (2a, 0), as shown. If D is the midpoint of AB, its coordinates are (b, c). If E is the midpoint of AC, its coordinates are (a b, c). The slope of BC is 0 and the slope of DE is 0, so they are parallel. The length of BC is 2a and the length of DE is a, 1 so DE BC. 2 The line segment joining the midpoints of two sides of a triangle is parallel to the third side and equal to one-half of it. y
Therefore AC BD. Note that positioning the rectangle makes the proof simple. y
A(2b, 2c)
D(b, c)
E(a + b, c)
B(0, 0)
x C(2a, 0)
B(a, b)
A(0, b)
D(0, 0)
C(a, 0)
x 9. Let the parallelogram be positioned as shown, and let
7. Name the vertices as in the diagram. Since we use
midpoints, we name vertices appropriately. Since W is the midpoint of A(0, 2b) and B(2a, 2b), its coordinates are (a, 2b). Similarly X has coordinates (2a, b), Y has coordinates (a, 0), and Z has coordinates (0, b). 2 Now WX (2 a a )2 (b 2b )2 a b2 2 XY (2 a a )2 (b 0 )2 a b2 2 YZ a b2
ZW (0 a )2 (b 2b )2 a2 b2. These are all equal. Then WXYZ is a rhombus.
the coordinates of A be (b, c), D be (0, 0), and C be (a, 0). Then since AB is equal and parallel to DC, the coordinates of B are (a b, c). Now AB2 BC2 CD2 DA2 (b2 c2) [(a b b) 2 02] [(a b a)2 c2] a2 b2 c2 a2 b2 c2 a2 2 (a2 b2 c2). Also AC2 BD2 (a b)2 c2 (a b)2 c2 2(a2 b2 c2). The sum of the squares of the sides is equal to the sum of the squares of the diagonals. y A(b, c)
B(a + b, c)
y A(0, 2b) W(a, 2b)
B(2a, 2b)
Z(0, b)
X(2a, b) x x
D(0, 0)
4
Y(a, 0)
C(2a, 0)
Chapter 1: Introduction to Proof
D(0, 0)
C(a, 0)
10. Let the triangle be positioned as shown, and let the
coordinates of B be (0, 0), C be (2a, 0), and A be (b, c). Since D is the midpoint of BC its coordinates are (a, 0). Then AB2 AC2 b2 c2 (2a b)2 c2 4a2 2b2 2c2 4ab 2 2 Also 2BD 2AD 2(a2) 2 ((b a)2 c2) 4a2 2b2 2c2 4ab. The statement is true. y
A(b, c)
B(0, 0) D(a, 0)
X
C
p
A
Y
D
12. Position the triangle as shown and let the coordinates
x C(2a, 0)
11. a. Let the rectangle be positioned as shown. Let P
(x, y) be any point in the interior. Then 0 x a and 0 y b. Now PA2 PC2 x2 y2 (a x) 2 (b y2) 2x2 2y2 a2 b2 2ax 2by. Also PB2 PD2 x2 (y b)2 (a x)2 y2 2x2 2y2 a2 b2 2ax 2by. Then PA2 PC2 PB2 PD2. C(a, b)
B(0, b)
B
of the vertices be A (0, a), B (–b, 0), and C (c, 0). Then the perpendicular from A meets BC at O (0, 0), and the equation of AO is x 0. a The slope of AB is , so the slope of CE is – b, and b a b the equation of CE is y 0 (x c). a The intersection of AO and CE is the point P and for the coordinates of P we set x 0 in the equation of CE. bc Then y . a bc P is the point 0, . a bc 0 a –– c . Then the slope of the line BF is –––––– 0b a
a The slope of AC is . c c a Now 1. a c
Then BF ⊥ AC, and the altitudes are concurrent. y
P(x, y)
A(0, a) F
A(0, 0)
D(a, 0)
E P
b. Through P draw XY ⊥ AD and BC, as shown.
Then AY BX and YD XC. Now PA2 PY2 AY2 and PC2 PX2 XC2 so PA2 PC2 PY2 PX2 AY2 XC2. Also PB2 PD2 PX2 BX2 PY2 YD2 PX2 AY2 PY2 XC2. 2 2 Then PA PC PB2 PD2.
B(–b, 0)
O
x C(c, 0)
Chapter 1: Introduction to Proof
5
13. Position the triangle as shown and let the coordinates
be A(2a, 2b), B(–2c, 0) and C(2c, 0). Then the median from A meets BC at D (0, 0). The median from C meets AB at E (a c, b). b The equation of AD is y x. a b The equation of CE is y 0 x 2c . a 3c These lines intersect at P, and for intersection b b x x 2c a a 3c (a 3c) x a (x 2c)
1. c(x, y) must lie on the right bisector of AB. The
equation of the right bisector is x 2. 2. In a convex hexagon there are six vertices at each of
which there is an interior and an exterior angle. The sum of the interior and exterior angles at a vertex is 180. Then the sum of the six interior and six exterior angles is 6 180 1080. The sum of the six interior angles is 180(6 2) 720. Therefore the sum of the exterior angles is 1080 720 360.
–3cx –2ac 2a x 3 b 2b Then y x . a 3
Chapter 1 Test
3. Let the trapezium be positioned as shown and let the
coordinates of the vertices be A(2a, 2b), B(2c, 2b), C(2e, 0), and D(0, 0). Then the coordinates of P, Q, R, S, are P(a, b), Q(c e, b), R(c, b), S(a e, b). All of these points are b units above the x-axis, so all of them lie on the line whose equation is y b. y
2a 2b P has coordinates , . 3 3 The median from B meets AC at F (a c, b). b The equation of BF is y x 2c . a 3c We check that P lies on BF. 2b b 2a For y , the right side is 2c 3 a 3c 3
b 2a 6c a 3c 3
A(2a, 2b)
•
P
2b . 3 Point P is on BF, and the medians are concurrent.
R
•
• S
A(2a, 2b)
E
C(2e, 0)
4. Let the equal parts of ∠ABC be x and let the equal
F
D
Q
x D(0, 0)
y
B(–2c, 0)
•
B(2c, 2b)
x C(2c, 0)
parts of ∠ACD be y. Now let ∠ACB be z. Because ∠BCD 180, 2y z 180, In ∆ ABC, 2x z 58 180 2x z 122 Then 2x 2y 2z 302 x y z 151. In ∆ EBC, ∠E x y z 180 Therefore ∠E 29. A E
58º
x B
6
Chapter 1: Introduction to Proof
x
Z
y C
y
D
5. Solution 1.
Let the three consecutive even numbers be 2k2, 2k, 2k2. Then (2k 2)2 (2k)2 (2k 2)2 4k2 8k 4 4k2 4k2 8k 4 12k2 8 4(3k2 2) 22(3k2 2). The expression is always divisible by 2, 2, and at least one other number.
ac a 2 2 The slope of DF is b b2 c2 ac 2 2b c 2 –-––––––––––––––––––––––––––––––––––––––––––––––– b2 b2 c2 ac b c ––––––––– ac c2 b
6. Let D be the midpoint of AB. D has coordinates
0, a2.
a The right bisector of AB has equation y . 2 Let E be the midpoint of BC. b c E has coordinates , . 2 2 c The slope of BC is . b Then the right bisector of BC has slope – b. c The equation of the right bisector of BC is c b b y x 2 c 2
b . ac b ca The product of these slopes is 1. ac b Then PF is the right bisector of AC. Therefore the right bisectors are concurrent. y •
•
b b2 c or y x c 2c 2
a b b2 c2 x 2 c 2c b b2 c2 a x c 2c 2
F • C(b, c)
P D
b b2 c2 x . c 2c These right bisectors intersect at P. For the coordinates of P solve the equations: a b b2 c2 y and y x 2 c 2c
• E x
B(0, 0)
7. a. D 92 4(8)(10)
239 Since D 0 the equation has imaginary roots.
b. For the equation (n 1) x2 nx (n 1) 0,
b2 c2 ac 2c b2 c2 ac x 2b 2 b c2 ac a P has coordinates , . 2b 2
A(0, a)
D n2 4(n 1)(n 1) n2 4(n2 1) 3n2 4. Since the coefficients are positive integers, n 2. Then D 0 for all n. Therefore the equation has imaginary roots for all n.
b ac We now show that if F , is the midpoint of 2 2 AC, then PF is the right bisector of AC. This requires only that PF ⊥ AC, or that the product of the slopes of PF and AC be –1. ca The slope of AC is . b
Chapter 1: Introduction to Proof
7
41 or 41 43, which is composite. Actually, if n 40, we have 402 40 41 40 (40 1) 41 412. The expression gives a prime number for n 1, 2, 3, … , 39, but not for 40.
6. The expression to be tested is n2 9, n even.
For n 4, n2 – 9 7, which is not composite. The statement is not true.
7. The expression to be examined is n2 9, n odd.
For n 5, n2 9 16, which is divisible by 8. For n 7, n2 9 40, which is divisible by 8. For n 9, n2 9 72. For n 13, n2 9 160. For n 15, n2 9 216. All are divisible by 8.
8. The expression to be tested is n2 3.
For n 5, n2 3 22. For n 6, n2 3 33. For n 7, n2 3 46. For n 8, n2 3 61, which is prime. The expression is not a composite number for all values of n.
3. n3 4n n(n2 4)
n(n 2)(n 2). If n is even, then set n 2k, k 1, 2, 3, …. Now n3 4n 2k(2k 2)(2k 2) 8k(k 1)(k 1). Since k 1, k, k 1 are three consecutive integers for all values of k, one of them is divisible by 2 and one (possibly the same one) is divisible by 3. Then k(k 1)(k 1) is divisible by 6. Therefore n3 4n is divisible by 8 6 48.
4. Every odd integer n can be written as n 2p1,
k 0, 1, 2, …. Then n2 4p2 4p 1 4p (p 1) 1. Now p and p 1 are consecutive integers, so one of them is even. Then p(p 1) 2k. Therefore n2 8k 1, where k is an integer.
5. Since 71 ends in 7, 72 ends in 49, 73 ends in 43, 74
ends in 01, then 75 71 74 ends in 07, 76 72 74 ends in 49, 77 73 74 ends in 43, and 78 74 74 ends in 01. Then 79 ends in 07, 710 ends in 49, 711 ends in 43, and 712 ends in 01. This cycle continues. Hence 74k+1 ends in 07, 74k+2 ends in 49, and so on. Since 201 4 50 1, 7201 ends in 07. 6. If x and y are integers, then 2x is even and 4y is even.
Hence the left side of the equation is divisible by 2, which is impossible if the right side is 5. Then it is impossible for x and y both to be integers.
Exercise 1.2 2. Every odd integer n can be written as 2p 1, where
p 1, 2, 3, …. Now, p itself can be either even or odd. If p is even, set p 2k, k 0, 1, 2, 3, …. If p is odd, set p 2k 1, k 0, 1, 2, 3, …. Then every odd number can be written as n 4k 1 or n 4k 3, k 0, 1, 2, 3, …. If n 4k 1, then n 7 4k 8 4(k 2). If n 4k 3, then n 5 4k 8 4(k 2). Hence one of n 5, n 7 is always divisible by 4.
7. n5 5n3 4n n(n4 5n2 4)
n(n2 1)(n 2 4) (n 2)(n 1)n(n 1)(n 2). This is the product of 5 consecutive integers, so one of them is divisible by 5, and at least one is divisible by each of 4, 3, and 2, provided that n 3. Then the expression is divisible by 5 4 3 2 120, for n 3 and an integer.
Chapter 1: Introduction to Proof
1
8. Since p and q are odd primes, then p q is even, so 2
divides p q. pq Now, if q p, then lies between p and q. It is 2 larger than p and smaller than q. But every number between p and q is composite, pq since p and q are consecutive odd primes, so 2 has at least two divisors. Then p q has at least three divisors.
9. Every integer can be written as 3k, 3k 1, or 3k 2.
Hence there are two possibilities. The first is that there are three of a1, a2, a3, a4, a5, of the same form. That is, three of them are of the form 3k1, 3k2, 3k3, or they are 3k1 1, 3k2 1, 3k3 1, or they are 3k1 2, 3k2 2, 3k3 2. In each case their sum is divisible by 3. The second case is that there are not three of the same form, so there must be at least one of each form, say 3k1, 3k2 1, 3k3 2. But now this sum is 3(k1 k2 k3 1), which is divisible by 3. Hence there is always at least one subset of three numbers whose sum is divisible by 3.
a b x y
c 4. Let ∠ABC be 2x and ∠ACB be 2y.
Then 2x 2y 90 180 (angles in a triangle). Therefore x y 45. Now ∠BDC x y 180 (angles in a triangle). Therefore ∠BDC 135. ∠BDC 135. B xx
D
11. Every odd integer n can be written as n 2k 1,
k 2, if we wish integers greater than 5. Now n2 25 (2k 1) 2 25 4k2 4k 24 4(k)(k 1) 24. Now k and k 1 are consecutive integers, so one of them is even. Then 4k(k 1) is divisible by 8. Since 24 is also divisible by 8, then n2 25 is divisible by 8 for n 5.
y y
A
10. Consider n2 4 (n 2)(n 2).
If n 3, then each of n 2 and n 2 is greater than 1, and n2 4 is a composite number.
d
C
5. In ∆ PQR, ∠PQR ∠PRQ (isosceles triangle).
In ∆ QRS, ∠QRS ∠QSR (isosceles triangle). Then ∠SQR 2 ∠QSR 180 (angles in a triangle). Also ∠SQR ∠PQR 180. Then ∠PQR 2 ∠QSR. Then ∠PQS 3 ∠QSR. P
Exercise 1.3 Q
3. Let the angles in any quadrilateral have measures a,
b, c, d, and let the exterior angles at opposite vertices have measures x and y, as shown. Then a b c d 360 (angles in a quadrilateral). Also b x 180 and d y 180 (straight angles). Then b x d y 360. Therefore a x c y 0 by subtraction and a c x y. The sum of the exterior angles at opposite vertices is equal to the sum of the interior angles at the other two vertices.
2
Chapter 1: Introduction to Proof
S
R
6. If the polygon has n sides and n angles each of size x,
180(n 2) nx 180n nx 360 360 n . 180 x Note that (180 x) must divide evenly into 360 if the polygon is constructable. 7. Mark angles as shown, using the given information.
Since AB AD, ∠ADB b. Since AB BE, ∠BEA a. Now 2x a 180 1 so x 90 – a. 2 Also 2y b 180 1 so y 90 – b. 2 1 In ∆ ABD, b b a 90 – a 2 a4b 1 In ∆ ABE, a a b 90 – b 2 4a b Adding 5a 5b 360 a b 72. In ∆ ABC, a b c 180 Then c 180 72 108. ∠BCA is 108.
8. Mark angles as shown.
Since b 2x 180 b x 90 – . 2 a b In ∆ ABD, b 90 – t 180 (angle sum) 2 2 1 t (a b) 90 2 2t (a b) 180 In ∆ ABC, c (a b) 180 (angle sum) Subtracting 2t c 0 1 t C 2 1 or ∠APB ∠ACB. 2 A a a 2 2
180 180. 180 B
180.
c
b x x
C
t A
P
x
a x
9. a. For the polygon ABCDEFG the sum of the exterior
c
B b y y
C
D
angles is 360. But each exterior angle occurs twice. Hence ∠P ∠Q ∠R∠S∠T ∠U∠V720 7 180. Then ∠P ∠Q ∠R ∠S ∠T ∠U ∠V 540. P B
A
V
Q
G
C
E
R U
D F
E
S
T
Chapter 1: Introduction to Proof
3
b. In general, if there are n sides then there are n
small triangles surrounding the inner polygon. Let the sum of the angles at the tips of the star be S. Then S 720 n 180 S 180(n 4).
Exercise 1.4 6. Name the vertices as in the diagram, using properties
of the rectangle. 2 Then AC a b2 2 and BD a b2
8. Let ∆ ABC be any triangle and let its coordinates be
(2b, 2c), (0, 0), and (2a, 0), as shown. If D is the midpoint of AB, its coordinates are (b, c). If E is the midpoint of AC, its coordinates are (a b, c). The slope of BC is 0 and the slope of DE is 0, so they are parallel. The length of BC is 2a and the length of DE is a, 1 so DE BC. 2 The line segment joining the midpoints of two sides of a triangle is parallel to the third side and equal to one-half of it. y
Therefore AC BD. Note that positioning the rectangle makes the proof simple. y
A(2b, 2c)
D(b, c)
E(a + b, c)
B(0, 0)
x C(2a, 0)
B(a, b)
A(0, b)
D(0, 0)
C(a, 0)
x 9. Let the parallelogram be positioned as shown, and let
7. Name the vertices as in the diagram. Since we use
midpoints, we name vertices appropriately. Since W is the midpoint of A(0, 2b) and B(2a, 2b), its coordinates are (a, 2b). Similarly X has coordinates (2a, b), Y has coordinates (a, 0), and Z has coordinates (0, b). 2 Now WX (2 a a )2 (b 2b )2 a b2 2 XY (2 a a )2 (b 0 )2 a b2 2 YZ a b2
ZW (0 a )2 (b 2b )2 a2 b2. These are all equal. Then WXYZ is a rhombus.
the coordinates of A be (b, c), D be (0, 0), and C be (a, 0). Then since AB is equal and parallel to DC, the coordinates of B are (a b, c). Now AB2 BC2 CD2 DA2 (b2 c2) [(a b b) 2 02] [(a b a)2 c2] a2 b2 c2 a2 b2 c2 a2 2 (a2 b2 c2). Also AC2 BD2 (a b)2 c2 (a b)2 c2 2(a2 b2 c2). The sum of the squares of the sides is equal to the sum of the squares of the diagonals. y A(b, c)
B(a + b, c)
y A(0, 2b) W(a, 2b)
B(2a, 2b)
Z(0, b)
X(2a, b) x x
D(0, 0)
4
Y(a, 0)
C(2a, 0)
Chapter 1: Introduction to Proof
D(0, 0)
C(a, 0)
10. Let the triangle be positioned as shown, and let the
coordinates of B be (0, 0), C be (2a, 0), and A be (b, c). Since D is the midpoint of BC its coordinates are (a, 0). Then AB2 AC2 b2 c2 (2a b)2 c2 4a2 2b2 2c2 4ab 2 2 Also 2BD 2AD 2(a2) 2 ((b a)2 c2) 4a2 2b2 2c2 4ab. The statement is true. y
A(b, c)
B(0, 0) D(a, 0)
X
C
p
A
Y
D
12. Position the triangle as shown and let the coordinates
x C(2a, 0)
11. a. Let the rectangle be positioned as shown. Let P
(x, y) be any point in the interior. Then 0 x a and 0 y b. Now PA2 PC2 x2 y2 (a x) 2 (b y2) 2x2 2y2 a2 b2 2ax 2by. Also PB2 PD2 x2 (y b)2 (a x)2 y2 2x2 2y2 a2 b2 2ax 2by. Then PA2 PC2 PB2 PD2. C(a, b)
B(0, b)
B
of the vertices be A (0, a), B (–b, 0), and C (c, 0). Then the perpendicular from A meets BC at O (0, 0), and the equation of AO is x 0. a The slope of AB is , so the slope of CE is – b, and b a b the equation of CE is y 0 (x c). a The intersection of AO and CE is the point P and for the coordinates of P we set x 0 in the equation of CE. bc Then y . a bc P is the point 0, . a bc 0 a –– c . Then the slope of the line BF is –––––– 0b a
a The slope of AC is . c c a Now 1. a c
Then BF ⊥ AC, and the altitudes are concurrent. y
P(x, y)
A(0, a) F
A(0, 0)
D(a, 0)
E P
b. Through P draw XY ⊥ AD and BC, as shown.
Then AY BX and YD XC. Now PA2 PY2 AY2 and PC2 PX2 XC2 so PA2 PC2 PY2 PX2 AY2 XC2. Also PB2 PD2 PX2 BX2 PY2 YD2 PX2 AY2 PY2 XC2. 2 2 Then PA PC PB2 PD2.
B(–b, 0)
O
x C(c, 0)
Chapter 1: Introduction to Proof
5
13. Position the triangle as shown and let the coordinates
be A(2a, 2b), B(–2c, 0) and C(2c, 0). Then the median from A meets BC at D (0, 0). The median from C meets AB at E (a c, b). b The equation of AD is y x. a b The equation of CE is y 0 x 2c . a 3c These lines intersect at P, and for intersection b b x x 2c a a 3c (a 3c) x a (x 2c)
1. c(x, y) must lie on the right bisector of AB. The
equation of the right bisector is x 2. 2. In a convex hexagon there are six vertices at each of
which there is an interior and an exterior angle. The sum of the interior and exterior angles at a vertex is 180. Then the sum of the six interior and six exterior angles is 6 180 1080. The sum of the six interior angles is 180(6 2) 720. Therefore the sum of the exterior angles is 1080 720 360.
–3cx –2ac 2a x 3 b 2b Then y x . a 3
Chapter 1 Test
3. Let the trapezium be positioned as shown and let the
coordinates of the vertices be A(2a, 2b), B(2c, 2b), C(2e, 0), and D(0, 0). Then the coordinates of P, Q, R, S, are P(a, b), Q(c e, b), R(c, b), S(a e, b). All of these points are b units above the x-axis, so all of them lie on the line whose equation is y b. y
2a 2b P has coordinates , . 3 3 The median from B meets AC at F (a c, b). b The equation of BF is y x 2c . a 3c We check that P lies on BF. 2b b 2a For y , the right side is 2c 3 a 3c 3
b 2a 6c a 3c 3
A(2a, 2b)
•
P
2b . 3 Point P is on BF, and the medians are concurrent.
R
•
• S
A(2a, 2b)
E
C(2e, 0)
4. Let the equal parts of ∠ABC be x and let the equal
F
D
Q
x D(0, 0)
y
B(–2c, 0)
•
B(2c, 2b)
x C(2c, 0)
parts of ∠ACD be y. Now let ∠ACB be z. Because ∠BCD 180, 2y z 180, In ∆ ABC, 2x z 58 180 2x z 122 Then 2x 2y 2z 302 x y z 151. In ∆ EBC, ∠E x y z 180 Therefore ∠E 29. A E
58º
x B
6
Chapter 1: Introduction to Proof
x
Z
y C
y
D
5. Solution 1.
Let the three consecutive even numbers be 2k2, 2k, 2k2. Then (2k 2)2 (2k)2 (2k 2)2 4k2 8k 4 4k2 4k2 8k 4 12k2 8 4(3k2 2) 22(3k2 2). The expression is always divisible by 2, 2, and at least one other number.
ac a 2 2 The slope of DF is b b2 c2 ac 2 2b c 2 –-––––––––––––––––––––––––––––––––––––––––––––––– b2 b2 c2 ac b c ––––––––– ac c2 b
6. Let D be the midpoint of AB. D has coordinates
0, a2.
a The right bisector of AB has equation y . 2 Let E be the midpoint of BC. b c E has coordinates , . 2 2 c The slope of BC is . b Then the right bisector of BC has slope – b. c The equation of the right bisector of BC is c b b y x 2 c 2
b . ac b ca The product of these slopes is 1. ac b Then PF is the right bisector of AC. Therefore the right bisectors are concurrent. y •
•
b b2 c or y x c 2c 2
a b b2 c2 x 2 c 2c b b2 c2 a x c 2c 2
F • C(b, c)
P D
b b2 c2 x . c 2c These right bisectors intersect at P. For the coordinates of P solve the equations: a b b2 c2 y and y x 2 c 2c
• E x
B(0, 0)
7. a. D 92 4(8)(10)
239 Since D 0 the equation has imaginary roots.
b. For the equation (n 1) x2 nx (n 1) 0,
b2 c2 ac 2c b2 c2 ac x 2b 2 b c2 ac a P has coordinates , . 2b 2
A(0, a)
D n2 4(n 1)(n 1) n2 4(n2 1) 3n2 4. Since the coefficients are positive integers, n 2. Then D 0 for all n. Therefore the equation has imaginary roots for all n.
b ac We now show that if F , is the midpoint of 2 2 AC, then PF is the right bisector of AC. This requires only that PF ⊥ AC, or that the product of the slopes of PF and AC be –1. ca The slope of AC is . b
Chapter 1: Introduction to Proof
7
