Intermediate Algebra Problem Solving ThugzMath
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February 15, 2011 Algebra, for a long time, is considered one of the most invincible foundation of mathematics. From our early mathematicians, we learned their theorems and proofs, causing the subject more rigid. Howeve However, r, when the mathematica mathematicall competitions petitions came out, not so long ago, our expectations expectations became became higher. As the difficulty of the problems of these math competitions rises, our comprehesion in problem-solving becomes wider, making the techniques more applicable to use. In this article, we will discuss some techniques to solve some algebra problems from different math competitions. We will tackle the topics that we use to handle them. Here are the examples.
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Examples
1)[iTest 1)[iTest 2008 TOC] Find the maximum value of x + y if x and y are positive real numbers such that x3 + y 3 + (x + y)3 + 36xy = 3456
. Solution: We make the substitutions a = x + y and b = xy. Then the original equation can be rewritten as a3
3
− 3ab + a + 36b = 3456 2a − 3ab + 36b − 3456 = 0 (1) This reduction comes from the identity ( x + y ) − 3xy (x + y) = x + y . Factoring 3
3
3
3
(1) we have
2a 3
3
− 3456 − (3ab − 36b) = 2(a − 1728) − 3b(a − 12) = 0 2(a − 12)(a + 12a + 144) − 3b(a − 12) = 0 (a − 12)(2a + 24a + 288 − 3b) = 0 2
2
(2)
Russelle Guadalupe (ThugzMath) is a third year student from Valenzuela City Science High School, and a finalist in the 13th Philippine Mathematical Olympiad. ∗
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Clearly from (2), we have a = x + y = 12 and 2a2 + 24a + 288 latter equation can be changed into
− 3b = 0. The
2(x + y )2 + 24(x + y) + 288 2
2x + 4xy + 2y
2
− 3xy = 0 + 24x + 24y + 288 − 3xy = 0
2x2 + xy + 2y 2 + 24x + 24y + 288 = 0
2x2 + x(y + 24) + 2y 2 + 24y + 288 = 0
(3)
For x and y to be positive real numbers, the discriminant of (3) must be nonnegative: (y + 24)2
15y 2
2
− 4(2)(2y + 24y + 288) ≥ 0 + 144y + 1728 ≤ 0 → 5y + 48y + 576 ≤ 0 2
Notice that the resulting inequality has no solution, since its discriminant is 482 4(5)(576) < 0. Thus the (maximum) value of x + y is 12.
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2)[NYCIML1 1975] Find all ordered pairs (a, b) of real numbers such that a + bi = x + x3 +
where i =
1
1 x
= 110
x3
√−1.
Solution: Knowing that
x+
3
1 x
3
=x +
1 x3
+3 x+
1 x
,
from the given system of equations, we have (a + bi)3 = 110 + 3(a + bi)
(4)
Collecting terms in (4) gives a3 + 3a2 bi + 3ab2 i2 + b3 i3
− 3a − 3bi = 110 a + 3a bi − 3ab − b i − 3a − 3bi = 110 a − 3ab − 3a + (3a b − b − 3b)i = 110 (5) From (5), it follows that a − 3ab − 3a = 110 and 3a b − b − 3b = 0. The latter equation can be factored as b(3a − b − 3) = 0. Thus, we need to consider two 3
2
3
2
3
2
3
2
2
2
2
2
cases. 1
3
New York City Interscholastic Math League
2
3
1. Case (1): b = 0 If b = 0, then we have a3 3a = 110 or a3 3a 110 = 0. By Rational Roots Theorem, 5 is the root of this cubic equation, so factoring we get (a 5)(a2 + 5a + 22) = 0. Note that a2 + 5a + 22 = 0 is irreducible, since its discriminant is 63 < 0. Thus, we get ( a, b) = (5,0) as the first real solution.
−
−
− −
−
2. Case (2): b = 0 If b = 0, then we have 3a2 3 = b2 . Plugging this equation to the equation a3 3ab2 3a = 110 gives
−
− − a − 3a(3a − 3) − 3a = 110 → a − 9a 3
2
3
3
8a
+ 9a 3
− 3a = 110
− 6a + 110 = 0 4a − 3a + 55 = 0 3
By Rational Roots Theorem, a = 4a 3
−
5 2
(6)
is a root of (6). Thus we have 2
− 3a + 55 = (2a + 5)(2a − 5a + 11) = 0
The quadratic factor of (6) is also irreducible, since its discriminant is 63 < 0. So a = 52 , which gives
−
−
√ 5 21 3 7 −1 =3 4 → = ± 2 =3 2 − − . and the other real solutions are ( ) = − 2
b
2
b
2
a, b
5 2
,
3
,
√
7
2
,
5 2
,
√
3
7
2
3)[AIME2 2005] An infinite geometric series has the sum 2005. A new series is formed such that its terms are the squares of the corresponding terms of original series and has the sum 20050. If the common ratio r = m , where m and n are n relatively prime positive integers, find m + n. Solution: The original infinite geometric series is given by a + ar + ar2 + ar3 + . . . ,
where a and r are the first term and common ratio, respectively. From the given problem, we have a
1
− r = 2005
A new series formed is a2 + a2 r2 + a2 r4 + a2 r6 + . . . , 2
American Invitational Math Examination
3
(7)
which has the sum a2 = 20050 1 r2
(8)
−
Dividing (8) and (7), we get a
= 10
1+r
(9)
From (7) and (9), it follows that 10 + 10r = 2005 2005r 1995 399 r= = 2015 403
−
so m + n = 399 + 403 = 802. 4)[AIME 2006 I] A sequence x0 , x1 , x2 , x3 , . . . , xn , . . . of positive integers satisfies x0 = 3 and xk = xk−1 + 3 for all positive integers k 1. Find the minimum possible value of
| | |
|
|x
1
Solution: We have for k
≥
+ x2 + x3 + . . . + x2006 .
|
≥ 1,
|x | = |x − k
k
1
+3
2
|→x
k
= x2k−1 + 6xk−1 + 9
It follows that from k = 1, 2, 3, . . . , 2007, x21 = x20 + 6x0 + 9 x22 = x21 + 6x1 + 9 x23 = x22 + 6x2 + 9
··················
x22007 = x22006 + 6x2006 + 9
Adding all equations gives 2007
2006 2
xi
i=1
Let S =
2006 i=0
=
2006 2
xi + 6
i=0
xi + 9(2007)
i=0
xi . With x0 = 0, we have 2007
i=1
2006 2
xi
−
x2i = 6S + 18063,
i=0
or, after simplifying, x22007 = 6 S + 18063
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In order for S to be small as possible, x22007 must be also small as possible. Since we have x22007 > 18063 and S is an integer, we must have x22007 = 135 so that
||
S =
18225
− 6
1352 = 6 S + 18063 18063 S = 27 = 27
→| |
5)[MTG3 2010] Given that 1 a= 2 determine the value of a2 +
√a
4
√
2+
1 8
− 18
√
2,
+ a + 1.
√
1 Solution: From a, we transpose 2 to the left side, square both sides and 8 collect terms:
√2
a2 =
4
(1
− a)
(10)
Note that a is a root of the quadratic equation (10). We square (10), getting a4 =
1 2 (a 8
2
− 2a + 1) = a8 − a4 + 18 ,
and adding a + 1, we have 4
a +a+1 =
=
a2
− a4 + 18 + a + 1
8 a2
+
3a 9 + 4 8
8 1 = (a2 + 6a + 9) 8 It follows that from (10) and (11), we get a
2
+ + a4
√2
(1 a) + (a2 + 6a + 9) 4 8 2 2 2 a+ = (a + 3) 4 4 4 2 = (1 + 3) = 2 4
a+1 =
√ √
3
1
−
− √
√
√
Mathematics Trainers’ Guild, Philippines
5
(11)
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Tips
Here are the tips in solving hard algebra problems: 1. Use the basic theorems and formulas, like the square of the binomial, the cube of the binomial, etc. whenever possible. 2. At the same manner, use these formulas to factor a polynomial (if it’s difficult to factor, apply Rational Roots Theorem or Vieta’s Formulas). 3. Be careful to what have you solved; sometimes if there’s no proper order in writing a solution, it’s confusing. 4. If the numbers a,b,c are in arithmetic progression, use substitutions a = x y, b = x, c = x + y or the formula 2b = a + c.
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5. Similarly, if the numbers a,b,c are in geometric progression, use substitutions a = xy , b = x, c = xy or the formula b2 = ac. 6. In solving systems of equations, use the methods like elimination, substitution, etc. 7. In solving system of symmetric equations, use substitutions a = x + y and b = xy , or x = u v and y = u + v .
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8. To have a nice solution, solve the problem first in a rough draft, then in a separate sheet of clean paper, rewrite it systematically. 9. Don’t let the problem ”scare” you. Don’t use brute force first. Although most of the problems in this article are difficult, there are techniques that govern their solution. Try to find those techniques, and use basic concepts and formulas. 10. ”Practice makes perfect”. The best way to be good in algebra is to solve problems. Your algebra problem-solving exposure makes your ability better, as the techniques/methods have a higher chance to handle the problem.
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Problems
The problems in this article came from various math competitions ((Mock)AIME, ARML4 , HMMT5 , PMO6 , etc.). These problems are arranged according to level of difficulty. Before you solve these problems, make sure that you have tried to solve the given examples. Here are the problems: 1. Find all ordered pairs ( a, b) of integers such that root of the quadratic equation x2 + ax + b = 0. 4
American Regional Math League Harvard-MIT Math Tournament 6 Philippine Mathematical Olympiad 5
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2010 + 2√2009 is a
2. Assume that x, y and z are positive real numbers satisfying the system of equations x + y + xy = 8 y + z + yz = 15 z + x + zx = 35
Determine the value of x + y + z + xyz . 3. Suppose that the roots of x3 + 3x2 + 4x 11 = 0 are a, b and c and that the roots of x3 + rx2 + sx + t = 0 are a + b, b + c and c + a. Find t.
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4. If a, b and c are the roots of x3
a2
1
−a
2
2
− 5x
b2
+ 4x 1
−b
2
− 1 = 0, determine the value of
c2
1
−c
2
.
5. Find x2 + y2 if x and y are positive integers such that x + y + xy = 71 x2 y + xy 2 = 880
6. Let x be a real number such that x3 + 4 x = 8. Determine the value of x7 + 64x2 . 7. The equation 2000x6 + 100x5 + 10x3 + x 2 =√ 0 has two real roots and one of them can be expressed in the form m+r n , where m, n and r are integers with r > 0 and m and r relatively prime. Find m + n + r.
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More Problems
Here are the difficult ones (Solve ’em, if you can) 7 : 1. If x, y and z are real numbers such that x+y+z =9 xy + yz + zx = 24
find the maximum value of z . 2. Suppose that x and y are real numbers such that x2 + 9y 2
− 4x + 6 y + 4 = 0 . Determine the maximum value of 4 x − 9y. 7
You may have some references from other math books/articles to get more information.
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3. Solve the equation8 log3x+4 (4x2 + 4x + 1) + log2x+1 (6x2 + 11x + 4) = 4. 4. If x and y are real numbers such that x+y 2
− xy = 155
x + y 2 = 325
find the value of x3
3
| − y |.
5. p , q , r are positive real numbers such that p2 + pq + q 2 = 211 q 2 + qr + r 2 = 259 r2 + rp + p2 = 307
Compute the value of pq + qr + rp. 6. A sequence of real numbers a0 , a1 , a2 , . . . is defined by a0 = 3 and for all integers n 0, (3 an+1 )(6 + an ) = 18.
≥
−
5
1 Find the value of 9
k=0
8 9
ak
.
from Bulgarian Mathematical Olympiad from China Mathematical Olympiad
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Hints
5.1
Problems
1. What can you say about the given quadratic surd (i.e
2010 + 2√2009)?
2. From the given system, it follows that (x +1)(y +1) = 9, (y +1)(z +1) = 16 and (z + 1)(x + 1) = 36. 3. Use Vieta’s Formulas. 4. Use Vieta’s Formulas. 5. Since the system is symmetric, substitute a = x + y and b = xy .
6. This problem is quite tricky. Make a chain of expressions by first isolating 4x to the given equation and multiplying the result by x. Repeat this until the degree of the left-hand term is 7. 7. Rearrange terms to get 2000 x6 2+100x5 +10x3 + x = 0, and then factor.
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5.2
More Problems
1. First, find the values of x + y and xy in terms of z . Then make a quadratic equation whose roots are x and y. Since x and y are real, the discriminant of this equation must be 0.
≥
2. Assume that the line 4x 9y = k intersects the curve x2 +9y2 4x+6y +4 = 0 at two points. These points will coincide if the given line is tangent to the given curve. Thus, the line is tangent to the curve if, after substitution, the resulting quadratic formula has discriminant 0.
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3. Use the properties of logarithms and the change-of-base formula: log b a = logx a . logx b 4. There are two answers. Like problem 5 in Section 3: Problems, substitute a = x + y and b = xy . Solve the resulting equations for a and b, and use it to find x and y . Use the identity x3 y 3 = (x y)3 + 3xy (x y ).
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−
−
5. Since 2(259) = 211 + 307, it follows that 2(q 2 + qr + r 2 ) = p2 + pq + q 2 + r2 + rp + p2 . Solve this equation in terms of q and r, and show that q, p and r are in arithmetic progression. 6. This problem is quite hard. Before tackling this problem, read first about recurrence formulas and how to solve them.
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