JHMT 2015

Algebra Test Solutions

14 February 2015

1. In a Super Smash Brothers tournament, 1 of the contestants play as Fox, 1 of the contestants
2
3
play as Falco, and 1 of the contestants play as Peach. Given that there were 40 more people
6
who played either Fox or Falco than who played Peach, how many contestants attended the
tournament?
Answer: 60
Solution: Let x denote the number of contestants in the tournament. Then 1 x + 1 x
2
3
Thus, 2 x = 40 and hence x = 60 contestants attended the tournament.
3

1
6x

= 40.

2. Find all pairs (x, y) that satisfy
x2 + y 2 = 1
x + 2y = 2

Answer: (0, 1) and ( 4 , 3 )
5 5
Solution: The second equation tells us that x = 2 2y. Substituting this into the first equation,
we have
(2

2y)2 + y 2 = 1

5y 2
(5y

Thus, y =

8y + 3 = 0
3)(y

1) = 0

3
5

or y = 1. Plugging in these values of y and solving for x, we find that the possible
✓
◆
4 3
solutions to the system are (0, 1),
,
.
5 5
3. Find the unique x > 0 such that
Answer:

p

1
9

x+

p
p
x + x = 1.

Solution: We solve
p

q
p
x+ x+ x=1
q
p
p
x+ x=1
x
p
p 2
x + x = (1
x)
p
p
x+ x=1+x 2 x
p
3 x=1
x=

4. Find the sum of all real roots of x5 + 4x4 + x3
Answer:

3

x2

1
9

4x

1.

JHMT 2015

Algebra Test Solutions

14 February 2015

Solution 1: Notice that x5 + 4x4 + x3 x2 4x 1 = x3 (x2 + 4x + 1) (x2 + 4x + 1) =
(x3 1)(x2 + 4x + 1). The only real root of x3 1 is 1, and the real roots of x2 + 4x + 1
p
are 2 ± p by the quadratic formula. Thus, the sum of all real roots of the polynomial is
3
p
1 + ( 2 + 3) + ( 2
3) = 3 .
Solution 2: By the rational root theorem, we quickly discover that 1 is a root of the quintic
polynomial. Factoring x 1 out, we are left with the quartic x4 + 5x3 + 6x2 + 5x + 1. This
1
1
polynomial is symmetric, so we may factor it as x2 ((x + x )2 + 5(x + x ) + 4). Thus, it su ces
1
1
1
to find the roots of (x + x )2 + 5(x + x ) + 4. Writing y = x + x , we have y 2 + 5y + 4 which has
1
roots y = 1, 4. Thus, the roots of the quartic are the solutions to the equations x + x = 1
1
and x + x = 4. Solving the first equation, we have
1
= 1
x
x2 + x + 1 = 0
x+

which has no real roots. The second equation gives
1
= 4
x
x2 + 4x + 1 = 0
x+

which has the roots x =

2±

p

3.

Therefore, the sum of all real roots of the polynomial is 1 + ( 2 +

p

3) + ( 2

p

3) =

3.

5. Let a, b, c, d be integers with no common divisor such that
p
p
a34+b32+c
1
p
= p
3
d
2 4+ 32+1
Compute a + b + c + d.
Answer: 26
Solution: If we let a, b, c, d be rational numbers, then the solution is defined up to a scaling
factor. Thus, we will first solve for rational a, b, c assuming d = 1 and then scale the solution
such that a, b, c, d are all integers with no common divisor.
p
p
Now, we wish to find rational a, b, c such that a 3 4 + b 3 2 + c = p 1p . Cross multiplying
2 3 4+ 3 2+1
and simplifying, we obtain the equation:
p
p
3
3
(a + b + 2c) 4 + (4a + b + c) 2 + (2a + 4b + c 1) = 0
p p
Since a, b, c are to be rational, the coe cients of 3 4, 3 2, and the constant term must all be
zero. This gives the system of equations:
a + b + 2c = 0
4a + b + c = 0
2a + 4b + c = 1
1
7
3
Solving this system, we find that a = 23 , b = 23 , and c = 23 . This was under the assumption
that d = 1, so if we scale everything up by 23, we find that a = 1, b = 7, c = 3, and d = 23.
Therefore, a + b + c + d = 26 .

JHMT 2015

Algebra Test Solutions

14 February 2015

Solution: There must exist polynomials p(x) = ax2 + bx + c and q(x) = dx + e so that
p(x)(2x2 + x + 1) + q(x)(x3 2) = 1 for all x, as the polynomials are relatively prime. We
expand the expression to obtain
(2a + d)x4 + (a + 2b + e)x3 + (a + b + 2c)x2 + (b + c

2d)x + (c

2e) = 1,

where all of these coe cients are equal to 0 except for the last one which is equal to 1. Thus, we
1
7
have a system of linear equations. However we solve the system, we find that a = 23 , b = 23 , c =
p
p
p
3
4+7 3 2 3
3
. When we plug in x = 3 2 into the polynomial expression, we get p 1p
=
3
3
23
23
2

so a =

1, b = 7, c =

4+

2+1

3, and d = 23. Therefore, a + b + c + d = 26 .

1
6. Let f (a, b) = a+b . Suppose that x, y, z are distinct integers such that x + y + z = 2015 and
f (f (x, y), z) = f (x, f (y, z)). Compute y.

Answer:

2015

Solution: Since f (f (x, y), z) = f (x, f (y, z)), we have that
1
1
=
1
+z
x + y+z

1
x+y

Simplifying, we have that
1
1
=
+z
y+z
x+y
x(y + z)(x + y) + (x + y) = (y + z) + z(x + y)(y + z)
x+

x ((y + z)(x + y) + 1) = z ((x + y)(y + z) + 1)
(x

z) ((y + z)(x + y) + 1) = 0

Since x and z are distinct, x

z 6= 0 so we may divide through by x

z to obtain

(y + z)(x + y) + 1 = 0
(y + z)(x + y) =
Since x + y + z = 2015, y + z = 2015
(2015

1

x and x + y = 2015
x)(2015

z) =

z so
1

Since x, y, z are integers, we have that x = 2014 and z = 2016 (or the other way around). In
either case, y = 2015 x z = 2015 2014 2016 = 2015 .
7. Compute all pairs (a, b) such that (x2 + ax + b)2 + a(x2 + ax + b)
number r.
Answer: (0, 0), 1,

b = (x

r)4 for some real

1
4

Solution: Let P (x) = x2 + ax + b and Q(x) = x2 + ax b. Then we are looking for a, b such
that Q(P (x)) has only a single real repeated root. The roots of Q(P (x)) are the solutions to the
equations P (x) = r1 and P (x) = r2 where r1 , r2 are the roots of Q. Thus, a necessary condition
is for Q(x) to have a repeated root, so we require that the discriminant a2 + 4b = 0. Then the
repeated root of Q is r = a .
2

JHMT 2015

Algebra Test Solutions

14 February 2015

Now, we require that P (x) = a have a repeated root, so the discriminant of P (x) + a must be
2
2
zero. Therefore, we require that a2 4(b + a ) = 0. From earlier, we know that a2 + 4b = 0, so
2
2
we may substitute in b = a . Hence, we have the equation
4

So a = 0, 1 with corresponding b = 0,

1
4.

4(

a2
+
4
a2

a
)=0
2
a=0

a(a

a2

1) = 0

✓
Thus, the desired pairs of a and b are (0, 0), 1,

1
4

◆

.

8. Let a, b, c, d satisfy
ab + cd = 11
ac + bd = 13
ad + bc = 17
abcd = 30
Find the greatest possible value of a.
p
Answer: 30
Solution: Note that ab, cd are roots of the quadratic equation x2 11x+30 because ab+cd = 11
and ab · cd = abcd = 30. But this clearly has roots 5, 6, thus {ab, cd} = {5, 6}. Similarly, we
must have that
{ab, cd} = {5, 6},

{ac, bd} = {3, 10},

{ad, bc} = {2, 15}.

But we have that a2 = ab·ac·ad is maximized when we maximize ab, ac, ad. Using our previous reabcd
sult, we set (ab, cd) = (6, 5), (ac, bd) = (10, 3), (ad, bc) = (15, ⇣ andq q that ⌘ maximum
2)
conclude q the
p
p
2 is 6·10·15 = 30, thus a 
of a
30. Note that (a, b, c, d) =
30, 6 , 10 , 15 satisfies the
30
5
3
2
p
2 , c2 , d2 were found in analogous ways to a2 ), thus we have a =
equation (b
30 is the greatest
possible value of a.
9. Given that real numbers x, y satisfy the equation x4 + x2 y 2 + y 4 = 72, what is the minimum
possible value of 2x2 + xy + 2y 2 ?
p
Answer: 6 6
Solution 1: Let a = x2 + xy + y 2 and b = x2 xy + y 2 . Then ab = x4 + x2 y 2 + y 4 = 72 and the
expression we are trying to minimize is a+b + a. Substituting b = 72 , the expression wo want to
2
a
q
p
p
3a 36
minimize becomes 2 + a . By AM-GM, this is greater than or equal to 2 3a · 36 = 2 54 = 6 6.
2
a
p
Finally, we show 6 6 is achievable. In AM-GM, equality is only achieved when 3a = 36 so
2
a
p p
p p
p
p
p
a = 2 6. If we let x = 2 6 and y =
2 6, then a = 2 6 so 6 6 is achievable. Hence, the
p
minimum possible value is 6 6 .
Solution 2: Let a = x2 + y 2 and b = xy. Then we are trying to minimize 2a + b. The expression
x4 + x2 y 2 + y 4 can be factored as (x2 + y 2 )2 (xy)2 = (a b)(a + b). Thus, we have the condition

JHMT 2015

Algebra Test Solutions

14 February 2015

(a b)(a + b) = 72. Now, notice that |x2 + y 2 | > |xy| for any choice of x, y, so both a
a + b are always positive.
Now, suppose a b = r. Then the condition (a
we can solve the system of equations

b)(a + b) = 72 tells us that a + b =

72
r .

b and
Hence,

a

b=r
72
a+b=
r
to obtain a =
2a + b =

108
r

+

36
r
r
2.

+

r
2

36
r

and b =

r
2.

The expression we are trying to minimize is thus
q
p
r
By AM-GM, this is greater than or equal to 2 108 · 2 = 6 6.
r

r
Finally, we show that this is achievable. Equality holds in AM-GM precisely when 108 = 2 or
r
p
p
p
when r = 6 6. Then a = 4 6 and b = 2 6. So this is achievable by finding x, y such that
p p
p p
p
p
x2 + y 2 = 4 6 and xy = 2 6. One such x, y is x = 2 6 and y =
2 6. Thus, the
p
minimum possible value is indeed 6 6 .

10. Consider a sequence defined recursively by an = 1 + (a0 + 1)(a1 + 1) · · · (an
2 < a0 < 1 such that
2015
X an
a0 + 4
=
2
a
1
a2 1
0
n=0 n
Answer: 3

1
22015

1

+ 1). Find

2

1

Solution: First, notice that for n > 1,
an = 1 + (a0 + 1)(a1 + 1) · · · (an
= 1 + (an
=

1)(an

1

1

2

+ 1)(an

1

+ 1)

+ 1)

a2 1
n
1, an = (a0 + 2)2

and that a1 = a0 + 2. Therefore, for n

n 1

.

Next, notice that for n > 0
1
an

1
1

an+1

1
1
(a0 + 1)(a1 + 1) · · · (an
an
=
(a0 + 1) · · · (an + 1)
an
=
(an 1)(an + 1)
an
= 2
an 1
=

1

1
(a0 + 1)(a1 + 1) · · · (an + 1)

+ 1)

Therefore, our sum telescopes as follows
a0
a2
0

1

+

2015
X
n=1

an
a2
n

1

=
=

a0
a2
0

1

+

2015 ✓
X
n=1

1
an

a0
1
+
a2 1 a1 1
0

1
1

an+1

1
a2016

1

1

◆

JHMT 2015

Algebra Test Solutions

Using the fact that a1 = a0 + 2 and a2016 = (a0 + 2)2
to
a0
a2
0

1

+

1
a0 + 1

2015

14 February 2015

, we can simplify the above expression

1
(a0 + 2)22015

1

If we let x = a0 + 2, then the condition in the problem statement becomes
x
(x

2
2)2

1

+

1
x

2x
1)(x

1
1

1

+
3) x 1
3x 3
(x 1)(x 3)
3
x 3

(x

x22015

3(x2

2015

x22015
1
22015
x
1
22015
x
1) (x
3x2

Since we want

1

1

2015

1
1

=

x+2
(x 2)2 1

=0
=0

=0
1
3) = 0
x=0

2 < a0 , 0 < x, so we can divide through by x to get
3x2

2015

1

1=0
x=3

And hence a0 = 3

1
22015

1

2.

1
22015

1