J.M.

Basilla
E.P. Bautista
I.J.L. Garces
J.A. Marasigan
A.R.L. Valdez

The PMO 2007-2008

Problems and Solutions of the Tenth
Philippine Mathematical Olympiad

DOST-SEI • MSP • HSBC • VPHI

 Preface and Introduction
This booklet contains the questions and answers in the Tenth Philippine
Mathematical Olympiad, which was held during School Year 2007-2008.
First held in 1984, the PMO was created as a venue for high school stu-
dents with interest and talent in mathematics to come together in the spirit of
friendly competition and sportsmanship. Its aims are: (1) to awaken greater
interest in and promote the appreciation of mathematics among students
and teachers; (2) to identify mathematically-gifted students and motivate
them towards the development of their mathematical skills; (3) to provide
a vehicle for the professional growth of teachers; and (4) to encourage the
involvement of both public and private sectors in the promotion and devel-
opment of mathematics education in the Philippines.
The PMO is the first part of the selection process leading to participation
in the International Mathematical Olympiad (IMO). It is followed by the
Mathematical Olympiad Summer Camp (MOSC), a five-phase program for
the twenty national finalists of PMO. The four selection tests given during
the second phase of MOSC determine the tentative Philippine Team to the
IMO. The final team is determined after the third phase of MOSC.
The PMO is a continuing project of the Department of Science and Tech-
nology - Science Education Institute (DOST-SEI), and is being implemented
by the Mathematical Society of the Philippines (MSP).
Though great effort was put in checking and editing the contents of this
booklet, some errors may have slipped from the eyes of the reviewers. Should
you find some errors, it would be greatly appreciated if these are reported to
one of the authors at the following e-mail address:
garces@math.admu.edu.ph

Quezon City The Authors

20 June 2008
The Problems
 Area Stage
24 November 2007

µ ¶−1
2−1 + 3−1
1. Simplify: .
2−1 − 3−1
2. If 2A99561 is equal to the product when 3 × (523 + A) is multiplied by
itself, find the digit A.

3. The perimeter of a square inscribed in a circle is p. What is the area

of the square that circumscribes the circle?

4. The sum of the first ten terms of an arithmetic sequence is 160. The
sum of the next ten terms of the sequence is 340. What is the first
term of the sequence?

5. It is given that 4CAB ∼

= 4EF D. If AC = x + y + z, AB = z + 6,
BC = x + 8z, EF = 3, DF = 2y − z, and DE = y + 2, find x2 + y 2 + z 2 .

6. Container A contained a mixture that is 40% acid, while container B

contained a mixture that is 60% acid. A chemist took some amount
from each container, and mixed them. To produce 100 liters of mixture
that is 17% acid, she needed to pour 70 liters of pure water to the
mixture she got from containers A and B. How many liters did she
take from container A?

7. If a and b are integers such that a log250 2 + b log250 5 = 3, what is the

value of a + 2b?

8. Find all real values of x satisfying the inequality

sµ ¶2
1
+ 1 ≥ 2.
2−x

2
 9. Find the polynomial of least degree,√ having
√ integral coefficients and
leading coefficient equal to 1, with 3 − 2 as a zero.

10. Let x = cos θ. Express cos 3θ in terms of x.

11. Solve the system of equations:

½ √
x + y + xy = 28
x2 + y 2 + xy = 336.

12. Let P be a point on the diagonal AC of the square ABCD. If AP is

one-fourth of the length of one side of the square and the area of the
quadrilateral ABP D is 1 square unit, find the area of ABCD.

13. A circle is inscribed in 4ABC with sides AB = 4, BC = 6, and

AC = 8. If P and Q are the respective points of tangency of AB and
AC with the circle, determine the length of chord P Q.
√ √
14. If 3 x + 5 − 3 x − 5 = 1, find x2 .

15. Let a, b, and c be real constants such that x2 + x + 2 is a factor of

25
ax3 + bx2 + cx + 5, and 2x − 1 is a factor of ax3 + bx2 + cx − 16 . Find
a + b + c.

16. Consider the function f defined by

2
f (x) = 1 + .
x
Find the roots of the equation

(f ◦ f ◦ · · · ◦ f )(x) = x,
| {z }
10 times

where “◦” denotes composition of functions.

3
17. How many ordered pairs (x, y) of positive integers, where x < y, satisfy
the equation
1 1 1
+ = .
x y 2007
−→
18. Let ABC be an equilateral triangle. Let AB be extended to a point D
such that B is the midpoint of AD. A variable point E is taken on the
same plane such that DE = AB. If the distance between C and E is
as large as possible, what is ∠BED?

19. For what values of k does the equation

|x − 2007| + |x + 2007| = k

have (−∞, −2007) ∪ (2007, +∞) as its solution set?

20. Find the sum of the maximum and minimum values of

1
.
1 + (2 cos x − 4 sin x)2

21. Let k be a positive integer. A positive integer n is said to be a k-flip

if the digits of n are reversed in order when it is multiplied by k. For
example, 1089 is a 9-flip because 1089 × 9 = 9801, and 21978 is a 4-flip
because 21978 × 4 = 87912. Explain why there is no 7-flip integer.

22. Let ABC be an acute-angled triangle. Let D and E be points on BC

and AC, respectively, such that AD ⊥ BC and BE ⊥ AC. Let P be
−−→
the point where AD meets the semicircle constructed outwardly on BC,
−−→
and Q the point where BE meets the semicircle constructed outwardly
on AC. Prove that P C = QC.

23. Two friends, Marco and Ian, are talking about their ages.
Ian says, “My age is a zero of a polynomial with integer coefficients.”

4
 Having seen the polynomial p(x) Ian was talking about, Marco ex-
claims, “You mean, you are seven years old? Oops, sorry I miscalcu-
lated! p(7) = 77 and not zero.”
“Yes, I am older than that,” Ian’s agreeing reply.
Then Marco mentioned a certain number, but realizes after a while
that he was wrong again because the value of the polynomial at that
number is 85.
Ian sighs, “I am even older than that number.”
Determine Ian’s age.

National Stage
Oral Competition
12 January 2008

15-Second Round

15.1. If 

 wxy = 10

wyz = 5

 wxz = 45

xyz = 12
what is w + y?

15.2. Simplify: (x − 1)4 + 4(x − 1)3 + 6(x − 1)2 + 4(x − 1) + 1.

15.3. By how much does the sum of the first 15 positive odd integers exceed
the sum of the first 10 positive even integers?
23
15.4. Solve for x: 161/8 + x1/4 = √ .
5− 2

5
 15.5. The area of a trapezoid is three times that of an equilateral triangle.
√
If the heights of the trapezoid and the triangle are both equal to 8 3,
what is the length of the median of the trapezoid?

15.6. If 1
2
sin2 x + C = − 14 cos 2x is an identity, what is the value of C?

15.7. If ABCDEF is a regular hexagon with each side of length 6 units,

what is the area of 4ACE?

15.8. Find the smallest positive integer x such that the sum of x, x + 3, x + 6,
x + 9, and x + 12 is a perfect cube.

15.9. The length of one side of the square ABCD is 4 units. A circle is drawn
tangent to BC and passing through the vertices A and D. Find the
area of the circle.

15.10. If f (x + y) = f (x) · f (y) for all positive integers x, y and f (1) = 2, find
f (2007).

15.11. It is given that 4ABC ∼ 4DEF . If the area of 4ABC is 32 times

that of 4DEF and AB = BC = AC = 2, what is the perimeter of
4DEF ?

15.12. For which real numbers x does the inequality

µ ¶
a+b
2 logx ≤ logx a + logx b
2

hold for all positive numbers a and b?

15.13. In Figure 1, what part of 4ABC is shaded?

15.14. In how many ways can the letters of the word SPECIAL be permuted
if the vowels are to appear in alphabetical order?

15.15. Graph theory’s Four-Color Theorem says that four colors are enough to
color the regions in a plane so that no two adjacent regions receive the

6
 A
1
2

2
1
B 3 C

Figure 1: Problem 15.13.

same color. The theorem was proved in 1976 by Kenneth Appel and
Wolfgang Haken, 124 years after the Four-Color Problem was posed.
Fermat’s Last Theorem in Number Theory was proved by Andrew
Wiles in 1995, after 358 years of attempts by generations of mathe-
maticians.
In 2003, Grigori Perelman completed the proof of a conjecture in topol-
ogy. Considered as one of the seven millennium prize problems, the
conjecture says that the sphere is the only type of bounded three-
dimensional surface that contains no holes. Mathematicians worked on
this conjecture for almost a century. What is the name of this conjec-
ture that earned Perelman the Fields Medal which he refused to accept
in 2006?

30-Second Round

30.1. What is the least 6-digit natural number that is divisible by 198?

30.2. Given that x + 2 and x − 3 are factors of p(x) = ax3 + ax2 + bx + 12,
what is the remainder when p(x) is divided by x − 1?

30.3. The graphs of x2 +y = 12 and x+y = 12 intersect at two points. What

is the distance between these points?

7
 30.4. In an arithmetic sequence, the third, fifth and eleventh terms are dis-
tinct and form a geometric sequence. If the fourth term of the arith-
metic sequence is 6, what is its 2007th term?

30.5. Let each of the characters A, B, C, D, E denote a single digit, and

ABCDE4 and 4ABCDE represent six-digit numbers. If

4 × ABCDE4 = 4ABCDE,

what is C?

30.6. Let ABC be an isosceles triangle with AB = AC. Let D and E be the
feet of the perpendiculars from B and C to AC and AB, respectively.
Suppose that CE and BD intersect at point H. If EH = 1 and
AD = 4, find DE.

30.7. Find the number of real roots of the equation

4 cos(2007a) = 2007a.

30.8. In 4ABC, ∠A = 15◦ and BC = 4. What is the radius of the circle

circumscribing 4ABC?

30.9. Find the largest three-digit number such that the number minus the
sum of its digits is a perfect square.

30.10. The integer x is the least among three positive integers whose product
is 2160. Find the largest possible value of x.

60-Second Round

60.1. Three distinct diameters are drawn on a unit circle such that√ chords
are drawn as shown in Figure 2. If the length of one chord is 2 units
and the other two chords are of equal lengths, what is the common
length of these chords?

8
 ?

Figure 2: Problem 60.1.

60.2. If a and b are positive real numbers, what is the minimum value of the
expression µ ¶
√ 1 1
a+b √ + √ ?
a b
60.3. What is the remainder when the sum

15 + 25 + 35 + · · · + 20075

is divided by 5?

60.4. Let ABCD be a square. Let M be the midpoint of DC, N the midpoint
of AC, and P the intersection of BM and AC. What is the ratio of
the area of 4M N P to that of the square ABCD?

60.5. Sharon has a chandelier containing n identical candles. She lights up

the candles for n consecutive Sundays in the following manner: the first
Sunday, she lights up one candle for one hour; the second Sunday, she
lights up two candles, conveniently chosen, for one hour; and continues
in the same fashion, increasing the number of candles lighted each
Sunday by one, until in the nth Sunday, she lights up all the n candles
for one hour. For what values of n is it possible for all the n candles to
be of equal lengths right after the nth Sunday?

9
 National Stage
Written Competition
12 January 2008

1. Prove that the set {1, 2, . . . , 2007} can be expressed as the union of
disjoint subsets Ai (i = 1, 2, . . . , 223) such that

(a) each Ai contains 9 elements, and

(b) the sum of all the elements in each Ai is the same.

2. Find the largest integer n such that

n2007 + n2006 + · · · + n2 + n + 1
n + 2007
is an integer.

3. Let P be a point outside a circle, and let the two tangent lines through
P touch the circle at A and B. Let C be a point on the minor arc
−→
AB, and let P C intersect the circle again at another point D. Let
L be the line that passes through B and is parallel to P A, and let L
−→ −−→
intersect AC and AD at points E and F , respectively. Prove that B is
the midpoint of EF .

4. Let f be the function defined by

20082x
f (x) = , x ∈ R.
2008 + 20082x
Prove that
µ ¶ µ ¶ µ ¶ µ ¶
1 2 2005 2006
f +f + ··· + f +f = 1003.
2007 2007 2007 2007

10
Answers and Solutions
 Area Stage
1
1.
5
µ ¶−1 1 1 1
2−1 + 3−1 2−1 − 3−1 2
− 3 6 1
= = 1 1 = 5 =
2−1 − 3−1 2−1 + 3−1 2
+ 3 6
5
2. 4
We are given that 2A99561 = [3 × (523 + A)]2 , which is equivalent to
2A99561 = 9 × (523 + A)2 . Since (523 + A)2 is an integer, it follows
that 2A99561 is divisible by 9. By the rule on divisibility by 9, after
adding all the digits of 2A99561, it suffices to find the digit A for which
A + 5 is divisible by 9, which yields A = 4.
p2
3.
8
The area of the square that circumscribes the circle is equal to the
square of the diameter of the circle. The side of the inner square has
length equal to p/4, so that the diameter of the circle (which is equal
to the length of the diagonal of the inner square) is given by
r³ ´ √
p 2 ³ p ´2 2p
+ = .
4 4 4
79
4.
10
Let a1 , a2 , . . . , a20 be the arithmetic sequence, and let d be its common
difference. Then a1 + a2 + · · · + a10 = 160 and a1 + a2 + · · · + a10 + a11 +
a12 + · · · + a20 = 160 + 340 = 500. Recalling the formula for the sum of
an arithmetic series involving the first term a1 and common difference
d, the first equation yields 5(2a1 +9d) = 160 or 2a1 +9d = 32, while the
second equation yields 10(2a1 + 19d) = 500 or 2a1 + 19d = 50. Thus,
we get a system of linear equations:
½
2a1 + 9d = 32
2a1 + 19d = 50.

12
 Solving the system gives the value of a1 .
5. 21
Since 4CAB ∼
= 4EF D, it follows that AC = EF , AB = F D, and
BC = ED. Thus, we need to solve the following system of linear
equations: 
 x+y+z = 3
z + 6 = 2y − z

x + 8z = y + 2.
Solving the system gives x = −2, y = 4, and z = 1.
6. 5
Let a be the amount (in liters) of mixture the chemist took from
container A, and b the amount she took from container B. Then
a + b + 70 = 100. On the other hand, computing the amount of
acid involved in the mixtures, we have 0.40a + 0.60b = 0.17(100) or
4a + 6b = 170. Solving for a in the following system of equations:
½
a + b + 70 = 100
4a + 6b = 170,
we get a = 5.
7. 21
Applying laws of logarithms to the given equation, we get
log250 (2a 5b ) = 3 or 2a 5b = 2503 = 23 59 .
Since a and b are integers and gcd(2, 5) = 1, we get a = 3 and b = 9,
so that a + 2b = 21.
8. [1, 2) ∪ (2, 7/3]
√
We recall that a2 = |a| for any a ∈ R. Thus, the given inequality is
equivalent to ¯ ¯
¯3 − x¯
¯ 2 − x ¯ ≥ 2,
¯ ¯

13
 which is further equivalent to the following compound inequality:
3−x 3−x
≥ 2 or ≤ −2. (?)
2−x 2−x
We solve the first inequality in (?).
3−x 3−x x−1
≥ 2 =⇒ − 2 ≥ 0 =⇒ ≥0
2−x 2−x 2−x
The last inequality gives x = 1 and x = 2 as critical numbers.
negative
.•........positive
...................... negative
................................................................................................................................................................................................................................................................
0 1 2 3

Thus, the solution set of the first inequality in (?) is [1, 2).
Solving the second inequality in (?), we get the interval (2, 7/3]. Thus,
the solution set of the original inequality is [1, 2) ∪ (2, 7/3].
9. x4 − 10x2 + 1
√ √
We let x = 3 − 2. We find the monic polynomial equation of least
degree in terms of x. Squaring, we get
³√ √ ´2 √ √
2
x = 3 − 2 = 5 − 2 6 or x2 − 5 = −2 6.

Squaring the last equation, we finally get

³ √ ´2
(x2 − 5)2 = −2 6 or x4 − 10x2 + 1 = 0.

10. 4x3 − 3x
cos 3θ = cos(2θ + θ)
= cos 2θ cos θ − sin 2θ sin θ
= (2 cos2 θ − 1) cos θ − 2 sin2 θ cos θ
= (2 cos2 θ − 1) cos θ − 2(1 − cos2 θ) cos θ
= (2x2 − 1)x − 2(1 − x2 )x
= 4x3 − 3x

14
11. (4, 16) and (16, 4)
√
After rewriting the first equation into x + y = 28 − xy, we square to
get
√
x2 + xy + y 2 = 784 − 56 xy.
Using the second given equation, the last equation becomes
√ √
336 = 784 − 56 xy or xy = 8.

Substituting this last equation to the first given equation, we get y =

20 − x, and the second given equation then becomes

x2 + (20 − x)2 + 64 = 336,

which yields x = 4 and x = 16. Knowing that xy = 64, the solutions

are the ordered pairs (4, 16) and (16, 4).
√
12. 4 2 square units
Let s be the length of one side of the square ABCD. Let (ABP )
and (ABP D) denote the areas of 4ABP and quadrilateral ABP D,
respectively. Then (ABP ) = 21 (ABP D) = 21 .
Since the diagonals of a square are perpendicular
√ to each other and the
1
length of each diagonal of ABCD is equal to 2 2s, we have
Ã√ ! √
1 1 ³s´ 2s 2s2
(ABP ) = (base × height) = = .
2 2 4 2 16
√
Since (ABP ) is also equal to 12 , we get s2 = 4 2 square units as the
area of the square ABCD.
√
3 10
13.
4
Applying the Law of Cosines to 4ABC, we get
42 + 8 2 − 6 2 11
cos A = = .
2·4·8 16
15
 Let AP = AQ = x, P B = y, and QC = z. Then we have the following
system of equations: 
 x+y = 4
x+z = 8

y + z = 6.
Adding the equations, we get 2x + 2y + 2z = 18 or x + y + z = 9, so
that x = (x + y + z) − (y + z) = 9 − 6 = 3, y = 1, and z = 5. Finally,
applying the Law of Cosines to 4AP Q, we have
11 45
P Q2 = AP 2 + AQ2 − 2AP · AQ cos A = 32 + 32 − 2 · 3 · 3 · = .
16 8

14. 52
By factoring with difference of two cubes, we have
³√ ´3 ¡ √ ¢3
3
x+5 − 3x−5
³√ √ ´ ³p p p ´
3 3 3 2 3 3 2
= x+5− x−5 (x + 5) + (x + 5)(x − 5) + (x − 5) ,

which can be simplified into

³p √
3
p ´
(x + 5) − (x − 5) = 3 (x + 5)2 + x2 − 25 + 3 (x − 5)2

or p √ p
3 3
(x + 5)2 + x2 − 25 + 3
(x − 5)2 = 10. (?)
On the other hand, squaring the given equation, we get
p3
√
3
p
(x + 5)2 − 2 x2 − 25 + 3 (x − 5)2 = 1. (??)

Subtracting (??) from (?), we obtain

√
3 x2 − 25 = 9 or x2 = 52.
3

16
 45
15.
11
Using long division, when ax3 + bx2 + cx + 5 is divided by x2 + x + 2, the
quotient is ax + (b − a) and the remainder is (c − a − b)x + 5 + 2a − 2b.
Since x2 +x+2 is a factor of ax3 +bx2 +cx+5, we must have c−a−b = 0
and 5 + 2a − 2b = 0. On the other hand, since 2x − 1 is a factor of
25
ax3 + bx2 + cx − 16 , by the Remainder Theorem, we must have
µ ¶3 µ ¶2 µ ¶
1 1 1 25
a +b +c − =0
2 2 2 26
or
a b c 25
+ + − = 0 or 2a + 4b + 8c − 25 = 0.
8 4 2 26
Solving the following system of equations:

 c−a−b = 0
5 + 2a − 2b = 0

2a + 4b + 8c − 25 = 0,
5 25 45 45
we get a = − 22 ,b= 11
, and c = 22
, so that a + b + c = 11
.
16. −1 and 2
Let f (n) (x) = (f ◦ f ◦ · · · ◦ f )(x). For allowed values of x, note that
| {z }
n times
f (n) (x) is of the form
an x + b n
f (n) (x) = ,
cn x + d n
where an , bn , cn , dn ∈ Z for all integers n ≥ 1. The equation
an x + b n
= x or cn x2 + (dn − an )x − bn = 0
cn x + d n
has at most two real roots. Since f (−1) = −1 and f (2) = 2, it follows
that f (n) (−1) = −1 and f (n) (2) = 2 for all n ≥ 1. Thus, the roots of
f (10) (x) = x are −1 and 2.

17
17. seven
We can rewrite the given equation into

(x − 2007)(y − 2007) = 20072 = 34 · 2232 .

Since x < y, we have x − 2007 < y − 2007. It follows that

−2007 < x − 2007 < 2007 or |x − 2007| < 2007.

Thus, we have |y − 2007| > 2007.

x − 2007 y − 2007
1 34 · 2232
3 33 · 2232
32 32 · 2232
33 3 · 2232
34 2232
223 34 · 223
3 · 223 33 · 223
For every pair of values of x − 2007 and y − 2007 in the above table,
there is a corresponding pair of x and y. Thus, there are seven such
ordered pairs.
18. 15◦
To make C and E as far as possible, C, D, E must be collinear in that
order.
With ∠ABC = 60◦ , we have ∠CBD = 120◦ . Since BC = BD, we
then have ∠CDB = 12 (180◦ − 120◦ ) = 30◦ . Finally, since BD = DE,
we have ∠BED = 21 · 30◦ = 15◦ .
19. k > 4014
If x ∈ (−∞, −2007), then
k
−(x − 2007) − (x + 2007) = k or x = − .
2
18
 If x ∈ [−2007, 2007], then

−(x − 2007) + (x + 2007) = k or k = 4014.

If x ∈ (2007, +∞), then

k
(x − 2007) + (x + 2007) = k or x = .
2
Thus, the given equation has (−∞, −2007) ∪ (2007, +∞) as its solution
set if and only if
k
> 2007 or k > 4014.
2
22
20.
21
Note that
µ ¶
√ 2 4
2 cos x − 4 sin x = 20 √ cos x − √ sin x .
20 20
2 4
Let ϕ be a real number such that cos ϕ = √ and sin ϕ = √ . We
20 20
obtain √
2 cos x − 4 sin x = 20 cos(x + ϕ).
Then
0 ≤ (2 cos x − 4 sin x)2 = 20 cos2 (x + ϕ) ≤ 20.
Note here that we can particularly choose a value of x so that 20 cos2 (x+
ϕ) = 20, and a value of x so that 20 cos2 (x + ϕ) = 0. Furthermore, we
get
1 ≤ 1 + (2 cos x − 4 sin x)2 ≤ 21,
and so
1 1
≤ ≤ 1.
21 1 + (2 cos x − 4 sin x)2
Thus, the sum of the maximum and minimum values of the given ex-
1 22
pression is 1 + 21 = 21 .

19
 Q

B C
D

Figure 3: Problem 22.

21. Suppose, by way of contradiction, that the number A . . . Z is a 7-flip.

Then A . . . Z × 7 = Z . . . A. So that there will be no carry-over in the
multiplication of the last digit, A should be 1. This will imply two
contradicting statements: (1) Z ≥ 7 and (2) the product 7Z should
have a units digit of 1, making Z equal to 3.

22. (This problem is taken from the British Mathematical Olympiad 2005.)
Refer to Figure 3. By the Pythagorean Theorem, we have QC 2 =
EQ2 + EC 2 . On the other hand, since 4AQC is right-angled at Q and
QE ⊥ AC, we have EQ2 = AE · EC. It follows that

QC 2 = EQ2 + EC 2 = AE · EC + EC 2 = EC(AE + EC) = EC · AC.

Similarly, we also have

P C 2 = DC · BC.

But since 4ADC ∼ 4BEC, we obtain

DC EC
= or DC · BC = EC · AC.
AC BC
Thus, P C 2 = QC 2 , which is equivalent to P C = QC.

20
23. Let a be Ian’s age. Then

p(x) = (x − a)q(x),

where q(x) is a polynomial with integer coefficients.

Since p(7) = 77, we have

p(7) = (7 − a)q(7) = 77 = 7 · 11.

Since q(7) is an integer and 7 − a < 0, we restrict

a − 7 ∈ {1, 7, 11, 77}. (?)

Let b be the second number mentioned by Marco. Since p(b) = 85, we

have
p(b) = (b − a)q(b) = 85 = 5 · 17.
Since q(b) is an integer and b − a < 0, we restrict

a − b ∈ {1, 5, 17, 85}. (??)

Finally, we know from algebra that b − 7 is a divisor of p(b) − p(7) =

85 − 77 = 8 = 23 . It follows that

b − 7 ∈ {1, 2, 4, 8}. (? ? ?)

Considering all the possibilities from (??) and (? ? ?), since a − 7 =

(a − b) + (b − 7), we get

a − 7 ∈ {2, 3, 5, 6, 7, 9, 13, 18, 19, 21, 25, 86, 87, 89, 93}.

Recalling (?), we get a − 7 = 7 or a = 14.

21
 National Stage
Oral Competition

19
15.1.
6
We multiply the four given equations.

(wxy)(wyz)(wxz)(xyz) = 10 · 5 · 45 · 12

(wxyz)3 = 23 33 53
wxyz = 2 · 3 · 5 = 30
wxyz 30 5 wxyz 30 2
w= = = , y= = =
xyz 12 2 wxz 45 3
5 2 19
w+y = + =
2 3 6
15.2. x4

(x − 1)4 + 4(x − 1)3 + 6(x − 1)2 + 4(x − 1) + 1 = [(x − 1) + 1]4 = x4

15.3. 115
We use the formula for the sum of an arithmetic series.
15 10
(2 + 14 · 2) − (4 + 9 · 2) = 152 − 10 · 11 = 115
2 2

15.4. 625
After rationalizing the denominator, we get
√
161/8 + x1/4 = 5 + 2.

It follows that x1/4 = 5 or x = 625.

22
15.5. 24
√
3
The height of an equilateral triangle is times the length of each
2
of its sides. Thus, the length of one side of the equilateral triangle is
2 ¡ √ ¢
√ 8 3 = 16, and its area is
3
1³ √ ´ √
8 3 (16) = 64 3.
2
Since the area of the trapezoid is three times that of the equilateral
triangle, we have
√
(height) × (median of the trapezoid) = 3 · 64 3
√ √
8 3 × (median of the trapezoid) = 3 · 64 3
median of the trapezoid = 24.
1
15.6. −
4
Since the equation is an identity, it is true for all x in the domain (which
is R) of the equation. To find C, we only set a particular value of x.
For convenience, when we let x = 0, we have C = − 41 .
√
15.7. 27 3 square units
Note that 4ACE is equilateral. Each interior angle of ABCDEF
measures 61 (6 − 2)(180◦ ) = 120◦ . Using a property of a 30◦ -60◦ -90◦
triangle, we have
√
1 3 √
AC = · 6 or AC = 6 3.
2 2
√
3 √
The height of 4ACE is · 6 3 = 9, so that
2
1 √ √
area of 4ACE = · 6 3 · 9 = 27 3.
2

23
 4

2 5 2

A
A

Figure 4: Problem 15.9.

15.8. 19

x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 5x + 30 = 5(x + 6)
To make up the least possible cube, we must have x + 6 = 52 or x = 19.
25π
15.9. square units
4
See Figure 4. Let R be the radius of the circle. Using the Extended
Law of Sines, we have
4 4 2
2R = = = 2 4 = 5,
sin 2A 2 sin A cos A 2
√ · √
5 2 5

giving R = 52 . Thus, we get

µ ¶2
5 25
area of the circle = π = π.
2 4

15.10. 22007
By induction, it can be shown that f (n) = [f (1)]n = 2n for all positive
integers n. Thus, f (2007) = 22007 .

24
 √
15.11. 2 6
We first define a notation. Let (XY Z) denote the area of 4XY Z.
µ ¶2 3
r
AB (ABC) (DEF ) 3 AB 3
= = 2 = =⇒ =
DE (DEF ) (DEF ) 2 DE 2
r √
2 2 6
DE = AB = = EF = DF
3 3
√
2 6 √
perimeter of 4DEF = 3 · =2 6
3
15.12. 0 < x < 1
The domain of the variable x in the inequality is (0, 1) ∪ (1, +∞). Use
properties of logarithms, the inequality can be rewritten into
µ ¶
a+b √
logx ≤ logx ab.
2

By the AM-GM Inequality, we know that

a+b √
≥ ab.
2
It follows that the function logx is decreasing, which only happens when
0 < x < 1.
2
15.13.
9
Refer to Figure 5. Triangle ABC is equilateral, which can be subdi-
vided into 9 smaller equilateral triangles of side 1 unit in length. The
shaded part comprises half the area of two smaller equilateral triangles.
Thus, it is 29 that of 4ABC.

15.14. 840

25
 A
1
2

2
1
B 3 C

Figure 5: Problem 15.13.

We first arrange the letter without restrictions. There are 7! such ar-
rangements. There are 3! ways to arrange the vowels into three partic-
ular positions, but only one of these is where the vowels are arranged
in alphabetical order. Thus, the desired number of arrangements is
7! ÷ 3! = 4 · 5 · 6 · 7 = 840.
15.15. Poincaré Conjecture
30.1. 100188
Since 198×500 = 99000 and 198×5 = 990, we have 198×505 = 99990.
Thus, the least six-digit natural number that is divisible by 198 is
99990 + 198 = 100188.
30.2. 18
Since x + 2 is a factor of p(x) = ax3 + ax2 + bx + 12, Factor Theorem
guarantees that
p(−2) = −8a + 4a − 2b + 12 = 0 or 2a + b = 6. (1?)
Similarly, we also have
p(3) = 27a + 9a + 3b + 12 = 0 or 12a + b = −4. (2?)
Solving the system of equations involving (1?) and (2?), we get a = −1
and b = 8. Thus, we have p(x) = −x3 − x2 + 8x + 12, so that p(1) = 18.

26
 √
30.3. 2
Solving the following system of equations using the elimination method:
½ 2
x + y = 12
x + y = 12,

we get the ordered pairs (0, 12) and (1, 11). That
p is, the given graphs in-
tersect at (0, 12) and (1, 11), whose distance is (0 − 1)2 + (12 − 11)2 =
√
2.

30.4. 6015
Let a and d be the first term and the common difference of the given
arithmetic sequence. Since there are distinct terms of the sequence, it
follows that d 6= 0. Since the third, fifth and eleventh terms form a
geometric sequence, we have
a + 4d a + 10d
= or a + d = 0. (1?)
a + 2d a + 4d
Since the fourth term of the sequence is 6, we also have

a + 3d = 6. (2?)

Solving the system of equations involving (1?) and (2?), we get a = −3

and d = 3. Thus, the 2007th term is −3 + 2006(3) = 6015.

30.5. 2
Let x = ABCDE. Then 4 × ABCDE4 = 4ABCDE implies

4(10x + 4) = 400000 + x,

which gives us x = 10256, so that C = 2.

√
8 17
30.6.
17

27
 A

4 4

E D
1 1
H
B C

Figure 6: Problem 30.6.

√
Refer to Figure 6. By Pythagorean Theorem, we get AH = 17.
Since ∠AEH + ∠ADH = 90◦ + 90◦ = 180◦ , the quadrilateral ADHE
is cyclic. By Ptolemy’s Theorem (in a cyclic quadrilateral, the product
of the lengths of the diagonals equals the sum of the products of the
lengths of the opposite sides), we have
√
DE · 17 = 1 · 4 + 1 · 4,
√
8 17
or DE = .
17
30.7. three
Let x = 2007a. Then the given equation becomes 4 cos x = x. The
graphs of the equations y = 4 cos x and y = x intersect at three points.
Thus, the equation 4 cos x = x has three roots. Consequently, the
equation 4 cos(2007a) = 2007a also has three roots.
p √ √ √
30.8. 4 2 + 3 or 2( 6 + 2)
We use the Extended Law of Sines. Let R be radius of the circle
circumscribing 4ABC.
4 2
2R = ◦
or R =
sin 15 sin 15◦
28
 Depending on the formula used (either half-angle formula or difference
of two angles), we have
p √ √ √
◦ 2− 3 ◦ 6− 2
sin 15 = or sin 15 = .
2 4
Then we solve for R using either of these values.

30.9. 919
Let abc be a three-digit number such that difference between the num-
ber and the sum of its digits is a perfect square; that is,

(100a + 10b + c) − (a + b + c) = 99a + 9b = 9(11a + b)

is a perfect square. To maximize the number 100a + 10b + c, we set

a = 9, b = 1, and c = 9.

30.10. 10
Note that 2160 = 24 · 33 · 5. By trial-and-error method, we will notice
that the set of three positive integers whose product is 2160 that will
have a maximum least integer is {10, 12, 18}.
p √
60.1. 2 − 2 units
Refer to Figure
√ 7. Let θ be the central angle subtended by the chord
of length 2, and α the central angle subtended by each of the chords
of equal lengths (and let x be this common length). By the Law of
Cosines, we have

x2 = 12 + 12 − 2 cos α = 2 − 2 cos α.

Since θ + 2α = 180◦ , we get

µ ¶ √
◦ θ θ 2
cos α = cos 90 − = sin = .
2 2 2
We can then solve for x.

29
 ?

Figure 7: Problem 60.1.

√
60.2. 2 2
By the AM-GM Inequality, we have
q
√ √ √
a + b ≥ 2 ab = 2(ab)1/4

and s
1 1 1 1 2
√ +√ ≥2 √ ·√ = ,
a b a b (ab)1/4
where both inequalities become equalities if and only if a = b. Multi-
plying the two inequalities, we get
µ ¶
√ 1 1 √
a+b √ + √ ≥ 2 2.
a b

60.3. 3
By Fermat’s Little Theorem, we have a5 ≡ a (mod 5) for any integer
a. Modulo 5, we have

15 +25 +35 +· · ·+20075 ≡ 1+2+3+· · ·+2007 = 2007·1004 ≡ 2·4 ≡ 3.

Thus, the desired remainder is 3.

30
 A B

N
P

D C
M

Figure 8: Problem 60.4.

60.4. 1 : 24
Refer to Figure 8. Notice that 4M N P ∼ 4BCP , so that
NP MN 1 MP MN 1
= = and = = .
PC BC 2 BP BC 2
Recall that the ratio of the areas of two triangles of equal altitudes is
equal to the ratio of the corresponding bases. With the notation (Z)
to mean the area of polygon Z.
· ¸ · ¸
1 1 1 2 1 1
(M N P ) = (M P C) = (BP C) = (M BC) = (ABCD) ,
2 4 4 3 6 4
and so (M N P ) : (ABCD) = 1 : 24.
60.5. all positive odd integers
Since the candles are identical and they are of equal lengths right after
the nth Sunday, they are used equal number of times for all the n
Sundays. The total number of times they are used for the n Sundays
is
n(n + 1)
1 + 2 + 3 + ··· + n = .
2
It follows that each candle is used 21 (n + 1) times for all the Sundays.
Thus, since 12 (n + 1) must be an integer, n must be odd.

31
 We need to exhibit a procedure to show that, when n is odd, it is
indeed possible for Sharon to carry out the lighting of the candles. We
first label the candles with the numbers 1, 2, . . . , n. We arrange these
numbers in triangular array, writing 1, 2, . . . , n consecutively, and goes
back to 1 after n, until we complete the n Sundays. The procedure is
illustrated below when n = 7:

1
2 3
4 5 6
7 1 2 3
4 5 6 7 1
2 3 4 5 6 7
1 2 3 4 5 6 7

National Stage
Written Competition

1. We first arrange the numbers 670, 671, . . . into 223 rows and 6 columns
in the following way:

670 1115 → 1116 1561 → 1562 2007

671 1114 1117 1560 1563 2006
672 1113 1118 1559 1564 2005
↓ ↑ ↓ ↑ ↓ ↑
890 895 1336 1341 1782 1787
891 894 1337 1340 1783 1786
892 → 893 1338 → 1339 1784 → 1785

32
Let Ci represent the set containing the numbers in the ith row of the
above arrangement. It is easy to check that the numbers in each Ci
add up to a constant sum.
We now need to arrange the numbers 1, 2, . . . , 669 into 223 rows and 3
columns in such a way the sum of the numbers in each row is the same
for all the rows:

1 335 669
2 336 667
3 337 665
↓ ↓ ↓
111 445 449
112 446 447

113 224 668

114 225 666
115 226 664
↓ ↓ ↓
221 332 452
222 333 450
223 334 448
Note that the sum of the numbers in the first and second columns
of one row is different from the sum of the numbers in the first and
second columns of another row. Since we expect that the sum of the
numbers in each row is (669)(670) ÷ (2)(223) = 1005, we choose the
number in the third column of a row to be the difference between 1005
and the sum of its numbers in the first and second columns. Thus, the
numbers in the third column are distinct. Since the sum of the numbers
in the first and second columns in each row ranges from 336 to 558,
we expect that the numbers in the third column are the numbers from
1005 − 558 = 447 to 1005 − 336 = 669. Let Bi be the set containing
the numbers in the ith row of the above arrangement.

33
 The desired decomposition of the set {1, 2, . . . , 2007} is Ai = Bi ∪ Ci ,
i = 1, 2, . . . , 223.

2. Let f (n) = n2007 + n2006 + · · · + n2 + n + 1. As a geometric series, we

have
n2008 − 1
f (n) = .
n−1
Using Division Algorithm, we can write

n2008 − 1
f (n) = = (n + 2007)g(n) + R,
n−1
20072008 − 1
where g(n) is integer-valued function, and R = f (−2007) = ,
−2008
which is an integer, so that

20072008 − 1
f (n) = (n + 2007)g(n) − .
2008
So that f (n) is divisible by n + 2007, we need n + 2007 to be a factor
of R. To find the largest integer n, we should have

20072008 − 1 20072008 − 1
n + 2007 = or n = − 2007.
2008 2008

3. Refer to Figure 9. We only need to see four pairs of similar triangles.

BE BC
4AEB ∼ 4ABC =⇒ = (1?)
AB AC
BF BD
4AF B ∼ 4ABD =⇒ = (2?)
AB AD
BC BD BC BP
4P BC ∼ 4P DB =⇒ = =⇒ = (3?)
BP DP BD DP
AC AD AC AP
4P AC ∼ 4P DA =⇒ = =⇒ = (4?)
AP DP AD DP

34
 A

P
C F

D
B
E

Figure 9: Problem 3.

Since AP = BP , from (3?) and (4?), we get

BC AC BC BD
= or = .
BD AD AC AD
Using this last proportion and applying transitivity property to (1?)
and (2?) yield
BE BF
= or BE = BF.
AB AB
4. We first show that the function satisfies the identity f (x)+f (1−x) = 1.
20082(1−x) 20082 2008−2x 2008
f (1 − x) = 2(1−x)
= 2 −2x
= 2x
2008 + 2008 2008 + 2008 2008 2008 + 2008
20082x 2008
f (x) + f (1 − x) = 2x
+ 2x
=1
2008 + 2008 2008 + 2008
Pairing off the terms of the left-hand side of the desired equality into
· µ ¶ µ ¶¸ · µ ¶ µ ¶¸
1 2006 1003 1004
f +f + ··· + f +f ,
2007 2007 2007 2007
and applying the above identity solve the problem.

35
10th Philippine Mathematical Olympiad
Area Stage: 24 November 2007
National Stage: 12 January 2008, UPNISMED
Project Director
Dr. Evangeline P. Bautista, Ateneo de Manila University
National Winners
First Place
Stephanie Anne A. Oliveros, Philippine Science High School (Diliman)
Second Place
Jillian Kristel G. Sy, Chiang Kai Shek College
Third Place
Diogo Miguel S. Moitinho de Almeida, Ateneo de Manila High School

49th International Mathematical Olympiad

10-22 July 2008, Madrid, Spain
Philippine IMO Team
Contestants
Jeffrey Kenneth L. Go, Xavier School
Diogo Miguel S. Moitinho de Almeida, Ateneo de Manila High School
Mark Benedict C. Tan, Xavier School
Deputy Leader
Dr. Julius M. Basilla, University of the Philippines (Diliman)
Leader
Dr. Ian June L. Garces, Ateneo de Manila University
 view.php3 (JPEG Image, 840x888 pixels) - Scaled (71%) https://mail.ateneo.net/horde/imp/view.php3?mailbox=INBOX&inde...

11th Philippine Mathematical Olympiad

Questions, Answers, and Hints

1 of 1 11/5/2008 5:02 PM
Questions

Qualifying Stage, 18 October 2008

Part I. Each correct answer is worth two points.

1. Simplify: 1 − 31 + 15 − 17 + 1
11
.
940 941 942 943
(a) 1155
(b) 1155
(c) 1155
(d) 1155

2. If 30% of p is q, and 20% of q is 12, what is 50% of p + q?

1 3
(a) 2
(q + 12) (b) 130 (c) 10
p + 15 q (d) 100
√
3
√
6
3. Which of the following is equal to 54 + 4?
√ √ √ √
(a) 3 128 (b) 3 56 (c) 9
54 · 4 (d) 6
112

4. Let ABCD be a square with each side of length 1 unit. If M is the intersection of its
diagonals and P is the midpoint of M B, what is the square of the length of AP ?
3 5 1 3
(a) 4
(b) 8
(c) 2
(d) 8

5. How many ordered pairs (x, y) of positive integers satisfy 2x + 5y = 100?

(a) 8 (b) 9 (c) 10 (d) 11

6. Find the area of the circle that circumscribes a right triangle whose legs are of lengths
6 cm and 10 cm.

(a) 34π cm2 (b) 68π cm2 (c) 102π cm2 (d) 136π cm2

2
7. How many real roots does the equation log(x2 −3x)3 4 = 3
have?

(a) 0 (b) 2 (c) 3 (d) 4

 8. Find the largest integer value of n such that 1 × 3 × 5 × 7 × · · · × 31 × 33 × 35 is divisible
by 3n .

(a) 6 (b) 7 (c) 8 (d) 9

9. Let 0 < x < 1. Which of the following has the largest value?

(a) x3 (b) x2 + x (c) x2 + x3 (d) x4

10. Find the sum of all the 4-digit positive numbers with no zero digit.

(a) 36644635 (b) 36644645 (c) 36445335 (d) 36446355

11. Find the number of real roots of the equation x5 − x4 + x3 − 4x2 − 12x = 0.

(a) 0 (b) 1 (c) 3 (d) 5

12. Let ABCD be a trapezoid with BCkAD and AB = BC = CD = 21 AD. Determine

∠ACD.

(a) 75◦ (b) 85◦ (c) 90◦ (d) 135◦

13. How many 4-digit positive numbers, whose digits are from the set {1, 2, 3, 4}, are
divisible by 4?

(a) 16 (b) 32 (c) 48 (d) 64

14. How many real numbers x satisfy the equation xx = x2 ?

(a) 1 (b) 2 (c) 4 (d) infinite

15. Let x be the solution of the equation

x+1 x+2 x+3 x + 2007 x + 2008
+ + + ··· + + = 2008.
1 2 3 2007 2008
Which of the following is true?

(a) x > 1 (b) x = 1 (c) 0 < x < 1 (d) x ≤ 0

Part II. Each correct answer is worth three points.

16. The roots of the quadratic equation x2 − 51x + k = 0 differ by 75, where k is a real
number. Determine the sum of the squares of the roots.

(a) 756 (b) 3825 (c) 4113 (d) 5625

17. Marco plans to give (not necessarily even) his eight marbles to his four friends. If each
of his friends receives at least one marble, in how many ways can he apportion his
marbles?

(a) 32 (b) 33 (c) 34 (d) 35

18. How many triangles (up to congruence) with perimeter 16 cm and whose lengths of its
sides are integers?

(a) 3 (b) 4 (c) 5 (d) 6

19. Which of the following is not satisfied by any solution of the system
 2
x − xy − 2y 2 = 4
x2 + 2xy + 3y 2 = 3 ?
(a) x = −2y (b) 9y = −x (c) x = −4y (d) y 2 = 1

20. If a, b, c, d are real numbers with a > b and c > d, which of the following is always
true?

(a) ac > bd (c) ac + bd > ad + bc

(b) a2 + c2 > b2 + d2 (d) ad + bc > ac + bd

a−b
21. Given that 0 < b < a and a2 + b2 = 6ab, what is the value of ?
a+b
√ √ √ √
(a) 2 (b) 1 + 2 (c) 12 2 (d) −1 + 2

n+3
22. How many values of n for which n and are both integers?
n−1
(a) 3 (b) 4 (c) 5 (d) 6
 23. In an isosceles triangle ABC, where AB = BC, there exists a point P on the segment
AC such that AP = 6, P C = 4, and BP = 2. Determine the perimeter of triangle
ABC.
√ √ √ √
(a) 10 + 2 7 (b) 10 + 4 7 (c) 12 + 2 7 (d) 12 + 4 7

24. Each side of the square ABCD is 12 meters long. The side AB is divided into three
equal segments: AE, EF , and F B. Segments CE and DF intersect at point H. Find
the area of triangle HCD.

(a) 48 m2 (b) 54 m2 (c) 60 m2 (d) 72 m2

25. An operation ∗ on the set of positive integers is defined by a ∗ b = (a + b)a−b . Evaluate

1024 ∗ (512 ∗ (256 ∗ (128 ∗ (64 ∗ (32 ∗ (16 ∗ (8 ∗ (4 ∗ (2 ∗ 1))))))))).

(a) 2027 (b) 2028 (c) 2047 (d) 2048

Part III. Each correct answer is worth six points.

26. If a3 + 12ab2 = 679 and 9a2 b + 12b3 = 978, find a2 − 4ab + 4b2 .

(a) 1 (b) 9 (c) 25 (d) 49

√ √ √
27. How many ordered pairs (x, y) of positive integers satisfy the equation y= 17+ x?

(a) none (b) 1 (c) 2 (d) infinite

28. Let f : R → R be a function such that f (a + b) = f (a) + f (b) and that f (2008) = 3012.
What is f (2009)?

(a) 3012.5 (b) 3013 (c) 3013.5 (d) 3014

29. The length of each side of the squares ABCD and DEF G (both labeled in counter-
clockwise direction) is 5 units. If AG is 8 units, how many units is BF ?

(a) 11 (b) 12 (c) 13 (d) 14

30. A point M is chosen inside the square ABCD in such a way that ∠M AC = ∠M CD =
x. Determine ∠ABM .

(a) 90 − 2x (b) 180 − 3x (c) 90 − x (d) 2x

 Area Stage, 22 November 2008

Part I. No solution is needed. All answers must be in simplest form. Each correct answer
merits two points.
√ √
1. For what integer a does the compound inequality a < 48 + 140 < a + 1 hold?
2 +2
2. For what real numbers x is the equation (x2 + 6x + 10)x = 1 true?

3. Each element of the arithmetic sequence 101, 111, 121, . . . , 201 is multiplied to each
element of the arithmetic sequence 212, 222, 232, . . . , 302. What is the sum when all
these products are added?

4. Let x be a real number such that csc x + cot x = 3. Evaluate csc x − cot x.

5. The sum of the areas of two similar polygons is 65 square units. If their perimeters are
12 units and 18 units, respectively, find the area of the larger polygon.
x
6. Find all positive real values of x for which xx = (xx )x .

7. In quadrilateral ABCD, it is given that AB = BC = 4 cm, ∠ABC = 100◦ , and

∠CDA = 130◦ . Find the length of BD.

8. How many polynomials are there of the form x3 − 10x2 + cx + d, where c and d are
integers, such that the three roots are distinct positive integers?

9. The bases of a trapezoid are 9 in and 15 in, respectively. Its altitude is 7 in. Find the
area of the quadrilateral formed by joining the midpoints of the adjacent sides of the
trapezoid.

10. If x ∈ R, what is the least possible value of the expression |x + 1| + |x − 2| + |5 − x|?

11. Chuckie was born before the year 2000. This year 2008, his age is exactly the sum of
the digits of his year of birth. How old is Chuckie now?

12. Let x, y, z ∈ R such that 6x2 + y 2 + z 2 = 2x(y + 2z). What is x + y + z?

13. In how many ways can the letters of the word OLYMPIAD be arranged if the vowels
must be in alphabetical order?

14. Find the largest integer that divides all terms of the sequence {an }, where an = n5 − n,
n ≥ 1.

15. A right triangle has sides of integral length and its area is equal to its perimeter. What
is the least possible length of one of its legs?

16. Give the prime factorization of 320 + 319 − 12.

17. How many integers between 2 and 10000 do not share a prime factor with 10000?
 18. In isosceles triangle ABC, the base angles at B and C measure 40◦ . The bisector of
angle B intersects AC at point D, and BD is extended to point E so that DE = AD.
Find ∠BEC.

19. At least how many 3-digit composite numbers should be chosen to ensure that at least
two of the chosen numbers are not relatively prime?

20. Let AB and BC be two consecutive sides of a regular pentagon inscribed in a unit
circle (that is, a circle of radius 1 unit). Find the value of (AB · AC)2 .

Part II. Show the solution to each problem. A complete and correct solution merits ten
points.

21. Consider the numbers 1, 10, 19, . . . , 2008, which form an arithmetic sequence. A num-
ber n is the sum of eleven distinct numbers from this sequence. How many (different)
possible values of n are there?

22. Let a, b, and c be nonnegative real numbers such that a + b + c = 1. Prove that
√ √ √ 1
a bc + b ac + c ab ≤ .
3

23. The bisector of ∠BAC intersects the circumcircle of 4ABC at a second point D.
Let AD and BC intersect at point E, and F be the midpoint of segment BC. If
AB 2 + AC 2 = 2AD2 , show that EF = DF .

National Stage, 24 January 2009

Oral Phase

15-Second Round. Each correct answer credits two points.

15.1 Define the operation “?” by a ? b = 4a − 3b + ab for all a, b ∈ R. For what real numbers
y does 12 = 3 ? y?

15.2 Six numbers from a list of nine integers are 4, 9, 3, 6, 7, and 3. What is the largest
possible value of the median of all nine integers in this list?

15.3 Triangle ABC has vertices with coordinates A(1, 3), B(4, 1), and C(3, −5). A point
D on AC is chosen so that the area of triangle ABD is equal to the area of triangle
CBD. Find the length of segment BD.

15.4 How many positive integers less than 2009 are divisible by 28 but not by 12?
3 +x2 −2x
15.5 Find all real numbers x that satisfy the equation (x3 − x)x = 0.
 15.6 What is the maximum number of points of intersection of the graphs of two differ-
ent fourth-degree polynomial functions y = P (x) and y = Q(x), each with leading
coefficient 1?
f (x)
15.7 Let f be a function satisfying f (xy) = for all positive real numbers x and y. If
y
f (2008) = 1, what is f (2009)?

15.8 Simplify the following expression √ √

2+ 6
p √ .
2+ 3
15.9 If the letters of the word MATHEMATICS are repeatedly and consecutively written,
what is the 2009th letter?

15.10 A rectangular piece of paper, 24 cm long by 18 cm wide, is folded once in such a way
that two diagonally opposite corners coincide. What is the length of the crease?

15.11 A chemist has x liters of sugar solution that is x% sugar. How many liters of sugar
must be added to this solution to yield a sugar solution that is 3x% sugar?

15.12 In how many ways can 60 students be distributed into 6 buses if a bus can contain
zero to 60 students?

15.13 If r1 , r2 , r3 , r4 are the roots of the equation 4x4 − 3x3 − x2 + 2x − 6 = 0, what is

1 1 1 1
+ + + ?
r1 r2 r3 r4
√
15.14 If sin x + cos x = 3 − 1, what is sin 2x?

15.15 The Philippines officially joined the International Mathematical Olympiad (IMO) in
1989. In what country was this IMO held?

30-Second Round. Each correct answer credits three points.

30.1 The first, fourth, and eighth terms of a nonconstant arithmetic sequence form a geo-
metric sequence. If its twentieth term is 56, what is its tenth term?
√
1 10
30.2 Let α and β be acute angles such that tan α = 7
and sin β = 10
. Find cos(α + 2β).

30.3 A point P is outside a circle and is 15 cm from the center. A secant through P cuts
the circle at Q and R so that the external segment P Q is 9 cm and QR is 8 cm. Find
the radius of the circle.

30.4 What is the coefficient of x5 in the polynomial expansion of (2 − x + x2 )4 ?

30.5 All the students in a geometry class took a 100-point test. Six students scored 100,
each student scored at least 50, and the mean score was 68. What is the smallest
possible number of students in the class?
 30.6 Let a ≥ b > 1. What is the largest possible value of
a b
loga + logb ?
b a

30.7 Let P (x) be a polynomial that, when divided by x − 19, has the remainder 99, and,
when divided by x − 99, has the remainder 19. What is the remainder when P (x) is
divided by (x − 19)(x − 99)?

30.8 In the plane, two concentric circles with radii 8 cm and 10 cm are given. The smaller
circle divides a chord of the larger circle into three equal parts. Find the length of the
chord.
12
X
30.9 Let f be a function such that f (2 − 3x) = 4 − x. Find the value of f (i).
i=1

30.10 The number 63 999 999 has exactly five prime factors. Find their sum.

60-Second Round. Each correct answer credits six points.

60.1 Equilateral triangle DEF is inscribed in equilateral triangle ABC with DE ⊥ BC. If
the area of 4DEF is 6 cm2 , what is the area of 4ABC?

60.2 For any positive integer n, define


log9 n if log9 n is rational,
f (n) =
0 otherwise.
2009
X
What is f (n)?
n=1

60.3 A point P is selected at random from the interior of the pentagon with vertices A =
(0, 2), B = (4, 0), C = (2π + 1, 0), D = (2π + 1, 4), and E = (0, 4). What is the
probability that ∠AP B is acute?

60.4 Let P (n) and S(n) denote the product and the sum, respectively, of the digits of the
positive integer n. Determine all two-digit numbers N that satisfy the equation

N = P (N ) + 2S(N ).

60.5 The vertices of a cube are each colored by either black or white. Two colorings of the
cube are said to be geometrically the same if one can be obtained from the other by
rotating the cube. In how many geometrically different ways can such coloring of the
cube be done?
 Written Phase

You are given three hours to solve all problems. Each item is worth eight points.

1. The sequence {a0 , a1 , a2 , . . .} of real numbers satisfies the recursive relation

n(n + 1)an+1 + (n − 2)an−1 = n(n − 1)an

for every positive integer n, where a0 = a1 = 1. Calculate the sum

a0 a1 a2008
+ + ··· + .
a1 a2 a2009

2. (a) Find all pairs (n, x) of positive integers that satisfy the equation 2n + 1 = x2 .
(b) Find all pairs (n, x) of positive integers that satisfy the equation 2n = x2 + 1.

3. Each point of a circle is colored either red or blue.

(a) Prove that there always exists an isosceles triangle inscribed in this circle such
that all its vertices are colored the same.
(b) Does there always exist an equilateral triangle inscribed in this circle such that
all its vertices are colored the same?

4. Let k be a positive real number such that

1 1 1
+ + ≤1
k+a k+b k+c
for any positive real numbers a, b, and c with abc = 1. Find the minimum value of k.

5. Segments AC and BD intersect at point P such that P A = P D and P B = P C. Let

E be the foot of the perpendicular from P to the line CD. Prove that the line P E
and the perpendicular bisectors of the segments P A and P B are concurrent.
 Answers and Hints
Qualifying Stage

1. b 6. a 11. c 16. c 21. c 26. b

2. b 7. b 12. c 17. d 22. d 27. d

3. a 8. d 13. d 18. c 23. b 28. c

4. b 9. b 14. b 19. c 24. b 29. d

5. b 10. d 15. d 20. c 25. c 30. a

Area Stage

1. 18 11. 23

2. −3 12. 0

3. 4268770 13. 6720

4. 1/3 14. 30

5. 45 15. 5

6. 1 and 2 16. 25 · 3 · 7 · 13 · 19 · 37 · 757

7. 4 17. 3999

8. 4 18. 80◦

9. 42 19. 12

10. 6 20. 5

21. Write n = 11 + 9(a1 + a2 + · · · + a11 ), where a1 , a2 , . . . , a11 are distinct elements of the
set {0, 1, 2, . . . , 223}. Let X = a1 + a2 + · · · + a11 . By showing that the integers from
55 to 2398 are possible values of X, there are 2398 − 55 + 1 = 2344 possible values of
X (and also of n).

22. The desired inequality follows from the following inequalities:

√ √ √
(i) 2a bc ≤ ab + ac, 2b ac ≤ ab + bc, 2c ab ≤ ac + bc
√ √ √
(ii) 4a bc ≤ a2 + a2 + b2 + c2 , 4b ac ≤ a2 + b2 + b2 + c2 , 4c ab ≤ a2 + b2 + c2 + c2 .

These inequalities follow from the AM-GM Inequality.

 23. By symmetry, we assume that AB < AC. Use Ptolemy’s Theorem and the Angle
Bisector Theorem to show that
BC(AC − AB)
DF = EF = .
2(AB + AC)

National Stage, Oral Phase

15.1. all real numbers 15.11. 2x2 /(100 − 3x) 30.6. 0

15.2. 7 15.12. 660 30.7. −x + 118

√ √
15.3. 2 2 15.13. 1/3 30.8. 9 2 cm
√
15.4. 48 15.14. 3 − 2 3 30.9. 66

15.5. −1 15.15. Germany 30.10. 577

15.6. 3 30.1. 36 60.1. 18 cm2

√
15.7. 2008/2009 30.2. 2/2 60.2. 21/2
√
15.8. 2 30.3. 6 2 cm 60.3. 11/16

15.9. A 30.4. −28 60.4. 14, 36, and 77

15.10. 22.5 cm 30.5. 17 60.5. 24

National Stage, Written Phase

1. It can be proved, by induction on n, that an−1 /an = n for every positive integer n.
Thus, we obtain
a0 a1 a2008 (2009)(2010)
+ + ··· + = 1 + 2 + · · · + 2009 = = 2019045.
a1 a2 a2009 2
2. (a) Rewrite the given equation into 2n = (x − 1)(x + 1). With the power-of-two
argument, the only pair that satisfies the equation is (3, 3).
(b) If n = 1, then x = 1. So, (1, 1) satisfies the equation. It is not possible that n ≥ 2.
Thus, (1, 1) is the only pair that satisfies the equation.
3. (a) By the Pigeonhole Principle, there are three vertices of the same color that form
an isosceles triangle.
(b) There is a coloring of the points of the circle such that no inscribed equilateral
triangle has three vertices of the same color.
4. After considering three cases (when k = 2, when k > 2, and when 0 < k < 2), one can
conclude that the minimum value of k is 2.
5. Let the perpendicular bisectors of P A and P B meet at O. Let OP meet CD at F . It
suffices to show that OF ⊥ CD.
 12th Philippine Mathematical Olympiad
Qualifying Stage
24 October 2009

Part I. Each correct answer is worth two points.

1. If 2009
| + 2009{z+ · · · + 2009} = 2009x , find the value of x.
2009 terms

(a) 2 (b) 3 (c) 2009 (d) 2010

2. What is the least positive difference between two three-digit numbers

if one number has all the digits 2, 4, and 6, while the other has all the
digits 1, 2, and 9?
(a) 72 (b) 54 (c) 48 (d) 27

3. Which of the following numbers is closest to one of the roots of the

equation x2 − 10000x − 10000 = 0?
(a) 10001 (b) 5000 (c) 100 (d) 25

4. The ratio of the areas of two squares is 3 : 4. What is the ratio of the
lengths of their corresponding diagonals?
√
(a) 1 : 2 (b) 3 : 4 (c) 2 : 3 (d) 3 : 2

5. In 4ABC, let P be a point on segment BC such that BP : P C = 1 : 4.

Find the ratio of the area of 4ACP to that of 4ABC.
(a) 1 : 4 (b) 1 : 5 (c) 3 : 4 (d) 4 : 5

6. In how many ways can three distinct numbers be selected from the set
{1, 2, 3, . . . , 9} if the product of these numbers is divisible by 21?
(a) 15 (b) 16 (c) 17 (d) 18

7. If |2x − 3| ≤ 5 and |5 − 2y| ≤ 3, find the least possible value of x − y.

(a) −5 (b) 0 (c) −1 (d) 5
 8. Define the operations ♣ and ♥ by

a ♣ b = ab − a − b and a ♥ b = a2 + b − ab.

What is the value of (−3 ♥ 4) − (−3 ♣ 4)?

(a) 38 (b) 12 (c) −12 (d) −38

9. Solve for x in the following system of equations:


 log x + log y = 2
log y + log z = 7
log z + log x = 3.


(a) 10 (b) 1 (c) 0.1 (d) 0.01

10. If a + 1 = b − 2 = c + 3 = d − 4, which is the smallest among the

numbers a, b, c, and d?
(a) a (b) b (c) c (d) d

11. Solve for x in the inequality 5x ≥ 252x .

(a) x ≤ 1 (b) x ≤ 0 (c) x ≥ 0 (d) x ≥ 1

12. Today, 24 October 2009, is a Saturday. On what day of the week will
10001 days from now fall?
(a) Saturday (b) Monday (c) Thursday (d) Friday

13. The lines 2x + ay + 2b = 0 and ax − y − b = 1 intersect at the point

(−1, 3). What is 2a + b?
(a) −6 (b) −4 (c) 4 (d) 6

14. Let x be a real number that satisfies the equation

2
16 (log9 x)4 = log3 x3 + 10.

Determine (log9 x)2 .

√ 5
√
5
(a) 10 (b) 10 (c) 2
(d) 2
 15. Let r and s be the roots of the equation x2 − 2mx − 3 = 0. If r + s−1
and s + r−1 are the roots of the equation x2 + px − 2q = 0, what is q?
2
(a) 1 (b) 3
(c) −3 (d) − 34

Part II. Each correct answer is worth three points.

16. On the blackboard, 1 is initially written. Then each of ten students,

one after another, erases the number he finds on the board, and write
its double plus one. What number is erased by the tenth student?
(a) 211 − 1 (b) 211 + 1 (c) 210 − 1 (d) 210 + 1
p √
17. For how many real numbers x is 2009 − x an integer?
(a) 0 (b) 45 (c) 90 (d) 2009

18. How many distinct natural numbers less than 1000 are multiples of 10,
15, 35, or 55?
(a) 145 (b) 146 (c) 147 (d) 148
√
19. Let x and y be nonnegative real numbers such that 2x+2y = 8 2. What
is the maximum possible value of xy?
√
(a) 8 2 (b) 49/4 (c) 49/32 (d) 1

20. In how many ways can ten people be divided into two groups?
(a) 45 (b) 511 (c) 637 (d) 1022

21. Let P be the point inside the square ABCD such that 4P CD is equi-
lateral. If AP = 1 cm, what is the area of the square?
√ √
(a) 3 + 3 cm2 (b) 2 + 3 cm2 (c) 94 cm2 (d) 2 cm2

22. Let x and y be real numbers such that 22x + 2x−y − 2x+y = 1. Which
of the following equations is always true?
(a) x + y = 0 (b) x = 2y (c) x + 2y = 0 (d) x = y

23. In 4ABC, M is the midpoint of BC, and N is the point on the bisector
of ∠BAC such that AN ⊥ N B. If AB = 14 and AC = 19, find M N .
(a) 1 (b) 1.5 (c) 2 (d) 2.5
 24. Seven distinct integers are randomly chosen from the set {1, 2, . . . , 2009}.
What is the probability that two of these integers have a difference that
is a multiple of 6?
7 2 1
(a) 2009
(b) 7
(c) 2
(d) 1

25. A student on vacation for d days observed that (1) it rained seven
times, either in the morning or in the afternoon, (2) there were five
clear afternoons, and (3) there were six clear mornings. Determine d.
(a) 7 (b) 8 (c) 9 (d) 10

Part III. Each correct answer is worth six points.

26. How many sequences containing two or more consecutive positive inte-
gers have a sum of 2009?
(a) 3 (b) 4 (c) 5 (d) 6

27. In 4ABC, let D, E, and F be points on the sides BC, AC, and AB,
respectively, such that BC = 4CD, AC = 5AE, and AB = 6BF . If
the area of 4ABC is 120 cm2 , what is the area of 4DEF ?
(a) 60 cm2 (b) 61 cm2 (c) 62 cm2 (d) 63 cm2

28. A function f is defined on the set of positive integers by f (1) = 1,

f (3) = 3, f (2n) = n, f (4n + 1) = 2f (2n + 1) − f (n), and f (4n + 3) =
3f (2n + 1) − 2f (n) for all positive integers n. Determine
X10
[f (4n + 1) + f (2n + 1) − f (4n + 3)].
n=1
(a) 55 (b) 50 (c) 45 (d) 40

29. A sequence of consecutive positive integers beginning with 1 is written

on the blackboard. A student came along and erased one number. The
7
average of the remaining numbers is 35 17 . What number was erased?
(a) 7 (b) 8 (c) 9 (d) 10

30. Let M be the midpoint of the side BC of 4ABC. Suppose that AB = 4

and AM = 1. Determine the smallest possible measure of ∠BAC.
(a) 60◦ (b) 90◦ (c) 120◦ (d) 150◦
 12th Philippine Mathematical Olympiad
Area Stage
21 November 2009

Part I. No solution is needed. All answers must be in simplest form. Each correct answer
merits two points.

1. If a = 2−1 and b = 23 , what is the value of (a−1 + b−1 )−2 ?

2. Find the sum of all (numerical) coefficients in the expansion of (x + y + z)3 .
3. A circle has radius 4 units, and a point P is situated outside the circle. A line through
P intersects the circle at points A and B. If P A = 4 units and P B = 6 units, how far
is P from the center of the circle?
4. Let y = (1 + ex )(ex − 6)−1 . If the values of x run through all real numbers, determine
the values of y.
5. The sum of the product and the sum of two integers is 95. The difference between the
product and the sum of these integers is 59. Find the integers.
6. Let A, B, C, D (written in the order from left to right) be four equally-spaced collinear
points. Let ω and ω 0 be the circles with diameters AD and BD, respectively.
√ A line
through A that is tangent to ω 0 intersects ω again at point E. If AB = 2 3 cm, what
is AE?
7. A certain high school offers its students the choice of two sports: football and basket-
ball. One fifth of the footballers also play basketball, and one seventh of the basketball
players also play football. There are 110 students who practice exactly one of the
sports. How many of them practice both sports?
√ √
8. Simplify: sin4 15◦ + 4 cos2 15◦ − cos4 15◦ + 4 sin2 15◦ .
9. Let a, b, and c be the roots of the equation 2x3 − x2 + x + 3 = 0. Find the value of
a3 − b 3 b 3 − c 3 c 3 − a3
+ + .
a−b b−c c−a

10. In 4ABC, let D, E, and F be points on sides BC, CA, and AB, respectively, so that
the segments AD, BE, and CF are concurrent at point P . If AF : F B = 4 : 5 and
the ratio of the area of 4AP B to that of 4AP C is 1 : 2, determine AE : AC.
11. A circle of radius 2 cm is inscribed in 4ABC. Let D and E be the points of tangency
of the circle with the sides AC and AB, respectively. If ∠BAC = 45◦ , find the length
of the minor arc DE.
12. Two regular polygons with the same number of sides have sides 48 cm and 55 cm
in length. What is the length of one side of another regular polygon with the same
number of sides whose area is equal to the sum of the areas of the given polygons?
 13. The perimeter of a right triangle is 90 cm. The squares of the lengths of its sides sum
up to 3362 cm2 . What is the area of the triangle?
14. Determine all real solutions (x, y, z) of the following system of equations:

2 2
x − y = z

y 2 − z = x2

 2
z − x = y2.

15. For what value(s) of k will the lines 2x + 7y = 14 and kx − y = k + 1 intersect in the
first quadrant?
16. For what real numbers r does the system of equations

x2 = y 2
(x − r) + y 2 = 1
2

have no solutions?
17. Determine the smallest positive integer n such that n is divisible by 20, n2 is a perfect
cube, and n3 is a perfect square.
p √
18. Find all pairs (a, b) of integers such that 2010 + 2 2009 is a solution of the quadratic
equation x2 + ax + b = 0.
19. Determine all functions f : (0, +∞) → R such that f (2009) = 1 and

 2009 
f (x)f (y) + f 2009
x
f y = 2f (xy)
for all positive real numbers x and y.
20. Find all pairs (k, r), where k is an integer and r is a rational number, such that the
equation r(5k − 7r) = 3 is satisfied.
Part II. Show the solution to each problem. A complete and correct solution merits ten
points.
21. Each of the integers 1, 2, 3, . . . , 9 is assigned to each vertex of a regular 9-sided polygon
(that is, every vertex receives exactly one integer from {1, 2, . . . , 9}, and two vertices
receive different integers) so that the sum of the integers assigned to any three consec-
utive vertices does not exceed some positive integer n. What is the least possible value
of n for which this assignment can be done?
22. Let E and F be points on the sides AB and AD of a convex quadrilateral ABCD such
that EF is parallel to the diagonal BD. Let the segments CE and CF intersect BD at
points G and H, respectively. Prove that if the quadrilateral AGCH is a parallelogram,
then so is ABCD.
23. Let p be a prime number. Let a, b, and c be integers that are divisible by p such that
the equation x3 + ax2 + bx + c = 0 has at least two different integer roots. Prove that
c is divisible by p3 .

- end of the problems -

 Answer Key

4 3
1. 49
11. 2
π cm

2. 33 12. 73 cm
√
3. 2 10 units 13. 180 cm2

4. (−∞, − 16 ) ∪ (1, +∞) 14. (0, 0, 0), (1, 0, −1), (0, −1, 1), (−1, 1, 0)

5. 11 and 7 15. (−∞, −3) ∪ ( 16 , +∞)

√ √
6. 9 cm 16. (−∞, − 2) ∪ ( 2, +∞)

7. 11 students 17. 1000000

√
8. 21 3 18. (−2, −2008)

9. −1 19. f (x) = 1 for all x ∈ (0, +∞)

10. 2 : 7 20. (2, 1), (−2, −1), (2, 37 ), (−2, − 37 )

Problem 21. There is an assignment of the integers 1, 2, 3, . . . , 9 to the vertices of the

regular nonagon that gives n = 16. Let S be the sum of all sums of the integers assigned to
three consecutive vertices. If there are integers assigned to three consecutive vertices whose
sum is at most 14, then S < 135, which is a contradiction. Thus, every sum of the integers
assigned to three consecutive vertices is equal to 15. Consider a, b, c, d. Then a + b + c = 15
and b + c + d = 15, which implies that a = d, a contradiction again. 2
Problem 22. Let EF intersect AG and AH at points I and J, respectively. Note that
EGHJ and F IGH are parallelograms. It follows that 4F HJ ∼ = 4IGE, which implies that
F J = EI. Using three pairs of similar triangles, F J = EI implies that BG = DH.
Let S be the midpoint of AC. Since AGCH is parallelogram, S is the midpoint of GH.
Finally, with BG = DH, we have BS = BG + GS = DH + HS = DS. This means that
the diagonals AC and BD bisect each other, and so ABCD is a parallelogram. 2
Problem 23. Let r and s be two different integral roots of x3 + ax2 + bx + c = 0; that is,
r3 + ar2 + br + c = 0 and s3 + as2 + bs + c = 0. Since p divides a, b, and c, it follows that p
divides both r3 and s3 . Being prime, p divides r and s.
Subtracting the above equations involving r and s, we get

r3 − s3 + a(r2 − s2 ) + b(r − s) = 0, or(r − s)(r2 + rs + s2 + a(r + s) + b) = 0.

Since r 6= s, the last equation becomes

r2 + rs + s2 + a(r + s) + b = 0.

Because the terms (other than b) are divisible by p2 , the last equation forces p2 to divide b.
Finally, the terms (other than c) of r3 + ar2 + br + c = 0 are divisible by p3 , it follows
that p3 divides c. 2
 12th Philippine Mathematical Olympiad
National Stage, 23 January 2010

Oral Phase

15.1. What is the smallest positive integral value of n such that n300 > 3500 ?

15.2. A figure consists of two overlapping circles that have radii 4 and 6. If
the common region of the circles has area 2π, what is the area of the
entire figure?
2010
15.3. Find all real values of x that satisfy the equation xx = x2010 .

15.4. Both roots of the quadratic equation x2 − 30x + 13k = 0 are prime
numbers. What is the largest possible value of k?

15.5. Let f : R → R be a function that satisfies the functional equation

f (x − y) = 2009f (x)f (y)

√
for all x, y ∈ R. If f (x) is never zero, what is f ( 2009)?

15.6. If the parabola y + 1 = x2 is rotated clockwise by 90◦ about its focus,

what will be the new coordinates of its vertex?

15.7. How many ways can you choose four integers from the set {1, 2, 3, . . . , 10}
so that no two of them are consecutive?

15.8. Let ABC be a triangle with AB = 12, BC = 16, and AC = 20.

Compute the area of the circle that passes through C and the midpoints
of AB and BC.

15.9. Which real numbers x satisfy the inequality |x − 3| ≥ |x|?

15.10. Let log14 16 be equal to a. Express log8 14 in terms of a.

15.11. Find the values of a and b such that ax4 +bx2 +1 is divisible by x2 −x−2.

15.12. What is the probability that a randomly chosen positive divisor of 2010
has two digits?

15.13. Let [[x]] denote the greatest integer less than or equal to the real number
x. What is the largest two-digit integral value of x[[x]]?

15.14. How many times does the graph of y + 1 = log1/2 |x| cross the x-axis?
15.15. Considered to be the most prolific mathematician of all time, he pub-
lished, in totality, the most number of mathematical pages in history.
Undertaken by the Swiss Society of Natural Sciences, the project of
publishing his collected works is still going on and will require more
than 75 volumes. Who is this great mathematician Switzerland has
produced?

30.1. The nonzero numbers x, y, and z satisfy the equations

xy = 2(x + y), yz = 4(y + z), and xz = 8(x + z).

Solve for x.

30.2. The positive integers are grouped as follows:

A1 = {1}, A2 = {2, 3, 4}, A3 = {5, 6, 7, 8, 9}, and so on.

In which group does 2009 belong to?

30.3. Triangle ABC is right-angled at C, and point D on AC is the foot of

the bisector of ∠B. If AB = 6 cm and the area of 4ABD is 4.5 cm2 ,
what is the length, in cm, of CD?

30.4. For each positive integer n, let Sn be the sum of the infinite geometric
1
series whose first term is n and whose common ratio is n+1 . Determine
the least value of n such that

S1 + S2 + · · · + Sn > 5150.

30.5. Let x and y be positive real numbers such that x + 2y = 8. Determine

the minimum value of
3 9
x+y+ + .
x 2y
30.6. Let d and n be integers such that 9n + 2 and 5n + 4 are both divisible
by d. What is the largest possible value of d?

30.7. Find all integers n such that 5n − 7, 6n + 1, and 20 − 3n are all prime
numbers.

30.8. When
(x2 + 2x + 2)2009 + (x2 − 3x − 3)2009
is expanded, what is the sum of the coefficients of the terms with odd
exponents of x?
 30.9. If 0 < θ < π/2 and 1 + sin θ = 2 cos θ, determine the numerical value
of sin θ.
30.10. For what real values of k does the system of equations

x − ky = 0
x2 + y = −1
have real solutions?
60.1. In 4ABC with BC = 24, one of the trisectors of ∠A is a median, while
the other trisector is an altitude. What is the area of 4ABC?
60.2. How many integral solutions does the equation
|x| + |y| + |z| = 9

60.3. Let X, Y , and Z be points on the sides BC, AC, and AB of 4ABC,
respectively, such that AX, BY , and CZ are concurred at point O.
The area of 4BOC is a. If BX : XC = 2 : 3 and CY : Y A = 1 : 2,
what is the area of 4AOC?
60.4. Find the only value of x in the open interval (−π/2, 0) that satisfies
the equation √
3 1
+ = 4.
sin x cos x
60.5. The incircle of a triangle has radius 4, and the segments into which one
side is divided by the point of contact with the incircle are of lengths
6 and 8. What is the perimeter of the triangle?

Written Phase
1. Find all primes that can be written both as a sum of two primes and
as a difference of two primes.
2. On a cyclic quadrilateral ABCD, there is a point P on side AD such
that the triangle CDP and the quadrilateral ABCP have equal perime-
ters and equal areas. Prove that two sides of ABCD have equal lengths.
3. Let R? be the set of all real numbers, except 1. Find all functions
f : R? → R that satisfy the functional equation
 
x + 2009
x + f (x) + 2f = 2010.
x−1
 4. There are 2008 blue, 2009 red, and 2010 yellow chips on a table. At
each step, one chooses two chips of different colors, and recolor both of
them using the third color. Can all the chips be of the same color after
some steps? Prove your answer.

5. Determine, with proof, the smallest positive integer n with the following
property: For every choice of n integers, there exist at least two whose
sum or difference is divisible by 2009.

Answers
Oral Phase

15.1. 7 15.11. a = 1/4, b = −5/4 30.6. 26

15.2. 50π 1
15.12. 4
30.7. only n = 6
√
15.3. 2010 √
2010, 15.13. 99 30.8. −1
− 2010 2010, 1
15.14. 4 30.9. 3/5
15.4. 17
1 15.15. Leonhard Euler 30.10. − 12 ≤ x ≤ 1
2
15.5. 2009
√
30.1. 16/3 60.1. 32 3
15.6. (− 34 , − 14 )

15.7. 35 30.2. A45 60.2. 326

15.8. 25π 30.3. 1.5 60.3. 3a

15.9. (−∞, 3/2] 30.4. 101 60.4. −4π/9

4
15.10. 3a
30.5. 8 60.5. 42

Oral Phase
1. Let p be a prime that can be written as a sum of two primes and as a
difference of two primes. Clearly, we have p > 2. Then p must be odd,
so that p = q + 2 = r − 2 for some odd primes q and r.

We consider three cases.

Case 1. Suppose that q ≡ 1 (mod 3). Then p is a multiple of 3,
implying that p = 3. It follows that p = 3, which means that q = 1, a
contradiction.
 Case 2. Suppose that q ≡ 2 (mod 3). Then r ≡ 0 (mod 3), which
implies that r = 3. This leads to p = 1, which is again a contradiction.
Case 3. Suppose that q ≡ 0 (mod 3). Then q = 3, and it follows that
p = 5 and r = 7.

From the above three cases, p = 5 is the only prime that is a sum of
two primes and a difference of two primes. 2

2. We denote by (XY Z) and (W XY Z) the areas of 4XY Z and quadrilat-

eral W XY Z, respectively. We use the labels depicted in the following
figure.

B
θ
a
b

A
x α
z
P C

y c

With equal perimeters, we get

a+b+z+x=c+y+z

or
a + b + x = c + y. (1?)
With equal areas, we get

(ABC) + (ACP ) = (CDP ). (2?)

Since 4ACP and 4CDP have the same altitude from C, we have

(ACP ) x x
= =⇒ (ACP ) = · (CDP ).
(CDP ) y y

With (2?), we have

 
x y−x
(ABC) = 1 − (CDP ) = · (CDP ). (3?)
y y
 On the other hand, since ABCD is cyclic, we know that ∠D = 180◦ −θ.
Then (ABC) = 21 ab sin θ and (CDP ) = 12 cy sin(180◦ − θ). After noting
that sin(180◦ − θ) = sin θ and applying (1?), equation (3?) reduces to

ab = c(a + b − c).

This last equation is equivalent to

(c − b)(c − a) = 0,

which implies that b = c or a = c. 2

x + 2009
3. Let g(x) = . Then the given functional equation becomes
x−1
x + f (x) + 2f (g(x)) = 2010. (1?)

Replacing x with g(x) in (1?), and after noting that g(g(x)) = x, we

get
g(x) + f (g(x)) + 2f (x) = 2010. (2?)
Eliminating f (g(x)) in (1?) and (2?), we obtain

x − 3f (x) − 2g(x) = −2010.

x + 2009
Solving for f (x) and using g(x) = , we have
x−1
x2 + 2007x − 6028
f (x) = .
3(x − 1)

It is not difficult to verify that this function satisfies the given functional
equation. 2

4. After some steps, suppose that there a blue, b red, and c yellow chips
on the table. We denote this scenario by the ordered triple (a, b, c).
Then the next step produces (a − 1, b − 1, c + 2), (a + 2, b − 1, c − 1), or
(a − 1, b + 2, c − 1). One crucial observation on these three possibilities
is the fact that

(a − 1) − (b − 1) ≡ (a + 2) − (b − 1) ≡ (a − 1) − (b + 2) ≡ a − b (mod 3);

that is, from one step to the next, the difference between the number
of blue chips and the number of red chips does not change modulo 3.
 Starting with (2008, 2009, 2010), we verify if we can end up with (6027, 0, 0),
(0, 6027, 0), or (0, 0, 6027). Since 2008 − 2009 ≡ 2 (mod 3), but

6027 − 0 ≡ 0 − 6027 ≡ 0 − 0 ≡ 0 (mod 3),

it follows that all the chips can never be of the same color after any
number of steps. 2

5. We show that the least integer with the desired property is 1006. We
write 2009 = 2 · 1004 + 1.

Consider the set {1005, 1006, . . . , 2009}, which contains 1005 integers.
The sum of every pair of distinct numbers from this set lies between
2011 and 4017, none of which is divisible by 2009. On the other hand,
the (absolute) difference between two distinct integers from this set lies
between 1 and 1004, none of which again is divisible by 2009. It follows
that the smallest integer with the desired property is at least 1006.

Let A be a set of 1006 integers. If there are two numbers in A that

have the same remainder when divided by 2009, then we are done.

Suppose, on the contrary, that all the 1006 remainders of the inte-
gers in A modulo 2009 are all different. Thus, the set of remainders
is a 1006-element subset of the set {0, 1, . . . , 2008}. One can also
consider the remainders as forming a 1006-element subset of the set
X = {−1004, −1003, . . . , −1, 0, 1, 2, . . . , 1004}. Every 1006-element
subset of X contains two elements whose sum is zero. Thus, A con-
tains two numbers whose sum is divisible by 2009. Since |A| = 1006,
we deduce that 1006 is the least integer with the desired property. 2
Problem 1. Find all nonempty finite sets X of real numbers with the following property:

x + |x| ∈ X for all x ∈ X.

Solution. Let X = {x1 , x2 , . . . , xn }, n ≥ 1, where x1 < x2 < · · · < xn .

If xn > 0, then xn + |xn | = 2xn ∈ X, which is a contradiction because xn < 2xn but xn
is the largest element of X.
The contradiction in the previous paragraph implies that xn ≤ 0. If x1 < 0, then
x1 + |x1 | = x1 − x1 = 0 ∈ X. Hence, we must have xn = 0, so that xi + |xi | = xi − xi = 0 ∈ X
for any xi ∈ X, i = 1, 2, . . . , n.
Hence, in order for the desired property to be satisfied, X must be a finite subset of the
interval (−∞, 0] and it must contain 0. On the other hand, such subsets satisfy the said
property.
The only nonempty finite sets that satisfy the desired property are those finite subsets
of (−∞, 0] containing 0. q.e.d.
Problem 2. In 4ABC, let X and Y be the midpoints of AB and AC, respectively. On
segment BC, there is a point D, different from its midpoint, such that ∠XDY = ∠BAC.
Prove that AD is perpendicular to BC.
Solution. Let Z be the midpoint of BC. Without loss of generality, we assume that D
is between B and Z. Then XY kBC, XZkAC, and Y ZkAB, and so ∠ABC = ∠XY Z and
∠XDY = ∠BAC = ∠XZY . It follows that quadrilateral XY ZD is cyclic.
A

X Y

B C
D Z

Since opposite angles of a cyclic quadrilateral are supplementary, we have

∠XDB = 180◦ − ∠XDZ = ∠XY Z = ∠ABC,

which implies that XA = XB = XD. Therefore, AB is a diameter of the circumcircle of

4ABD, and so ∠ADB = 90◦ . q.e.d.
Problem 3. The 2011th prime number is 17483, and the next prime is 17489.
Does there exist a sequence of 20112011 consecutive positive integers that contains exactly
2011 prime numbers? Prove your answer.
Solution. Let N = 20112011 . Since N > 17489, there are more than 2011 primes in the
sequence 1, 2, 3, . . . , N .
Claim. There exists a sequence of N consecutive positive integers that are all composite.
Proof of the Claim. The sequence

(N + 1)! + 2, (N + 1)! + 3, (N + 1)! + 4, . . . , (N + 1)! + N, (N + 1)! + (N + 1)

consists of N consecutive composite numbers, and so the claim follows.

Back to the solution of the problem, let

a, a + 1, a + 2, . . . , a + N − 1 (?)

be a sequence of N consecutive positive integers with no prime. We do repeatedly the

following operation to the numbers in (?). Delete the far-right number a + N − 1, and
append to the far-left the number a − 1. The resulting sequence

a − 1, a, a + 1, . . . , a + N − 2

has at most one prime. Repeating this operation until we reach the sequence 1, 2, 3, . . . , N ,
which has more than 2011 primes.
Performing such operation either retains, increases by one, or decreases by one the number
of primes of the previous sequence. Since the starting sequence has no prime at all, while
the last sequence has more than 2011 primes, there exists a sequence (after applying the
operation a number of times) that contains exactly 2011 primes. q.e.d.
Problem 4. Find all (if there is one) functions f : R → R that satisfy the following
functional equation:
f (f (x)) + xf (x) = 1 for all x ∈ R. (1?)

Solution. We prove that no such function exists.

Suppose there exists a function f that satisfies (1?). Let f (0) = a.
Set x = 0 in (1?) to get
f (a) = 1, (2?)
which implies, after setting x = a in (1?), that

f (1) = 1 − a. (3?)

This last equation and setting x = 1 in (1?) yield

f (1 − a) = f (f (1)) = 1 − f (1) = a.

Further, with the preceding equation and setting x = 1 − a in (1?), we have

f (a) = f (f (1 − a)) = 1 − (1 − a)f (1 − a) = 1 − (1 − a)a = 1 − a + a2 .

This last equation and (2?) give a = 0 or a = 1.

If a = 0, then f (0) = 0, which contradicts (2?).
If a = 1, then setting x = 0 in (1?) gives f (1) = 1, which contradicts (3?).
Problem 5. The chromatic number of the (infinite) plane, denoted by χ, is the smallest
number of colors with which we can color the points on the plane in such a way that no two
points of the same color are one unit apart.
Prove that 4 ≤ χ ≤ 7.
Solution. Suppose χ ≤ 3. Consider the following configuration, where each segment has
unit length. Then the points A, B, and G must receive different colors, and so are the points
A, E, and F . This will force points C and D to receive the same color as A, which is a
contradiction. Thus, we obtain χ ≥ 4.
A

B E

F G

C D

On the other hand, we will exhibit a coloring of the points on the plane using 7 colors in
such a way that points one unit apart have different colors. We first tile the plane by regular
hexagons with unit sides. Now, we color one hexagon with color 1, and its six neighbors
with colors 2, 3, . . . , 7, as highlighted in the following diagram.

7 3 2
4 1 7 3 2
5 6 4 1 7
3 2 5 6 4
7 3 2 5 6
1 7 3 2 5
6 4 1 7 3
2 5 6 4 1
3 2 5 6 4
7 3 2 5
1 7 3
4 1

The union of the seven highlighted hexagons forms a symmetric polygon P of 18 sides.
Translates of P also tile the plane and determine how we color the plane using 7 colors.
It is easy to compute√ that each color does not have monochromatic segments of any
length d, where 2 < d < 7. Thus, if we proportionally shrink the configuration by a factor
of, say, 2.1, we will get 7-coloring that has no monochromatic segments of unit length, which
implies that χ ≤ 7. q.e.d.
 13th Philippine Mathematical Olympiad
Qualifying Stage
23 October 2010

Part I. Each correct answer is worth two points.

1. What is the sum of the roots of x2 − 2009x − 2010 = 0?

(a) 2010 (b) 2009 (c) 2011 (d) −2010

√
q p
2. Find the value of 2 2 2 2 · · ·.
√ √
(a) 2 (b) 2 (c) 4 (d) 2 2
x
3. If 22 = 43 , what is x?
(a) log2 6 (b) log4 6 (c) log6 2 (d) log6 4

4. For what values of a does the system

 2
x − y2 = 0
(x − a)2 + y 2 = 0

have a unique solution?

(a) a = −1 (b) a = 0 (c) a = 1 (d) a = 2

5. If x + y = 4 and x2 + y 2 = 10, what is the value of x4 + y 4 ?

(a) 84 (b) 100 (c) 68 (d) 82

6. Let f be a function defined on the set of integers such that f (1) = 5

and f (x + 1) = 2f (x) + 1 for all integers x. What is the value of
f (7) − f (0)?
(a) 380 (b) 189 (c) 191 (d) 381

7. There are k zeros at the end of 34! = 34 · 33 · 32 · · · · 4 · 3 · 2 · 1. What

is the value of k?
(a) 7 (b) 4 (c) 6 (d) 5
 8. Find the sum cos 1◦ + cos 3◦ + cos 5◦ + · · · + cos 177◦ + cos 179◦ .
√
2 (b) 1 (c) 0 1
(a) (d)
2 2
18x + 7y 2 x
9. If = , what is the value ?
12y + 5x 3 y
57 44 46 3
(a) (b) (c) (d)
46 3 57 44
10. A 4 by 6 inch paper is folded so that its upper right corner touches
the midpoint of an opposite side and such that the fold obtained is the
longer one. Find the length of the fold.
√ √ 5
(a) 2 13 in (b) 5 in (c) 65 in (d) 5 in
24
11. If a − b + c = 1, b − 2c = 0, 2a + c = 5, what is the sum a + b + c?
(a) 3 (b) 4 (c) 5 (d) 0

12. A triangle is formed inside a circle by connecting the center C to two

points A and B on the circle. If ∠ACB = 30◦ , what is the ratio of the
areas of the circle to the triangle?
(a) 6π : 1 (b) 9 : 1 (c) 4π : 1 (d) 9π : 2

13. A ball rebounds each time to a height which is half that of the previous
one. If the total distance traveled before coming to rest is 72 meters,
from how high was the ball dropped?
(a) 24 meters (b) 18 meters (c) 36 meters (d) 12 meters
π x + π −x
14. Let f be the function defined by f (x) = . Find f (2p) if
π x − π −x
f (p) = 2.
1 3 5 (d) 4
(a) (b) (c)
4 4 4
15. If x > 0, find the solution set of log x ≥ log 2 + log(x − 1).
√
(a) (1, 2] (b) (−∞, 2] (c) (0, 1] (d) ( 2, 1]
Part II. Each correct answer is worth three points.

1. Solve for (x, y) in the system (ex + 2)2 − y = 3, 4(ex + 2) − y = −1.

√ √
(a) ( 2, 3) (c) (ln 2, 3)
√ √ √ √
(b) (ln 2 2, 9 + 8 2) (d) (ln 2, 2 + 4 2)

2. Mica has six differently colored crayons. She can use one or more
colors in her painting. What is the likelihood that she will use only her
favorite color?
1 1 1 1
(a) (b) (c) (d)
24 48 81 63
1 1 − bn
3. If b1 = and bn+1 = , for n ≥ 2, find b2010 − b2009 .
3 1 + bn
1 1 1 1
(a) (b) − (c) (d) −
2 3 6 6
4. Let
1
x=1− 1 .
2− 1− 1
1
2− 1−...

Find (2x − 1)2 .

(a) 4 (b) −4 (c) 8 (d) −8

5. cos 15◦ is equal to

r √ r √ √ √ √ √
2− 3 2− 3 6− 2 6+ 2
(a) (b) (c) (d)
2 4 4 4
(log5 x)2 − 4
6. Solve for x in the equation + 2log5 x = −1.
(log5 x)2 + log5 x4 + 4
(a) x = 1 (b) x = −1 (c) x = 2 (d) x = 3

7. A line with y-intercept 5 and positive slope is drawn such that this line
intersects x2 + y 2 = 9. What is the least slope of such a line?
1 (b) 1 5 7
(a) (c) (d)
3 6 6
8. A metal bar bent into a square is to be painted. How many distinct
ways can one color the metal bar using four distinct colors on the edges
using red, white, blue, and yellow.
(a) 8 (b) 24 (c) 3 (d) 4
 √
9. If 92x − 92x−1 = 8 3, find (2x − 1)2x .
√ √
2 2 1 1
(a) (b) (c) (d)
8 4 4 8
10. In how many ways can the letters of the word MURMUR be arranged
without letting two letters which are the same be adjacent?
(a) 54 (b) 24 (c) 45 (d) 36

PART III. Each correct answer is worth six points.

1. Let 
n + 1, if n is odd
f (n) =
n − 1, if n is even
be a function whose domain is the set of positive integers. Then
f ((n2 + 1)2 + (n2 − 1)2 ) =
(a) 2n4 − 1 (b) 2n4 (c) 2n4 + 1 (d) 2n4 + 2
2. Find all polynomials p(x) where xp(x − 1) = (x − 5)p(x) and p(6) = 5!
 
x(x − 1)(x − 2)(x − 3)(x − 4)(x − 5)
(a) , 120x
6
 
x(x − 1)(x − 2)(x − 3)(x − 4)
(b)
6
(c) {x(x − 1)(x − 2)(x − 3)(x − 4)}
 
x(x − 1)(x − 2)(x − 3)(x − 4)
(d) , 24x
6
3. Let n = 231 319 . How many positive divisors of n2 are less than n but
do not divide n?
(a) 588 (b) 560 (c) 561 (d) 589

4. Four spheres, each of radius 1.5, are placed in a pile with three at
the base and the other on top. If each sphere touches the other three
spheres, give the height of the pile.
√ √ √ √
(a) 3 + 3 (b) 3 + 6 (c) 6 (d) 6 3

5. Let ABC be a 3-digit number such that its digits A, B, and C form
an arithmetic sequence. The largest integer that divides all numbers of
the form ABCABC is
(a) 11 (b) 101 (c) 1001 (d) 3003