22nd PMO Area Stage With Solutions
22nd PMO Area Stage With Solutions
22nd PMO Area Stage With Solutions
PART I. Give the answer in the simplest form that is reasonable. No solution is needed. Figures
are not drawn to scale. Each correct answer is worth three points.
1. If the sum of the first 22 terms of an arithmetic progression is 1045 and the sum of the next
22 terms is 2013, find the first term.
4. Determine the number of ordered quadruples (a, b, c, d) of odd positive integers that satisfy the
equation a + b + c + d = 30.
log √
3 (log3 x) + log3 (log27 x) + log27 (log √
3 3 x) = 1.
3
6. Let f (x) = x2 + 3. How many positive integers x are there such that x divides f (f (f (x)))?
8. Find the largest three-digit integer for which the product of its digits is 3 times the sum of its
digits.
9. A wooden rectangular brick with dimensions 3 units by a units by b units is painted blue on
all six faces and then cut into 3ab unit cubes. Exactly 1/8 of these unit cubes have all their
faces unpainted. Given that a and b are positive integers, what is the volume of the brick?
10. In square ABCD with side length 1, E is the midpoint of AB and F is the midpoint of BC.
The line segment EC intersects AF and DF at G and H, respectively. Find the area of
quadrilateral AGHD.
11. A sequence {an }n≥1 of positive integers satisfies the recurrence relation an+1 = nban /nc + 1
for all integers n ≥ 1. If a4 = 34, find the sum of all the possible values of a1 .
12. Let (0, 0), (10, 0), (10, 8), and (0, 8) be the vertices of a rectangle on the Cartesian plane. Two
lines with slopes −3 and 3 pass through the rectangle and divide the rectangle into three regions
with the same area. If the lines intersect above the rectangle, find the coordinates of their point
of intersection.
2019
n
X
13. For a positive integer x, let f (x) be the last two digits of x. Find f 77 .
n=1
p
14. How many positive rational numbers less than 1 can be written in the form , where p and q
q
are relatively prime integers and p + q = 2020?
1 8
2 1
15. The constant term in the expansion of ax − + 2 is 210a5 . If a > 0, find the value of
x x
a.
16. Let A = {n ∈ Z | |n| ≤ 24}. In how many ways can two distinct numbers be chosen (simulta-
neously) from A such that their product is less than their sum?
17. Points A, B, C, and D lie on a line ` in that order, with AB = CD = 4 and BC = 8. Circles
Ω1 , Ω2 , and Ω3 with diameters AB, BC, and CD, respectively, are drawn. A line through A
and tangent to Ω3 intersects Ω2 at the two points X and Y . Find the length of XY .
18. A musical performer has three different outfits. In how many ways can she dress up for seven
different performances such that each outfit is worn at least once? (Assume that outfits can be
washed and dried between performances.)
√ √
19. In 4P M O, P M = 6 3, P O = 12 3, and S is a point on M O such that P S is the angle
bisector of ∠M P O. Let T be the reflection of S across P M . If P O is parallel to M T , find the
length of OT .
20. A student writes the six complex roots of the equation z 6 + 2 = 0 on the blackboard. At every
step, he randomly chooses two numbers a and b from the board, erases them, and replaces
them with 3ab − 3a − 3b + 4. At the end of the fifth step, only one number is left. Find the
largest possible value of this number.
PART II. Show your solution to each problem. Each complete and correct solution is worth ten
points.
1. Consider all the subsets of {1, 2, 3, . . . , 2018, 2019} having exactly 100 elements. For each subset,
take the greatest element. Find the average of all these greatest elements.
1. Consider all the subsets of {1, 2, 3, . . . , 2018, 2019} having exactly 100 elements. For each subset,
take the greatest element. Find the average of all these greatest elements.
Solution 1: Let M be the average that we are computing. First, there are 2019
100 ways to
a 100-element subset. Next, if x is the largest element, then x ≥ 100, and there are
choose
x−1
99 subsets having x as the largest element. Hence
2019
X x−1
x
99
M = x=100
.
2019
100
But note that
x−1 x
x = 100 .
99 100
Using this fact and the hockey stick identity, we have
2019
X x
100
100
x=100
M =
2019
100
2020
100
101
=
2019
100
100 · 2020
=
101
= 2000
Solution 2: As in Solution 1, the required average M can be written as
2019
X x−1
x
99
x=100
M =
2019
100
99 100 2018
100 + 101 + · · · + 2019
99 99 99
= .
2019
100
We can simplify this expression by grouping the terms in the numerator in specific ways,
applying the hockey stick identity to each group of terms, and then applying the hockey stick
identity again to the resulting terms. This can be done in several ways, two of which are shown
next:
where the last two lines follow from the hockey stick identity.
2020
101 2020
Hence, M = 2020 − = 2020 − = 2000 .
2019 101
100
Thus,
100 101 2018
M X = 100X + 1919X − + + ··· +
100 100 100
2019
= 2019X −
101
which gives
2019
101 2020
M = 2019 − = 2020 − = 2000 .
2019 101
100
Solution 1: Let {Fn }∞ n=1 = {1, 1, 2, 3, 5, 8, . . .} be the sequence of Fibonacci numbers. We first
claim that an = 2Fn + 1 for all n ∈ N. Clearly, this is true for n = 1, 2. Let k ∈ N and suppose
that the claim is true for n = k and for n = k + 1. Then
By strong induction, the claim is proved. Therefore, we now find the remainder when 2F2020 + 1
is divided by 22. To this end, it is easier to find residues modulo 2 and modulo 11 and process
them to get the residue modulo 22 (e.g. through Chinese Remainder Theorem).
Clearly, 2F2020 is even, i.e., 2F2020 + 1 ≡ 1 (mod 2). We will see later that 2F2020 + 1 ≡ 0 (mod
11). Therefore, by Chinese Remainder Theorem, a2020 = 2F2020 + 1 ≡ 11 (mod 22) .
There are several ways to find the residue of 2F2020 + 1 (mod 11).
Way 1.1: By Fermat’s Little Theorem, 210 ≡ 1 (mod 11). This prompts us to consider the
sequence of residues of Fn (mod 10) in order to find F2020 (mod 10):
We see that the sequence is cyclic with period 60. Therefore, since 2020 = 60(33) + 40, we
obtain F2020 ≡ 5 (mod 10). Consequently, for some k ∈ Z,
k
2F2020 + 1 = 210k+5 + 1 ≡ 210 25 + 1 ≡ 33 ≡ 0 (mod 11). (2)
Way 1.2: Another way to find F2020 (mod 10) is to get the residues of F2020 modulo 2 and 5
and process them to find the residue modulo 10. Again, we list down the sequence of residues
modulo 2 and 5: {Fn (mod 2)}∞ n=1 = {1, 1, 0, ...cycle} which has period 3, and
which has period 20. Since, 2020 ≡ 1 (mod 3) and 2020 ≡ 0 (mod 20), then F2020 ≡ 1 (mod 2)
and F2020 ≡ 0 (mod 5). Therefore, F2020 ≡ 5 (mod 10) by Chinese Remainder Theorem. Thus,
(2) holds.
Since the sequence is cyclic with period 60, and 2020 ≡ 40 (mod 60), then a2020 ≡ 11 (mod 22).
3. In 4ABC, AB = AC. A line parallel to BC meets sides AB and AC at D and E, respectively.
The angle bisector of ∠BAC meets the circumcircle of 4ABC and 4ADE at points X and
Y , respectively. Let F and G be the midpoints of BY and XY , respectively. Let T be the
intersection of lines CY and DF . Prove that the circumcenter of 4F GT lies on line XY .
Solution 1: Let F 0 be the reflection of F over line XY . Observe that, by symmetry, we get
∠ADY = ∠AEY . As quadrilateral ADY E is cyclic, both angles must be right, and hence
∠BDY is right as well.
Thus F is the circumcenter of triangle BDY , so ∠F DB = ∠F BD. By reflecting over XY , we
also get ∠F BD = ∠F 0 CE. Thus
]T CA = ]F 0 CE = ]DBF = ]F DB = ]T DA,
the last step following from F G k BX and F 0 G k CX. Thus the points F , T , G, and F 0 are
concyclic. Their circumcenter must lie on the perpendicular bisector of F F 0 , which is line XY .
But this is also the circumcenter of triangle F GT , as desired.
which implies that F F 0 GT is cyclic. The logic in Solution 1 finishes the proof.
Remark: The directed angles are necessary here due to configuration issues.