USA Mathematical Talent Search
Round 2 Solutions
Year 33 — Academic Year 2021–2022
www.usamts.org
1/2/33. A 5 × 5 Latin Square is a 5 × 5 grid of squares in which each square contains one
of the numbers 1 through 5 such that every number appears exactly once in each row and
column. A partially completed grid (with numbers in some of the squares) is puzzle-ready
if there is a unique way to fill in the remaining squares to complete a Latin Square.
Below is a partially completed grid with seven squares filled in and an additional three
squares shaded. Determine what numbers must be filled into the shaded squares to make
the grid (now with ten squares filled in) puzzle-ready, and then complete the Latin Square.
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There is a unique solution, but you do not need to prove that your answer is the only
one possible. You merely need to find an answer that satisfies the constraints above. (Note:
In any other USAMTS problem, you need to provide a full proof. Only in this problem is
an answer without justification acceptable.)
1
13
23 5
5
Solution
1 4 5 2 3
5 1 3 4 2
2 3 4 1 5
3 2 1 5 4
4 5 2 3 1
USA Mathematical Talent Search
Round 2 Solutions
Year 33 — Academic Year 2021–2022
www.usamts.org
2/2/33. Let n be a fixed positive integer. Which is greater?
1. The number of n-tuples of integers whose largest value is 7 and whose smallest value
is 0; or
2. The number of ordered triples (A, B, C) that satisfy the following property: A, B, C
are subsets of {1, 2, 3, . . . , n}, and neither C ⊆ A ∪ B, nor B ⊆ A ∪ C.
Your answer can be: (1), (2), the two counts are equal, or it depends on n.
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Solution
The answer is that the two sets are the same size. This can be proven by directly
counting each and comparing the results. Here, we will demonstrate a bijection between the
two sets.
Note that sets A, B, and C break {1, 2, 3, . . . , n} up into 8 total regions. We will label
each element in the set based on which region it is in. We will call the region in B but not
in A or C region 0. The region in C but not in A or B will be region 7. The other regions
will be labeled arbitrarily.
7
4 1
2 6
5 3 0
A B
Given this Venn Diagram, any n-tuple of numbers from 0 to 7 can be used to construct
sets A, B, and C. We simply put the k-th element in the region defined by the number
assigned to it. So, for example, if the first element were labeled 1, we would put it in both
sets B and C but not in A.
Now, let’s look at conditions (1) and (2) in the problem statement.
We’ll start with (1). If an assigned n-tuple has a maximum of 7, then that means the
corresponding subsets have at least one element that is in C but not in A or B. That means
C 6⊆ A ∪ B. Similarly, if it has a minimum of 0, then that means the corresponding subsets
have an element in B that is not in A or C. That means B 6⊆ A ∪ C. Hence, every n-tuple
from part (1) will map to A, B, and C with the properties in (2).
USA Mathematical Talent Search
Round 2 Solutions
Year 33 — Academic Year 2021–2022
www.usamts.org
Likewise, if a set of subsets satisfies (2), then since B 6⊆ A ∪ C, that means region 0 is
occupied, or at least one number is assigned to region 0. Hence, our n-tuple from (1) will
use the 0. Similarly, since C 6⊆ A ∪ B, we know that there is at least one number in region
7.
Thus, we have established a bijection and we see that (1) and (2) have the same number
of elements! Although, of course, this solution makes absolutely no claim as to what that
number of elements is! We’ll leave that to you!
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USA Mathematical Talent Search
Round 2 Solutions
Year 33 — Academic Year 2021–2022
www.usamts.org
3/2/33. Let x and y be distinct real numbers such that
√ p
x2 + 1 + y 2 + 1 = 2021x + 2021y.
Find, with proof, the value of
√ p
(x + x2 + 1)(y + y 2 + 1).
Solution
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Subtracting x + y from both sides, we get
√ p
x2 + 1 − x + y 2 + 1 − y = 2020(x + y).
√ √
Because a2 + 1 − a a + a2 + 1 = (a2 + 1) − a2 = 1 for all real a, we have
1 1
√ + p = 2020(x + y).
x + x2 + 1 y + y 2 + 1
Writing the left-hand side under a common denominator, we get
√ p
x + x2 + 1 + y + y 2 + 1
√ p = 2020(x + y).
x + x2 + 1 y + y 2 + 1
√ p
But, since x2 + 1 + y 2 + 1 = 2021(x + y), we have
2022(x + y)
√ p = 2020(x + y).
x+ 2
x +1 y+ y +1 2
√ p
Because x2 + 1 + y 2 + 1 ≥ 1 + 1 > 0, we have 2021(x + y) > 0, so x + y > 0. Thus, we
may write
√ p 2022(x + y) 1011
x + x2 + 1 y + y 2 + 1 = = .
2020(x + y) 1010
USA Mathematical Talent Search
Round 2 Solutions
Year 33 — Academic Year 2021–2022
www.usamts.org
4/2/33. Let ABC be a scalene triangle, and let X, Y , Z be points on sides BC, CA, AB,
respectively. Let I and O denote the incenter and circumcenter, respectively, of triangle
ABC. Suppose that
BX − CX CY − AY AZ − BZ
= = .
BA − CA CB − AB AC − BC
Prove that there exists a point P on line IO such that P X ⊥ BC, P Y ⊥ CA, and P Z ⊥
AB.
Solution
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O
I X’
B DX M C
To avoid lengthy discussion of configurations, let us now take all ratios of lengths to be
directed. Hence we may write the given condition as
BX − XC CY − Y A AZ − ZB
α= = = .
|AB| − |AC| |BC| − |BA| |CA| − |CB|
Observe that if α = 0, we may select P ≡ O, and if α = 1, we may select P ≡ I. Thus it
suffices to consider the case where α 6= 0, 1.
Since ABC is scalene, the points I and O are distinct, and IO is not perpendicular to
BC. Let M be the midpoint of BC and let D be the tangency point of the incircle onto side
BC.
The perpendicular to BC to X is not parallel to line IO, so the two intersect at some
point X 0 . Evidently
OX 0 MX
= .
OI MD
However, notice that
CX − XB
MX =
2
and
|AC| − |AB|
MD =
2
so
MX
= α.
MD
USA Mathematical Talent Search
Round 2 Solutions
Year 33 — Academic Year 2021–2022
www.usamts.org
Hence if we define Y 0 and Z 0 similarly, we find that X 0 ≡ Y 0 ≡ Z 0 , so the required P exists
and lies on line IO.
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USA Mathematical Talent Search
Round 2 Solutions
Year 33 — Academic Year 2021–2022
www.usamts.org
5/2/33. For a finite nonempty set A of positive integers, A = {a1 , a2 , . . . , an }, we say the
calamitous complement of A is the set of all positive integers k for which there do not
exist nonnegative integers w1 , w2 , . . . , wn with
k = a1 w 1 + a2 w 2 + · · · + an w n .
The calamitous complement of A is denoted cc(A). For example,
cc({5, 6, 9}) = {1, 2, 3, 4, 7, 8, 13}.
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Find all pairs of positive integers a, b with 1 < a < b for which there exists a set G satisfying
all of the following properties:
1. G is a set of at most three positive integers,
2. cc({a, b}) and cc(G) are both finite sets, and
3. cc(G) = cc({a, b}) ∪ {m} for some m not in cc({a, b}).
Solution
The answers are (a, b) = (2, 2k + 1), (a, b) = (3, 3k + 1), or (a, b) = (3, 3k + 2) where k is
any positive integer. It will suffice to prove that a ∈ {2, 3} and gcd(a, b) = 1.
We cite the well-known Frobenius Coin problem to assert that cc({a, b}) is finite if and
only if gcd(a, b) = 1.
First, we show that the pairs we listed in our answer work. We will use G = {a, a + b, 2b}.
It is clear that cc(G) ⊆ cc({a, b}). We show that cc(G) = cc({a, b}) ∪ {b}. This means that
for any x, y ≥ 0, we need to write xa + yb as a nonnegative linear combination of a, a + b, 2b
except the case (x, y) = (0, 1). That (x, y) = (0, 1) is impossible follows from the fact that
out of {a, a + b, 2b}, only a is less than b, and gcd(a, b) = 1.
If y = 0, then xa is a linear combination of as only. If y = 1 and x > 0, then we use 1
term a + b and x − 1 terms a. So it remains to consider y ≥ 2.
If a = 2, then 3b = 1(a + b) + (b − 1)a, and if a = 3, then 3b = ba. Either way, this means
3b ∈
/ cc(G). Next, observe that all multiples of b greater than b can be written as a sum of
terms of the form 2b and 3b. This means that all numbers of the form xa+yb for x ≥ 0, y ≥ 2
are not in cc(G). That finishes our proof of the claim that cc(G) = cc({a, b}) ∪ {b}.
USA Mathematical Talent Search
Round 2 Solutions
Year 33 — Academic Year 2021–2022
www.usamts.org
Now we need to show no other pairs work. The Frobenius Coin problem tells us imme-
diately that gcd(a, b) > 1 fails, so it remains to prove that a < 4. Suppose for the sake of
contradiction that a, b ≥ 4 and that a set G exists satisfying the properties (1), (2), (3) for
some a, b. Note that we will remove the restriction that a < b for this argument.
First, we show that the m in property (3) is one of a or b. If m is neither of them, then
since a, b ∈ / cc({a, b}), we must have a, b ∈ / cc(G) by (3). But then a, b are nonnegative linear
combinations of elements of G, which means that cc({a, b}) ⊆ cc(G). This means that no
such m can exist. This contradiction means that m ∈ {a, b}. Without loss of generality, we
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will assume m = b.
We now claim that two of the elements of G must be a and a + b. Let H be the set
of positive integers not in cc(G), which means H is all positive integers that are a nonneg-
ative linear combination of elements of G. The smallest element in H not a multiple of b
is a. Since gcd(a, b) = 1, a cannot be a linear combination of smaller elements of H, so
a must be in G. Likewise, a is the only element in H that is smaller than a + b and not
a multiple of b. Because b cannot be in H by assumption, this means a + b must also be in G.
We now know that G = {a, a + b, x} for some x. The smallest element of H that is not
a nonnegative linear combination of a, a + b is 2b. This is because all elements of H are of
the form xa + yb for x, y nonnegative integers and (x, y) 6= (0, 1), and adding terms of the
form a, a + b will obtain all such numbers when y is 0 or 1. Furthermore, 2b is not a linear
combination of a, a + b since gcd(a, b) = 1 and a ≥ 4.
So we must have G = {a, a + b, 2b}. We now show 3b is not a linear combination of any
elements of G, which obtains our final contradiction. 3b is not a sum of terms of the form 2b,
obviously. If we involve the other terms, then we have xa + yb for x > 0. Since gcd(a, b) = 1,
this can only be a multiple of b if x is a multiple of b. So x ≥ b, but a ≥ 4 meaning that
xa + yb ≥ 4b > 3b, a contradiction.
Problems by Ryan Kuroyama, Michael Tang, Evan Chen, and USAMTS Staff.
c 2021 Art of Problem Solving Initiative, Inc.
