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Problem 3: Page 1
Marcos Pertierra
10890
modularmarc101
USAMTS Year 21 Round 3
3/3/21
We will prove that we can draw a circle of diameter 1 with a white interior by
proving that there exists an area where we can place the center of the circle such
that the circle does not overlap or is tangent to any unit square or the 20 x 25
piece of paper.
First of all, since the circle has radius , the center has to be more than unit
away from each side of the 20 x 25 paper. If not, the circle will be tangent to a
side of the rectangle, or have a part of it outside the rectangle. So the area where
we can place the center is now restricted to a 19 x 24 rectangle.
Now take any unit square. Tracing a line around the unit square that is unit
away from its perimeter, we have the area around the square where we can not
place the center. Firstly, it obviously can not be inside the unit square, because
its interior will be blue. Secondly, if it is inside the line that we traced, the circle
will be tangent to the perimeter of the unit square, or overlap with the square.
The area around each unit square where we can not place the center of the circle
will look like this:
Therefore, we must place the center outside this area so the interior is white.
Problem 3: Page 2
Marcos Pertierra
10890
modularmarc101
USAMTS Year 21 Round 3
In the case where the surrounding areas of each unit square do not intersect,
there will be a minimal area to place the center of the circle. Hence, by proving
that it’s possible to place the center of the circle in that case, it proves that we
can draw it the circle in any case.
The area of each unit square along with it’s surrounding area is equal to
(area of unit square) + 4(area of 1 x 1/2 rectangle) + 4(area of 90 degree sector
of circle with radius 1/2).
So the area is
.
Now the area of where we can place the center of the circle is equal to
(area of 19 x 24 rectangle) – 120(area of unit square with surrounding area).
So it suffices to prove that
Which is clearly true because
Therefore, since there exists an area where we can place the center of the circle,
we can draw the circle on the remaining paper such that the entire interior of the
circle is white.
Problem 2: Page 1
Marcos Pertierra
10890
modularmarc101
USAMTS Year 21 Round 3
2/3/21
We begin by noting that
,
which is equal to the number of ways we can choose 500 objects out of
objects. Hence, the result will be a positive integer for any positive integer
we choose.
that
Let all the primes under 500 be
and let
be the smallest, positive integers such that
where
.
We will show that
satisfies the problem. Firstly, notice that if
, then
has the same number of factors of as . We can
show this by letting
(
and are relatively prime), and since
, the greatest power of we can factor out of
is .
Since we showed that
know that
does not divide
we know that
where
and
have the same number of factors of
. This holds for
and
has no factors of
.
, we
, so
Problem 2: Page 2
Marcos Pertierra
10890
modularmarc101
USAMTS Year 21 Round 3
Therefore, we can conclude that
is a possible value for , where
are all the primes under 500 and where
are the smallest,
positive integers, such that
is greater than 500.
Problem 4: Page 1
Marcos Pertierra
10890
modularmarc101
USAMTS Year 21 Round 3
4/3/21
We begin by showing that
contradiction; let
have that
and
have that
can not be written as
Hence,
.
and are relatively prime. We proceed by
for some positive integer greater than 1. Then we
for positive integers
and . Furthermore, we
, so there are infinitely many numbers that
(those that are not divisible by ); contradiction.
The Chicken McNugget Theorem states that for any two relatively prime positive
integers and , the greatest number that can not be written as
(where
and are nonnegative integers) is
.
Now we will consider two cases for and . They can not be both even because
then they will not be relatively prime.
Case 1:
and are odd
Let
and
. By the Chicken McNugget Theorem, the
greatest number that is not expressible as
is
.
Notice that all even positive integers less than
. Hence, there are a total of
that are not expressible.
Case 2:
are not expressible as
numbers
is even and is odd
Let
and
. By the Chicken McNugget Theorem, the greatest
number that is not expressible as
is
Problem 4: Page 2
Marcos Pertierra
10890
modularmarc101
USAMTS Year 21 Round 3
Therefore, there are a total of
that are not expressible.
numbers
Knowing this, we have the following equation:
From this equation, we can deduce that
solutions for
. These are
.
is even and is odd, and the possible
, and
We can eliminate the solutions
and
because and must be
relatively prime. We can eliminate the solutions
because 2, 8 and 3 divide 1776, which would make 1776 expressible as
.
This leaves
as the only possible values for
and . Hence, we have that
.
Note: The fact that the greatest integer that can not be written as
and that there are
is
positive integers that are not
expressible that way is mentioned in pages 42-44 of this Number Theory paper,
by Naoki Sato, in this link:
http://www.artofproblemsolving.com/Resources/Papers/SatoNT.pdf
Problem 1: Page 1
Marcos Pertierra
10890
modularmarc101
USAMTS Year 21 Round 3
1/3/21
Take any right triangle with right angle C. Drop an altitude from C to
AB. We have that
Also, by the Pythagorean Theorem, we have that
Now we can apply this to the four right triangles in our quadrilateral : ABP, BCP,
CDP, and DAP. Let AP = a, BP = b, CP = c, DP = d, and let the distance from P
to AD be x. We have the following equations:
Solving two equations for
and the two equations for
, we get:
Problem 1: Page 2
Marcos Pertierra
10890
modularmarc101
USAMTS Year 21 Round 3
By equating each pair of equations and clearing denominators, we get that
Solving each of these for
, we get that
Again, we equate these two equations, clear denominators, and simplify (divide
by ) to get
Problem 5: Page 1
Marcos Pertierra
10890
modularmarc101
USAMTS Year 21 Round 3
5/3/21
Firstly, let
. Notice that
From which we get that
.
Also, notice that since
and
, we have that
.
Now to the inequality; we can remove the absolute value by rewriting the
inequality as
By Arithmetic Mean – Geometric Mean, we have that