Found problems: 248
2015 USAMTS Problems, 2
A net for a polyhedron is cut along an edge to give two [b]pieces[/b]. For example, we may cut a cube net along the red edge to form two pieces as shown.
[asy]
size(5.5cm);
draw((1,0)--(1,4)--(2,4)--(2,0)--cycle);
draw((1,1)--(2,1));
draw((1,2)--(2,2));
draw((1,3)--(2,3));
draw((0,1)--(3,1)--(3,2)--(0,2)--cycle);
draw((2,1)--(2,2),red+linewidth(1.5));
draw((3.5,2)--(5,2));
filldraw((4.25,2.2)--(5,2)--(4.25,1.8)--cycle,black);
draw((6,1.5)--(10,1.5)--(10,2.5)--(6,2.5)--cycle);
draw((7,1.5)--(7,2.5));
draw((8,1.5)--(8,2.5));
draw((9,1.5)--(9,2.5));
draw((7,2.5)--(7,3.5)--(8,3.5)--(8,2.5)--cycle);
draw((11,1.5)--(11,2.5)--(12,2.5)--(12,1.5)--cycle);
[/asy]
Are there two distinct polyhedra for which this process may result in the same two pairs of pieces? If you think the answer is no, prove that no pair of polyhedra can result in the same two pairs of pieces. If you think the answer is yes, provide an example; a clear example will suffice as a proof.
2023 USAMTS Problems, 2
Malmer Pebane's apartment uses a six-digit access code, with leading zeros allowed. He noticed that his fingers leave that reveal which digits were pressed. He decided to change his access code to provide the largest number of possible combinations for a burglar to try when the digits are known. For each number of distinct digits that could be used in the access code, calculate the number of possible combinations when the digits are known but their order and frequency are not known. For example, if there are smudges on $3$ and $9,$ two possible codes are $393939$ and $993999.$ Which number of distinct digits in the access code offers the most combinations?
2012 USAMTS Problems, 1
Fill in each of the ten boxes with a 3-digit number so that the following conditions are satisfied.
[list=1]
[*]Every number has three distinct digits that sum to $15$. $0$ may not be a leading digit. One digit of each number has been given to you.
[*]No two numbers in any pair of boxes use the same three digits. For example, it is not allowed for two different boxes to have the numbers $456$ and $645$.
[*]Two boxes joined by an arrow must have two numbers that share an equal hundreds digit, tens digit, or ones digit. Also, the smaller number must point to the larger.[/list]
You do not need to prove that your configuration is the only one possible; you merely need to find a configuration 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.)
[asy]
size(200);
defaultpen(linewidth(0.8));
path arrow;
pair squares[]={(2,4),(6,4),(10,4),(0,0),(4,0),(8,0),(12,0),(2,-4),(6,-4),(10,-4)};
pair horizarrows[]={(4,4),(2,0),(6,0),(10,0),(4,-4),(8,-4)};
bool isLeft[]={false,false,true,false,false,false};
pair diagarrows[]={(1,2),(7,2),(9,2),(1,-2),(5,-2),(11,-2)};
bool isDown[]={true,false,true,false,false,true};
for(int i=0;i<=9;i=i+1)
{
draw(box(squares[i]-(1,1),squares[i]+(1,1)));
label("$"+(string)i+"$",squares[i]);
}
for(int j=0;j<=5;j=j+1)
{
if(isLeft[j])
arrow=(horizarrows[j].x-1,horizarrows[j].y)--(horizarrows[j].x+1,horizarrows[j].y);
else
arrow=(horizarrows[j].x+1,horizarrows[j].y)--(horizarrows[j].x-1,horizarrows[j].y);
draw(arrow,BeginArrow(size=7));
}
for(int k=0;k<=5;k=k+1)
{
if(isDown[k])
arrow=(diagarrows[k].x-1/3,diagarrows[k].y-1)--(diagarrows[k].x+1/3,diagarrows[k].y+1);
else
arrow=(diagarrows[k].x-1/3,diagarrows[k].y+1)--(diagarrows[k].x+1/3,diagarrows[k].y-1);
draw(arrow,BeginArrow(size=7));
}
[/asy]
2013 USAMTS Problems, 5
Let $S$ be a planar region. A $\emph{domino-tiling}$ of $S$ is a partition of $S$ into $1\times2$ rectangles. (For example, a $2\times3$ rectangle has exactly $3$ domino-tilings, as shown below.)
[asy]
import graph; size(7cm);
pen dps = linewidth(0.7); defaultpen(dps);
draw((0,0)--(3,0)--(3,2)--(0,2)--cycle, linewidth(2));
draw((4,0)--(4,2)--(7,2)--(7,0)--cycle, linewidth(2));
draw((8,0)--(8,2)--(11,2)--(11,0)--cycle, linewidth(2));
draw((1,0)--(1,2));
draw((2,1)--(3,1));
draw((0,1)--(2,1), linewidth(2));
draw((2,0)--(2,2), linewidth(2));
draw((4,1)--(7,1));
draw((5,0)--(5,2), linewidth(2));
draw((6,0)--(6,2), linewidth(2));
draw((8,1)--(9,1));
draw((10,0)--(10,2));
draw((9,0)--(9,2), linewidth(2));
draw((9,1)--(11,1), linewidth(2));
[/asy]
The rectangles in the partition of $S$ are called $\emph{dominoes}$.
(a) For any given positive integer $n$, find a region $S_n$ with area at most $2n$ that has exactly $n$ domino-tilings.
(b) Find a region $T$ with area less than $50000$ that has exactly $100002013$ domino-tilings.
2005 USAMTS Problems, 1
$\overline{AB}$ is a diameter of circle $C_1$. Point $P$ is on $C_1$ such that $AP>BP$. Circle $C_2$ is centered at $P$ with radius $PB$. The extension of $\overline{AP}$ past $P$ meets $C_2$ at $Q$. Circle $C_3$ is centered at $A$ and is externally tangent to $C_2$. Circle $C_4$ passes through $A$, $Q$, and $R$. Find, with proof, the ratio between the area of $C_4$ and the area of $C_1$, and show that this ratio is the same for all points $P$ on $C_1$ such that $AP>BP$.
2014 USAMTS Problems, 5:
Let $a_0,a_1,a_2,\dots$ be a sequence of nonnegative integers such that $a_2=5$, $a_{2014}=2015$, and $a_n=a_{a_{n-1}}$ for all positive integers $n$. Find all possible values of $a_{2015}$.
2009 USAMTS Problems, 1
Archimedes planned to count all of the prime numbers between $2$ and $1000$ using the Sieve of Eratosthenes as follows:
(a) List the integers from $2$ to $1000$.
(b) Circle the smallest number in the list and call this $p$.
(c) Cross out all multiples of $p$ in the list except for $p$ itself.
(d) Let $p$ be the smallest number remaining that is neither circled nor crossed out. Circle $p$.
(e) Repeat steps $(c)$ and $(d)$ until each number is either circled or crossed out.
At the end of this process, the circled numbers are prime and the crossed out numbers are composite.
Unfortunately, while crossing out the multiples of $2$, Archimedes accidentally crossed out two odd primes in addition to crossing out all the even numbers (besides $2$). Otherwise, he executed the algorithm correctly. If the number of circled numbers remaining when Archimedes finished equals the number of primes from $2$ to $1000$ (including $2$), then what is the largest possible prime that Archimedes accidentally crossed out?
2004 USAMTS Problems, 4
Find, with proof, all integers $n$ such that there is a solution in nonnegative real numbers $(x,y,z)$ to the system of equations
\[2x^2+3y^2+6z^2=n\text{ and }3x+4y+5z=23.\]
2002 USAMTS Problems, 4
Two overlapping triangles could divide a plane into up to eight regions, and three overlapping triangles could divide the plane into up to twenty regions. Find, with proof, the maximum number of regions into which six overlapping triangles could divide the plane. Describe or draw an arrangement of six triangles that divides the plane into that many regions.
1999 USAMTS Problems, 2
Let $a$ be a positive real number, $n$ a positive integer, and define the [i]power tower[/i] $a\uparrow n$ recursively with $a\uparrow 1=a$, and $a\uparrow(i+1)=a^{a\uparrow i}$ for $i=1,2,3,\ldots$. For example, we have $4\uparrow 3=4^{(4^4)}=4^{256}$, a number which has $155$ digits. For each positive integer $k$, let $x_k$ denote the unique positive real number solution of the equation $x\uparrow k=10\uparrow (k+1)$. Which is larger: $x_{42}$ or $x_{43}$?
2011 USAMTS Problems, 5
In the game of Tristack Solitaire, you start with three stacks of cards, each with a different positive integer number of cards. At any time, you can double the number of cards in any one stack of cards by moving cards from exactly one other, larger, stack of cards to the stack you double. You win the game when any two of the three stacks have the same number of cards.
For example, if you start with stacks of $3$, $5$, and $7$ cards, then you have three possible legal moves:
[list]
[*]You may move $3$ cards from the $5$-card stack to the $3$-card stack, leaving stacks of $6$, $2$, and $7$ cards.
[*]You may move $3$ cards from the $7$-card stack to the $3$-card stack, leaving stacks of $6,$ $5$, and $4$ cards.
[*]You may move $5$ cards from the $7$-card stack to the $5$-card stack, leaving stacks of $3$, $10$, and $2$ cards.[/list]
Can you win Tristack Solitaire from any starting position? If so, then give a strategy for winning. If not, then explain why.
2009 USAMTS Problems, 2
Alice has three daughters, each of whom has two daughters, each of Alice's six grand-daughters has one daughter. How many sets of women from the family of $16$ can be chosen such that no woman and her daughter are both in the set? (Include the empty set as a possible set.)
2015 USAMTS Problems, 3
Let $P$ be a convex n-gon in the plane with vertices labeled $V_1,...,V_n$ in counterclockwise order. A point $Q$ not outside $P$ is called a balancing point of $P$ if, when the triangles the blue and green regions are the same. Suppose $P$ has exactly one balancing point/ Show that the balancing point must be a vertex of $P$
2005 USAMTS Problems, 5
Sphere $S$ is inscribed in cone $C$. The height of $C$ equals its radius, and both equal $12+12\sqrt2$. Let the vertex of the cone be $A$ and the center of the sphere be $B$. Plane $P$ is tangent to $S$ and intersects $\overline{AB}$. $X$ is the point on the intersection of $P$ and $C$ closest to $A$. Given that $AX=6$, find the area of the region of $P$ enclosed by the intersection of $C$ and $P$.
2020 USAMTS Problems, 3:
[b]3/1/32.[/b] The bisectors of the internal angles of parallelogram $ABCD$ determine a quadrilateral with the same area as $ABCD$. Given that $AB > BC$, compute, with proof, the ratio $\frac{AB}{BC}$.
2024 USAMTS Problems, 5
Let $f(x) = x^2 + bx + 1$ for some real number $b$. Across all possible values of $b$, find all
possible values for the number of integers $x$ that satisfy $f(f(x) + x) < 0$.
2004 USAMTS Problems, 1
Determine with proof the number of positive integers $n$ such that a convex regular polygon with $n$ sides has interior angles whose measures, in degrees, are integers.
2014 USAMTS Problems, 1:
Fill in each blank unshaded cell with a positive integer less than 100, such that every consecutive group of unshaded cells within a row or column is an arithmetic sequence. 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.)
[asy]
size(9cm);
for (int x=0; x<=11; ++x)
draw((x, 0) -- (x, 5), linewidth(.5pt));
for (int y=0; y<=5; ++y)
draw((0, y) -- (11, y), linewidth(.5pt));
filldraw((0,4)--(0,3)--(2,3)--(2,4)--cycle, gray, gray);
filldraw((1,1)--(1,2)--(3,2)--(3,1)--cycle, gray, gray);
filldraw((4,1)--(4,4)--(5,4)--(5,1)--cycle, gray, gray);
filldraw((7,0)--(7,3)--(6,3)--(6,0)--cycle, gray, gray);
filldraw((7,4)--(7,5)--(6,5)--(6,4)--cycle, gray, gray);
filldraw((8,1)--(8,2)--(10,2)--(10,1)--cycle, gray, gray);
filldraw((9,4)--(9,3)--(11,3)--(11,4)--cycle, gray, gray);
draw((0,0)--(11,0)--(11,5)--(0,5)--cycle);
void foo(int x, int y, string n)
{
label(n, (x+0.5, y+0.5));
}
foo(1, 2, "10");
foo(4, 0, "31");
foo(5, 0, "26");
foo(10, 0, "59");
foo(0, 4, "3");
foo(7, 4, "59");
[/asy]
2021 USAMTS Problems, 3
Let $x$ and $y$ be distinct real numbers such that
\[ \sqrt{x^2+1}+\sqrt{y^2+1}=2021x+2021y. \]
Find, with proof, the value of
\[ \left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right). \]
2020 USAMTS Problems, 3:
Find, with proof, all positive integers $n$ with the following property: There are only finitely many positive multiples of $n$ which have exactly $n$ positive divisors
1999 USAMTS Problems, 3
Triangle $ABC$ has angle $A$ measuring $30^\circ$, angle $B$ measuring $60^\circ$, and angle $C$ measuring $90^\circ$. Show four different ways to divide triangle $ABC$ into four triangles, each similar to triangle $ABC$, but with one quarter of the area. Prove that the angles and sizes of the smaller triangles are correct.
2009 USAMTS Problems, 2
The ordered pair of four-digit numbers $(2025, 3136)$ has the property that each number in the pair is a perfect square and each digit of the second number is $1$ more than the corresponding digit of the fi rst number. Find, with proof, all ordered pairs of fi ve-digit numbers and ordered pairs of six-digit numbers with the same property: each number in the pair is a perfect square and each digit of the second number is $1$ more than the corresponding digit of the first number.
2015 USAMTS Problems, 4
Find all polynomials $P(x)$ with integer coefficients such that, for all integers $a$ and $b$, $P(a+b) - P(b)$ is a multiple of $P(a)$.
2011 USAMTS Problems, 4
Let $ABCDEF$ and $ABC'D'E'F'$ be regular planar hexagons in three-dimensional space with side length $1$, such that $\angle EAE'=60^{\circ}$. Let $P$ be the convex polyhedron whose vertices are $A$, $B$, $C$, $C'$, $D$, $D'$, $E$, $E'$, $F$, and $F'$.
(a) Find the radius $r$ of the largest sphere that can be enclosed in polyhedron $P$.
(b) Let $S$ be a sphere enclosed in polyhedron $P$ with radius $r$ (as derived in part (a)). The set of possible centers of $S$ is a line segment $\overline{XY}$. Find the length $XY$.
2013 USAMTS Problems, 4
An infinite sequence $(a_0,a_1,a_2,\dots)$ of positive integers is called a $\emph{ribbon}$ if the sum of any eight consecutive terms is at most $16$; that is, for all $i\ge0$,
\[a_i+a_{i+1}+\dots+a_{i+7}\le16.\]A positive integer $m$ is called a $\emph{cut size}$ if every ribbon contains a set of consecutive elements that sum to $m$; that is, given any ribbon $(a_0,a_1,a_2,\dots)$, there exist nonnegative integers $k\le l$ such that
\[a_k+a_{k+1}+\dots+a_l=m.\]Find, with proof, all cut sizes, or prove that none exist.