Found problems: 248
1999 USAMTS Problems, 4
In $\triangle PQR$, $PQ=8$, $QR=13$, and $RP=15$. Prove that there is a point $S$ on line segment $\overline{PR}$, but not at its endpoints, such that $PS$ and $QS$ are also integers.
[asy]
size(200);
defaultpen(linewidth(0.8));
pair P=origin,Q=(8,0),R=(7,10),S=(3/2,15/7);
draw(P--Q--R--cycle);
label("$P$",P,W);
label("$Q$",Q,E);
label("$R$",R,NE);
draw(Q--S,linetype("4 4"));
label("$S$",S,NW);
[/asy]
1998 USAMTS Problems, 1
Determine the leftmost three digits of the number
\[1^1+2^2+3^3+...+999^{999}+1000^{1000}.\]
2015 USAMTS Problems, 3
For all positive integers $n$, show that:
$$ \dfrac1n \sum^n _{k=1} \dfrac{k \cdot k! \cdot {n\choose k}}{n^k} = 1$$
2014 USAMTS Problems, 3:
Let $P$ be a square pyramid whose base consists of the four vertices $(0, 0, 0), (3, 0, 0), (3, 3, 0)$, and $(0, 3, 0)$, and whose apex is the point $(1, 1, 3)$. Let $Q$ be a square pyramid whose base is the same as the base of $P$, and whose apex is the point $(2, 2, 3)$. Find the volume of the intersection of the interiors of $P$ and $Q$.
2014 USAMTS Problems, 4:
Nine distinct positive integers are arranged in a circle such that the product of any two non-adjacent numbers in the circle is a multiple of $n$ and the product of any two adjacent numbers in the circle is not a multiple of $n$, where $n$ is a fixed positive integer. Find the smallest possible value for $n$.
2012 USAMTS Problems, 4
Let $n$ be a positive integer. Consider an $n\times n$ grid of unit squares. How many ways are there to partition the horizontal and vertical unit segments of the grid into $n(n + 1)$ pairs so that the following properties are satisfi ed?
(i) Each pair consists of a horizontal segment and a vertical segment that share a common endpoint, and no segment is in more than one pair.
(ii) No two pairs of the partition contain four segments that all share the same endpoint.
(Pictured below is an example of a valid partition for $n = 2$.)
[asy]
import graph; size(2.6cm);
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);
pen dotstyle = black;
draw((-3,4)--(-3,2));
draw((-3,4)--(-1,4));
draw((-1,4)--(-1,2));
draw((-3,2)--(-1,2));
draw((-3,3)--(-1,3));
draw((-2,4)--(-2,2));
draw((-2.8,4)--(-2,4), linewidth(2));
draw((-3,3.8)--(-3,3), linewidth(2));
draw((-1.8,4)--(-1,4), linewidth(2));
draw((-2,4)--(-2,3.2), linewidth(2));
draw((-3,3)--(-2.2,3), linewidth(2));
draw((-3,2.8)--(-3,2), linewidth(2));
draw((-3,2)--(-2.2,2), linewidth(2));
draw((-2,3)--(-2,2.2), linewidth(2));
draw((-1,2)--(-1.8,2), linewidth(2));
draw((-1,4)--(-1,3.2), linewidth(2));
draw((-2,3)--(-1.2,3), linewidth(2));
draw((-1,2.8)--(-1,2), linewidth(2));
dot((-3,2),dotstyle);
dot((-1,4),dotstyle);
dot((-1,2),dotstyle);
dot((-3,3),dotstyle);
dot((-2,4),dotstyle);
dot((-2,3),dotstyle);[/asy]
1998 USAMTS Problems, 1
Determine the unique pair of real numbers $(x,y)$ that satisfy the equation
\[(4 x^2+ 6 x + 4)(4 y^2 - 12 y + 25 ) = 28 .\]
2002 USAMTS Problems, 1
The sequence of letters [b]TAGC[/b] is written in succession 55 times on a strip, as shown below. The strip is to be cut into segments between letters, leaving strings of letters on each segment, which we call words. For example, a cut after the first G, after the second T, and after the second C would yield the words [b]TAG[/b], [b]CT[/b] and [b]AGC[/b]. At most how many distinct words could be found if the entire strip were cut? Justify your answer.
\[\boxed{\textbf{T A G C T A G C T A G}}\ldots\boxed{\textbf{C T A G C}}\]
2015 USAMTS Problems, 1
Fill in each space of the grid with either a $0$ or a $1$ so that all $16$ strings of four consecutive numbers across and down are distinct.
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]
draw((8,0)--(8,4)--(1,4)--(1,9)--(0,9) -- (0,5) -- (5,5)--(5,0)--(9,0)--(9,1)--(4,1)--(4,8)--(0,8));
draw((0,6)--(4,6));
draw((0,7)--(4,7));
draw((4,3)--(8,3));
draw((4,2)--(8,2));
draw((2,4)--(2,8));
draw((3,4)--(3,8));
draw((6,0)--(6,4));
draw((7,0)--(7,4));
label("0",(0.5, 8.5));
label("",(0.5, 7.5));
label("0",(0.5, 6.5));
label("1",(0.5, 5.5));
label("1",(1.5, 7.5));
label("",(1.5, 6.5));
label("",(1.5, 5.5));
label("0",(1.5, 4.5));
label("0",(2.5, 7.5));
label("1",(2.5, 6.5));
label("",(2.5, 5.5));
label("",(2.5, 4.5));
label("",(3.5, 7.5));
label("",(3.5, 6.5));
label("0",(3.5, 5.5));
label("1",(3.5, 4.5));
label("",(4.5, 4.5));
label("",(4.5, 3.5));
label("",(4.5, 2.5));
label("0",(4.5, 1.5));
label("0",(5.5, 3.5));
label("",(5.5, 2.5));
label("",(5.5, 1.5));
label("",(5.5, 0.5));
label("",(6.5, 3.5));
label("",(6.5, 2.5));
label("",(6.5, 1.5));
label("",(6.5, 0.5));
label("",(7.5, 3.5));
label("0",(7.5, 2.5));
label("",(7.5, 1.5));
label("1",(7.5, 0.5));
label("",(8.5, 0.5));
[/asy]
1998 USAMTS Problems, 5
In $\triangle A B C$, let $D, E$, and $F$ be the midpoints of the sides of the triangle, and let $P, Q,$ and $R$ be the midpoints of the corresponding medians, $AD ,B E,$ and $C F$, respectively, as shown in the figure at the right. Prove that the value of
\[\frac{AQ^2 + A R^2 + B P^2 + B R^2 + C P^2+ C Q^2 }{A B^2 + B C^2 + C A^2}\]
does not depend on the shape of $\triangle A B C$ and find that value.
[asy]
defaultpen(linewidth(0.7)+fontsize(10));size(200);
pair A=origin, B=(14,0), C=(9,12), D=midpoint(C--B), E=midpoint(C--A), F=midpoint(A--B), R=midpoint(C--F), P=midpoint(D--A), Q=midpoint(E--B);
draw(A--B--C--A, linewidth(1));
draw(A--D^^B--E^^C--F);
draw(B--R--A--Q--C--P--cycle, dashed);
pair point=centroid(A,B,C);
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));
label("$E$", E, dir(point--E));
label("$F$", F, dir(point--F));
label("$P$", P, dir(40)*dir(point--P));
label("$Q$", Q, dir(40)*dir(point--Q));
label("$R$", R, dir(40)*dir(point--R));
dot(P^^Q^^R);[/asy]
2004 USAMTS Problems, 1
The numbers 1 through 9 can be arranged in the triangles labeled $a$ through $i$ illustrated below so that the numbers in each of the $2\times2$ triangles sum to the value $n$; that is \[a+b+c+d=b+e+f+g=d+g+h+i=n.\] For each possible sum $n$, show an arrangement, labeled with the sum as shown below. Prove that there are no possible arrangements for any other values of $n$.
[asy]
size(150);
defaultpen(linewidth(0.7)+fontsize(12)); picture p = new picture;
draw(p,(-3,-3^.5)/2--(3,-3^.5)/2^^(-1,0)--(1,0)^^(-1,3^.5)/2--(1,3^.5)/2); add(p); add(rotate(120)*p); add(rotate(240)*p);
string[] hexlbl = {'d','c','b','f','g','h'}, trilbl = {'a','e','i'};
for(int i = 0; i < hexlbl.length; ++i) label('$'+hexlbl[i]+'$',dir(30+60*i)/3^.5);
for(int i = 0; i < trilbl.length; ++i) label('$'+trilbl[i]+'$',dir(90+120*i)*2/3^.5);[/asy]
2012 Romania Team Selection Test, 1
Let $m$ and $n$ be two positive integers greater than $1$. Prove that there are $m$ positive integers $N_1$ , $\ldots$ , $N_m$ (some of them may be equal) such that \[\sqrt{m}=\sum_{i=1}^m{(\sqrt{N_i}-\sqrt{N_i-1})^{\frac{1}{n}}.}\]
2021 USAMTS Problems, 4
Let ABC be a triangle whose vertices are inside a circle $\Omega$. Prove that we can choose two of the vertices of ABC such that there are infinitely many circles $\omega$ that satisfy the following properties:
1. $\omega$ is inside of $\Omega$,
2. $\omega$ passes through the two chosen vertices, and
3. the third vertex is in the interior of $\omega$ .
2014 USAMTS Problems, 4:
A point $P$ in the interior of a convex polyhedron in Euclidean space is called a [i]pivot point[/i] of the polyhedron if every line through $P$ contains exactly $0$ or $2$ vertices of the polyhedron. Determine, with proof, the maximum number of pivot points that a polyhedron can contain.
2012 USAMTS Problems, 4
Denote by $\lfloor x\rfloor$ the greatest positive integer less than or equal to $x$. Let $m\ge2$ be an integer, and let $s$ be a real number between $0$ and $1$. Defi ne an infi nite sequence of real numbers $a_1, a_2, a_3,\ldots$ by setting $a_1 = s$ and $ak = ma_{k-1}-(m-1)\lfloor a_{k-1}\rfloor$ for all $k\ge2$. For example, if $m = 3$ and $s = \tfrac58$, then we get $a_1 = \tfrac58$, $a_2 = \tfrac{15}8$, $a_3 = \tfrac{29}8$, $a_4 = \tfrac{39}8$, and so on.
Call the sequence $a_1, a_2, a_3,\ldots$ $\textbf{orderly}$ if we can find rational numbers $b, c$ such that $\lfloor a_n\rfloor = \lfloor bn + c\rfloor$ for all $n\ge1$. With the example above where $m = 3$ and $s = \tfrac58$, we get an orderly sequence since $\lfloor a_n\rfloor = \left\lfloor\tfrac{3n}2-\tfrac32\right\rfloor$ for all $n$.
Show that if $s$ is an irrational number and $m\ge2$ is any integer, then the sequence $a_1, a_2, a_3,\ldots$ is $\textbf{not}$ an orderly sequence.
2012 USAMTS Problems, 3
The $\textbf{symmetric difference}$, $\triangle$, of a pair of sets is the set of elements in exactly one set. For example, \[\{1,2,3\}\triangle\{2,3,4\}=\{1,4\}.\] There are fifteen nonempty subsets of $\{1,2,3,4\}$. Assign each subset to exactly one of the squares in the grid to the right so that the following conditions are satisfied.
(i) If $A$ and $B$ are in squares connected by a solid line then $A\triangle B$ has exactly one element.
(ii) If $A$ and $B$ are in squares connected by a dashed line then the largest element of $A$ is equal to the largest element of $B$.
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(150);
defaultpen(linewidth(0.8));
draw((0,1)--(0,3)--(3,3)^^(2,3)--(2,2)--(3,2)--(3,1)--(1,1)--(1,2)--(0,2)^^(2,1)--(2,0)--(0,0));
draw(origin--(0,1)^^(1,0)--(3,2)^^(1,1)--(0,2)^^(1,2)--(0,3)^^(1,3)--(2,2),linetype("4 4"));
real r=1/4;
path square=(r,r)--(r,-r)--(-r,-r)--(-r,r)--cycle;
int limit;
for(int i=0;i<=3;i=i+1)
{
if (i==0)
limit=2;
else
limit=3;
for(int j=0;j<=limit;j=j+1)
filldraw(shift(j,i)*square,white);
}
[/asy]
2009 USAMTS Problems, 1
Fill in the circles in the picture at right with the digits $1-8$, one digit in each circle with no digit repeated, so that no two circles that are connected by a line segment contain consecutive digits. In how many ways can this be done?
[asy]
defaultpen(linewidth(1)); real r = 0.2; pair[] dots = {(0,1),(1,0),(0,0),(-1,0),(1,-1),(0,-1),(-1,-1),(0,-2)}; draw((0,1)--(-1,0)--(-1,-1)--(0,-2)--(1,-1)--(1,0)--(0,1)--(0,-2)); draw((-1,0)--(1,0)--(0,-1)--cycle); draw((-1,-1)--(1,-1)); for(int i = 0; i < dots.length; ++i) filldraw(circle(dots[i], r), white);[/asy]
2024 USAMTS Problems, 3
Let $a, b$ be positive integers such that $a^2 \ge b$. Let
$x = \sqrt{a+\sqrt{b}} - \sqrt{a-\sqrt{b}}$
(a) Prove that for all integers $a \ge 2$, there exists a positive integer $b$ such that $x$ is also
a positive integer.
(b) Prove that for all sufficiently large $a$, there are at least two $b$ such that $x$ is a positive
integer.
$\textbf{Note}$: We’ve received some questions about what is meant by “for all sufficiently large
$a$.” To give a simple example of this phrasing, it is true that for all sufficiently large positive
integers $n$, we have $n^2 \ge 100$. Specifically, this is true for all $n \ge 10$.
1999 USAMTS Problems, 3
Suppose that the 32 computers in a certain network are numbered with the 5-bit integers $00000, 00001, 00010, ..., 11111$ (bit is short for binary digit). Suppose that there is a one-way connection from computer $A$ to computer $B$ if and only if $A$ and $B$ share four of their bits with the remaining bit being $0$ at $A$ and $1$ at $B$. (For example, $10101$ can send messages to $11101$ and to $10111$.) We say that a computer is at level $k$ in the network if it has exactly $k$ 1’s in its label $(k = 0, 1, 2, ..., 5)$. Suppose further that we know that $12$ computers, three at each of the levels $1$, $2$, $3$, and $4$, are malfunctioning, but we do not know which ones. Can we still be sure that we can send a message from $00000$ to $11111$?
2014 USAMTS Problems, 5:
Find the smallest positive integer $n$ that satisfies the following:
We can color each positive integer with one of $n$ colors such that the equation $w + 6x = 2y + 3z$ has no solutions in positive integers with all of $w, x, y$ and $z$ having the same color. (Note that $w, x, y$ and $z$ need not be distinct.)
2020 USAMTS Problems, 2:
[b]2/1/32.[/b] Is it possible to fill in a $2020$ x $2020$ grid with the integers from $1$ to $4,080,400$ so that the sum of each row is $1$ greater than the previous row?
2002 USAMTS Problems, 5
Prove that if the cross-section of a cube cut by a plane is a pentagon, as shown in the figure below, then there are two adjacent sides of the pentagon such that the sum of the lengths of those two sides is greater than the sum of the lengths of the other three sides. (For ease of grading, please use the names of the points from the figure below in your solution.)
[asy]
import three;
defaultpen(linewidth(0.8));
currentprojection=orthographic(1,3/5,1/2);
draw(unitcube, white, thick(), nolight);
draw(O--(1,0,0)^^O--(0,1,0)^^O--(0,0,1), linetype("4 4")+linewidth(0.7));
triple A=(1/3, 1, 1), B=(2/3, 1, 0), C=(1, 1/2, 0), D=(1, 0, 1/2), E=(2/3, 0, 1);
draw(E--A--B^^C--D);
draw(B--C^^D--E, linetype("4 4")+linewidth(0.7));
label("$A$", A, dir(85));
label("$B$", B, SE);
label("$C$", C, S);
label("$D$", D, W);
label("$E$", E, NW);[/asy]
2013 USAMTS Problems, 4
An infinite sequence of real numbers $a_1,a_2,a_3,\dots$ is called $\emph{spooky}$ if $a_1=1$ and for all integers $n>1$,
\[\begin{array}{c@{\;\,}c@{\;\,}c@{\;\,}c@{\;\,}c@{\;\,}c@{\;\,}c@{\;\,}c@{\;\,}c@{\;\,}c@{\;\,}c@{\;\,}c@{\;\,}c}
na_1&+&(n-1)a_2&+&(n-2)a_3&+&\dots&+&2a_{n-1}&+&a_n&<&0,\\
n^2a_1&+&(n-1)^2a_2&+&(n-2)^2a_3&+&\dots&+&2^2a_{n-1}&+&a_n&>&0.
\end{array}\]Given any spooky sequence $a_1,a_2,a_3,\dots$, prove that
\[2013^3a_1+2012^3a_2+2011^3a_3+\cdots+2^3a_{2012}+a_{2013}<12345.\]
1997 Hungary-Israel Binational, 1
Is there an integer $ N$ such that $ \left(\sqrt{1997}\minus{}\sqrt{1996}\right)^{1998}\equal{}\sqrt{N}\minus{}\sqrt{N\minus{}1}$?
2016 USAMTS Problems, 1:
Fill in each cell of the grid with one of the numbers 1, 2, or 3. After all numbers are filled in, if a row, column, or any diagonal has a number of cells equal to a multiple of 3, then it must have the same amount of 1’s, 2’s, and 3’s. (There are 10 such diagonals, and they are all marked in the grid by a gray dashed line.) Some numbers have been given to you.
[asy]
defaultpen(linewidth(0.45));
real[][] arr = {
{0, 2, 1, 0, 0, 0, 0, 0, 0},
{3, 0, 0, 2, 0, 0, 0, 0, 0},
{0, 0, 0, 2, 0, 0, 3, 2, 0},
{0, 2, 1, 0, 0, 1, 0, 0, 3},
{3, 0, 0, 0, 0, 3, 0, 0, 3},
{2, 0, 0, 0, 0, 0, 2, 3, 0},
{3, 2, 3, 2, 0, 2, 0, 0, 3},
{0, 0, 0, 0, 0, 3, 0, 0, 1},
{0, 0, 0, 0, 0, 0, 1, 3, 0}};
for (int i=0; i<9; ++i){
for (int j=0; j<9; ++j){
draw((i,j)--(i+1,j)--(i+1, j+1)--(i,j+1)--cycle);
if(arr[8-j][i] != 0){
label((string) arr[8-j][i], (i+0.5, j+0.5));
}
}
}
draw((3,0)--(0,3), linetype(new real[] {4,4})+grey);
draw((6,0)--(0,6), linetype(new real[] {4,4})+grey);
draw((9,0)--(0,9), linetype(new real[] {4,4})+grey);
draw((3,9)--(9,3), linetype(new real[] {4,4})+grey);
draw((6,9)--(9,6), linetype(new real[] {4,4})+grey);
draw((6,0)--(9,3), linetype(new real[] {4,4})+grey);
draw((3,0)--(9,6), linetype(new real[] {4,4})+grey);
draw((0,0)--(9,9), linetype(new real[] {4,4})+grey);
draw((0,3)--(6,9), linetype(new real[] {4,4})+grey);
draw((0,6)--(3,9), linetype(new real[] {4,4})+grey);
[/asy]
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.)