Found problems: 85335
Problems 15, 16, and 17 all refer to the following:
In the very center of the Irenic Sea lie the beautiful Nisos Isles. In 1998 the number of people on these islands is only 200, but the population triples every 25 years. Queen Irene has decreed that there must be at least 1.5 square miles for every person living in the Isles. The total area of the Nisos Isles is 24,900 square miles.
17. In how many years, approximately, from 1998 will the population of Nisos be as much as Queen Irene has proclaimed that the islands can support?
$ \text{(A)}\ 50\text{ yrs.}\qquad\text{(B)}\ 75\text{ yrs.}\qquad\text{(C)}\ 100\text{ yrs.}\qquad\text{(D)}\ 125\text{ yrs.}\qquad\text{(E)}\ 150\text{ yrs.} $
The plane shows $2020$ straight lines in general position, that is, there are none three intersecting at one point but no two parallel. Let's say, that the drawn line $a$ [i]detaches [/i] the drawn line $b$ if all intersection points of line $b$ with the other drawn lines lie in one half plane wrt to line $a$ (given the most straightforward $a$). Prove that you can be guaranteed find two drawn lines $a$ and $b$ that $a$ detaches $b$, but $b$ does not detach $a$.
Let $P$ be a point inside a triangle $ABC$ such that $\angle PBC = \angle PCA < \angle PAB$. The line $PB$ meets the circumcircle of triangle $ABC$ at a point $E$ (apart from $B$). The line $CE$ meets the circumcircle of triangle $APE$ at a point $F$ (apart from $E$). Show that the ratio $\frac{\left|APEF\right|}{\left|ABP\right|}$ does not depend on the point $P$, where the notation $\left|P_1P_2...P_n\right|$ stands for the area of an arbitrary polygon $P_1P_2...P_n$.
For each non-negative integer $n$ find the sum of all $n$-digit numbers with the digits in a decreasing sequence.
[I]Proposed by P. Kozhevnikov[/I]
Let $f : N \to R$ be a function, satisfying the following condition:
for every integer $n > 1$, there exists a prime divisor $p$ of $n$ such that $f(n) = f \big(\frac{n}{p}\big)-f(p)$.
If $f(2^{2007}) + f(3^{2008}) + f(5^{2009}) = 2006$, determine the value of $f(2007^2) + f(2008^3) + f(2009^5)$
If $ a$, $ b$, $ c$, $ d$, and $ e$ are constants such that every $ x > 0$ satisfies
\[ \frac{5x^4 \minus{} 8x^3 \plus{} 2x^2 \plus{} 4x \plus{} 7}{(x \plus{} 2)^4}
\equal{} a \plus{} \frac{b}{x \plus{} 2} \plus{} \frac{c}{(x \plus{} 2)^2}
\plus{} \frac{d}{(x \plus{} 2)^3} \plus{} \frac{e}{(x \plus{} 2)^4} \, ,\]
then what is the value of $ a \plus{} b \plus{} c \plus{} d \plus{} e$?
In order to study a certain ancient language, some researchers formatted its discovered words into expressions formed by concatenating letters from an alphabet containing only two letters. Along the study, they noticed that any two distinct words whose formatted expressions have an equal number of letters, greater than $ 2, $ differ by at least three letters.
Prove that if their observation holds indeed, then the number of formatted expressions that have $ n\ge 3 $ letters is at most $ \left[ \frac{2^n}{n+1} \right] . $
Find $\lim_{n\to\infty} \frac{1}{(\ln n)^2}\sum_{k=3}^n \frac{\ln k}{k}.$
The circle is inscribed in a triangle, inscribed in a semicircle. Find the marked angle $a$.
[img]https://cdn.artofproblemsolving.com/attachments/8/e/334c8662377155086e9211da3589145f460b52.png[/img]
In a circle of radius $ 5$ units, $ CD$ and $ AB$ are perpendicular diameters. A chord $ CH$ cutting $ AB$ at $ K$ is $ 8$ units long. The diameter $ AB$ is divided into two segments whose dimensions are:
$ \textbf{(A)}\ 1.25, 8.75 \qquad\textbf{(B)}\ 2.75,7.25 \qquad\textbf{(C)}\ 2,8 \qquad\textbf{(D)}\ 4,6$
$ \textbf{(E)}\ \text{none of these}$
In the regular hexagon $ABCDEF$ on the line $AF$, the point $X$ is taken so that the angle $XCD$ is $45^o$. Find the angle $\angle FXE$.
(Kiev Olympiad)
Does there exist a polyhedron (not necessarily convex) which could have the following complete list of edges?
$AB, AC, BC, BD, CD, DE, EF, EG, FG, FH, GH, AH$.
[img]http://1.bp.blogspot.com/-wTdNfQHG5RU/XVk1Bf4wpqI/AAAAAAAAKhA/8kc6u9KqOgg_p1CXim2LZ1ANFXFiWgnYACK4BGAYYCw/s1600/TOT%2B1982%2BAutum%2BS2.png[/img]
Let $ABCD$ be a convex quadrilateral. Extend line $CD$ past $D$ to meet line $AB$ at $P$ and extend line $CB$ past $B$ to meet line $AD$ at $Q$. Suppose that line $AC$ bisects $\angle BAD$. If $AD = \frac{7}{4}$, $AP = \frac{21}{2}$, and $AB = \frac{14}{11}$ , compute $AQ$.
Let $f(x)=x^5-3x^4+2x^3+6x^2+x-14=a(x-1)^5+b(x-1)^4+c(x-1)^3+d(x-1)^2+e(x-1)+f,$ for some real constants $a,b,c,d,e,f.$ Determine the value of $ab+bc+cd+de+ad+be.$
Anton and Britta play a game with the set $M=\left \{ 1,2,\dots,n-1 \right \}$ where $n \geq 5$ is an odd integer. In each step Anton removes a number from $M$ and puts it in his set $A$, and Britta removes a number from $M$ and puts it in her set $B$ (both $A$ and $B$ are empty to begin with). When $M$ is empty, Anton picks two distinct numbers $x_1, x_2$ from $A$ and shows them to Britta. Britta then picks two distinct numbers $y_1, y_2$ from $B$. Britta wins if
$(x_1x_2(x_1-y_1)(x_2-y_2))^{\frac{n-1}{2}}\equiv 1\mod n$
otherwise Anton wins. Find all $n$ for which Britta has a winning strategy.
Let $ABC$ be a triangle with $AB<AC$, let $G,H$ be its centroid and otrhocenter. Let $D$ be the otrhogonal projection of $A$ on the line $BC$, and let $M$ be the midpoint of the side $BC$. The circumcircle of $ABC$ crosses the ray $HM$ emanating from $M$ at $P$ and the ray $DG$ emanating from $D$ at $Q$, outside the segment $DG$. Show that the lines $DP$ and $MQ$ meet on the circumcircle of $ABC$.
$a+b+c \leq 3000000$ and $a\neq b \neq c \neq a$ and $a,b,c$ are naturals.
Find maximum $GCD(ab+1,ac+1,bc+1)$
Let $ABCD$ be a convex quadrilateral whose sides $AD$ and $BC$ are not parallel. Suppose that the circles with diameters $AB$ and $CD$ meet at points $E$ and $F$ inside the quadrilateral. Let $\omega_E$ be the circle through the feet of the perpendiculars from $E$ to the lines $AB,BC$ and $CD$. Let $\omega_F$ be the circle through the feet of the perpendiculars from $F$ to the lines $CD,DA$ and $AB$. Prove that the midpoint of the segment $EF$ lies on the line through the two intersections of $\omega_E$ and $\omega_F$.
[i]Proposed by Carlos Yuzo Shine, Brazil[/i]
The prime numbers $p$ and $q$ and the integer $a$ are chosen such that $p> 2$ and $a \not\equiv 1$ (mod $q$), but $a^p \equiv 1$ (mod $q$). Prove that $(1 + a^1)(1 + a^2)...(1 + a^{p - 1})\equiv 1$ (mod $q$) .
Jeffrey rolls fair three six-sided dice and records their results. The probability that the mean of these three numbers is greater than the median of these three numbers can be expressed as $\frac{m}{n}$ for relatively prime positive integers $m$ and $n$. Compute $m+n$.
[i]Proposed by Nathan Xiong[/i]
A square piece of paper is folded twice into four equal quarters, as shown below, then cut along the dashed line. When unfolded, the paper will match which of the following figures?
[asy]
//kante314
size(11cm);
filldraw((0,0)--(29,0)--(29,29)--(0,29)--cycle,mediumgray);
draw((36,29/2)--(54,29/2),EndArrow(size=7));
draw((36,29/2)--(52.5,29/2),linewidth(1.5));
filldraw((61,22)--(63,22)--(63,6)--cycle,mediumgray);
fill((63,6+1*17/16)--(80,6+1*17/16)--(80,6+2*17/16)--(63,6+2*17/16)--cycle,lightgray);
fill((63,6+3*17/16)--(80,6+3*17/16)--(80,6+4*17/16)--(63,6+4*17/16)--cycle,lightgray);
fill((63,6+5*17/16)--(80,6+5*17/16)--(80,6+6*17/16)--(63,6+6*17/16)--cycle,lightgray);
fill((63,6+7*17/16)--(80,6+7*17/16)--(80,6+8*17/16)--(63,6+8*17/16)--cycle,lightgray);
fill((63,6+9*17/16)--(80,6+9*17/16)--(80,6+10*17/16)--(63,6+10*17/16)--cycle,lightgray);
fill((63,6+11*17/16)--(80,6+11*17/16)--(80,6+12*17/16)--(63,6+12*17/16)--cycle,lightgray);
fill((63,6+13*17/16)--(80,6+13*17/16)--(80,6+14*17/16)--(63,6+14*17/16)--cycle,lightgray);
fill((63,6+15*17/16)--(80,6+15*17/16)--(80,6+16*17/16)--(63,6+16*17/16)--cycle,lightgray);
draw((63,6)--(63,23)--(68,23)--(69,12)--(80,6)--cycle);
filldraw((69,12)--(69,27)--(67,28)--cycle,mediumgray);
filldraw((69,12)--(69,29)--(80,23)--(80,6)--cycle,white);
fill((69,12+1*15/13)--(80,6+1*15/13)--(80,6+2*15/13)--(69,12+2*15/13)--cycle,lightgray);
fill((69,12+3*15/13)--(80,6+3*15/13)--(80,6+4*15/13)--(69,12+4*15/13)--cycle,lightgray);
fill((69,12+5*15/13)--(80,6+5*15/13)--(80,6+6*15/13)--(69,12+6*15/13)--cycle,lightgray);
fill((69,12+7*15/13)--(80,6+7*15/13)--(80,6+8*15/13)--(69,12+8*15/13)--cycle,lightgray);
fill((69,12+9*15/13)--(80,6+9*15/13)--(80,6+10*15/13)--(69,12+10*15/13)--cycle,lightgray);
fill((69,12+11*15/13)--(80,6+11*15/13)--(80,6+12*15/13)--(69,12+12*15/13)--cycle,lightgray);
fill((69,12+13*15/13)--(80,6+13*15/13)--(80,6+14*15/13)--(69,12+14*15/13)--cycle,lightgray);
draw((69,12)--(69,29)--(80,23)--(80,6)--cycle);
draw((87,29/2)--(105,29/2),EndArrow(size=7));
draw((87,29/2)--(102.5,29/2),linewidth(1.5));
fill((112,6+1*17/16)--(129,6+1*17/16)--(129,6+2*17/16)--(112,6+2*17/16)--cycle,lightgray);
fill((112,6+3*17/16)--(129,6+3*17/16)--(129,6+4*17/16)--(112,6+4*17/16)--cycle,lightgray);
fill((112,6+5*17/16)--(129,6+5*17/16)--(129,6+6*17/16)--(112,6+6*17/16)--cycle,lightgray);
fill((112,6+7*17/16)--(129,6+7*17/16)--(129,6+8*17/16)--(112,6+8*17/16)--cycle,lightgray);
fill((112,6+9*17/16)--(129,6+9*17/16)--(129,6+10*17/16)--(112,6+10*17/16)--cycle,lightgray);
fill((112,6+11*17/16)--(129,6+11*17/16)--(129,6+12*17/16)--(112,6+12*17/16)--cycle,lightgray);
fill((112,6+13*17/16)--(129,6+13*17/16)--(129,6+14*17/16)--(112,6+14*17/16)--cycle,lightgray);
fill((112,6+15*17/16)--(129,6+15*17/16)--(129,6+16*17/16)--(112,6+16*17/16)--cycle,lightgray);
draw((112,6)--(129,6)--(129,23)--(112,23)--cycle);
draw((112+17/2,6)--(129,6+17/2),dashed+linewidth(.3));
draw((111.7,6.7)--(111.7,23.3)--(128.3,23.3),linewidth(1));
draw((111.75,6.6)--(111.75,6.3));
draw((128.4,23.25)--(128.7,23.25));
[/asy]
[asy]
//kante314
size(11cm);
label(scale(.85)*"\textbf{(A)}", (2,55));
filldraw((7,31)--(13,31)--(19.5,37)--(26,31)--(32,31)--(32,37)--(26,43.5)--(32,50)--(32,56)--(26,56)--(19.5,50)--(13,56)--(7,56)--(7,50)--(13,43.5)--(7,37)--cycle,mediumgray);
label(scale(.85)*"\textbf{(B)}", (44,55));
filldraw((49,31)--(55,31)--(61.5,37)--(68,31)--(74,31)--(74,37)--(74,50)--(74,56)--(68,56)--(61.5,50)--(55,56)--(49,56)--(49,50)--(49,37)--cycle,mediumgray);
label(scale(.85)*"\textbf{(C)}", (86,55));
filldraw((91,31)--(116,31)--(116,56)--(91,56)--cycle,mediumgray);
filldraw((91+25/4,31+25/4)--(116-25/4,31+25/4)--(116-25/4,56-25/4)--(91+25/4,56-25/4)--cycle,white);
label(scale(.85)*"\textbf{(D)}", (2,24));
filldraw((7,0)--(32,0)--(32,25)--(7,25)--cycle,mediumgray);
filldraw((7+25/4,25/2)--(32-25/4,25/2)--(7+25/2,25-25/4)--cycle,white);
label(scale(.85)*"\textbf{(E)}", (44,24));
filldraw((49,0)--(74,0)--(74,25)--(49,25)--cycle,mediumgray);
filldraw((49+25/4,25/2)--(49+25/2,25/4)--(74-25/4,25/2)--(49+25/2,25-25/4)--cycle,white);
[/asy]
A house and store were sold for $ \$12000$ each. The house was sold at a loss of $ 20\%$ of the cost, and the store at a gain of $ 20\%$ of the cost. The entire transaction resulted in:
$ \textbf{(A)}\ \text{no loss or gain} \qquad\textbf{(B)}\ \text{loss of } \$1000 \qquad\textbf{(C)}\ \text{gain of } \$1000 \qquad\textbf{(D)}\ \text{gain of }\$2000 \qquad\textbf{(E)}\ \text{none of these}$
Esmeralda chooses two distinct positive integers \(a\) and \(b\), with \(b > a\), and writes the equation
\[
x^2 - ax + b = 0
\]
on the board. If the equation has distinct positive integer roots \(c\) and \(d\), with \(d > c\), she writes the equation
\[
x^2 - cx + d = 0
\]
on the board. She repeats the procedure as long as she obtains distinct positive integer roots. If she writes an equation for which this does not occur, she stops.
a) Show that Esmeralda can choose \(a\) and \(b\) such that she will write exactly 2024 equations on the board.
b) What is the maximum number of equations she can write knowing that one of the initially chosen numbers is 2024?
Let $n$ be a positive integer. Starting with the sequence $1,\frac{1}{2}, \frac{1}{3} , \cdots , \frac{1}{n}$, form a new sequence of $n -1$ entries $\frac{3}{4}, \frac{5}{12},\cdots ,\frac{2n -1}{2n(n -1)}$, by taking the averages of two consecutive entries in the first sequence. Repeat the averaging of neighbors on the second sequence to obtain a third sequence of $n -2$ entries and continue until the final sequence consists of a single number $x_n$. Show that $x_n < \frac{2}{n}$.
Find the number of pairs of integers $x, y$ with different parities such that $\frac{1}{x}+\frac{1}{y} = \frac{1}{2520}$.