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Difference between revisions of "2015 AIME I Problems"

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{{AIME Problems|year=2015|n=I}}
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==Problem 1==
 
==Problem 1==
 
The expressions <math>A</math> = <math> 1 \times 2 + 3 \times 4 + 5 \times 6 + \cdots + 37 \times 38 + 39 </math> and <math>B</math> = <math> 1 + 2 \times 3 + 4 \times 5 + \cdots + 36 \times 37 + 38 \times 39 </math> are obtained by writing multiplication and addition operators in an alternating pattern between successive integers.  Find the positive difference between integers <math>A</math> and <math>B</math>.
 
The expressions <math>A</math> = <math> 1 \times 2 + 3 \times 4 + 5 \times 6 + \cdots + 37 \times 38 + 39 </math> and <math>B</math> = <math> 1 + 2 \times 3 + 4 \times 5 + \cdots + 36 \times 37 + 38 \times 39 </math> are obtained by writing multiplication and addition operators in an alternating pattern between successive integers.  Find the positive difference between integers <math>A</math> and <math>B</math>.
  
 
[[2015 AIME I Problems/Problem 1|Solution]]
 
[[2015 AIME I Problems/Problem 1|Solution]]
 +
 +
  
 
==Problem 2==
 
==Problem 2==
−
The nine delegates to the Economic Cooperation Conference include <math>2</math> officials from Mexico, <math>3</math> officials from Canada, and <math>4</math> officials from the United States. During the opening session, three of the delegates fall asleep. Assuming that the three sleepers were determined randomly, the probability that exactly two of the sleepers are from the same country is <math>\frac{m}{n}</math>, where m and n are relatively prime positive integers. Find <math>m+n</math>.
+
The nine delegates to the Economic Cooperation Conference include <math>2</math> officials from Mexico, <math>3</math> officials from Canada, and <math>4</math> officials from the United States. During the opening session, three of the delegates fall asleep. Assuming that the three sleepers were determined randomly, the probability that exactly two of the sleepers are from the same country is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
 
[[2015 AIME I Problems/Problem 2|Solution]]
 
[[2015 AIME I Problems/Problem 2|Solution]]
 +
 +
  
 
==Problem 3==
 
==Problem 3==
Line 13: Line 19:
  
 
[[2015 AIME I Problems/Problem 3|Solution]]
 
[[2015 AIME I Problems/Problem 3|Solution]]
 +
 +
  
 
==Problem 4==
 
==Problem 4==
 +
Point <math>B</math> lies on line segment <math>\overline{AC}</math> with <math>AB=16</math> and <math>BC=4</math>. Points <math>D</math> and <math>E</math> lie on the same side of line <math>AC</math> forming equilateral triangles <math>\triangle ABD</math> and <math>\triangle BCE</math>. Let <math>M</math> be the midpoint of <math>\overline{AE}</math>, and <math>N</math> be the midpoint of <math>\overline{CD}</math>. The area of <math>\triangle BMN</math> is <math>x</math>. Find <math>x^2</math>.
 +
 +
[[2015 AIME I Problems/Problem 4|Solution]]
 +
 +
  
 
==Problem 5==
 
==Problem 5==
−
In a drawer Sandy has <math>5</math> pairs of socks, each pair a different color.  On Monday Sandy selects two individual socks at random from the <math>10</math> socks in the drawer.  On Tuesday Sandy selects <math>2</math> of the remaining <math>8</math> socks at random and on Wednesday two of the remaining <math>6</math> socks at random.  The probability that Wednesday is the first day Sandy selects matching socks is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers,  Find <math>m+n</math>.
+
In a drawer Sandy has <math>5</math> pairs of socks, each pair a different color.  On Monday, Sandy selects two individual socks at random from the <math>10</math> socks in the drawer.  On Tuesday Sandy selects <math>2</math> of the remaining <math>8</math> socks at random, and on Wednesday two of the remaining <math>6</math> socks at random.  The probability that Wednesday is the first day Sandy selects matching socks is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
 
[[2015 AIME I Problems/Problem 5|Solution]]
 
[[2015 AIME I Problems/Problem 5|Solution]]
 +
 +
  
 
==Problem 6==
 
==Problem 6==
 +
Point <math>A,B,C,D,</math> and <math>E</math> are equally spaced on a minor arc of a circle. Points <math>E,F,G,H,I</math> and <math>A</math> are equally spaced on a minor arc of a second circle with center <math>C</math> as shown in the figure below. The angle <math>\angle ABD</math> exceeds <math>\angle AHG</math> by <math>12^\circ</math>. Find the degree measure of <math>\angle BAG</math>.
 +
 +
<asy>
 +
pair A,B,C,D,E,F,G,H,I,O;
 +
O=(0,0);
 +
C=dir(90);
 +
B=dir(70);
 +
A=dir(50);
 +
D=dir(110);
 +
E=dir(130);
 +
draw(arc(O,1,50,130));
 +
real x=2*sin(20*pi/180);
 +
F=x*dir(228)+C;
 +
G=x*dir(256)+C;
 +
H=x*dir(284)+C;
 +
I=x*dir(312)+C;
 +
draw(arc(C,x,200,340));
 +
label("$A$",A,dir(0));
 +
label("$B$",B,dir(75));
 +
label("$C$",C,dir(90));
 +
label("$D$",D,dir(105));
 +
label("$E$",E,dir(180));
 +
label("$F$",F,dir(225));
 +
label("$G$",G,dir(260));
 +
label("$H$",H,dir(280));
 +
label("$I$",I,dir(315));</asy>
 +
 +
[[2015 AIME I Problems/Problem 6|Solution]]
 +
 +
  
 
==Problem 7==
 
==Problem 7==
 
In the diagram below, <math>ABCD</math> is a square. Point <math>E</math> is the midpoint of <math>\overline{AD}</math>. Points <math>F</math> and <math>G</math> lie on <math>\overline{CE}</math>, and <math>H</math> and <math>J</math> lie on <math>\overline{AB}</math> and <math>\overline{BC}</math>, respectively, so that <math>FGHJ</math> is a square. Points <math>K</math> and <math>L</math> lie on <math>\overline{GH}</math>, and <math>M</math> and <math>N</math> lie on <math>\overline{AD}</math> and <math>\overline{AB}</math>, respectively, so that <math>KLMN</math> is a square. The area of <math>KLMN</math> is 99. Find the area of <math>FGHJ</math>.
 
In the diagram below, <math>ABCD</math> is a square. Point <math>E</math> is the midpoint of <math>\overline{AD}</math>. Points <math>F</math> and <math>G</math> lie on <math>\overline{CE}</math>, and <math>H</math> and <math>J</math> lie on <math>\overline{AB}</math> and <math>\overline{BC}</math>, respectively, so that <math>FGHJ</math> is a square. Points <math>K</math> and <math>L</math> lie on <math>\overline{GH}</math>, and <math>M</math> and <math>N</math> lie on <math>\overline{AD}</math> and <math>\overline{AB}</math>, respectively, so that <math>KLMN</math> is a square. The area of <math>KLMN</math> is 99. Find the area of <math>FGHJ</math>.
  
−
INSERT DIAGRAM HERE
+
<asy>
 +
pair A,B,C,D,E,F,G,H,J,K,L,M,N;
 +
B=(0,0);
 +
real m=7*sqrt(55)/5;
 +
J=(m,0);
 +
C=(7*m/2,0);
 +
A=(0,7*m/2);
 +
D=(7*m/2,7*m/2);
 +
E=(A+D)/2;
 +
H=(0,2m);
 +
N=(0,2m+3*sqrt(55)/2);
 +
G=foot(H,E,C);
 +
F=foot(J,E,C);
 +
draw(A--B--C--D--cycle);
 +
draw(C--E);
 +
draw(G--H--J--F);
 +
pair X=foot(N,E,C);
 +
M=extension(N,X,A,D);
 +
K=foot(N,H,G);
 +
L=foot(M,H,G);
 +
draw(K--N--M--L);
 +
label("$A$",A,NW);
 +
label("$B$",B,SW);
 +
label("$C$",C,SE);
 +
label("$D$",D,NE);
 +
label("$E$",E,dir(90));
 +
label("$F$",F,NE);
 +
label("$G$",G,NE);
 +
label("$H$",H,W);
 +
label("$J$",J,S);
 +
label("$K$",K,SE);
 +
label("$L$",L,SE);
 +
label("$M$",M,dir(90));
 +
label("$N$",N,dir(180)); </asy>
 +
 
 +
[[2015 AIME I Problems/Problem 7|Solution]]
 +
 
 +
 
  
 
==Problem 8==
 
==Problem 8==
 
For positive integer <math>n</math>, let <math>s(n)</math> denote the sum of the digits of <math>n</math>.  Find the smallest positive integer satisfying <math>s(n) = s(n+864) = 20</math>.
 
For positive integer <math>n</math>, let <math>s(n)</math> denote the sum of the digits of <math>n</math>.  Find the smallest positive integer satisfying <math>s(n) = s(n+864) = 20</math>.
 +
 +
[[2015 AIME I Problems/Problem 8|Solution]]
 +
 +
  
 
==Problem 9==
 
==Problem 9==
 +
Let <math>S</math> be the set of all ordered triple of integers <math>(a_1,a_2,a_3)</math> with <math>1 \le a_1,a_2,a_3 \le 10</math>. Each ordered triple in <math>S</math> generates a sequence according to the rule <math>a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |</math> for all <math>n\ge 4</math>. Find the number of such sequences for which <math>a_n=0</math> for some <math>n</math>.
 +
 +
[[2015 AIME I Problems/Problem 9|Solution]]
  
 
==Problem 10==
 
==Problem 10==
 +
 +
Let <math>f(x)</math> be a third-degree polynomial with real coefficients satisfying
 +
<cmath>|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.</cmath> Find <math>|f(0)|</math>.
 +
 +
[[2015 AIME I Problems/Problem 10|Solution]]
 +
 +
  
 
==Problem 11==
 
==Problem 11==
 +
Triangle <math>ABC</math> has positive integer side lengths with <math>AB=AC</math>. Let <math>I</math> be the intersection of the bisectors of <math>\angle B</math> and <math>\angle C</math>. Suppose <math>BI=8</math>. Find the smallest possible perimeter of <math>\triangle ABC</math>.
 +
 +
[[2015 AIME I Problems/Problem 11|Solution]]
 +
 +
  
 
==Problem 12==
 
==Problem 12==
−
Consider all 1000-element subsets of the set {1, 2, 3, ... , 2015}.  From each such subset choose the least element.  The arithmetic mean of all of these least elements is <math> \frac{p}{q} </math>, where <math>p</math> and <math>q</math> are relatively prime positive integers.  Find <math>p + q</math>.
+
Consider all 1000-element subsets of the set <math> \{ 1, 2, 3, ... , 2015 \} </math>.  From each such subset choose the least element.  The arithmetic mean of all of these least elements is <math> \frac{p}{q} </math>, where <math>p</math> and <math>q</math> are relatively prime positive integers.  Find <math>p + q</math>.
 +
 
 +
[[2015 AIME I Problems/Problem 12|Solution]]
 +
 
 +
 
  
 
==Problem 13==
 
==Problem 13==
 +
With all angles measured in degrees, the product <math>\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n</math>, where <math>m</math> and <math>n</math> are integers greater than 1. Find <math>m+n</math>.
 +
 +
[[2015 AIME I Problems/Problem 13|Solution]]
 +
 +
  
 
==Problem 14==
 
==Problem 14==
 +
 +
For each integer <math>n \ge 2</math>, let <math>A(n)</math> be the area of the region in the coordinate plane defined by the inequalities <math>1\le x \le n</math> and <math>0\le y \le x \left\lfloor \sqrt x \right\rfloor</math>, where <math>\left\lfloor \sqrt x \right\rfloor</math> is the greatest integer not exceeding <math>\sqrt x</math>. Find the number of values of <math>n</math> with <math>2\le n \le 1000</math> for which <math>A(n)</math> is an integer.
 +
 +
[[2015 AIME I Problems/Problem 14|Solution]]
 +
 +
  
 
==Problem 15==
 
==Problem 15==
 
A block of wood has the shape of a right circular cylinder with radius <math>6</math> and height <math>8</math>, and its entire surface has been painted blue.  Points <math>A</math> and <math>B</math> are chosen on the edge of one of the circular faces of the cylinder so that <math>\overarc{AB}</math> on that face measures <math>120^\text{o}</math>.  The block is then sliced in half along the plane that passes through point <math>A</math>, point <math>B</math>, and the center of the cylinder, revealing a flat, unpainted face on each half.  The area of one of these unpainted faces is <math>a\cdot\pi + b\sqrt{c}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are integers and <math>c</math> is not divisible by the square of any prime.  Find <math>a+b+c</math>.
 
A block of wood has the shape of a right circular cylinder with radius <math>6</math> and height <math>8</math>, and its entire surface has been painted blue.  Points <math>A</math> and <math>B</math> are chosen on the edge of one of the circular faces of the cylinder so that <math>\overarc{AB}</math> on that face measures <math>120^\text{o}</math>.  The block is then sliced in half along the plane that passes through point <math>A</math>, point <math>B</math>, and the center of the cylinder, revealing a flat, unpainted face on each half.  The area of one of these unpainted faces is <math>a\cdot\pi + b\sqrt{c}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are integers and <math>c</math> is not divisible by the square of any prime.  Find <math>a+b+c</math>.
 +
 +
<asy>
 +
import three; import solids;
 +
size(8cm);
 +
currentprojection=orthographic(-1,-5,3);
 +
 +
picture lpic, rpic;
 +
 +
 +
size(lpic,5cm);
 +
draw(lpic,surface(revolution((0,0,0),(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8),Z,0,120)),gray(0.7),nolight);
 +
draw(lpic,surface(revolution((0,0,0),(-3*sqrt(3),-3,8)..(-6,0,4)..(-3*sqrt(3),3,0),Z,0,90)),gray(0.7),nolight);
 +
draw(lpic,surface((3,3*sqrt(3),8)..(-6,0,8)..(3,-3*sqrt(3),8)--cycle),gray(0.7),nolight);
 +
draw(lpic,(3,-3*sqrt(3),8)..(-6,0,8)..(3,3*sqrt(3),8));
 +
draw(lpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0),dashed);
 +
draw(lpic,(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0)--(-3,3*sqrt(3),0)..(-3*sqrt(3),3,0)..(-6,0,0),dashed);
 +
draw(lpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(-3*sqrt(3),-3,0)..(-6,0,0));
 +
draw(lpic,(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),0)--(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),8));
 +
 +
size(rpic,5cm);
 +
draw(rpic,surface(revolution((0,0,0),(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0),Z,230,360)),gray(0.7),nolight);
 +
draw(rpic,surface((-3,3*sqrt(3),0)..(6,0,0)..(-3,-3*sqrt(3),0)--cycle),gray(0.7),nolight);
 +
draw(rpic,surface((-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--cycle),white,nolight);
 +
draw(rpic,(-3,-3*sqrt(3),0)..(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0));
 +
draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)..(-3,3*sqrt(3),0),dashed);
 +
draw(rpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8));
 +
draw(rpic,(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)..(3*sqrt(3),3,8)..(6,0,8));
 +
draw(rpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(0,-6,4)..(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(3*sqrt(3),-3,8)..(6,0,8));
 +
draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)--(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),8));
 +
label(rpic,"$A$",(-3,3*sqrt(3),0),W);
 +
label(rpic,"$B$",(-3,-3*sqrt(3),0),W);
 +
 +
add(lpic.fit(),(0,0));
 +
add(rpic.fit(),(1,0)); </asy>
 +
 +
[[2015 AIME I Problems/Problem 15|Solution]]
 +
 +
 +
{{AIME box|year=2015|n=I|before=[[2014 AIME II Problems]]|after=[[2015 AIME II Problems]]}}
 +
{{MAA Notice}}

Latest revision as of 16:03, 9 October 2024

2015 AIME I (Answer Key)
Printable version | AoPS Contest Collections • PDF

Instructions

  1. This is a 15-question, 3-hour examination. All answers are integers ranging from $000$ to $999$, inclusive. Your score will be the number of correct answers; i.e., there is neither partial credit nor a penalty for wrong answers.
  2. No aids other than scratch paper, graph paper, ruler, compass, and protractor are permitted. In particular, calculators and computers are not permitted.
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Problem 1

The expressions $A$ = $1 \times 2 + 3 \times 4 + 5 \times 6 + \cdots + 37 \times 38 + 39$ and $B$ = $1 + 2 \times 3 + 4 \times 5 + \cdots + 36 \times 37 + 38 \times 39$ are obtained by writing multiplication and addition operators in an alternating pattern between successive integers. Find the positive difference between integers $A$ and $B$.

Solution


Problem 2

The nine delegates to the Economic Cooperation Conference include $2$ officials from Mexico, $3$ officials from Canada, and $4$ officials from the United States. During the opening session, three of the delegates fall asleep. Assuming that the three sleepers were determined randomly, the probability that exactly two of the sleepers are from the same country is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution


Problem 3

There is a prime number $p$ such that $16p+1$ is the cube of a positive integer. Find $p$.

Solution


Problem 4

Point $B$ lies on line segment $\overline{AC}$ with $AB=16$ and $BC=4$. Points $D$ and $E$ lie on the same side of line $AC$ forming equilateral triangles $\triangle ABD$ and $\triangle BCE$. Let $M$ be the midpoint of $\overline{AE}$, and $N$ be the midpoint of $\overline{CD}$. The area of $\triangle BMN$ is $x$. Find $x^2$.

Solution


Problem 5

In a drawer Sandy has $5$ pairs of socks, each pair a different color. On Monday, Sandy selects two individual socks at random from the $10$ socks in the drawer. On Tuesday Sandy selects $2$ of the remaining $8$ socks at random, and on Wednesday two of the remaining $6$ socks at random. The probability that Wednesday is the first day Sandy selects matching socks is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution


Problem 6

Point $A,B,C,D,$ and $E$ are equally spaced on a minor arc of a circle. Points $E,F,G,H,I$ and $A$ are equally spaced on a minor arc of a second circle with center $C$ as shown in the figure below. The angle $\angle ABD$ exceeds $\angle AHG$ by $12^\circ$. Find the degree measure of $\angle BAG$.

[asy] pair A,B,C,D,E,F,G,H,I,O; O=(0,0); C=dir(90); B=dir(70); A=dir(50); D=dir(110); E=dir(130); draw(arc(O,1,50,130)); real x=2*sin(20*pi/180); F=x*dir(228)+C; G=x*dir(256)+C; H=x*dir(284)+C; I=x*dir(312)+C; draw(arc(C,x,200,340)); label("$A$",A,dir(0)); label("$B$",B,dir(75)); label("$C$",C,dir(90)); label("$D$",D,dir(105)); label("$E$",E,dir(180)); label("$F$",F,dir(225)); label("$G$",G,dir(260)); label("$H$",H,dir(280)); label("$I$",I,dir(315));[/asy]

Solution


Problem 7

In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\overline{AD}$. Points $F$ and $G$ lie on $\overline{CE}$, and $H$ and $J$ lie on $\overline{AB}$ and $\overline{BC}$, respectively, so that $FGHJ$ is a square. Points $K$ and $L$ lie on $\overline{GH}$, and $M$ and $N$ lie on $\overline{AD}$ and $\overline{AB}$, respectively, so that $KLMN$ is a square. The area of $KLMN$ is 99. Find the area of $FGHJ$.

[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,dir(90)); label("$F$",F,NE); label("$G$",G,NE); label("$H$",H,W); label("$J$",J,S); label("$K$",K,SE); label("$L$",L,SE); label("$M$",M,dir(90)); label("$N$",N,dir(180)); [/asy]

Solution


Problem 8

For positive integer $n$, let $s(n)$ denote the sum of the digits of $n$. Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$.

Solution


Problem 9

Let $S$ be the set of all ordered triple of integers $(a_1,a_2,a_3)$ with $1 \le a_1,a_2,a_3 \le 10$. Each ordered triple in $S$ generates a sequence according to the rule $a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |$ for all $n\ge 4$. Find the number of such sequences for which $a_n=0$ for some $n$.

Solution

Problem 10

Let $f(x)$ be a third-degree polynomial with real coefficients satisfying \[|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.\] Find $|f(0)|$.

Solution


Problem 11

Triangle $ABC$ has positive integer side lengths with $AB=AC$. Let $I$ be the intersection of the bisectors of $\angle B$ and $\angle C$. Suppose $BI=8$. Find the smallest possible perimeter of $\triangle ABC$.

Solution


Problem 12

Consider all 1000-element subsets of the set $\{ 1, 2, 3, ... , 2015 \}$. From each such subset choose the least element. The arithmetic mean of all of these least elements is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$.

Solution


Problem 13

With all angles measured in degrees, the product $\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n$, where $m$ and $n$ are integers greater than 1. Find $m+n$.

Solution


Problem 14

For each integer $n \ge 2$, let $A(n)$ be the area of the region in the coordinate plane defined by the inequalities $1\le x \le n$ and $0\le y \le x \left\lfloor \sqrt x \right\rfloor$, where $\left\lfloor \sqrt x \right\rfloor$ is the greatest integer not exceeding $\sqrt x$. Find the number of values of $n$ with $2\le n \le 1000$ for which $A(n)$ is an integer.

Solution


Problem 15

A block of wood has the shape of a right circular cylinder with radius $6$ and height $8$, and its entire surface has been painted blue. Points $A$ and $B$ are chosen on the edge of one of the circular faces of the cylinder so that $\overarc{AB}$ on that face measures $120^\text{o}$. The block is then sliced in half along the plane that passes through point $A$, point $B$, and the center of the cylinder, revealing a flat, unpainted face on each half. The area of one of these unpainted faces is $a\cdot\pi + b\sqrt{c}$, where $a$, $b$, and $c$ are integers and $c$ is not divisible by the square of any prime. Find $a+b+c$.

[asy] import three; import solids; size(8cm); currentprojection=orthographic(-1,-5,3); picture lpic, rpic; size(lpic,5cm); draw(lpic,surface(revolution((0,0,0),(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8),Z,0,120)),gray(0.7),nolight); draw(lpic,surface(revolution((0,0,0),(-3*sqrt(3),-3,8)..(-6,0,4)..(-3*sqrt(3),3,0),Z,0,90)),gray(0.7),nolight); draw(lpic,surface((3,3*sqrt(3),8)..(-6,0,8)..(3,-3*sqrt(3),8)--cycle),gray(0.7),nolight); draw(lpic,(3,-3*sqrt(3),8)..(-6,0,8)..(3,3*sqrt(3),8)); draw(lpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0),dashed); draw(lpic,(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0)--(-3,3*sqrt(3),0)..(-3*sqrt(3),3,0)..(-6,0,0),dashed); draw(lpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(-3*sqrt(3),-3,0)..(-6,0,0)); draw(lpic,(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),0)--(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),8)); size(rpic,5cm); draw(rpic,surface(revolution((0,0,0),(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0),Z,230,360)),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(6,0,0)..(-3,-3*sqrt(3),0)--cycle),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--cycle),white,nolight); draw(rpic,(-3,-3*sqrt(3),0)..(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)..(-3,3*sqrt(3),0),dashed); draw(rpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)); draw(rpic,(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)..(3*sqrt(3),3,8)..(6,0,8)); draw(rpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(0,-6,4)..(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(3*sqrt(3),-3,8)..(6,0,8)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)--(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),8)); label(rpic,"$A$",(-3,3*sqrt(3),0),W); label(rpic,"$B$",(-3,-3*sqrt(3),0),W); add(lpic.fit(),(0,0)); add(rpic.fit(),(1,0)); [/asy]

Solution


2015 AIME I (Problems • Answer Key • Resources)
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2014 AIME II Problems 		Followed by
2015 AIME II Problems
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