Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2010 AIME I Problems/Problem 11"

m (will writeup later)
 
(by myself)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
Let <math>\mathcal{R}</math> be the region consisting of the set of points in the coordinate plane that satisfy both <math>|8 - x| + y \le 10</math> and <math>3y - x \ge 15</math>. When <math>\mathcal{R}</math> is revolved around the line whose equation is <math>3y - x = 15</math>, the volume of the resulting solid is <math>\frac {m\pi}{n\sqrt {p}}</math>, where <math>m</math>, <math>n</math>, and <math>p</math> are positive integers, <math>m</math> and <math>n</math> are relatively prime, and <math>p</math> is not divisible by the square of any prime. Find <math>m + n + p</math>.
+
Let <math>\mathcal{R}</math> be the region consisting of the set of points in the [[coordinate plane]] that satisfy both <math>|8 - x| + y \le 10</math> and <math>3y - x \ge 15</math>. When <math>\mathcal{R}</math> is revolved around the [[line]] whose equation is <math>3y - x = 15</math>, the [[volume]] of the resulting [[solid]] is <math>\frac {m\pi}{n\sqrt {p}}</math>, where <math>m</math>, <math>n</math>, and <math>p</math> are positive integers, <math>m</math> and <math>n</math> are [[relatively prime]], and <math>p</math> is not divisible by the square of any prime. Find <math>m + n + p</math>.
  
 
== Solution ==
 
== Solution ==
 +
<center><asy>size(280);
 +
import graph; real min = 2, max = 12; pen dark = linewidth(1);
 +
 +
real P(real x) { return x/3 + 5; }
 +
real Q(real x) { return 10 - abs(x - 8); }
 +
path p = (2,P(2))--(8,P(8))--(12,P(12)), q = (2,Q(2))--(12,Q(12));
 +
pair A = (8,10), B = (4.5,6.5), C= (9.75,8.25), F=foot(A,B,C), G=2*F-A; fill(A--B--C--cycle,rgb(0.9,0.9,0.9));
 +
draw(graph(P,min,max),dark);
 +
draw(graph(Q,min,max),dark);
 +
draw(Arc((8,7.67),A,G,CW),dark,EndArrow(8)); draw(B--C--G--cycle,linetype("4 4"));
 +
 +
label("$y \ge x/3 + 5$",(max,P(max)),E,fontsize(10));
 +
label("$y \le 10 - |x-8|$",(max,Q(max)),E,fontsize(10));
 +
label("$\mathcal{R}$",(6,Q(6)),NW);
 +
 +
/* axes */
 +
Label f; f.p=fontsize(8);
 +
xaxis(0, max, Ticks(f, 6, 1));
 +
yaxis(0, 10, Ticks(f, 5, 1));
 +
</asy></center>
 +
 +
The [[inequality|inequalities]] are equivalent to <math>y \ge x/3 + 5, y \le 10 - |x - 8|</math>. We can set them equal to find the two points of intersection, <math>x/3 + 5 = 10 - |x - 8| \Longrightarrow |x - 8| = 5 - x/3</math>. This implies that one of <math>x - 8, 8 - x = 5 - x/3</math>, from which we find that <math>(x,y) = \left(\frac 92, \frac {13}2\right), \left(\frac{39}{4}, \frac{33}{4}\right)</math>. The region <math>\mathcal{R}</math> is a [[triangle]], as shown above. When revolved about the line <math>y = x/3+5</math>, the resulting solid is the union of two right [[cone]]s that share the same base and axis.
 +
 +
<center><asy>size(200);
 +
import three; currentprojection = perspective(0,0,10); defaultpen(linewidth(0.7)); pen dark=linewidth(1.3);
 +
pair Fxy = foot((8,10),(4.5,6.5),(9.75,8.25));
 +
triple A = (8,10,0), B = (4.5,6.5,0), C= (9.75,8.25,0), F=(Fxy.x,Fxy.y,0), G=2*F-A, H=(F.x,F.y,abs(F-A)),I=(F.x,F.y,-abs(F-A));
 +
real theta1 = 1.2, theta2 = -1.7,theta3= abs(F-A),theta4=-2.2;
 +
 +
triple J=F+theta1*unit(A-F)+(0,0,((abs(F-A))^2-(theta1)^2)^.5 ),K=F+theta2*unit(A-F)+(0,0,((abs(F-A))^2-(theta2)^2)^.5 ),L=F+theta3*unit(A-F)+(0,0,((abs(F-A))^2-(theta3)^2)^.5 ),M=F+theta4*unit(A-F)-(0,0,((abs(F-A))^2-(theta4)^2)^.5 );
 +
draw(C--A--B--G--cycle,linetype("4 4")+dark); draw(A..H..G..I..A); draw(C--B^^A--G,linetype("4 4")); draw(J--C--K); draw(L--B--M);
 +
dot(B);dot(C);dot(F);
 +
 +
label("$h_1$",(B+F)/2,SE,fontsize(10));
 +
label("$h_2$",(C+F)/2,S,fontsize(10));
 +
label("$r$",(A+F)/2,E,fontsize(10));
 +
</asy></center>
 +
 +
Let <math>h_1,h_2</math> denote the height of the left and right cones, respectively (so <math>h_1 > h_2</math>), and let <math>r</math> denote their common radius. The volume of a cone is given by <math>\frac 13 Bh</math>; since both cones share the same base, then the desired volume is <math>\frac 13 \cdot \pi r^2 \cdot (h_1 + h_2)</math>. The [[distance]] from the point <math>(8,10)</math> to the line <math>x - 3y + 15 = 0</math> is given by <math>\left|\frac{(8) - 3(10) + 15}{\sqrt{1^2 + (-3)^2}}\right| = \frac{7}{\sqrt{10}}</math>. The distance between <math>\left(\frac 92, \frac {13}2\right)</math> and <math>\left(\frac{39}{4}, \frac{33}{4}\right)</math> is given by <math>h_1 + h_2 = \sqrt{\left(\frac{18}{4} - \frac{39}{4}\right)^2 + \left(\frac{26}{4} - \frac{33}{4}\right)^2} = \frac{7\sqrt{10}}{4}</math>. Thus, the answer is <math>\frac{343\sqrt{10}\pi}{120} = \frac{343\pi}{12\sqrt{10}} \Longrightarrow 343 + 12 + 10 = \boxed{365}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 14:44, 17 March 2010

Problem

Let $\mathcal{R}$ be the region consisting of the set of points in the coordinate plane that satisfy both $|8 - x| + y \le 10$ and $3y - x \ge 15$. When $\mathcal{R}$ is revolved around the line whose equation is $3y - x = 15$, the volume of the resulting solid is $\frac {m\pi}{n\sqrt {p}}$, where $m$, $n$, and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m + n + p$.

Solution

[asy]size(280); import graph; real min = 2, max = 12; pen dark = linewidth(1); real P(real x) { return x/3 + 5; } real Q(real x) { return 10 - abs(x - 8); } path p = (2,P(2))--(8,P(8))--(12,P(12)), q = (2,Q(2))--(12,Q(12)); pair A = (8,10), B = (4.5,6.5), C= (9.75,8.25), F=foot(A,B,C), G=2*F-A; fill(A--B--C--cycle,rgb(0.9,0.9,0.9)); draw(graph(P,min,max),dark); draw(graph(Q,min,max),dark); draw(Arc((8,7.67),A,G,CW),dark,EndArrow(8)); draw(B--C--G--cycle,linetype("4 4")); label("$y \ge x/3 + 5$",(max,P(max)),E,fontsize(10)); label("$y \le 10 - |x-8|$",(max,Q(max)),E,fontsize(10)); label("$\mathcal{R}$",(6,Q(6)),NW); /* axes */ Label f; f.p=fontsize(8); xaxis(0, max, Ticks(f, 6, 1)); yaxis(0, 10, Ticks(f, 5, 1)); [/asy]

The inequalities are equivalent to $y \ge x/3 + 5, y \le 10 - |x - 8|$. We can set them equal to find the two points of intersection, $x/3 + 5 = 10 - |x - 8| \Longrightarrow |x - 8| = 5 - x/3$. This implies that one of $x - 8, 8 - x = 5 - x/3$, from which we find that $(x,y) = \left(\frac 92, \frac {13}2\right), \left(\frac{39}{4}, \frac{33}{4}\right)$. The region $\mathcal{R}$ is a triangle, as shown above. When revolved about the line $y = x/3+5$, the resulting solid is the union of two right cones that share the same base and axis.

[asy]size(200); import three; currentprojection = perspective(0,0,10); defaultpen(linewidth(0.7)); pen dark=linewidth(1.3); pair Fxy = foot((8,10),(4.5,6.5),(9.75,8.25)); triple A = (8,10,0), B = (4.5,6.5,0), C= (9.75,8.25,0), F=(Fxy.x,Fxy.y,0), G=2*F-A, H=(F.x,F.y,abs(F-A)),I=(F.x,F.y,-abs(F-A)); real theta1 = 1.2, theta2 = -1.7,theta3= abs(F-A),theta4=-2.2; triple J=F+theta1*unit(A-F)+(0,0,((abs(F-A))^2-(theta1)^2)^.5 ),K=F+theta2*unit(A-F)+(0,0,((abs(F-A))^2-(theta2)^2)^.5 ),L=F+theta3*unit(A-F)+(0,0,((abs(F-A))^2-(theta3)^2)^.5 ),M=F+theta4*unit(A-F)-(0,0,((abs(F-A))^2-(theta4)^2)^.5 ); draw(C--A--B--G--cycle,linetype("4 4")+dark); draw(A..H..G..I..A); draw(C--B^^A--G,linetype("4 4")); draw(J--C--K); draw(L--B--M); dot(B);dot(C);dot(F); label("$h_1$",(B+F)/2,SE,fontsize(10)); label("$h_2$",(C+F)/2,S,fontsize(10)); label("$r$",(A+F)/2,E,fontsize(10)); [/asy]

Let $h_1,h_2$ denote the height of the left and right cones, respectively (so $h_1 > h_2$), and let $r$ denote their common radius. The volume of a cone is given by $\frac 13 Bh$; since both cones share the same base, then the desired volume is $\frac 13 \cdot \pi r^2 \cdot (h_1 + h_2)$. The distance from the point $(8,10)$ to the line $x - 3y + 15 = 0$ is given by $\left|\frac{(8) - 3(10) + 15}{\sqrt{1^2 + (-3)^2}}\right| = \frac{7}{\sqrt{10}}$. The distance between $\left(\frac 92, \frac {13}2\right)$ and $\left(\frac{39}{4}, \frac{33}{4}\right)$ is given by $h_1 + h_2 = \sqrt{\left(\frac{18}{4} - \frac{39}{4}\right)^2 + \left(\frac{26}{4} - \frac{33}{4}\right)^2} = \frac{7\sqrt{10}}{4}$. Thus, the answer is $\frac{343\sqrt{10}\pi}{120} = \frac{343\pi}{12\sqrt{10}} \Longrightarrow 343 + 12 + 10 = \boxed{365}$.

See also

2010 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems/Problem_11&oldid=33931"