Difference between revisions of "2010 AIME I Problems/Problem 11"
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== 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 15:44, 17 March 2010
Problem
Let be the region consisting of the set of points in the coordinate plane that satisfy both and . When is revolved around the line whose equation is , the volume of the resulting solid is , where , , and are positive integers, and are relatively prime, and is not divisible by the square of any prime. Find .
Solution
The inequalities are equivalent to . We can set them equal to find the two points of intersection, . This implies that one of , from which we find that . The region is a triangle, as shown above. When revolved about the line , the resulting solid is the union of two right cones that share the same base and axis.
Let denote the height of the left and right cones, respectively (so ), and let denote their common radius. The volume of a cone is given by ; since both cones share the same base, then the desired volume is . The distance from the point to the line is given by . The distance between and is given by . Thus, the answer is .
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 |