https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Josu&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T06:44:46ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=Cone&diff=74920Cone2016-01-30T21:19:47Z<p>Josu: /* Problems */</p>
<hr />
<div>A '''cone''' (or ''circular cone'') is a three-dimensional [[solid]]. It consists of a [[circle | circular]] base, a [[point]] (called the ''vertex''), and all the points that lie on [[line segment]]s connecting the vertex to the base. Thus, the cone is the special case of the [[pyramid]] in which the base is circular.<br />
<br />
<center><!-- NOTE: two different versions --><asy> size(120);<br />
import three; currentprojection = perspective(0,-3,1); defaultpen(linewidth(0.7)); triple vertex = (0,0,1.5);<br />
<br />
path3 rightanglemark(triple A, triple B, triple C, real s=8) { // olympiad package<br />
triple P,Q,R; <br />
P=s*markscalefactor*unit(A-B)+B; <br />
R=s*markscalefactor*unit(C-B)+B; <br />
Q=P+R-B; <br />
return P--Q--R; <br />
} //documentation in second version<br />
path3 unitc=(1,0,0)..(0,1,0)..(-1,0,0)..(0,-1,0)..cycle;<br />
draw(unitc);dot(vertex);draw((1,0,0)--vertex--(-1,0,0));draw(vertex--(vertex.x,vertex.y,0)--(1,0,0));draw(rightanglemark(vertex,(vertex.x,vertex.y,0),(1,0,0),2)); label("$h$",(vertex.x,vertex.y,vertex.z/2),W);label("$r$",(0.5,0,0),S); label("$s$",((1+vertex.x)/2,(1+vertex.y)/2,vertex.z/2),NE);<br />
</asy>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <!-- VERSION 2 --><br />
<asy> size(120);<br />
import three; currentprojection = perspective(0,-3,1); defaultpen(linewidth(0.7));triple vertex = (0.4,0.6,1.5);<br />
<br />
path3 rightanglemark(triple A, triple B, triple C, real s=8) { // <br />
triple P,Q,R; <br />
P=s*markscalefactor*unit(A-B)+B; <br />
R=s*markscalefactor*unit(C-B)+B; <br />
Q=P+R-B; <br />
return P--Q--R; <br />
} <br />
// draw cone<br />
path3 unitc=(1,0,0)..(0,1,0)..(-1,0,0)..(0,-1,0)..cycle;<br />
draw(unitc);dot(vertex);draw((1,0,0)--vertex--(-1,0,0));draw(vertex--(vertex.x,vertex.y,0)--(1,0,0));draw(rightanglemark(vertex,(vertex.x,vertex.y,0),(1,0,0),2));<br />
<br />
// draw radius<br />
dot((0,0,0)); draw((0,0,0)--(1,0,0)); draw((0,0,0)--vertex,dotted);<br />
<br />
// label<br />
label("$r$",(0.5,0,0),S); label("$h$",(vertex.x,vertex.y,vertex.z/2),E);<br />
</asy></center><br />
<br />
== Terminology ==<br />
The distance from the vertex to the [[plane]] containing the base is the ''height'' of the cone, and is frequently denoted <math>h</math>. The [[radius]] of the base is called the radius of the cone and is frequently denoted <math>r</math>. If the vertex lies directly above the center of the base, we call the cone a '' '''right''' circular cone'' (or ''right cone'' for short). In this case, the vertex is the same distance from every point on the boundary of the base; this distance is called the ''slant height'' of the cone, and is sometimes denoted <math>s</math> or <math>\ell</math>. If a cone is not a right cone (that is, if the vertex is not directly above the center of the base), we call it an ''oblique cone''.<br />
<br />
== Properties ==<br />
* A cone with radius <math>r</math> and height <math>h</math> has [[volume]] <math>V = \frac{1}{3} \cdot \pi r^2 \cdot h</math>. This is a special case of the general formula for the volume of a pyramid, <math>V = \frac{1}{3} \cdot B \cdot h</math>, where <math>V</math> is the volume, <math>B</math> is the area of the base and <math>h</math> is the height.<br />
* A right cone of radius <math>r</math> and slant-height <math>s</math> has [[surface area]] <math>\pi r^2 + \pi r s</math> (the [[lateral area]] is <math>\pi rs</math>, and the area of the base is <math>\pi r^2</math>). <br />
<br />
<center><!-- NOTE: two different versions --><asy> size(120);<br />
import three; currentprojection = perspective(0,-3,1); defaultpen(linewidth(0.7)); triple vertex = (0,0,1.5); <br />
real top = 0.75;<br />
<br />
path3 rightanglemark(triple A, triple B, triple C, real s=8) { // olympiad package<br />
triple P,Q,R; <br />
P=s*markscalefactor*unit(A-B)+B; <br />
R=s*markscalefactor*unit(C-B)+B; <br />
Q=P+R-B; <br />
return P--Q--R; <br />
} //documentation in second version<br />
path3 unitc=(1,0,0)..(0,1,0)..(-1,0,0)..(0,-1,0)..cycle;<br />
draw(unitc, dashed);dot(vertex);draw((1,0,0)--vertex--(-1,0,0),dashed);<br />
draw(vertex--(vertex.x,vertex.y,0)--(1,0,0),dashed);<br />
draw(rightanglemark(vertex,(vertex.x,vertex.y,0),(1,0,0),2)); <br />
<br />
// labeling<br />
label("$h$",(vertex.x,vertex.y,vertex.z/3),W);label("$r$",(0.5,0,0),S); label("$s$",((1+vertex.x)/2,(1+vertex.y)/2,vertex.z/2+0.35),NE);<br />
<br />
// lifting effect<br />
triple A = (1.15,0,top), C = (-1.02,0,top/2);<br />
path3 liftc=A..(0,1.06,top*3/4)..C..(0,-1,top/4)..(1,0,0); <br />
draw(liftc); draw(A--vertex--C); draw(vertex--(1,0,0));<br />
<br />
</asy> &nbsp;&nbsp; The lateral surface can be laid out to become a portion of a circular disk.&nbsp;&nbsp; <asy><br />
<br />
size(110); defaultpen(linewidth(0.7));<br />
draw(arc((0,0),1,0,250));draw(expi(250/180*pi)--(0,0)--(1,0)); label("$s$",(.5,0),S);label("$2\pi r$",expi(30*pi/180),NE);</asy></center><br />
<br />
== Problems ==<br />
*An ice cream [[cone]] consists of a [[sphere]] of vanilla ice cream and a right circular cone that has the same diameter as the sphere. If the ice cream melts, it will exactly fill the cone. Assume that the melted ice cream occupies <math>75\%</math> of the volume of the frozen ice cream. What is the ratio of the cone’s height to its [[radius]]? ([[2003 AMC 12B Problems/Problem 13]])<br />
<br />
<br />
*A right circular [[cone]] has base radius <math>r</math> and height <math>h</math>. The cone lies on its side on a flat table. As the cone rolls on the surface of the table without slipping, the point where the cone's base meets the table traces a circular arc centered at the point where the vertex touches the table. The cone first returns to its original position on the table after making <math>17</math> complete rotations. The value of <math>h/r</math> can be written in the form <math>m\sqrt {n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m + n</math>. ([[2008 AIME I Problems/Problem 5]])<br />
<br />
<br />
*A container in the shape of a right circular [[cone]] is <math>12</math> inches tall and its base has a <math>5</math>-inch [[radius]]. The liquid that is sealed inside is <math>9</math> inches deep when the cone is held with its [[point]] down and its base horizontal. When the liquid is held with its point up and its base horizontal, the height of the liquid is <math>m - n\sqrt [3]{p},</math> from the base where <math>m,</math> <math>n,</math> and <math>p</math> are positive integers and <math>p</math> is not divisible by the cube of any prime number. Find <math>m + n + p</math>. ([[2000 AIME I Problems/Problem 8]])<br />
<br />
<br />
*A [[solid]] in the shape of a right circular [[cone]] is 4 inches tall and its base has a 3-inch radius. The entire [[surface]] of the cone, including its base, is painted. A [[plane]] [[parallel]] to the base of the cone divides the cone into two solids, a smaller cone-shaped solid <math> C </math> and a [[frustum]]-shaped solid <math> F, </math> in such a way that the [[ratio]] between the [[area]]s of the painted surfaces of <math> C </math> and <math> F </math> and the ratio between the [[volume]]s of <math> C </math> and <math> F </math> are both equal to <math> k</math>. Given that <math> k=\frac m n, </math> where <math> m </math> and <math> n </math> are [[relatively prime]] [[positive integer]]s, find <math> m+n. </math> ([[2004 AIME I Problems/Problem 11]])<br />
<br />
<br />
{{stub}}<br />
<br />
[[Category:Geometry]]</div>Josuhttps://artofproblemsolving.com/wiki/index.php?title=Cone&diff=74919Cone2016-01-30T21:19:23Z<p>Josu: /* Problems */</p>
<hr />
<div>A '''cone''' (or ''circular cone'') is a three-dimensional [[solid]]. It consists of a [[circle | circular]] base, a [[point]] (called the ''vertex''), and all the points that lie on [[line segment]]s connecting the vertex to the base. Thus, the cone is the special case of the [[pyramid]] in which the base is circular.<br />
<br />
<center><!-- NOTE: two different versions --><asy> size(120);<br />
import three; currentprojection = perspective(0,-3,1); defaultpen(linewidth(0.7)); triple vertex = (0,0,1.5);<br />
<br />
path3 rightanglemark(triple A, triple B, triple C, real s=8) { // olympiad package<br />
triple P,Q,R; <br />
P=s*markscalefactor*unit(A-B)+B; <br />
R=s*markscalefactor*unit(C-B)+B; <br />
Q=P+R-B; <br />
return P--Q--R; <br />
} //documentation in second version<br />
path3 unitc=(1,0,0)..(0,1,0)..(-1,0,0)..(0,-1,0)..cycle;<br />
draw(unitc);dot(vertex);draw((1,0,0)--vertex--(-1,0,0));draw(vertex--(vertex.x,vertex.y,0)--(1,0,0));draw(rightanglemark(vertex,(vertex.x,vertex.y,0),(1,0,0),2)); label("$h$",(vertex.x,vertex.y,vertex.z/2),W);label("$r$",(0.5,0,0),S); label("$s$",((1+vertex.x)/2,(1+vertex.y)/2,vertex.z/2),NE);<br />
</asy>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <!-- VERSION 2 --><br />
<asy> size(120);<br />
import three; currentprojection = perspective(0,-3,1); defaultpen(linewidth(0.7));triple vertex = (0.4,0.6,1.5);<br />
<br />
path3 rightanglemark(triple A, triple B, triple C, real s=8) { // <br />
triple P,Q,R; <br />
P=s*markscalefactor*unit(A-B)+B; <br />
R=s*markscalefactor*unit(C-B)+B; <br />
Q=P+R-B; <br />
return P--Q--R; <br />
} <br />
// draw cone<br />
path3 unitc=(1,0,0)..(0,1,0)..(-1,0,0)..(0,-1,0)..cycle;<br />
draw(unitc);dot(vertex);draw((1,0,0)--vertex--(-1,0,0));draw(vertex--(vertex.x,vertex.y,0)--(1,0,0));draw(rightanglemark(vertex,(vertex.x,vertex.y,0),(1,0,0),2));<br />
<br />
// draw radius<br />
dot((0,0,0)); draw((0,0,0)--(1,0,0)); draw((0,0,0)--vertex,dotted);<br />
<br />
// label<br />
label("$r$",(0.5,0,0),S); label("$h$",(vertex.x,vertex.y,vertex.z/2),E);<br />
</asy></center><br />
<br />
== Terminology ==<br />
The distance from the vertex to the [[plane]] containing the base is the ''height'' of the cone, and is frequently denoted <math>h</math>. The [[radius]] of the base is called the radius of the cone and is frequently denoted <math>r</math>. If the vertex lies directly above the center of the base, we call the cone a '' '''right''' circular cone'' (or ''right cone'' for short). In this case, the vertex is the same distance from every point on the boundary of the base; this distance is called the ''slant height'' of the cone, and is sometimes denoted <math>s</math> or <math>\ell</math>. If a cone is not a right cone (that is, if the vertex is not directly above the center of the base), we call it an ''oblique cone''.<br />
<br />
== Properties ==<br />
* A cone with radius <math>r</math> and height <math>h</math> has [[volume]] <math>V = \frac{1}{3} \cdot \pi r^2 \cdot h</math>. This is a special case of the general formula for the volume of a pyramid, <math>V = \frac{1}{3} \cdot B \cdot h</math>, where <math>V</math> is the volume, <math>B</math> is the area of the base and <math>h</math> is the height.<br />
* A right cone of radius <math>r</math> and slant-height <math>s</math> has [[surface area]] <math>\pi r^2 + \pi r s</math> (the [[lateral area]] is <math>\pi rs</math>, and the area of the base is <math>\pi r^2</math>). <br />
<br />
<center><!-- NOTE: two different versions --><asy> size(120);<br />
import three; currentprojection = perspective(0,-3,1); defaultpen(linewidth(0.7)); triple vertex = (0,0,1.5); <br />
real top = 0.75;<br />
<br />
path3 rightanglemark(triple A, triple B, triple C, real s=8) { // olympiad package<br />
triple P,Q,R; <br />
P=s*markscalefactor*unit(A-B)+B; <br />
R=s*markscalefactor*unit(C-B)+B; <br />
Q=P+R-B; <br />
return P--Q--R; <br />
} //documentation in second version<br />
path3 unitc=(1,0,0)..(0,1,0)..(-1,0,0)..(0,-1,0)..cycle;<br />
draw(unitc, dashed);dot(vertex);draw((1,0,0)--vertex--(-1,0,0),dashed);<br />
draw(vertex--(vertex.x,vertex.y,0)--(1,0,0),dashed);<br />
draw(rightanglemark(vertex,(vertex.x,vertex.y,0),(1,0,0),2)); <br />
<br />
// labeling<br />
label("$h$",(vertex.x,vertex.y,vertex.z/3),W);label("$r$",(0.5,0,0),S); label("$s$",((1+vertex.x)/2,(1+vertex.y)/2,vertex.z/2+0.35),NE);<br />
<br />
// lifting effect<br />
triple A = (1.15,0,top), C = (-1.02,0,top/2);<br />
path3 liftc=A..(0,1.06,top*3/4)..C..(0,-1,top/4)..(1,0,0); <br />
draw(liftc); draw(A--vertex--C); draw(vertex--(1,0,0));<br />
<br />
</asy> &nbsp;&nbsp; The lateral surface can be laid out to become a portion of a circular disk.&nbsp;&nbsp; <asy><br />
<br />
size(110); defaultpen(linewidth(0.7));<br />
draw(arc((0,0),1,0,250));draw(expi(250/180*pi)--(0,0)--(1,0)); label("$s$",(.5,0),S);label("$2\pi r$",expi(30*pi/180),NE);</asy></center><br />
<br />
== Problems ==<br />
*An ice cream [[cone]] consists of a [[sphere]] of vanilla ice cream and a right circular cone that has the same diameter as the sphere. If the ice cream melts, it will exactly fill the cone. Assume that the melted ice cream occupies <math>75\%</math> of the volume of the frozen ice cream. What is the ratio of the cone’s height to its [[radius]]? ([[2003 AMC 12B Problems/Problem 13]])<br />
<br />
*A right circular [[cone]] has base radius <math>r</math> and height <math>h</math>. The cone lies on its side on a flat table. As the cone rolls on the surface of the table without slipping, the point where the cone's base meets the table traces a circular arc centered at the point where the vertex touches the table. The cone first returns to its original position on the table after making <math>17</math> complete rotations. The value of <math>h/r</math> can be written in the form <math>m\sqrt {n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m + n</math>. ([[2008 AIME I Problems/Problem 5]])<br />
<br />
*A container in the shape of a right circular [[cone]] is <math>12</math> inches tall and its base has a <math>5</math>-inch [[radius]]. The liquid that is sealed inside is <math>9</math> inches deep when the cone is held with its [[point]] down and its base horizontal. When the liquid is held with its point up and its base horizontal, the height of the liquid is <math>m - n\sqrt [3]{p},</math> from the base where <math>m,</math> <math>n,</math> and <math>p</math> are positive integers and <math>p</math> is not divisible by the cube of any prime number. Find <math>m + n + p</math>. ([[2000 AIME I Problems/Problem 8]])<br />
<br />
*A [[solid]] in the shape of a right circular [[cone]] is 4 inches tall and its base has a 3-inch radius. The entire [[surface]] of the cone, including its base, is painted. A [[plane]] [[parallel]] to the base of the cone divides the cone into two solids, a smaller cone-shaped solid <math> C </math> and a [[frustum]]-shaped solid <math> F, </math> in such a way that the [[ratio]] between the [[area]]s of the painted surfaces of <math> C </math> and <math> F </math> and the ratio between the [[volume]]s of <math> C </math> and <math> F </math> are both equal to <math> k</math>. Given that <math> k=\frac m n, </math> where <math> m </math> and <math> n </math> are [[relatively prime]] [[positive integer]]s, find <math> m+n. </math> ([[2004 AIME I Problems/Problem 11]])<br />
<br />
<br />
{{stub}}<br />
<br />
[[Category:Geometry]]</div>Josu