# Difference between revisions of "Cone"

A cone (or circular cone) is a three-dimensional solid. It consists of a circular base, a point (called the vertex), and all the points that lie on line segments connecting the vertex to the base. Thus, the cone is the special case of the pyramid in which the base is circular.

$[asy] size(120); import three; currentprojection = perspective(0,-3,1); defaultpen(linewidth(0.7)); triple vertex = (0,0,1.5); path3 rightanglemark(triple A, triple B, triple C, real s=8) { // olympiad package triple P,Q,R; P=s*markscalefactor*unit(A-B)+B; R=s*markscalefactor*unit(C-B)+B; Q=P+R-B; return P--Q--R; } //documentation in second version path3 unitc=(1,0,0)..(0,1,0)..(-1,0,0)..(0,-1,0)..cycle; draw(unitc);dot(vertex);draw((1,0,0)--vertex--(-1,0,0));draw(vertex--(vertex.x,vertex.y,0)--(1,0,0));draw(rightanglemark(vertex,(vertex.x,vertex.y,0),(1,0,0),2)); label("h",(vertex.x,vertex.y,vertex.z/2),W);label("r",(0.5,0,0),S); label("s",((1+vertex.x)/2,(1+vertex.y)/2,vertex.z/2),NE); [/asy]$      $[asy] size(120); import three; currentprojection = perspective(0,-3,1); defaultpen(linewidth(0.7));triple vertex = (0.4,0.6,1.5); path3 rightanglemark(triple A, triple B, triple C, real s=8) { // triple P,Q,R; P=s*markscalefactor*unit(A-B)+B; R=s*markscalefactor*unit(C-B)+B; Q=P+R-B; return P--Q--R; } // draw cone path3 unitc=(1,0,0)..(0,1,0)..(-1,0,0)..(0,-1,0)..cycle; draw(unitc);dot(vertex);draw((1,0,0)--vertex--(-1,0,0));draw(vertex--(vertex.x,vertex.y,0)--(1,0,0));draw(rightanglemark(vertex,(vertex.x,vertex.y,0),(1,0,0),2)); // draw radius dot((0,0,0)); draw((0,0,0)--(1,0,0)); draw((0,0,0)--vertex,dotted); // label label("r",(0.5,0,0),S); label("s",((1+vertex.x)/2,(1+vertex.y)/2,vertex.z/2),ENE); label("h",(vertex.x,vertex.y,vertex.z/2),E); [/asy]$

## Terminology

The distance from the vertex to the plane containing the base is the height of the cone, and is frequently denoted $h$. The radius of the base is called the radius of the cone and is frequently denoted $r$. 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 $s$ or $\ell$. 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.

## Properties

• A cone with radius $r$ and height $h$ has volume $V = \frac{1}{3} \cdot \pi r^2 \cdot h$. This is a special case of the general formula for the volume of a pyramid, $V = \frac{1}{3} \cdot B \cdot h$, where $V$ is the volume, $B$ is the area of the base and $h$ is the height.
• A right cone of radius $r$ and slant-height $s$ has surface area $\pi r^2 + \pi r s$ (the lateral area is $\pi rs$, and the area of the base is $\pi r^2$).
$[asy] size(120); import three; currentprojection = perspective(0,-3,1); defaultpen(linewidth(0.7)); triple vertex = (0,0,1.5); real top = 0.75; path3 rightanglemark(triple A, triple B, triple C, real s=8) { // olympiad package triple P,Q,R; P=s*markscalefactor*unit(A-B)+B; R=s*markscalefactor*unit(C-B)+B; Q=P+R-B; return P--Q--R; } //documentation in second version path3 unitc=(1,0,0)..(0,1,0)..(-1,0,0)..(0,-1,0)..cycle; draw(unitc, dashed);dot(vertex);draw((1,0,0)--vertex--(-1,0,0),dashed); draw(vertex--(vertex.x,vertex.y,0)--(1,0,0),dashed); draw(rightanglemark(vertex,(vertex.x,vertex.y,0),(1,0,0),2)); // labeling label("h",(vertex.x,vertex.y,vertex.z/3),W);label("r",(0.5,0,0),S); label("s",((1+vertex.x)/2,(1+vertex.y)/2,vertex.z/2+0.35),NE); // lifting effect triple A = (1.15,0,top), C = (-1.02,0,top/2); path3 liftc=A..(0,1.06,top*3/4)..C..(0,-1,top/4)..(1,0,0); draw(liftc); draw(A--vertex--C); draw(vertex--(1,0,0)); [/asy]$    The lateral surface can be laid out to become    $[asy] size(110); defaultpen(linewidth(0.7)); draw(arc((0,0),1,0,250));draw(expi(250/180*pi)--(0,0)--(1,0)); label("s",(.5,0),S);label("2\pi r",expi(30*pi/180),NE);[/asy]$

## Problems

• 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 $75\%$ 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)
• A right circular cone has base radius $r$ and height $h$. 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 $17$ complete rotations. The value of $h/r$ can be written in the form $m\sqrt {n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n$. (2008 AIME I Problems/Problem 5)
• A container in the shape of a right circular cone is $12$ inches tall and its base has a $5$-inch radius. The liquid that is sealed inside is $9$ 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 $m - n\sqrt [3]{p},$ from the base where $m,$ $n,$ and $p$ are positive integers and $p$ is not divisible by the cube of any prime number. Find $m + n + p$. (2000 AIME I Problems/Problem 8)
• 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 $C$ and a frustum-shaped solid $F,$ in such a way that the ratio between the areas of the painted surfaces of $C$ and $F$ and the ratio between the volumes of $C$ and $F$ are both equal to $k$. Given that $k=\frac m n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$ (2004 AIME I Problems/Problem 11)