What is the radius of the base of the cone ?

Question

Prepp Team · Accepted Answer

This problem involves a geometric setup where a right conical cap covers two spheres of different radii placed one above the other on a table. The cone touches both spheres, and we are given the radii of the spheres (r for smaller, R for bigger) and the vertical angle of the cone (2$\theta$). We need to find the radius of the base of this cone.

Analyzing the Conical Cap and Sphere Geometry

Imagine a vertical cross-section passing through the axis of the cone. This cross-section shows an isosceles triangle representing the cone and two circles representing the spheres inside it. The circles are tangent to the sides of the triangle (the slant edges of the cone) and tangent to each other vertically. The larger sphere is placed on the table, which we assume is the plane containing the base of the cone.

Let the vertex of the cone be V, the center of the base be O, and the height of the cone be H = VO. The radius of the base is B = OA (where A is a point on the circumference of the base). The vertical angle of the cone is 2$\theta$, so the semi-vertical angle (the angle between the axis VO and the slant edge VA) is $\theta$.

Let C1 be the center of the smaller sphere (radius r) and C2 be the center of the larger sphere (radius R). Since the spheres are placed one above the other along the axis of the cone, C1 and C2 lie on the line segment VO. As the larger sphere rests on the table (level of O), its center C2 is located at a height R above the base O. So, OC2 = R.

The smaller sphere is placed directly above the larger sphere and touches it. The distance between the centers of two tangent spheres is the sum of their radii. Thus, the distance C1C2 = R + r. Since C2 is R above O, and C1 is R+r above C2, the height of C1 above the base O is OC1 = OC2 + C2C1 = R + (R+r) = 2R + r. So, C1 is at a height 2R+r above the base O.

Relating Sphere Radii, Cone Angle, and Height

Consider the cross-section. The slant edge of the cone is a line tangent to both spherical cross-sections (circles). The distance from the center of a sphere to a tangent line is equal to the radius of the sphere.

Let's use a coordinate system with the base center O at the origin (0,0) and the vertex V on the positive y-axis at (0, H). The axis of the cone is the y-axis. The centers of the spheres are C2 at (0, R) and C1 at (0, 2R+r).

A slant edge of the cone passes through the vertex (0, H) and makes an angle $\theta$ with the y-axis. The equation of such a line in the xy-plane can be written as $x = \tan \theta (H-y)$ or $x \cot \theta + y - H = 0$.

The distance from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$. For the slant line $x \cot \theta + y - H = 0$ (where $A=\cot \theta$, $B=1$, $C=-H$), the distance is $\frac{|x_0 \cot \theta + y_0 - H|}{\sqrt{\cot^2 \theta + 1}} = \frac{|x_0 \cot \theta + y_0 - H|}{\sqrt{\csc^2 \theta}} = |x_0 \cot \theta + y_0 - H| \sin \theta$.

For the center C2(0, R): The distance to the slant line is R.

$R = |0 \cdot \cot \theta + R - H| \sin \theta = |R - H| \sin \theta$

Since the center C2 is below the vertex V, R < H. So $|R-H| = H-R$.

$R = (H-R) \sin \theta \quad (*)$

For the center C1(0, 2R+r): The distance to the slant line is r.

$r = |0 \cdot \cot \theta + (2R+r) - H| \sin \theta = |2R+r - H| \sin \theta$

Since the center C1 is below the vertex V, 2R+r < H. So $|2R+r - H| = H - (2R+r)$.

$r = (H - 2R - r) \sin \theta \quad (**)$

Finding the Cone Height and Angle Relation

From equation $(*)$, $H-R = \frac{R}{\sin \theta} = R \csc \theta$. This gives the height $H = R + R \csc \theta$.

From equation $(**)$, $H - 2R - r = \frac{r}{\sin \theta} = r \csc \theta$. This gives the height $H = 2R + r + r \csc \theta$.

Equating the two expressions for H:

$R + R \csc \theta = 2R + r + r \csc \theta$

$R \csc \theta - r \csc \theta = 2R + r - R$

$(R-r) \csc \theta = R+r$

$\csc \theta = \frac{R+r}{R-r}$

This equation gives the relationship between the semi-vertical angle $\theta$ and the radii R and r.

Calculating the Base Radius of the Cone

The radius of the base B is the x-coordinate of the point where the slant line intersects the plane y=0 (the base). Substituting (B, 0) into the slant line equation $x \cot \theta + y - H = 0$:

$B \cot \theta + 0 - H = 0$

$B = H \tan \theta$

Now, substitute the expression for H we found: $H = R(1 + \csc \theta)$.

$B = R(1 + \csc \theta) \tan \theta$

Substitute the value of $\csc \theta = \frac{R+r}{R-r}$:

$B = R \left(1 + \frac{R+r}{R-r}\right) \tan \theta$

Simplify the term in the parenthesis:

$1 + \frac{R+r}{R-r} = \frac{R-r}{R-r} + \frac{R+r}{R-r} = \frac{(R-r) + (R+r)}{R-r} = \frac{R-r+R+r}{R-r} = \frac{2R}{R-r}$

So, the base radius B is:

$B = R \left(\frac{2R}{R-r}\right) \tan \theta = \frac{2R^2 \tan \theta}{R-r}$

This formula gives the radius of the base of the cone in terms of the radii of the spheres and the tangent of the semi-vertical angle.

Step-by-Step Derivation Summary

Set up the geometry of the cone covering two stacked, tangent spheres on a table.
Place the base of the cone on the table and determine the vertical positions of the sphere centers C1 and C2 relative to the base O.
Use the property that the distance from a sphere's center to the cone's slant edge equals the sphere's radius.
Derive two equations relating the cone height H, sphere radii R, r, and the semi-vertical angle $\theta$.
Solve these equations to find a relationship for $\csc \theta$ in terms of R and r, and an expression for H in terms of R and $\csc \theta$.
Use the formula relating base radius B, height H, and angle $\theta$ ($B = H \tan \theta$).
Substitute the expression for H and the relationship for $\csc \theta$ into the formula for B and simplify.

Quantity	Formula
Semi-vertical angle relation	$\csc \theta = \frac{R+r}{R-r}$
Cone Height (H)	$H = R(1 + \csc \theta) = \frac{2R^2}{R-r}$
Base Radius (B)	$B = H \tan \theta = \frac{2R^2 \tan \theta}{R-r}$

Revision Table: Key Formulas

Additional Information: Cone and Sphere Geometry

This problem is a classic example of tangency in 3D geometry, often simplified by looking at a 2D cross-section. The key concepts used are the distance from a point (sphere center) to a line (cone slant edge) and the geometry of stacked tangent spheres. The relation $(R-r) \csc \theta = R+r$ is a fundamental result when a cone just covers two spheres tangent to the cone and to each other. The height of the cone and its base radius are then determined by the position of the spheres relative to the cone's base.

Prepp Team · Answer

$\frac{2 r^2 \tan \theta}{R-r}$

Prepp Team · Answer

$\frac{2\left(\mathrm{r}^2+\mathrm{R}^2\right) \tan \theta}{\mathrm{R}-\mathrm{r}}$

Prepp Team · Answer

$\frac{\left(\mathrm{r}^2+\mathrm{R}^2\right) \tan \theta}{\mathrm{R}-\mathrm{r}}$

Analyzing the Conical Cap and Sphere Geometry

Relating Sphere Radii, Cone Angle, and Height

Finding the Cone Height and Angle Relation

Calculating the Base Radius of the Cone

Step-by-Step Derivation Summary

Revision Table: Key Formulas

Additional Information: Cone and Sphere Geometry

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Quantity	Formula
Semi-vertical angle relation	\(\csc \theta = \frac{R+r}{R-r}\)
Cone Height (H)	\(H = R(1 + \csc \theta) = \frac{2R^2}{R-r}\)
Base Radius (B)	\(B = H \tan \theta = \frac{2R^2 \tan \theta}{R-r}\)