Question 168·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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A right circular cone has base radius $r$ and height $h$ , where $h=2r$ . A sphere is inscribed in the cone so that it is tangent to the base of the cone and to the lateral surface of the cone.

Which choice is the ratio of the volume of the sphere to the volume of the cone?

An inscribed sphere in a cone is easier to study in a slice through the cone’s center: the slice is a triangle with a circle touching all three sides. For a volume ratio, choose a convenient cone radius, find the sphere’s radius from that slice, and compare the volume formulas. The cone’s radius is not the sphere’s radius.

Hints

Hint 1
A volume ratio stays the same if you scale every length: both volumes are multiplied by the same cubed factor. What convenient value could you give the cone’s radius?
Hint 2
Slice through the cone’s center and tip. Half of the resulting triangle is a right triangle; its legs are the cone’s radius and height. Use those legs to find the sloping side.
Hint 3
The sphere becomes a circle touching all three sides of the slice. A radius drawn to a touching point is perpendicular to that side. How can you use those radii as the heights of three smaller triangles?

Step-by-step

Approach 1: Use a slice and compare volumes
Step 1
Choose a convenient cone size
A volume ratio doesn’t change when every length is scaled, because both volumes scale by the same cubed factor. So set the cone’s base radius to $1$ . The given $h=2r$ makes its height $2$ .
Step 2
Find the sloping side of the slice
Slice through the cone’s center and tip. You get a triangle with base $2$ and height $2$ ; half of it is a right triangle with legs $1$ and $2$ . By the Pythagorean theorem, type $\sqrt{1^2+2^2}$ in Desmos. It shows about $2.236$ , so each sloping side is exactly $\sqrt5$ .
Step 3
Find the slice’s area
The triangle’s base is $2$ and its perpendicular height is $2$ . Type $\frac12(2)(2)$ in Desmos; it shows area $2$ . This is the area you’ll also express using the circle inside the triangle.
Step 4
Connect the sphere’s radius to that area
Call the sphere’s radius $s$ . In this slice, the sphere becomes a circle touching the base and both sloping sides. Draw a radius to each touching point: it meets that side at a right angle. The three smaller triangles have height $s$ and bases totaling $2+2\sqrt5$ . So the big triangle’s area is half the circle’s radius times its perimeter:
$2=\frac12s(2+2\sqrt5).$
Combine the terms:
$2=s(1+\sqrt5).$
The sloping side $\sqrt5$ is one of those bases, not the sphere’s radius.
Step 5
Find the sphere’s radius
Type $2\sim s_1(1+\sqrt5)$ in Desmos. Here $s_1$ stands for the sphere’s radius, and $\sim$ asks Desmos to find the value that makes the equation work. Under PARAMETERS, it shows $s_1\approx0.618034$ .
Step 6
Set up the requested volume ratio
Use the volume formulas with cone radius $1$ , cone height $2$ , and sphere radius $s$ :
$V_{\text{sphere}}=\frac43\pi s^3,\qquad V_{\text{cone}}=\frac13\pi(1)^2(2)=\frac23\pi.$
Divide and cancel $\pi$ and the thirds:
$\frac{V_{\text{sphere}}}{V_{\text{cone}}}=\frac{\frac43\pi s^3}{\frac23\pi}=2s^3.$
Don’t stop at $s^3$ : the two volume formulas leave a factor of $2$ .
Step 7
Evaluate and match the exact choice
Type $2s_1^3$ in Desmos; it shows about $0.472136$ . Then type $2\sqrt5-4$ ; Desmos shows the same value. So the sphere-to-cone volume ratio is $2\sqrt5-4$ . Choice C.

Approach 2: Check the match exactly
Step 1
Write the sphere’s radius as a radical
Instead of comparing decimals, start from $2=s(1+\sqrt5)$ . Divide by $1+\sqrt5$ :
$s=\frac{2}{1+\sqrt5}.$
The conjugate $\sqrt5-1$ makes the denominator a difference of squares. Multiply top and bottom by it:
$s=\frac{2(\sqrt5-1)}{(1+\sqrt5)(\sqrt5-1)}.$
Evaluate the denominator:
$s=\frac{2(\sqrt5-1)}{4}.$
Reduce the fraction:
$s=\frac{\sqrt5-1}{2}.$
Step 2
Cube the exact radius
The volume ratio uses $s^3$ . Cube the radius:
$s^3=\frac{(\sqrt5-1)^3}{8}.$
Square the first two factors:
$s^3=\frac{(6-2\sqrt5)(\sqrt5-1)}{8}.$
Distribute:
$s^3=\frac{8\sqrt5-16}{8}.$
Divide both terms by $8$ :
$s^3=\sqrt5-2.$
Step 3
Include the volume-formula factor
The sphere-to-cone ratio is $2s^3$ , not $s^3$ . Multiply the exact cube by $2$ :
$2s^3=2(\sqrt5-2).$
Distribute $2$ :
$2s^3=2\sqrt5-4.$
So the sphere’s volume divided by the cone’s volume is $2\sqrt5-4$ . Choice C.

Question 168·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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