A right circular cone is sliced by a plane parallel to its base, resulting in a smaller cone at the top and a frustum below. The volume of the original cone is $27$ times that of the smaller cone. When the lateral surface area of the smaller cone is unfolded, it forms a sector of a circle with a central angle of $120^\circ$. Determine the value of $H/l$, where $H$ is the height of the original cone and $l$ is the slant height of the smaller cone.

We are given a right circular cone that is sliced by a plane parallel to its base, creating a smaller cone at the top and a frustum below. The volume of the original cone is 27 times that of the smaller cone. Additionally, the lateral surface area of the smaller cone, when unfolded, forms a sector of a circle with a central angle of $120^\circ$. We are tasked with determining the ratio $H/l$, where $H$ is the height of the original cone and $l$ is the slant height of the smaller cone.

### Step 1: Set up the relationship for volumes

Let the radius and height of the original cone be $R$ and $H$, respectively. The radius and height of the smaller cone (after slicing) will be $r$ and $h$, respectively. The volume of a cone is given by the formula:
\[
V = \frac{1}{3} \pi r^2 h.
\]
Thus, the volume of the original cone is:
\[
V_{\text{original}} = \frac{1}{3} \pi R^2 H,
\]
and the volume of the smaller cone is:
\[
V_{\text{small}} = \frac{1}{3} \pi r^2 h.
\]
We are told that the volume of the original cone is 27 times that of the smaller cone, so:
\[
V_{\text{original}} = 27 V_{\text{small}}.
\]
Substituting the expressions for the volumes:
\[
\frac{1}{3} \pi R^2 H = 27 \times \frac{1}{3} \pi r^2 h.
\]
Canceling the common terms of $\frac{1}{3} \pi$, we get:
\[
R^2 H = 27 r^2 h.
\]
Since the smaller cone is similar to the original cone (as they are both right circular cones), the corresponding dimensions of the cones are proportional. That is:
\[
\frac{r}{R} = \frac{h}{H} = k \quad \text{for some constant } k.
\]
Thus, we can express $r = kR$ and $h = kH$. Substituting these into the volume equation:
\[
R^2 H = 27 (kR)^2 (kH),
\]
which simplifies to:
\[
R^2 H = 27 k^3 R^2 H.
\]
Canceling $R^2 H$ from both sides (assuming $R, H \neq 0$):
\[
1 = 27 k^3,
\]
which gives:
\[
k^3 = \frac{1}{27},
\]
so:
\[
k = \frac{1}{3}.
\]

### Step 2: Relate the lateral surface areas

Next, we use the information about the lateral surface area of the smaller cone. The lateral surface area of a cone is given by:
\[
A = \pi r l,
\]
where $l$ is the slant height. The slant height of the original cone is $L$, and the slant height of the smaller cone is $l$. We are told that when the lateral surface area of the smaller cone is unfolded, it forms a sector of a circle with a central angle of $120^\circ$.

The lateral surface area of the smaller cone is $A_{\text{small}} = \pi r l$, and the radius of the sector is $l$, so the arc length of the sector corresponds to the circumference of the base of the smaller cone, which is $2\pi r$. The formula for the arc length of a sector is:
\[
\text{Arc length} = \frac{\theta}{360^\circ} \times 2\pi l,
\]
where $\theta = 120^\circ$. Thus:
\[
2\pi r = \frac{120^\circ}{360^\circ} \times 2\pi l,
\]
which simplifies to:
\[
2\pi r = \frac{1}{3} \times 2\pi l.
\]
Canceling $2\pi$ from both sides:
\[
r = \frac{1}{3} l.
\]
Substituting $r = kR = \frac{1}{3} R$, we have:
\[
\frac{1}{3} R = \frac{1}{3} l,
\]
which gives:
\[
l = R.
\]

### Step 3: Find the ratio $H/l$

We now know that $k = \frac{1}{3}$ and $l = R$. Since $k = \frac{h}{H}$, we have $h = \frac{1}{3} H$. The ratio $H/l$ is:
\[
\frac{H}{l} = \frac{H}{R}.
\]
Using the fact that $k = \frac{1}{3}$, we find:
\[
\frac{H}{R} = 3.
\]
Thus, the value of $H/l$ is:
\[
\boxed{3}.
\]

A right circular cone is sliced by a plane parallel to its base, resulting in a smaller cone at the top and a frustum below. The volume of the original cone is 27 times that of the smaller cone. When the lateral surface area of the smaller cone is unfolded, it forms a sector of a circle with a central angle of 120°. Determine the value of H/l, where H is the height of the original cone and l is the slant height of the smaller cone.

This problem involves a right circular cone sliced to create a smaller cone and a frustum. The volume of the original cone is 27 times that of the smaller cone. The lateral surface area of the smaller cone forms a sector of a circle with a 120° central angle. Learn how to find the ratio H/l, where H is the height of the original cone and l is the slant height of the smaller cone. Suitable for Grades 10-12.

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