A circular cone has vertex $I$, a base with radius 1 , and a slant height of 4 . Point $A$ is on the circumference of the base and point $R$ is on the line segment $I A$ with $I R=3$. Shahid draws the shortest possible path starting at $R$, travelling once around the cone, and ending at $A$. If $P$ is the point on this path that is closest to $I$, what is the length $I P$ ?

![](https://cdn.mathpix.com/cropped/2024_04_30_3011c729c030a190170fg-4.jpg?height=474&width=333&top_left_y=381&top_left_x=950)
We cut the lateral surface of the cone along line segment $I A$.

This forms a sector of a circle with radius $I A=4$.

We label the second endpoint of the arc of this sector as $A^{\prime}$ (it previously connected with $A$ ). Since the circumference of the base of the original cone is $2 \pi(1)=2 \pi$, then the length of the arc of this sector is $2 \pi$.

Since a full circle with radius 4 has circumference $8 \pi$ and length of the arc of this sector is $2 \pi$, then this sector is $\frac{1}{4}$ of a circle, so has central angle $\angle A I A^{\prime}=90^{\circ}$.

Point $R$ still lies on $I A$ with $I R=3$.

The shortest path starting at $R$, travelling once around the cone, and ending at $A$ is now the shortest path in this sector beginning at $R$ and ending at $A^{\prime}$. This is the straight line segment from $R$ to $A^{\prime}$.

Since $I R=3, I A^{\prime}=4$, and $\angle R I A^{\prime}=90^{\circ}$, then $R A^{\prime}=\sqrt{3^{2}+4^{2}}=\sqrt{25}=5$, by the Pythagorean Theorem.

The point $P$ will be the point on $R A^{\prime}$ for which $I P$ is perpendicular to $R A^{\prime}$.

![](https://cdn.mathpix.com/cropped/2024_04_30_75c1cf70fbe3735bfd92g-14.jpg?height=447&width=483&top_left_y=969&top_left_x=867)

We can then calculate the area of $\triangle R I A^{\prime}$ in two different ways

$$
\text { Area }=\frac{1}{2} \cdot I R \cdot I A^{\prime}=\frac{1}{2} \cdot R A^{\prime} \cdot I P
$$

to obtain $I R \cdot I A^{\prime}=R A^{\prime} \cdot I P$ or $3 \cdot 4=5 \cdot I P$ and so $I P=\frac{12}{5}$.

Therefore, the length of $I P$ is $\frac{12}{5}$.

(We could also calculate the length $I P$ from $\triangle R I A^{\prime}$ using trigonometry or similar triangles.)

ANSWER: $\frac{12}{5}$