Circular Cone

**Ilsa** · November 20th 2009, 06:52 AM

Consider a right circular cone with the vertex V and O the center of the base. the radius of the base is 1 and the height VO is square root of 35. consider also the diameter AB of the base and C the midpoint of VA. Comput the length of the shortest path, on the surface of the cone from B to C.

**Soroban** · November 20th 2009, 08:33 AM

Hello, Ilsa!

Consider a right circular cone with the vertex $V$ and the center of the base $O$ .
The radius of the base is 1 and the height $VO$ is $\sqrt{35}$
Consider also the diameter $AB$ of the base and $C$ the midpoint of $VA$ .

Compute the length of the shortest path, on the surface of the cone from $B$ to $C.$

Code:

            V
            *
           /|\
          / | \
         /  |  \
      C * __|   \ 6
       / √35|    \
      /     |     \ 
     /      |      \
  A * - - - * - - - * B
            O   1

We have a right triangle with legs 1 and $\sqrt{35}.$
The hypotenuse (slant height of the cone) is 6.

If we "unroll" half the cone and press it flat,
. . we have the diagram below.

Code:

                    V
                    *
                   / \ 
                  /30°\ 3
                 /     \  C
              6 /       * 
               /     *   \
              /   *       \ 3
             / *           \
          B * - - - - - - - *
                  * * *
                    π

The circumference of the base of the cone is: . $2\pi r \:=\:2\pi(1) \:=\:2\pi$
Half the circumference is: $\pi$

We have: . $s \:=\:r\theta \quad\Rightarrow\quad \theta \:=\:\frac{s}{r}$
. . Hence: . $\theta \:=\:\frac{\pi}{6}$
The vertex angle is 30°.

The shortest distance between $B$ and $C$ is a straight line.

In $\Delta BVC$ , Law of Cosines:
. . $BC^2 \:=\:6^2 + 3^2 - 2(6)(3)\cos30^o \;=\;45 - 36\left(\frac{\sqrt{3}}{2}\right) \;=\;9(5 - 2\sqrt{3})$

Therefore: . $BC \;=\;3\sqrt{5-2\sqrt{3}} \;\approx\;3.718$

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