A right circular cone has radius 3 and height 3. If A and B are two points on its surface, what is the maximum possible straight-line distance between A and B?
I fixed $A$ to be a point on the bottom left corner of the cone.
From $A$, the diameter (Maximum horizontal distance) is 6, and the slant distance (Max distance to the vertex)is $\sqrt{18} = 4.24…$
I suspected that there may be a point $B$ from A, on the right hand half of the cone, along a slant line that intersects with the base $\pi$ radians from $A$ relative to the circumference to the base of the cone, that may be the maximum distance.
I chose the mid-point, and constructed a triangle with base 6, and a right hand side of $\frac{\sqrt{18}}{2}$ at an angle of $\pi/4$ to the base of length 6. The distance from $A$ was found to be $\sqrt{22.5} = 4.74$
This way, with $\sqrt{18}$ being the distance to the vertex, $\sqrt{22.5}$ being the distance to the point $B$, and 6 being the maximum distance to the other end of the cone horizontally, I concluded that 6 was the maximum distance between two points along the surface.
Is this conclusion correct? Could this be approached with a quicker method?
Best Answer
Your approach is very hand-wavy. Here is a rigorous proof.
Take any two points $A,B$ on the cone with apex $O$.
By symmetry we can assume that $A$ is closer to the base, and we can use a scaling centred at $O$ to bring $A$ to the base. Both of $A,B$ still remain on the surface of the cone, and are now further apart since the scale factor is more than $1$. Thus it suffices to consider only pairs of points with at least one on the base.
Now draw a line from $O$ through $B$ to intersect the base at $C$. Then $|AB| \le \max(|AC|,|AO|)$. $|AO| = 3\sqrt{2}$ as you have computed. Also $|AC| \le 2(3)$ since the diameter is the longest chord in any circle. Thus $|AB| \le 6$.
Finally, any two points diametrically opposite the base of the cone really have distance $6$, so the upper bound is achieved.