- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,689

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,689

- Admin
- #2

Now, orienting the right angle at the origin of the $xy$-plane, we find that the equation of the line along which the hypotenuse runs is:

\(\displaystyle y=-\frac{h}{a}x+h\)

We require that the perpendicular distance from the center of the circle to this line be $r$, hence:

\(\displaystyle r=\frac{|h-r|}{\sqrt{\left(\frac{h}{a} \right)^2+1}}\)

Since $h>r$, and solving for $r$, we find:

\(\displaystyle r=\frac{a\left(\sqrt{h^2+a^2}-a \right)}{h}\)

Now, the apothem $a$ is given by:

\(\displaystyle a=\frac{s}{2}\tan\left(\frac{\pi(n-2)}{2n} \right)\)

- Thread starter
- Admin
- #3

- Feb 14, 2012

- 3,689

HiHere is my solution:

Consider a cross-section of the pyramid-sphere system from the axis of symmetry running along an apothem $a$ of the base:

View attachment 1083

Now, orienting the right angle at the origin of the $xy$-plane, we find that the equation of the line along which the hypotenuse runs is:

\(\displaystyle y=-\frac{h}{a}x+h\)

We require that the perpendicular distance from the center of the circle to this line be $r$, hence:

\(\displaystyle r=\frac{|h-r|}{\sqrt{\left(\frac{h}{a} \right)^2+1}}\)

Since $h>r$, and solving for $r$, we find:

\(\displaystyle r=\frac{a\left(\sqrt{h^2+a^2}-a \right)}{h}\)

Now, the apothem $a$ is given by:

\(\displaystyle a=\frac{s}{2}\tan\left(\frac{\pi(n-2)}{2n} \right)\)

Also, I see that you have used one of the formulas that you posted here http://www.mathhelpboards.com/f49/finding-distance-between-point-line-2952/. I think MHB is truly a great place to search for challenging problems to solve and it also provides handy formulas for us to use!