# Thread: Spheres, cones and triangles.

1. ## Spheres, cones and triangles.

I got 2 math problems can you help me ?
1. Suppose that a sphere of radius 1 is inscribed in a right circular cone with radius r and height h.

(1) Express r in terms of h !
(2) Find the minimum of the volume of such a right circular one !

2.Let Triangle ABC be a right angled such that $angle A=90^0$, AB=AC and let M be the mid point of the side AC. Take the point P on the side BC so that AP is vertical to BM. Let H be the intersection point of AP and BM

(1) Find the ratio of the areas of the two triangles $\Delta{ABH} : \Delta{AHM}$!

(2) Find the ratio BP : PC !

2. Originally Posted by wizard654zzz
I got 2 math problems can you help me ?
1. Suppose that a sphere of radius 1 is inscribed in a right circular cone with radius r and height h.

(1) Express r in terms of h !
(2) Find the minimum of the volume of such a right circular one !

In the attachment, there are similar triangles,
due to both having A, B and 90 degrees.

Using Pythagoras theorem

$x^2=1+y^2,\ y=\sqrt{x^2-1}$

$tanA=\frac{x+1}{r}$

$tanA=\sqrt{x^2-1}$

$r=\frac{x+1}{\sqrt{x^2-1}}$

$r^2=\frac{(x+1)(x+1)}{x^2-1}=\frac{(x+1)(x+1)}{(x+1)(x-1)}=\frac{x+1}{x-1}=\frac{h}{h-2}$

since h=x+1, x-1=x+1-2

$V_{cone}=\frac{{\pi}r^2h}{3}=\frac{{\pi}h^2}{3(h-2)}$

Differentiate this with respect to h and set the result =0.
This gives the height of the cone of minimum volume from which it's radius can be found.

3. Originally Posted by wizard654zzz
I got 2 math problems can you help me ?
1. Suppose that a sphere of radius 1 is inscribed in a right circular cone with radius r and height h.

(1) Express r in terms of h !
(2) Find the minimum of the volume of such a right circular one !

2.Let Triangle ABC be a right angled such that $angle A=90^0$, AB=AC and let M be the mid point of the side AC. Take the point P on the side BC so that AP is vertical to BM. Let H be the intersection point of AP and BM

(1) Find the ratio of the areas of the two triangles $\Delta{ABH} : \Delta{AHM}$!

(2) Find the ratio BP : PC !

For Q2,

if we draw a sketch, we will again find similar triangles.

In the attachment, since the red lines cross at right-angles

$Z+V=90^o$

$\theta+V=90^o\ \Rightarrow\ \theta=Z$

and so

$Q=V$

The large triangle ABH is a magnified version of the smaller one AHM,
as we can re-orientate them as shown.

Since $|BA|=2|AM|$

then all linear measurements of the larger triangle are twice those of the smaller one.

Hence the height of the large triangle is twice that of the smaller one.

As area of a triangle is 0.5(base)(height), we can now write the ratio of their areas.

4. Originally Posted by wizard654zzz
I got 2 math problems can you help me ?
1. Suppose that a sphere of radius 1 is inscribed in a right circular cone with radius r and height h.

(1) Express r in terms of h !
(2) Find the minimum of the volume of such a right circular one !

2.Let Triangle ABC be a right angled such that $angle A=90^0$, AB=AC and let M be the mid point of the side AC. Take the point P on the side BC so that AP is vertical to BM. Let H be the intersection point of AP and BM

(1) Find the ratio of the areas of the two triangles $\Delta{ABH} : \Delta{AHM}$!

(2) Find the ratio BP : PC !

For Q2 (2)

the attachment shows how to obtain pairs of similar triangles.

Triangles EPC and BPA are equiangular,

hence if you rotate triangle EPC through 180 degrees,
it's a scaled-down version of triangle BPA, with side CE corresponding to side BA.

Hence, the sides of triangle EPC are half the length of the corresponding sides of triangle BPA.

Therefore, as side PC corresponds to side BP

$BP:PC=2:1$

For your Q1, if you continued to the end you should have had

$V_{cone}=\frac{{\pi}}{3}\frac{h^2}{h-2}$

Differentiate this with respect to h, using the quotient rule,
and set the derivative equal to zero to find where the tangent to the curve is horizontal,
which locates minimum volume

$\frac{dV}{dh}=\frac{{\pi}}{3}\left[\frac{(h-2)2h-h^2(1)}{(h-2)^2}\right]=0$

hence the numerator is zero, so

$\Rightarrow\ 2h^2-4h-h^2=0$

$h^2-4h=0\ \Rightarrow\ h(h-4)=0$

$h=4$

$r^2=\frac{h}{h-2}=2$

$V_{min}=\frac{{\pi}r^2h}{3}$ for r and h corresponding to min V

$=\frac{{\pi}(2)(4)}{3}=\frac{8{\pi}}{3}$

Also, the final answer to Q2 (1)

$[Area\ of\ ABH]:[Area\ of\ AHM]=\frac{kh}{2}:\frac{\frac{k}{2}\frac{h}{2}}{2}=kh: \frac{kh}{4}=1:\frac{1}{4}=4:1$