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Math Help - Spheres, cones and triangles.

  1. #1
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    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 !

    Thanks for your attention
    Last edited by mr fantastic; April 20th 2010 at 07:24 AM. Reason: Re-titled post.
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  2. #2
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    Quote Originally Posted by wizard654zzz View Post
    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 !

    Thanks for your attention
    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.
    Attached Thumbnails Attached Thumbnails Spheres, cones and triangles.-sphere-cone.jpg  
    Last edited by Archie Meade; April 20th 2010 at 10:33 AM.
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  3. #3
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    Quote Originally Posted by wizard654zzz View Post
    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 !

    Thanks for your attention
    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.
    Attached Thumbnails Attached Thumbnails Spheres, cones and triangles.-similar-triangles.jpg  
    Last edited by Archie Meade; April 21st 2010 at 04:49 AM.
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  4. #4
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    Quote Originally Posted by wizard654zzz View Post
    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 !

    Thanks for your attention
    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
    Attached Thumbnails Attached Thumbnails Spheres, cones and triangles.-similar-triangles-3.jpg  
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