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Math Help - A sphare in a triangle

  1. #1
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    A sphare in a triangle

    Hi!
    Today we had a test at school and I had this interesting question ,that I didn't know how to answer.
    So it was said: A sphere is touching all 3 edges of a right triangle, its edges are 9 and 12 cm long. The distance from the sphere's center to the triangle plane is 10cm. How long is the radius of the sphere?

    I don't know if I made the picture right, but I think it should look like this:
    Somewhere I heard that if it both edges were the same size (for example, 9 and 9) then the distance and radius would be the same size.
    How could I get the radius? The answer should be 12.25 cm, but not sure.
    Last edited by regdude; January 26th 2011 at 05:14 AM.
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  2. #2
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    Here's how I would do that- it may not be suitable for you as it uses "coordinate geometry". Set up a coordinate system so that the origin is at the right angle, the positive y axis along the side of the triangle of length 9 and the positive x-axis along the side of length l. Let the center be at (a, b). We can immediately say that a= b since a is the distance from the center to the y axis, b is the distance from the center to the x axis, and those are both radii of the sphere. That is, the center of the sphere is at (a, a).

    The hypotenuse runs from (12, 0) to (0, 9) and so has slope (9- 0)/(0- 12)= -9/12= -3/4. The equation of the hypotenuse is, in fact, \frac{x}{12}+ \frac{y}{9}= 1 or y= 9- (3/4)x. A line perpendicular to that will have slope 4/3 and, in particular, a line perpendicular to the hypotenuse through (a, a) will have equation y= (4/3)(x- a)+ a= (4/3)x- (1/3)a. That line intersects the line when y= 9- (3/4)x= (4/3)x- (1/3)a. Solve for x and y to find point where the radius touches the hypotenuse, in terms of a, then calculate the distance between that point and (a, a), again in terms of a. Finally, set that distance equal to a, since we know the other two given radii have length a, and solve for a, the radius of the sphere.

    By the way, this is only true if the sphere is completely in side the triangle- that is, that it is symmetric about the triangle. Since the sphere is three-dimensional and the triangle two-dimensional, you could have part of the sphere on one side of the triangle and a larger part on the other side. Then there would be an infinite number of possible solutions. In my opinion this problem would be better stated as a circle inside a triangle.
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  3. #3
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    Yeah, this is "bit" too hard for me to solve. Always wanted to learn how to use coordinate geometry, see some good solutions using it with a computer
    But if it was a circle then wouldn't the distance be the same as radius?
    I don't seem to understand where is the triangle plane...
    Thanks for the hint, maybe I will find the right picture and from that will see some known figures.
    Seems now that the only thing unknown in my picture is a small part of the hypotenuse.
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  4. #4
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    Um... I'm thinking something else.

    I'm picturing the picture as:



    To find the centre of the circle (the 'slice' of sphere in the plane of the triangle), I would use some knoledge of locii.

    The lines bisecting two angles are equidistant from each two lines making the angle and when two meets, the intersection gives the centre of the circle. Then using the coordinate system, you get lines with equations:

    y = x

    y = -\dfrac13 x + c (you can get c by using the point (12, 0)

    y = -2x+9

    [I used arctan(9/12) to get the angle, then divide be two, then did tan of that resulting angle to get the gradients]

    You take any two and solve simultaneously, to get the centre of that circle.

    Now, we use trigonometry.

    You must now picture your drawing another way, so that the plane containing the triangle becomes horizontal flat, and you see it as a line. From the centre of the sphere, to the centre of the previous circle to the point the sphere touches the triangle, you have a new right triangle, with one side having length 10, the other side having length equal to the centre of the circle.



    Use Pythagoras' Theorem now to get the radius of the sphere as 10.4 cm. (I just hope I didn't do any mistakes in my calculations... )
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  5. #5
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    Quote Originally Posted by regdude View Post
    Hi!
    Today we had a test at school and I had this interesting question ,that I didn't know how to answer.
    So it was said: A sphere is touching all 3 edges of a right triangle, its edges are 9 and 12 cm long. The distance from the sphere's center to the triangle plane is 10cm. How long is the radius of the sphere?

    I don't know if I made the picture right, but I think it should look like this:
    Somewhere I heard that if it both edges were the same size (for example, 9 and 9) then the distance and radius would be the same size.
    How could I get the radius? The answer should be 12.25 cm, but not sure.
    Here is a slightly different approach:

    1. The sides of the triangle have the lengths a = 12, b = 9, c = 15
    The vertices of the triangle are placed in a plane.

    2. The center of the sphere is located perpendicularly over the incenter of the triangle. The distance of the center of the sphere to the plane of the triangle is h = 10 cm.

    3. The height h, the radius of the incircle of the triangle and the radius of the sphere form a right triangle.
    Let A denote the area of the right triangle then the radius of the incircle is calculated by:
    r = \dfrac{2A}{a+b+c}

    That means r = 3.

    4. Therefore the radius R of the sphere is calculated by:

    R^2 = r^2+h^2

    With the given values R = \sqrt{109}\approx 10.44
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  6. #6
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    Ahh, the sphere isn't fully inside the triangle that way it is clear to me
    Seems that the book says the answer is 12.5, but it could be a mistake, won't be the first time.
    Thanks!
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  7. #7
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    Quote Originally Posted by regdude View Post
    ...
    Thanks for the hint, maybe I will find the right picture and from that will see some known figures.
    Seems now that the only thing unknown in my picture is a small part of the hypotenuse.
    I've attached a sketch. Maybe it is the right picture ... (?)
    Attached Thumbnails Attached Thumbnails A sphare in a triangle-kuglindreieck.png  
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  8. #8
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    I was about to answer this but see that earboth beat me to it.I calculated the triangle area = 54 and then summed the area of the three triangle areas comprising the total in terms of r . r x18 =54 r=3 and then to rad109 as the sphere R

    bjh
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  9. #9
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    Quote Originally Posted by regdude View Post
    Ahh, the sphere isn't fully inside the triangle that way it is clear to me
    Seems that the book says the answer is 12.5, but it could be a mistake, won't be the first time.
    Thanks!
    This is the answer to a different question: If the 3 vertices of the given right triangle are located on the surface of a sphere whose center is 10 cm above the plane of the triangle, then the center of the sphere is perpendicularly above the center of the circumcircle of the triangle whose radius is 7.5 cm.

    The height h above the triangle's plane, the radius r_c of the circumcircle and the radius R of the sphere form a right triangle:

    R^2 = r_c^2+h^2

    With the given values:

    R = \sqrt{7.5^2+10^2}=12.5
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