Tough Math Question

[JPtux]

Okay, I have a math question that I would like help with. My teacher and I are unable to solve this problem for some reason:

In 3-dimensional space line m intersect plane P in point T, and m is perpendicular to P. The points in space which are 2 inches or 4 inches or 6 inches from plane P are also 5 inches or 7 inches or 9 inches from point T lie on circles. Find the sum of the squares of the radius lengths of the circles.

 

Our answer is 616 but the answer is supposed to be 596. Can anyone solve this?

P.S. I'm sorry if I posted this in the wrong forum, I'm new.

NOTE: Not a HW problem.

Thanks in advance!

[Mr Skeptic]

First, assume that point T is the origin, plane P is the xy plane, and line m is the z axis (since these are nice and the problem doesn't specify their locations, only the relation between them)

 

Now make triangles, the distance from plane P is along the line m (the z axis) or a line parallel to it that intersects the circle, the distance from T to the circle is another line, and the distance from a point on m to the circle the third line. I assume the second set of distances is supposed to be the hypotenuse? The circles are the points a given distance from a the plane and from T? This should be a right triangle, with the distance from T a hypotenuse and the distance from the plane one side. The third side is the radius of the circles.

 

Anyhow, show the work you and your teacher did and we will tell you where you went wrong and help you to get the right answer.

[doG]

I too looked at this and found myself needing to make assuptions because the description wasn't clear enough. From what the OP described I envisioned 3 planes parallel to P 2", 4" and 6 inches from P that contained the 3 circles such that all of the points on those circles would lie 2", 4" or 6" from P. I then imagined 3 rays, 5", 7" and 9" long originating from point t, and revolving about m to make the circles on the planes where each circle would be the base of a cone on each of the respective planes and the rays would be the hypotenuse of the triangles described in Mr. Skeptic's post. This scenario does not lead to either of the answers given by the OP.

[JPtux]

I too looked at this and found myself needing to make assuptions because the description wasn't clear enough. From what the OP described I envisioned 3 planes parallel to P 2", 4" and 6 inches from P that contained the 3 circles such that all of the points on those circles would lie 2", 4" or 6" from P. I then imagined 3 rays, 5", 7" and 9" long originating from point t, and revolving about m to make the circles on the planes where each circle would be the base of a cone on each of the respective planes and the rays would be the hypotenuse of the triangles described in Mr. Skeptic's post. This scenario does not lead to either of the answers given by the OP.

 

That is pretty much what I did. You have to remember that the 5" ray is impossible in the plane that is 6" away from P. Furthermore, there would be 6 planes, the ones that you mentioned, and the ones on the opposite side of P.

[doG]

That is pretty much what I did. You have to remember that the 5" ray is impossible in the plane that is 6" away from P. Furthermore, there would be 6 planes, the ones that you mentioned, and the ones on the opposite side of P.

 

You would not need the planes on both side simply to calculate the radii of the circles. The second set of planes on the other side of P would be symmetric to the others. I also assumed there are only 3 circles, one drawn by the 5" ray on the plane 2" from P, one drawn by the 7" ray on the plane 4" from P and one drawn by the 9" ray on the plane 6" from P. As you said, the 5" ray would obviously not reach the 6" plane. The sum of the squares of the 3 circles I described is 99 so it seems like there is some missing information.

[the tree]

doG, three circles? I made it 16, with 3 planes on either side of P and 3 circles in each of them save for the pair of impossible circles.

 

(I would upload some images or something but it turns out the Wifi on trains isn't that great)

 

So I figured that the sum of the squares should be:

[math]2 \times \Bigg( \sum_{b \in \{ 2,4,6 \}} \sum_{c \in \{ 5,7,9 \}}

\begin{cases}

c^2 - b^2 & \mbox{if } c>b \\

0 & \mbox{o/w}

\end{cases} \Bigg\} \Bigg) = 616 [/math]

Which is the answer the OP got. So either we're both wrong in (probably) the same way, or the textbook is wrong in a different way.

 

Mr. Skeptic, did you get 596 somehow? If so, how?

[doG]

Which is the answer the OP got. So either we're both wrong in (probably) the same way, or the textbook is wrong in a different way.

 

Your answer matches the answer that's supposed to be the one given. I don't see how c=5, b=6 can be part of a real solution so it must be assumed that all of the possible real circles is what was intended by the text book. I get the same answer as you with that set.

[D H]

I'm with doG here: 99. The universal answer, 42, just doesn't seem to fit.

 

The question as written in the text would be nice.

[JPtux]

I'm with doG here: 99. The universal answer, 42, just doesn't seem to fit.

 

The question as written in the text would be nice.

 

I copied it exactly as it is.

[Mr Skeptic]

the tree got the answer folks.

 

Mr. Skeptic, did you get 596 somehow? If so, how?

 

No, I just set up for JPtux to find the radius of the circles. Since I did not realize that the problem meant all possible circles, I was very confused as my answer would have been under 300 using just 3 circles. So I didn't bother to calculate it. I figured JPtux would clarify how he got a bigger number, but you beat him to it.