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The tetrahedron BCDH is cut off from the cube and is then placed on top of the solid ABDEFGH as shown in the figure. The face BCD of the tetrahedron coincides with the face BAD of the solid ABDEFGH such that vertex H of the tetrahedron moves to the position of V and vertex C coincides with A. The faces BHD and BVD of the new solid lie on the same plane.

What's the projection of point F on the plane BDH?

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Just a placeholder to let you know that I am looking at this problem, and will reply soon (1-2 days). If you need a hint more quickly than that, just mention so, and I'll try to speed up a bit.

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I don't need a quick hint, just do it in your leisure time.

Your reply inspire really much. If you don't reply, I think I will probably forget it in 1-2 days.

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I'm not very sure what is meant by "the projection" of F onto the plane, but I'm assuming that question to be something like if you were to draw a line perpendicular to the plane and intersects F, where would that line intersect with the plane. If this is the question then the answer seems pretty straight forward. If it is not I must apologize for misunderstanding.

Assuming myunderstanding is correct however, you would draw a plane that is perpendicular to plane VDHB, and colinear with VH. You will get yourself a right triangle with verticle side of 2a, horizontal side of sqrt(2)a, and hypotenuse of sqrt(6)a. Putting the a's asside as just a unit figure, we can find all the angles of this right triangle. Draw a line from F to intersect with the line VH, and assume that it intersects as a right angle. Using the knowledge that all triangles have a 180 degree internal angle sum, you can find all the other angles as well. From here, it's very basic geometry, and I will leave it to you.

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My method:

Draw a line perpendicular from F to BH at I. Extention of BH may be needed.

Then, Draw a perpendicular line from BH in the plane BVDH. Name the line IZ.

Then, draw a line perpendicular from F to IZ cut at P.

Then, P is the projection.

Is this a correct method? Wait a minute, I show you the picture in order to help you to get fully understood.

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What softwares do you suggest?

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MS Paint? ^^v

I didn't understand your explanation above by the way. You used alot of unreferenced variables (they're not on the diagrams), but your method already seems unnecessarily complicated to say the least.

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so what is the projection in your method of F?

Projection of a point on a plane refers to the line joining the point and the projection(a point) is perpendicular to all line passing through the plane.

what's the distance from the point F to that plane>?

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I've explained it in my first reply. Follow the steps carefully and draw it out. All the variables are referenced on the diagram you provided.

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The plane that is perpendicular to plane VDHB is VFH?

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Yup. The plane I'm talking about has the lines VF, VH, and FH on it. The point is you want to get a plane to work on. Now this place must be perpendicular to the VDHB right? since that's your objective, to find the perpendicular line. The next part is to ensure that this plane intersects with point F, your point of interest, and lastly, to make it possible to solve, it should also intersect other points on the cube.

Once you identify this plane, you can draw out the plane on a 2D piece of paper. The right triangle with points V, F, and H. VF=2 units, FH = sqrt 2, VH = sqrt 6. Since you know all sides of this right triangle, you can figure out the angles using sine and cosine. Now draw a line from F to intersect with VH. Make the intersection 90 degrees. Now use geometry to work out the angles and the length.

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• 1 month later...

Well, I worked at this problem for a while and I believe to possess the (right I hope) solution.

I'm not at all enthusiastic about the prospect for my message, containing only the answer without the according steps, to be deleted ...It already happened to me once in another forum...that's why this question:

Do you wish the answer first, so that we can compare our results, or should I write it all down and place the answer at the end, or should I give a hint?

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I think I have the answer as well.

[hide]F projects to itself, as the two are on exactly the same plane. from H to M goes up one, over 1/2, and over the other way 1/2. From M to F goes up one, over 1/2, and over the other way 1/2. So the three are collinear, so F is on the plane, so its projection is itself.[/hide]

-Uncool-

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There is a method, including no angle calculations, and needing only the information about the inner diagonal FC of the cube.

The tetrahedron is nothing but the solid cut off the cube and placed on its rest, so I will use only the left figure and its symbols.

1. We will find the diagonal FC: $FC=\sqrt{(FG^2+GH^2)+HC^2}=a\sqrt{3}$ ;

2. Next we prove that F' lies on the segment FC:

Take the plane FACH. The points F and C both lie in this plane, then the straight line and, hence, the segment FC lie in this plane. Take the segment MH. M an H both lie in FACH, then the segment MH lies in this same plane. Since F and C lie in the opposite halfplanes in respect to MH, then FC crosses MH. Because of the fact that MH lies in both FACH and BDH, it is the only intersection line, so any line of FACH crosses BDH at a point lying on MH. The point of intersection is exactly F'.

3. Construct the height from M to the base FGHE, let it be N. Then the parallel segments NA and MH divide the segment FC in proption 1:1:1, that is in three equal parts.

Since the two parallel lines NA and MH divide AC in propotion $AM:MC=1:1$ then the same lines divide FC in the same propotion (see Fales theorem). The same is valid for NA,MH and the lines FH and FC. Since we operate with equal segments FN=CM=AM=HN, we can conclude that FC is split in three equal segment too.

So FF':F'C=2:1 and it follows $FF'=\frac{2}{3}FC=\frac{2}{3}a\sqrt{3}$

That's all. There is actually very little maths needed. Cheers!

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