Diagonal in Space
Three unit squares are each divided into two pieces (A and B) by cutting along the line connecting the midpoints of two adjacent sides. The pieces are then joined to a regular hexagon (C), as indicated in the figure below. The net is then folded up to form a polyhedron.
Find the volume of this polyhedron.
Give your answer to 2 decimal places.
The answer is 0.5.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Brilliant and exceptionally solution, but as always, sir. What intrigues me is that I tried the problem with a different approach. Will you please help me in pointing out where I am wrong?
When I visualised the formed polyhedron, I thought we could easily form a rectangular triangular pyramid. If we try to understand by your diagram it turns out if we choose any one of the two portions and for each of the three corresponding corners of the cube, we extend the edges of the cube and that of the hexagon, then two edges of hexagon and one of the cube will be concurrent and thus we will have formed a rectangular triangular pyramid formed from cube of side (3/2) units. Now that we can easily find out volume of a rectangular triangular pyramid, we can find the volume of original polyhedron by visualising only addition of three smaller pyramids.
It turns out the answer doesn't match. Please aid me in telling where I stand wrong?
Log in to reply
Okay, let me see if I follow you. Assuming that squares A have a side length of 2 , we divide the solid into 4 parts, 3 of them which are right pyramids with sides 1 , 1 , 2 , which have a total volume of
3 ⋅ 3 1 ⋅ 2 ⋅ 2 1 ( 1 ⋅ 1 ) = 1
and then one pyramid with a hexagon base of side length 2 and a height of 3 = 1 2 + 2 2 − ( 2 ) 2 , which has a volume of
3 1 ⋅ 3 ⋅ ( 2 3 3 ⋅ ( 2 ) 2 ) = 3
for a total volume of 4 which is half the volume of a cube of side length 2
Log in to reply
Thanks, though I meant something else, extending the lines along the edge's of cube and extending sides of hexagon, but your solution helped me correct mine. Thanks we have now another good solution.
Oh wow, how did you conceive that?
Log in to reply
Well, how many different ways 3 right-angle corners can fit together? It's an engineer's recipe for building a rigid box. And then, since the hexagon has to be regular (see comment above), where would we find a regular hexagon in a cube? It's like your problem about cutting a tetrahedron, where the cross section was a regular square.
This does suggest a class of problems, what other kinds of convex polyhedra exhibiting symmetries that non-trivially yields cross sections other than regular squares and hexagons? Is there one that will non-trivially yield a regular pentagon, for instance? Let me think about that.
Is this a unique solution?
Log in to reply
W Rose made sure of that by specifying that the cut is between midpoints of the sides of the squares. Otherwise, the solution wouldn't be unique.
WOW!Pure brilliance!
Two of these make up an unit cube.