3D Object
* Draw Square 1 with sides of length x. Draw a smaller square, inside square 1, with sides of length x/2. Align the smaller square such that one of it's sides lies on top of, and in the center of, the lower side of square 1. (lower side of square 1 refers to the side closest to the bottom edge of the paper upon which you are drawing)
* Repeat the above instructions and label the second drawing as Square 2
* Square 1 and Square 2 are identical.
* Square 1 is a plan view (top view) of a 3 dimensional object.
* Square 2 is the elevation (front view) of the same object.
* Draw a 3 dimensional representation of the object.
* DESCRIBE the OBJECT.
-- Zaux
* Repeat the above instructions and label the second drawing as Square 2
* Square 1 and Square 2 are identical.
* Square 1 is a plan view (top view) of a 3 dimensional object.
* Square 2 is the elevation (front view) of the same object.
* Draw a 3 dimensional representation of the object.
* DESCRIBE the OBJECT.
-- Zaux
Labels: geometrick, mathemagic, SharedPuzzle
posted by Rajesh Lal at
10:24 PM
29 Comments:
I think you might be describing two cubes, with the smaller having one half the side length the side of the other.
The small cube would be inside the larger one, pushed over to the middle of one side of the larger one.
See my blog at
Ragknot.blogspot.com
But there are other 3D objects (besides two cubes) that could give the exact same plan and front views.
What if the side view looked like a right triangle with a line crossing at 1/2 the height?
The solid is a ramp with a "cube like" protuberance. With my poor english it's very hard to describe it, but the (left) side view are two rectangle triangles:
1st with the vertices in (0;0), (x;0) and (0;y) and the other with the vertices (x;0), (x;x/2); (x/2;x/2).
hi ragknot ...
your 2 cube explanation works only if the object is transparent with non-transparent edges. This object is an opaque composite of 2 objects ... but not cubes.
I can think of no other opaque objects, fitting this description, with the same views, plan and elevation.
hi Miguel ...
sorry for the language difficulty ... but, help me understand: what is a "rectangle triangle"?
Ragnot's first comment seems pretty accurate other than to add that either the larger cube is constructed of transparent material (glass maybe?) or is not solid.
Otherwise the smaller shape would not be visisble through it.
Opps, sorry I missed Zaux's comment re Ragnot's attempt.
How about this.
The large square refers to a half cube - cut so that it slopes from the top rear edge, to the bottom front edge.
The second smaller cube is the same shape, upside down, resting on the bottom of the first half cube.
I hopr that makes sense.
hi Anonymous ...
you're right ... for Ragknot's solution to work, the object would have to be transparent with non-transparent edges. However, it is an opaque object.
See here for my attempt at drawing what I have tried to describe.
http://www.dualaspect-photography.co.uk/Shape-1.jpg
It's a square hole/tunnel on a 45 degree plane.
But we need a side elevation to be able to make an accurate description, otherwise our interpretations are going to make a lot of suppositions...
Karl
Zaux, its a triangle with a 90º angle, sry for that.
The solid is made by 2 prisms:
1st with triangular base (triangle with 2 sides size x and 90º between them) and hight x, perpendicular to base.
2nd is the same, but the triangle sides are x/2.
The solid has 12 vertices:
V1(0;0;x)
V2(0;0;0)
V3(x;0;0)
V4(0;x;x)
V5(0;x;0)
V6(x;x;0)
V7(x/2;x/4;x/2)
V8(x/2;3x/4;x/2)
V9(x;x/4;x/2)
V10(x;3x/4;x/2)
V11(x;x/4;0)
V12(x;3x/4;0)
and 17 edges:
E1(V1;V2)
E2(V2;V3)
E3(V3;V1)
E4(V1;V4)
E5(V4;V5)
E6(V5;V2)
E7(V4;V6)
E8(V5;V6)
E9(V6;V3)
E10(V7;V8)
E11(V8;V10)
E12(V10;V12)
E13(V12;V8)
E14(V7;V9)
E15(V9;V11)
E16(V11;V7)
E17(V7;V9)
Didn't double-check, no time... :P
The question says it's a 3D object..... it does not say it is a "Solid".
A wire framed object can be a 3D object, but not a soild.
Ok, I forgot the faces... but from the wireframe it's not difficult to place them ;)
Ragknot ....
sorry ... you're exactly right ... a wire framed object is certainly a 3D object ... but, in this instance, the desired solution is a 3D solid, opaque object.
Miguel ...
wow ... lots of coordinates ...
you and Anonymous seem to have similiar thougths about the object
Has anyone graphed Miguels' coordinates to what it is? I've not done that ... yet
Hi Karl ...
an interesting thought ...
but the square hole tunnel, on a 45degree plane, solution is clever ... but it won't work, because the plan and front elevation views would look similiar, but the dimensions would appear differest. In the desired object, the views and dimensions are identical.
oops ... typo...
meant to type "has anyone graphed Miguel's coordinates to see what they produce?"
Karl ...
I have pondered your solution ... it's works except for one small detail. The front elevation would be right ... however, the plan view would not have a continuous line along the edge of the tunneling. But it is a really clever approach ...
Friday, Dec. 4th, I will post a descriptive solution and a link so that a visual solution may be viewed.
Please review my revised picture on
ragknot.blogspot.com
Someone said a ramp, so I drew a ramp with a small cube subtracted.
Ragknot ...
nice pic ... fullfills all the criteria except one... if you look at the plan view of your object... there is no line across the front where the section was removed... remember the plan view is a complete square ...
I took the 12 verticies of Miguel Tato. and constructed the solids I thought they gave.
Check this pic on
ragknot.blogspot.com
For those who know AutoCad, it is the same solids as the other, but ended with a "Union" instead of a "Subtract". (a union of two solids rather that subtracting the small from the large)
If this one is correct, I want to thank Miguel.... He was right, I just filled his coordinates.
Ragknot ...
I was going to wait 'til Friday... but Miguel's coordinates are exactly correct ... also the second post by Anonymous describes the same object. He even suggests a site where he drew a conceptual image of the object (not to scale).
This problem is normally presented as a visual exercise ... by making it a word problem, it became a little more difficult.
If you examine the plan and front elevation of the object you plotted, it is clearly the same view.
Thanks for everyones participation ... I enjoyed it.
Zaux
Ragknot ...
I was going to wait 'til Friday... but Miguel's coordinates are exactly correct ... also the second post by Anonymous describes the same object. He even suggests a site where he drew a conceptual image of the object (not to scale).
This problem is normally presented as a visual exercise ... by making it a word problem, it became a little more difficult.
If you examine the plan and front elevation of the object you plotted, it is clearly the same view.
Thanks for everyones participation ... I enjoyed it.
Zaux
sorry for the double post... the first delayed so long I thought it was not goint ot post ...
Here's the descriptive solution I wrote after wording the puzzle:
The Object consists of a wedge, with a smaller wedge protruding from the thin edge of the larger wedge.
Consider a cube with sides of length x. Slice the cube diagonally to form 2 wedges. Keep one of the wedges and discard the other. The wedge has a square base, each side having a length x. The height of the thick side of the wedge is also x. Just to clarify ... the side view is an isosceles triangle, having 2 sides of length x. Call this Wedge 1.
Consider a smaller cube with sides of length x/2. Also slice this cube diagonally and discard one of the halves. This wedge has a square base with sides of length
x/2. Call this Wedge 2.
Imagine wedge 1 sitting on its base. Turn wedge 2 over such that its base is on top. Place wedge 2 on wedge 1 such that the hypotenuse planes of each wedge are touching. The thickest side of wedge 2 would be on top of the thinnest edge of wedge 1. Center wedge 2 along the thinnest edge of wedge 1. Now consider the 2 wedges as 1 piece and you have the OBJECT ... the plan view and front elevation will be identical.
Here's a link to my conceptual (not to scale) drawing:
http://www.blogger.com/post-edit.g?blogID=5698395345020257257&postID=4542619959218370090
Thanks again for the participation,
Xaux
My final comment never posted ... goint to try again ... sorry if it double posts:
December 1, 2009 9:21 PM
Zaux said...
Here's the descriptive solution I wrote after wording the puzzle:
The Object consists of a wedge, with a smaller wedge protruding from the thin edge of the larger wedge.
Consider a cube with sides of length x. Slice the cube diagonally to form 2 wedges. Keep one of the wedges and discard the other. The wedge has a square base, each side having a length x. The height of the thick side of the wedge is also x. Just to clarify ... the side view is an isosceles triangle, having 2 sides of length x. Call this Wedge 1.
Consider a smaller cube with sides of length x/2. Also slice this cube diagonally and discard one of the halves. This wedge has a square base with sides of length
x/2. Call this Wedge 2.
Imagine wedge 1 sitting on its base. Turn wedge 2 over such that its base is on top. Place wedge 2 on wedge 1 such that the hypotenuse planes of each wedge are touching. The thickest side of wedge 2 would be on top of the thinnest edge of wedge 1. Center wedge 2 along the thinnest edge of wedge 1. Now consider the 2 wedges as 1 piece and you have the OBJECT ... the plan view and front elevation will be identical.
Here's a link to my conceptual (not to scale) drawing:
http://www.blogger.com/post-edit.g?blogID=5698395345020257257&postID=4542619959218370090
Thanks again for the participation,
Xaux
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