We cut a prism having a 6-gon regular polygon with side length 1 as base (see the following figure).If the distances of the vertices of the oblique cutter plane from the base are $2,3,x,y,11,z$ calculate $x+y+z$.
I really have no idea for this!
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Hamid Reza Ebrahimi
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Let $ABCDEF$ be the base hexagon, and let $A'B'C'D'E'F'$ be the oblique hexagon where $$AA'=2,\quad BB'=3,\quad CC'=x,\quad DD'=y,\quad EE'=11,\quad FF'=z$$
Let $M$ be the intersection point of the oblique hexagon with the line parpendicular to the base passing through the center of the base hexagon.
Since the base is a regular hexagon, we can see that the midpoint of $A'D'$, the midpoint of $B'E'$, and the midpoint of $C'F'$ exist at $M$.
It follows from this that $$\dfrac{2+y}{2}=\dfrac{3+11}{2}=\dfrac{x+z}{2}\implies \color{red}{x+y+z=26}$$
mathlove
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Very good,bravo – Hamid Reza Ebrahimi Jun 10 '17 at 06:12
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Is the oblique hexagon regular too? – Hamid Reza Ebrahimi Jun 10 '17 at 06:14
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@HamidRezaEbrahimi: No, it is not necessarily regular. You can see that by considering the trapezoids of the sides of the prism. – mathlove Jun 10 '17 at 06:17
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@HamidRezaEbrahimi: We can prove that it is not regular. We have $A'B'=\sqrt 2$. In order for the oblique hexagon to be regular, we have to have $B'C'=C'D'=D'E'=\sqrt 2$ which implies $x=4,y=5,EE'=6$. This contradicts that $EE'=11$. – mathlove Jun 10 '17 at 06:26
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Let's position the base hexagon on the $XY$ plane as shown below:
Assuming the vertice heights are ordered counterclockwise, we can give the three vertices in the cutter plane with known heights the coordinates $A' = (\frac 1 2, -\frac{\sqrt 3}{2}, 2)$, $B' = (1, 0, 3)$ and $C' = (-1, 0, 11)$. We can then find the equation of the plane passing through these three points, which is $$4 \sqrt{3}x - 6y + \sqrt{3}z = 7 \sqrt{3}$$
The missing vertice heights can then be found by finding the intersection between the plane above and the lines passing through each vertice which are perpendicular to the $XY$ plane. The equations for these lines are: $$D: x=\frac 1 2, y= \frac{\sqrt 3}{2}, z=t$$ $$E: x=-\frac 1 2, y= \frac{\sqrt 3}{2}, z=t$$ $$F: x=-\frac 1 2, y= -\frac{\sqrt 3}{2}, z=t$$
Inserting the values for each line into the equation for the plane and solving for $t$ we find that $D' = (\frac 1 2, \frac{\sqrt 3}{2}, 8)$, $E' = (-\frac 1 2, \frac{\sqrt 3}{2}, 12)$ and $F' = (-\frac 1 2, -\frac{\sqrt 3}{2}, 6)$. I.e., in the terms of your question, $x=8, y=12, z=6$. Hence $x+y+z=26$.
Below is a figure showing the resulting shape:
Jens
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Correct,do you think it might have a non-analytic solution? – Hamid Reza Ebrahimi Jun 10 '17 at 05:41
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Thank you for your correct answer,but unfortunately just one answer could be accepted... – Hamid Reza Ebrahimi Jun 10 '17 at 06:34