No, it should not be. The prism is hollow. The rod goes in from the top, turns the corner inside the prism and leaves from the bottom right. The rod linear and not bending. Also assume that top and right-hand side prisms (small one) are long enough - as seen.

This is a hard geometry problem. Solving Fox 93 gives you an idea on this solution. 93 should be posted here: http://www.8foxes.com/Home/93

newzad, Your answer is posted above. You're on right path. But the answers to both foxes (93 and 94) are exact! Not an approximation. There may be a minor mistake in your approach. We have a different answer. We would insists on solving Fox 93 first: http://www.8foxes.com/Home/93

Then you can add another dimension and try solving Fox 94. Thanks.

sin(2alpha)=1/a cos(2alpha)=8/b Length of the rod = L = a+b = f(alpha) gIVEN by O. alpha=q L=1/sin(2q) + 8/cos(2q) L'=-(cos(2q)*2)/sin^2(2q) + sin(2q)*2*8/cos^2(2q) = 0 the min value of L can go tru the corner. i am skipping 2nd order condition (L''>0)

Hey binary: That's the solution for Fox 93. Not this one. Fox 94 can be solved with the same logic, but a little more work is needed, We'll post Fox 93 here as well.

newzad: the rod must touch the following points at the same time: 1. upper-left corner at the back AND 2. common line segment between upper prism and lower prism AND 3. lower-right corner in the front

i don't have the solution yet. no time to solve actually.something similar to ur good solution to 93 would work. do this: drop a normal line from common segment line (point 2 in my previous post) also draw the projection of the rod on the base.there should be pithagoras and congruent right triangles. then right the length of the rod in terms of one of the lengths (or an angle). then we may get a function to differantiate. will try myself here. but later. -binary descartes

it can be infinite

ReplyDelete?

No, it should not be.

ReplyDeleteThe prism is hollow. The rod goes in from the top, turns the corner inside the prism and leaves from the bottom right.

The rod linear and not bending.

Also assume that top and right-hand side prisms (small one) are long enough - as seen.

This is a hard geometry problem. Solving Fox 93 gives you an idea on this solution. 93 should be posted here:

http://www.8foxes.com/Home/93

Good luck.

then the answer is 12

ReplyDeleteBut i still didn't understand, rod can touch the bottom at infinity while it touch the corner

ReplyDeleteNo, the connected prism is an L-shaped tube. The only enterance is from the TOP, and it leaves from the RIGHT-HAND-SIDE. It's walls are rigid.

ReplyDeleteAs you say the rod can go to infinity while touching the corner, BUT, it can not come to that position by entering from the TOP.

Have you ever carried a long piece of wood or pipe around an L-shaped corridor?

but think about a infinit length rod entering from the top

ReplyDeleteAnd how is it gonna turn around the corner???

ReplyDeletemy english is not good

ReplyDeletemaybe i am missing something

by meaning go around do you mean entering from top and leaving from bottom?

ReplyDeleteEntering from Top, leaving from Bottom-Right.

ReplyDeleteThe answer is the same if it enters from Bottom-right and leaves from the Top.

i think the answer is C

ReplyDeletebut i guess it is B

http://i38.tinypic.com/2py59ts.gif

newzad,

ReplyDeleteYour answer is posted above. You're on right path. But the answers to both foxes (93 and 94) are exact! Not an approximation. There may be a minor mistake in your approach. We have a different answer. We would insists on solving Fox 93 first:

http://www.8foxes.com/Home/93

Then you can add another dimension and try solving Fox 94.

Thanks.

For Fox 93:

ReplyDeletesin(2alpha)=1/a

cos(2alpha)=8/b

Length of the rod = L = a+b = f(alpha)

Minimum value of f(alpha) gives the maximum length of the rod (why?)

Fox 94 may have a different solution.

This comment has been removed by the author.

ReplyDeleteI remember seeing a solution here. what happened?

ReplyDeletesin(2alpha)=1/a

cos(2alpha)=8/b

Length of the rod = L = a+b = f(alpha)

gIVEN by O. alpha=q

L=1/sin(2q) + 8/cos(2q)

L'=-(cos(2q)*2)/sin^2(2q) + sin(2q)*2*8/cos^2(2q) = 0 the min value of L can go tru the corner.

i am skipping 2nd order condition (L''>0)

2cos^3(2q)=16sin^3(2q) gives the answer.

binary descartes

Hey binary:

ReplyDeleteThat's the solution for Fox 93. Not this one.

Fox 94 can be solved with the same logic, but a little more work is needed, We'll post Fox 93 here as well.

newzad: the rod must touch the following points at the same time:

ReplyDelete1. upper-left corner at the back AND

2. common line segment between upper prism and lower prism AND

3. lower-right corner in the front

intuitively thats what i see.

-binary descartes

yes,

ReplyDeleteand how will you solve?

can you show your solution?

I have to go for now tomorrow i have an exam

ReplyDeletei will be here for a few hours later

post your answer

i don't have the solution yet. no time to solve actually.something similar to ur good solution to 93 would work. do this:

ReplyDeletedrop a normal line from common segment line (point 2 in my previous post) also draw the projection of the rod on the base.there should be pithagoras and congruent right triangles. then right the length of the rod in terms of one of the lengths (or an angle). then we may get a function to differantiate. will try myself here. but later.

-binary descartes

İ think i have to say again

ReplyDeletefor this problem you have to add 16

fox 93 the answer was 5sqrt(5)=sqrt(125)

for this question there is a more dimension different from fox 93

and this dimension doesn't effect the rod while going through the corner

so for the biggest length this dimension must be 16

a^2+b^2+c^2 must be the biggest

a^2+b^2=125 which proved at fox 93

c^2=16

ROD=sqrt(125+16)=sqrt(141)

it is solution is similar to fox 93

here is the solution

ReplyDelete