## A non-trivial problem

Anything goes, but keep it seemly...

### A non-trivial problem

We have a 4x4x8 brick (whatever measuring unit), with an ant at corner A:

Which point on the brick takes the longest travelling distance for the ant to reach?
udosuk

Posts: 2698
Joined: 17 July 2005

If the ant moves along the edges of the brick, then the corner diametrically opposite is 16 units from A.

Are you talking about the shortest direct path that is longest from point A?

MCC
MCC

Posts: 1275
Joined: 08 June 2005

MCC wrote:If the ant moves along the edges of the brick, then the corner diametrically opposite is 16 units from A.

Nope... Obviously the ant is allow to move on the surface (including the edges) of the block... Why would you think the ant could only move along the edges but not the faces!?

Are you talking about the shortest direct path that is longest from point A?

Yes... But the ant cannot go "inside" the brick... i.e. the brick is a piece of completely solid concrete without any holes... I think the word distance itself implies the shortest possible length...
udosuk

Posts: 2698
Joined: 17 July 2005

From A, up front face to half way along top edge, then along top face to dimetrically opposite corner.

2 x the square root of 32 = approximately 11.3137 units

MCC
MCC

Posts: 1275
Joined: 08 June 2005

That's the correct approach for the solvers to take.

But MCC's answer is not right... It is the trivial wrong answer which 90% of the population are supposed to come up with...
udosuk

Posts: 2698
Joined: 17 July 2005

What if the brick is lying on the ground?

then:

a. There is a whole side of the brick that cannot be reached by the ant.

b. The shortest route to the opposite corner of the bottom rectangle is 12 units. Longer than ~11.3.

c. The farthest point would lie somewhere in the middle of the opposite vertical edge.

Ruud.
Ruud

Posts: 664
Joined: 28 October 2005

Ok, how about if the ant ascends 4 units then travels diagonally across the top face to the corner.

4 + square root of 48 = 10.9282 units.

MCC
MCC

Posts: 1275
Joined: 08 June 2005

Anyway your destination is the incorrect point.
You should find a point that takes a longer minimal travelling distance for the ant to get to...

Ruud, good thinking... I didn't consider that before... Which is why when the original question was presented to me, the brick was "standing" on the square face instead of laid down like it is here...

But your insights then create another problem to solve...

For my version let's assume the ant can get between the ground and the bottom of the brick...
udosuk

Posts: 2698
Joined: 17 July 2005

Here we go again.

A picture tells more than a 1000 words.

Let's deflate, cut open and iron the brick. Then we can look at this problem in 2d.

I have drawn 4 copies of the opposite 4x4 square.

The circle crosses the opposite point, given by MCC as the answer.

When we overlap all 4 circle parts (as shown in the smaller square) you can see that there is still a tiny spot outside the circle. The point farthest from those segments is the true answer.

A real mathematician can now calculate the answer.

Any takers?

Ruud.
Ruud

Posts: 664
Joined: 28 October 2005

Ruud, once again great graphical analysis...

Also, worked out the variant problem culminated from your limitation that "the ant cannot go under the brick"...
Let's say the answer is a bit surprising too...

Furthermore, I have another extension to this problem:

Pick 2 points on the brick which have the longest (minimal) travelling distance from each other.
It's a theoretical question so no limitation of ant movement under brick whatsoever...

Hint: the centres of the 2 squares are 12 unit from each other no matter what. Can you do better?
udosuk

Posts: 2698
Joined: 17 July 2005

the answer to the first question:

1.5278640450004206071816526625374 Units Down from the corner top & opposite

That is 14.472135954999579392818347337463 units travelling distance

tarek

tarek

Posts: 2699
Joined: 05 January 2006

What was I thinking........

The answer is x units down from the corner top & opposite where:

SQR((4-x)^2+64)-x-8.944271909999158785636694674925=0

which will give a travelling distance of 12.944271909999158785636694674925 +x units

as you can see, the answer is 0.... so the ant would travel the longest distance if the destination was the top & opposite corner, travelling distance of 12.944271909999158785636694674925 units.

It doesn't matter if the ant can't go under the brick, as one side of the brick is a square.....

It would have been one good equation if the dimensions were 8*4*3 rather than 8x4x4

tarek

tarek

Posts: 2699
Joined: 05 January 2006

OK,

here is the graphical representation of the "ant cannot go under the brick" variation. The orange part cannot be traveled by the ant.

The 8x4 side opposite of a is shown twice. The outer border (red) shows the distance to the opposite corner again.

The smaller image shows how all outlines match up.

The farthest point must be on the opposite edge, ~1.5 units from the top.

Ruud.
Ruud

Posts: 664
Joined: 28 October 2005

Rethinking........

Looking at Ruud's Illustration....... the answer seems to be easier......

The corner top & opposite is 11.313708498984760390413509793678 units away........

making the corner opposite on same surface 12 units away......

so the point is somewhere over that edge

back to the equation

SQR(x^2+64)-x-4=0

& x is the number of units from the oppsite same level corner going up

tarek

tarek

Posts: 2699
Joined: 05 January 2006

Rethinking again,,,

Thanx for Ruud for this....

The answer should be 2.33333333 units up the edge of the corner on the same level & opposite (or 1.6666666 down the edge from the corner on the top & opposite...

The distance covered is 12.292725943057183080003032375684 units

tarek

tarek

Posts: 2699
Joined: 05 January 2006

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