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An ant had to travel from one corner of a room to diagonally
opposite corner. It can walk on walls, floor and roof only.
How will you find the shortest path for this ant ?
- 9 Answers
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Answer / kelley w.
The room is a rectangular prism (like a cube only
rectangular). The 4 floor corners will be represented by
A,B, C, and D. The ceiling corners will be represented by
E, F, G, and H, with E being directly above A, F being
directly above B, Gbeing directly above C, and H being
directly above D. (Draw a picture, it helps) the ant needs
to go from the floor corner A to the ceiling corner G (if
I’m not mistaken). The shortest path would be a diagonal
line from A to G, but seeing as the ant cannot walk on air,
we have to find the quickest path using the walls, floor
and ceiling.
F__________G
/| /|
/B|________/_|C
E/_________H/ /
| / | /
|/_________| /
A D
The ant has four logical choices:
1) He could cross diagonally from A to C, then crawl
up the wall to G. We would have to find the length of
diagonal AC and the height of the room, which would be CG.
(He could also crawl up from A to E, and then diagonally
from E to G, but this would be the same distance as the
aforementioned choice.)
2) He could cross along the wall from A to D, then
crawl diagonally up the wall from D to G. We would have to
find the width of the room, which would be AD, and the
length of diagonal DG.
(He could also go diagonally from A to F, and then crawl
along the ceiling from F to G, but that would be the same
distance as the aforementioned choice.)
3) He could cross along the floor from A to B, then
crawl diagonally from B to G. We would have to find the
length of the room, which would be AB, and the length of
diagonal BG.
(He could also go diagonally from A to H, and then crawl
along the ceiling from H to G, but that would be the same
distance as the aforementioned choice.)
4) He could cross along the wall from A to the
midpoint of DH, then continue from the midpoint of DH to G.
We’ll call the midpoint of DH “M”. We would have to find
the width of the room, which would be AD, and half the
height of the room, which would be DM and HM. We would then
have to find the length of the diagonal AM, and then the
length of the diagonal MG.
(He could also go diagonally from A to the midpoint of BF
and then diagonally to G, but this would be the same
distance as the aforementioned choice.)
Using common sense, we can see that choice 4 will be the
quickest route for the ant. If you have trouble seeing how
this option is quickest, make up fake dimensions for the
room and find the values I mentioned above for each choice
to determine the shortest distance.
(For example, pretend the room is 5 feet high, 10 feet
wide, and 20 long. AE=5, AD=10, and AB=20.)
| Is This Answer Correct ? | 43 Yes | 12 No |
Answer / hemant from india
1. sqrt((a+b)^2+c^2),
2. sqrt((b+c)^2+a^2)
3. sqrt((c+a)^2+b^2)
which one is smaller in these is right answer
cos ant travels only thorough plane so , just unfold cuboid
u will find u have to add 2 dimensions and third dimension
in perpendicular direction and gets this ans
| Is This Answer Correct ? | 36 Yes | 8 No |
Answer / dhruv saksena
The ant will walk to the midpoint of the edge touching the corner opposite diagonally on the floor. From the mid point it will travel directly to its destination corner. this is the shortest distance possible for it to travel.
This can be calculated using a combination of differentiation and the Pythagoras theorem, it was in my test.
| Is This Answer Correct ? | 10 Yes | 5 No |
Answer / guest
Shortest path is always a straight line.
So do flat the room walls and floor and draw a straight line
between the opposite corners of the room. If room is not
cubical then that line will not pass through any edge of room.
| Is This Answer Correct ? | 24 Yes | 23 No |
Answer / jawahir kashim
Unfold the 3D cuboid into 2D. Let (A B C D,E F G H) be the edges of the cuboid.Such that A is below E and others edges resp.Unfold such that point A and its diagonal are in the same plane. The straight line joining A and G ie the diagonal in 2D is the shortest line.
| Is This Answer Correct ? | 2 Yes | 4 No |
Answer / avinash singh
shortest distance in a cube is diognaly. but ant cant fly. she can travel in a flor or roof diognaly from one corner to another one. and one side.
ex= if side is given 3. so diognaly on floor distance is 3 under root 2. and add one side is 3. so total distance is 3(1+under root 2).
| Is This Answer Correct ? | 3 Yes | 5 No |
Answer / suda
Insteatd of moving on wall and roof,If the ant moves in a
straight way in floor,it can reach the opp.corner
diagonally and it would be the shortest path for the ant.
| Is This Answer Correct ? | 24 Yes | 28 No |
Answer / devangini
There is no restriction on the way the ant can walk.
Consider the cuboid given in answer 3 and let X be the
midpoint of the surface ABCD. now the ant can walk from A to
X, X to D and D to G.
| Is This Answer Correct ? | 1 Yes | 6 No |
Answer / guru
I tried with above 10,20, and % dimensions.But the shortest
is to move diagonally across the 2 longest dimensioned wall
(10 & 10) and then move straight..(5)
So its sqrt(10^2+20^2)+5 = 27.36
if you take any other method its beyond 30.2
So A to C and then G is shortest
| Is This Answer Correct ? | 1 Yes | 10 No |
Four men - Abraham, Bobby, Clinton and Denial - are standing in a straight line. 1. One man is fair, handsome and unscarred. 2. Two men who are not fair, are each standing next to Abraham. 3. Bobby is the only man standing next to exactly one handsome man. 4. Clinton is the only man not standing next to exactly one scarred man. Who is fair, handsome and unscarred?
how do i improve my logical questions. is there any website?
u use any operation solve this problem 45-45=45
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