Nov 6, 2009

Four Ants - HC Verma

Solved a more difficult problem in HC Verma 5-6 years back.. :)

Four bugs are on the corners of a 1 meter square. Each bug always faces the next bug (on the next clockwise corner). If they all walk forward at the same speed until they meet, how far does each bug travel?

Solution:
Highlight the part between the * symbols for the answer.
*Since the bugs always move perpendicular to each other, so when they meet at the center they have traveled a distance equivalent to their initial separation i.e. 1 m.

If some of you remember, in the HC Verma problem, the three ants were moving in a equilateral triangle of side-size 1. The three ants were such that 1 is following 2, 2 is following three and three is following 1. By symmetry, they would meet at the centre. The time taken was 2/3v as you had to take components along the line joining the two ants. So, v(1+cos60) was the real relative speed along the line joining two ants. So, time taken was 2/(3v). So, the distance traveled was 2/3. :)*

1. what is the solution? is it really 1!!!

the answer I got is :
1- distance travelled by each ant is Pi/4
2- Straight line distance from A to B is 1/2

2. We want the distance traveled by each ant. Solution goes as follows.

The ants keep constant speed throughout their motion. Their trajectory is extremely complex. So, geometrical solution will not help. We calculate the time taken for the ants to meet. Now, we see that ant 1 s always moving towards ant 2 and ant 2 towards ant 3. They initially form a square. By symmetry, they always form a square. So, ant 1 is always moving perpendicular to the line joining ant 2 and ant 3. This means, at all times, the velocity of ant 1 is perpendicular to that of ant 2. So, the relative velocity of ant 1 towards ant 2 is always v. So, the time taken to make this distance zero is 1/v.

So, the distance traveled total is 1 m. :)

Hope that helps!!

3. i remember having solved the HC Verma problem myself, and perhaps would have managed this one too (considering its simpler) but feeling too lazy to exercise my brains, i saw the solution. nice.

4. by same arguments relative velocity of ant 1 with respect to ant 3 is 2v and pointing towards 3 .
in fact trajectory of ant 1 in reference frame of ant 3 is straight line with speed 2v..
so time taken to meet is (square root 2)/2v. hence answer is (square root 2)/2..i don't know which one is right .

5. in the reference frame of ant 3 ,ant 1's trajectory is straight line with speed 2v.
(because velocity of ant1 is perpendicular to ant2 velocty and velocity of ant2 is perpendicular to ant 3 velocty so ant 1 velocity and ant 3 velocity are in opposite direction. )
hence the answer is square root(2)/2.
i don't know why the answers are not matching..please do sort out this issue .

Fraction Brainteaser

Source: Sent to me by Gaurav Sinha Problem: Siddhant writes a Maths test and correctly answers 5 out of 6 Arithmetic questions and 20...