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. :)*

Comments

  1. Unknown6:41 PM, November 06, 2009

    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

    ReplyDelete
  2. Pratik Poddar8:28 PM, November 06, 2009

    The answer is 1. :)

    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!!

    ReplyDelete
  3. suyash12:35 PM, November 08, 2009

    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.

    ReplyDelete
  4. Unknown5:25 PM, September 29, 2012

    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 .

    ReplyDelete
  5. Unknown5:30 PM, September 29, 2012

    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 .

    ReplyDelete

Post a Comment

Popular posts from this blog

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 out of 28 Geometry questions.  In total, Siddhant scores 25 out of 34.  Vaibhav writes another Maths test and correctly answers 20 out of 25 Arithmetic questions and 6 out of 9 Geometry questions. in total, Vaibhav scores 26 out of 34. Note that a) Vaibhav scores more than Siddhant b) Siddhant score better than Vaibhav in both individual topics -  5/6 > 20/25 and 20/28 > 6/9 How is it possible? 
Read more

Buying Dimsums

Source: Alok Goyal (Stellaris VP, Ex-Helion VC) puzzle blog Problem: A fast food restaurant sells dimsums in boxes of 7 and 3. What’s the greatest number of dimsums a person cannot buy. Generalize it for p and q where p and q are relatively prime. I loved the puzzle. Hope you enjoy it too.
Read more

Consecutive Heads

Let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process. 1) Calculate P{X>Y} 2) What's the expected value of X 3) What's the expected value of Y This is probably the hardest puzzle I have ever put on my blog. Hence, I will post its solution in the post directly rather than on comment. Solution: 1) (Solved by me finally after 13 months :)) Make a state diagram. Let the state be (a,b) where a is the number of consecutive heads streak "A" is on currently and b is the number of consecutive heads streak "B" is on currently. So, (0,3) (1,3) are final accepted states and (2,0) (2,1) (2,2) (2,3) are failure states. If you get tails, your contribution to the state reaches to "0" f(State) = P(X>Y | "State" configuration initially)
Read more