Math Problem Help

[Chaos Sorcerer]

So the Centre for Education in Mathematics and Computing at the University of Waterloo sends out math problems every Thursday at a few different levels. I am a student in the eighth grade who has also subscribed to the Gr. 9/10 problems. Today, I ran into this one:

Ali, Bill and Carl are lined up such that Ali is 100 m west of Bill and Carl is 160 m east of Bill. At noon, Carl begins to walk north at a constant rate of 41 m/min and Ali walks south at a constant rate of 20 m/min. (Bill does not move.)

At what time will the distance between Carl and Bill be twice the distance between Ali and Bill?

So I created a formula to help me find the number of minutes past noon it would take to arrive at this difference. According to the Pythagorean Theorem, a^2 + b^2 = c^2. Let x be the number of minutes.

(41x)^2 + 160^2 = 2 {(20x)^2 + 100^2}

According to the numerical distributive property, a (b + c) = ab + ac, so I simplified this formula down to (41x)^2 + 25600 = 2 (20x)^2 + 20000, then to (41x)^2 + 25600 - 20000 = 2 (20x)^2 + 20000 - 20000, then (41x)^2 + 5600 = (80x)^2.

Then I used the "check and guess" strategy and experimented with all sorts of numbers to find the value of x, and none of them seemed to work! Of course, I could try every number from 0 to 60, but that would take a long time.

Could someone let me know of a way to simplify this any further? Is there a way to simplify the powers? If so, what would be the time when the distance between Carl and Bill is twice that between Ali and Bill?Of course, I could make (41x)^2 = (80x)^2 - 5600, but would that be any more effective? The solutions to this problem should be emailed out to me (along with a new problem for the week), but I am really curious as to where this could go. Thank you.

 

[alexoiu]

The formula is:
(41x)^2 + 160^2 = 4 {(20x)^2 + 100^2} (the distance between Carl and Bill be twice the distance between Ali and Bill), so the distance between Carl and Bill ^2 = 4 x the distance between Ali and Bill ^2.
Start from here and you should get the result: 9x = 120, x = 13 mins and 20 secs

 

[Chaos Sorcerer]

The formula is:
(41x)^2 + 160^2 = 4 {(20x)^2 + 100^2} (the distance between Carl and Bill be twice the distance between Ali and Bill), so the distance between Carl and Bill ^2 = 4 x the distance between Ali and Bill ^2.
Start from here and you should get the result: 9x = 120, x = 13 mins and 20 secs

Thank you! So it was really my formula that was messed up.

 

[alexoiu]

You're welcome.
Not messed up entirely, just 2 instead of 4. It happens.
By the way, you can post the problems here, if you wish.

 

[Chaos Sorcerer]

You're welcome.
Not messed up entirely, just 2 instead of 4. It happens.
By the way, you can post the problems here, if you wish.

What do you mean? Like, the problems that come out every week? Or just the ones that I'm having difficulties with?

Here is the link if you want to subscribe.

 

[alexoiu]

Just the ones that you're having difficulties with.
Thank you for the link.

 

[Chaos Sorcerer]

Just the ones that you're having difficulties with.
Thank you for the link.

So this was this week's problem. I worked through it, but I am not sure if I got the right answer. Of course, the solutions come out next week, but I wanted to know beforehand. I would like your help in this.

99 is the greatest value for a, because a + 3c can't be greater than 99. In this case, c is 0 and therefore 3c is 0.

33 is the greatest value for c, because 33 x 3 = 99. In this case, a is 0. It can be any value from 0 to 33 because a is a variable. Thus, there are 34 possible combinations for c. Since it is a value in an ordered pair, the total number of ordered pairs that satisfy the equation is 34.

 

[alexoiu]

My result is a bit different.
If you solve the equation, the result is a=36 (a is not variable). Meaning that for any value of c (positive, but different of 0 - the fractions 2/a and 2/c have no sense when a or c = 0) and a=36, we have (a/c+a/2+1)/(2/a+2/c+1)=18 (you can observe that in their example a=36 and c=5).
The value of c is given by the restriction a+3c not greater than 99.
so 36+3c not greater than 99 meaning c not greater than 21.
Therefore for any pair (a=36, c =between 1 and 21) we have both equation and restriction satisfied.
The answer to the problem is 21 pairs.

Start by solving the equation and see if you get a=36.

 

[Chaos Sorcerer]

My result is a bit different.
If you solve the equation, the result is a=36 (a is not variable). Meaning that for any value of c (positive, but different of 0 - the fractions 2/a and 2/c have no sense when a or c = 0) and a=36, we have (a/c+a/2+1)/(2/a+2/c+1)=18 (you can observe that in their example a=36 and c=5).
The value of c is given by the restriction a+3c not greater than 99.
so 36+3c not greater than 99 meaning c not greater than 21.
Therefore for any pair (a=36, c =between 1 and 21) we have both equation and restriction satisfied.
The answer to the problem is 21 pairs.

Start by solving the equation and see if you get a=36.

Oh...I think I misinterpreted the question. I thought it was asking for the number of possibilities for a and c in just a+3c is equal or lesser than 99. I didn't think it was asking for them to comply to the original equation (a/c+a/2+1)/(2/a+2/c+1)=18...I thought it was just an example. My bad.

Then yes, I think your solution is correct. Thank you. But how did you arrive at the conclusion that 36 is the only possible value for a? So we know that a can't be greater than 99. We also know that it can be 36. Did you test out every value from 0 to 99? Did you simplify it? If so, what was the equation's most simplified form? I apologize if these are stupid questions. I can only simplify it as far as (2a/2c + a + 1) x (a/4 + c/4) = 36, and I'm not even sure if that was correctly simplified.

 

[alexoiu]

No problem.
I simplified the equation:
(a/c+a/2+1)/(2/a+2/c+1) = [(2a+ac+2c)/2c] / [(2c+2a+ac)/ac]= [(2a+ac+2c)/2c] x [ac/(2c+2a+ac)]= ac/2c = a/2 = 18
So a =36.

 

[popatim]

I'm afraid I read it same as Chaos. I had to re-read the problem to see that was a restriction and not a new problem. LoL