This person in the question needs a job ><
Hey guys! I got challenged by my teacher to do a certain problem for extra credit... i'm past the date to submit it anyway but it's completely wrecking my brain. Since we're studying arithmetic and geometric sequences, and their sums, I believe that we're supposed to use them. Here's the problem:
A person is out for a walk. He takes 1 step to his right, then turns right and takes 2 steps, then turns right and takes 3 steps, then turns right and takes 4 steps, then turns right and takes 5 steps, etc. After he has taken a total of 2000 steps he receives a phone call that tells him to return. Instead of retracing his steps, he returns in a straight line to his starting point. How many steps will he require?
I've tried to approach this problem in many ways. I'm going to try to scan my work in for you guys to see since I did it in paper, but I'll take a while since I'll have to scan it in and edit the mess.
Also, a note: I believe the doesn't get to the 2000 steps, since 2000 would be between the 62nd and 63rd iterations I believe (from my own calculations) - but it is accepted to stop after the 62nd iteration, at step 1953 I believe.
Thanks!
Indeed. I've tried separating it into 4 different sequences, for Up, Down, Left, and Right, so I could then subtract down from up and right from left to get some sort of length to which I could then apply the pythagorean theorem, but that didn't work out.
Here's my attempt. I hope it's clear enough without any diagrams
So the plan is to find a triangle and then calculate the hypotenuse, which will hopefully be the answer...
The first thing is to work out where the 63rd walking line is gonna be, since that's where he'll be when he gets the call. Unless I'm mistaken it'll be on the right side, going down. We can already calculate one side of the triangle, because he's gonna stop on exactly the 2000th step, which should be is the 47th step of the 63rd line.
By picking a point on the 63rd line (let's say point A) perpendicular with the center we can get a right angle, so point A is gonna be 31 steps down the 63rd line.
Let's take where he stops as point B, so that's 47 steps down the 63rd line.
Therefore we know the side AB = 16.
Next we want to find the length of the side from the center (point C) to A (halfway down 63rd line). This can be calculated by working out how many lines you have to cross walking in a straight line from C to A, then multiplying it by 2 (because each line is separated from the next by a distance of 2 steps). Thus 16*2 = 32 steps, so line AC = 32.
Now time to find the hypotenuse (i.e. his walk home):
AB^2 + AC^2 = BC^2
16^2 + 32^2 = BC^2
1280 = BC^2
BC = 35.777...
And rounding up the answer comes out as 36 steps. How does that sound?
bubbletea999 - I did that too, that's how I got that number. ( )
You have a mistake. "because he's gonna stop on exactly the 2000th step, which should be is the 47th step of the 62nd line." No, is the 47th step of the 63th line. (1+2+...+62=1953; 1953+47=2000)
Up to here I managed to work it out on my own and got the same numbers
Could you please explain what you're trying to do here? By center, do you mean the point from which he started?By picking a point halfway down the 63rd line (let's say point A) we can get a right angle with the center
Your arithmetic, with veileen's correction, seems correct to me, and I have gotten answers in that ballpark so it seems correctAnd rounding up the answer comes out as 36 steps. How does that sound?
Yep, by center I mean where he started off (and what I'm calling point C). The only reason it seems like a good idea to pick a point halfway down the 63rd line is because that gives you a right angle at the intersection between AB and AC. The right angle is useful because you can then use Pythagoras' theorum to find the hypotenuse, which is his walk home.
Hmm, yep, I think you're right. So the final equation at the end would be:
15.5^2 + 32^2 = BC^2
1264.25 = BC^2
BC = 35.5562...
Thankfully this still rounds up to 36 steps. It's weird how veileen managed to get the same slightly wrong answer as me then. I'm probably missing something too...
Also - and sorry if I'm being a little annoying - but how do you get 16 here?:
Never mind, I figured it out by putting it all on excel:This can be calculated by working out how many lines you have to cross walking in a straight line from C to A
"By picking a point halfway down the 63rd line (let's say point A)" I didn't see that... If A is the middle of the 63rd line then CA is not perpendicular on it. I don't know what draw you have but mine says something else. For being perpendicular A should be at 31 steps from right corner and 32 from left corner.
This is how I did it, not sure if there's a better way.
Firstly, imagine that you've drawn out all his walking lines and labelled them 1st, 2nd, 3rd...62nd, 63rd, etc.
If you start from the center point C and go in a straight line towards point A, the first line you hit will be the 3rd line. From here, each line you meet will be labelled n+4. So: 3, 7, 11, 15, ...all the way to 63. We only really want to count the gaps between the lines, since we know that it's 2 steps each time.
Because the values only start rising in a regular way (n+4 each time) from the 3rd line onwards, it's easier to work it out from there (rather than the center point), then simply +1 at the end.
So, in order to calculate the number of lines between the 3rd line and the 63rd line, you just take the difference between 3 and 63, which is 60, then divide 60 by 4 (because each line has a difference of 4), giving you 15 lines. Then you have to add one for the 3rd line which we didn't count first time around.
Thus, travelling from point C to point A you'll cross 16 gaps between the lines, each of which are 2 steps in length, giving 32.