Lenny Conundrum 186
Thu Oct 05, 2006 10:00 pm
The Conundrum Lenny wrote:
Gilbert the Gelert farmer (He's a Gelert who happens to be a farmer, not a farmer who grows Gelerts!) grows crops on a square field, 400 metres on a side. He has divided this field into seven different sections, such that each section has the same area, and each section meets at a point in the exact center of the field.
If he wants to build a fence along the perimeter of the two sections (the perimeter outlined in red), how many metres of fence is required? Round up to the nearest whole number, and please only submit a number as your answer.
Oooh... *evil*. No hints here yet.
[edit: 6:17 EDT] Ooh, wait, one hint: the central angles are not all equal!
[edit: 9:53 EDT]
Sum of digits in answer: 12.
Remember your triangle formulas, and there are fewer unknowns than you think.
Fri Oct 06, 2006 7:34 pm
i got it, not that i expect a prize. i'm just proud. here's a hint. it's symmetrical
Sat Oct 07, 2006 1:49 am
remember to round UP and not down.
Mon Oct 09, 2006 3:59 pm
Wow. That was a fun one. It was just the same kind of triangle geometry over and over again. I'm not sure I had to calculate everything I did, but I'm fairly sure I got the right answer (I started in the opposite corner from the area of interest for some strange reason). Too bad my internet was down when it came out; I'm pretty sure I'm not in the top 250. Unless fewer than 250 people get it right...
Mon Oct 09, 2006 6:32 pm
A second way to check your answer - the drawing is very accurate (but your screen might not be).
Wed Oct 11, 2006 6:37 pm
the prizes are out, the answer is 921, its phytagoras + some variables replacement
somebody did post the answer, but I think mod deleted the post
somebody did post the answer, but I think mod deleted the post
Thu Oct 12, 2006 2:28 am
Here's the really really long solution. (I don't *think* this will stretch the board - the width is 1024 now right? - but feel free to change these into links instead, if that'd still allow the images to be accessed)
And this was the way I checked my answer (and what I was talking about as a second way to check, since I'd already posted the sum-of-digits thing as a first check):
[edit, 10:46 PM] skipped a paragraph by accident, now fixed
And this was the way I checked my answer (and what I was talking about as a second way to check, since I'd already posted the sum-of-digits thing as a first check):
[edit, 10:46 PM] skipped a paragraph by accident, now fixed
Thu Oct 12, 2006 6:11 am
sorry for not able to draw as nice as you
Notice the area marked with X ? its because they have same "h" which is s/2 (s is the side), and the same area (s*s/7) (explanation : the square area divided by 7 same area).
X = (s*s/7) / (0.5*s/2)
After that we work with AI, trapezoid (border in blue).
AI area = the small triangle area (with border "a") + the 1/7 farm area.
(z+s/2)*s/2*0.5 = s*s/7 + x-s/2 * s/2 * 0.5
replace the variables and we obtain z
then we can obtain "b" by applying the phytagoras theorem to the triangle with green line.
c can be obtained using the phytagoras theorem as well.
I think this way is simpler, since phytagoras is thing that almost remembered by everyone, and I already forget law of cosine
since that was like 6 years ago for me
Notice the area marked with X ? its because they have same "h" which is s/2 (s is the side), and the same area (s*s/7) (explanation : the square area divided by 7 same area).
X = (s*s/7) / (0.5*s/2)
After that we work with AI, trapezoid (border in blue).
AI area = the small triangle area (with border "a") + the 1/7 farm area.
(z+s/2)*s/2*0.5 = s*s/7 + x-s/2 * s/2 * 0.5
replace the variables and we obtain z
then we can obtain "b" by applying the phytagoras theorem to the triangle with green line.
c can be obtained using the phytagoras theorem as well.
I think this way is simpler, since phytagoras is thing that almost remembered by everyone, and I already forget law of cosine
since that was like 6 years ago for me