Win a copy of Cross-Platform Desktop Applications: Using Node, Electron, and NW.js this week in the JavaScript forum!
programming forums Java Java JSRs Mobile Certification Databases Caching Books Engineering Languages Frameworks Products This Site Careers Other all forums
this forum made possible by our volunteer staff, including ...
Marshals:
Sheriffs:
Saloon Keepers:
Bartenders:

# Anybody for Geometry ?

Peter Chang
Greenhorn
Posts: 18
A certain city has a circular wall around it.
The wall has four gates pointing north, south, east and west.
A flag-post stands outside the city, three kms north of the north gate,
and it can just be seen from a point nine kms east of the South Gate.
What is the diameter of the wall that surrounds the city?

David O'Meara
Rancher
Posts: 13459
Dammit, HTML denied. I have an answer, but I'll wait for others.
[ January 27, 2005: Message edited by: David O'Meara ]

David O'Meara
Rancher
Posts: 13459
My answer is more than 10 and less than 50

Peter van de Riet
Ranch Hand
Posts: 112

a point nine kms east of the South Gate.

Is this right?
[ January 27, 2005: Message edited by: Peter van de Riet ]

Varun Khanna
Ranch Hand
Posts: 1400
Pretty easy to solve if you give me options

I just checked for the closest number satisfying the Pythagoras:
(3+x)square + (9)square = (Hypotenuse)square

[12*12 + 9*9 = 15*15]

David O'Meara
Rancher
Posts: 13459
Nope, you need a difference of 6, not 3...

David O'Meara
Rancher
Posts: 13459
What about a general solution for distance 'a' from the north gate, 'b' from the east gate?

Varun Khanna
Ranch Hand
Posts: 1400
Originally posted by David O'Meara:
Nope, you need a difference of 6, not 3...

why?
[ January 27, 2005: Message edited by: K Varun ]

Varun Khanna
Ranch Hand
Posts: 1400
Originally posted by David O'Meara:
What about a general solution for distance 'a' from the north gate, 'b' from the east gate?

Well i remmeber there use to be some equation for tangents meeting at some external point, dont remember it now ..
Also tangents from a same point to a circle are always equal ...

but not sure wht pythagoras wont work here. I just feel 9 should be an answer.

Wats ur reasoning ?
[ January 27, 2005: Message edited by: K Varun ]

David O'Meara
Rancher
Posts: 13459
I can tell you're still imagining the curve of the wall. I was able to come up with an answer fairly quickly once I drew it and recognised the trick. It's all about where you put the circle

David O'Meara
Rancher
Posts: 13459
Originally posted by K Varun:

why?

[ January 27, 2005: Message edited by: K Varun ]

Attempted ascii art:

Hopefully you'll see that the answer is x^2 = (r+3)^2 + (r+9)^2
ie the two sides have a difference of 6.
Also notice that you need to find r, not the lengths of the sides, and you need to find some way to define x to solve the equation...

Varun Khanna
Ranch Hand
Posts: 1400

Here is how my diagram would be.
CB and CA are tangents to the circle. CA will just hit the circle somewhere around K

ABC is a right angled triangle, with B=90 degrees.

Now if you agree that B should be the south gate, you can see your diagram isn't satisfying the condition "it can just be seen from a point nine kms east of the South Gate"

David O'Meara
Rancher
Posts: 13459
Originally posted by K Varun:
[QBit can just be seen from a point nine kms east of the South Gate[/QB]

Yes, but based on the elegant slution, I hope this is a typo. Still waiting on a response from the original poster.

Although I'm pretty sure I have a solution to your case, I just haven't solved for the final answer.

David O'Meara
Rancher
Posts: 13459
Well, since no one else is taking part, I'll drop my answer to the first and second types.

The first is the quetion as I drew it, assuming there is a typo in the question. Obviously the problem forms a triangle, from the center of town to the two flag poles. Less obvious, since it is right-angle triangle you can draw a circle containing these three points on its circumference, then the three sides must be (r+3), (r+9) and (2r). Solving a^2 + b^2 = c^2 for r gives two solutions, -3 and 15. I chose r=15.

To solve the second case, imagine it as the first case, using (r+m) (for some number m) instead of (r+9), then use similar triangles to remove m and solve for r. It should be solvable, but ends up with a fourth-order quadratic equation. No thanks.

Varun Khanna
Ranch Hand
Posts: 1400
eh just tried to google and here you go

Barry Gaunt
Ranch Hand
Posts: 7729
I've just come back from a teabreak after working out the answer 9. Why did you have to give a link?

Ok I have not look at the link yet, so I'll tell you how I did it anyway.

I found an angle for which I could write its tangent in two ways using the radius of the city as the unknown variable. I got it around to a simple cubic equation which I solved by guessing at a few possible roots. r = 4.5 solves it. So the diameter of the city is 9.

OK now I'll take a look at the link.

Barry Gaunt
Ranch Hand
Posts: 7729
They used Pythagoras to get a quartic?

With my approach you end up with a much simpler:

r * r * ( 2 * r + 3 ) = 243

for which 9 / 2 is the solution.

Varun Khanna
Ranch Hand
Posts: 1400

Nick George
Ranch Hand
Posts: 815
What did the seed say when it grew up?

Gee, I'm a tree!

 Don't get me started about those stupid light bulbs.