This week's book giveaway is in the Agile and other Processes forum. We're giving away four copies of The Mikado Method and have Ola Ellnestam and Daniel Brolund on-line! See this thread for details.
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?
What about a general solution for distance 'a' from the north gate, 'b' from the east gate?
Varun Khanna
Ranch Hand
Joined: May 30, 2002
Posts: 1400
posted
0
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
Joined: May 30, 2002
Posts: 1400
posted
0
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 ]
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
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
Joined: May 30, 2002
Posts: 1400
posted
0
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"
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
Joined: May 30, 2002
Posts: 1400
posted
0
eh just tried to google and here you go 9 is the answer.
Barry Gaunt
Ranch Hand
Joined: Aug 03, 2002
Posts: 7729
posted
0
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.
0
why?
I've just come back from a teabreak after working out the answer 9. Why did you have to give a 
link?
