# Anybody for Geometry ?

Peter Chang

Greenhorn

Posts: 18

posted 12 years ago

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?

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?

Varun Khanna

Ranch Hand

Posts: 1400

Varun Khanna

Ranch Hand

Posts: 1400

Varun Khanna

Ranch Hand

Posts: 1400

posted 12 years ago

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 ]

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 ]

- Varun

posted 12 years ago

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...

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

posted 12 years ago

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

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"**

- Varun

posted 12 years ago

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.

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.

posted 12 years ago

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.

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

Barry Gaunt

Ranch Hand

Posts: 7729

posted 12 years ago

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.

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.

Ask a Meaningful Question and HowToAskQuestionsOnJavaRanch

Getting someone to think and try something out is much more useful than just telling them the answer.

Barry Gaunt

Ranch Hand

Posts: 7729

posted 12 years ago

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.

With my approach you end up with a much simpler:

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

for which 9 / 2 is the solution.

Getting someone to think and try something out is much more useful than just telling them the answer.