# Anybody for Geometry ?

Peter Chang

Greenhorn

Posts: 18

posted 11 years ago

- 0

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 11 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 ]

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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 11 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...

- 0

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 11 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

- 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"**

- Varun

posted 11 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.

- 0

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 11 years ago

- 0

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 11 years ago

- 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.

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.

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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 11 years ago

- 0

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.

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