Tuesday Puzzle

Half The Diameter???

Uhhhh

I guess that is true, but not quite what I was looking for ;-)

Answer, semi-hidden

Is the wall about 28 1/4 miles long?

Like my table top problem, my solution to this involved a 4th degree polynomial. Surely there is a better way of looking at it?

yes and yes ;-)

hint

You got the answer, but for an easier way think properties of tangents and our old Greek friend.

Re: hint

Am I doing this right? I need 4 times the circumference of quarter circle that touches the hypotenuse of a right triangle with a length of 9.49 miles

Do You Mean Mr. Pythagareon?

I have the same r^^4th equation

but I can't recall how to solve them.

I looked out at a couple of tangents, and I know that's the answer, but I also couldn't recall enough math to finish it with that approach...

I used to be able to do this stuff in my head!

Steve

yes

solving a 4th degree polynomial

Inorganic solutions:

1) "solve for" in spreadsheet program

2) Draw the problem in CAD, and experiment with different size circles to get one that fits.

I think that to get the root of the 4th degree polynomial, you have to factor it, but that takes too much grey matter.

But Dan has indicated that there is an easier way to see it. I'll have to cogitate on that some tonight.

?

Will have to think about that for awhile to see what it means ;-)

Funny thing

Long ago, I had a final in a thermodynamics class. One particular problem on the exam ended up having a trick to it that I did not see, so I did it the hard way. If you saw the trick, it was fairly simple to do, but if not, it turned into a triple integral. I ended up solving it and getting the correct answer, but the professor marked it wrong because he couldn't figure out how I did it. I had to take him to the head of the math department and explain what I had done to him in order to get credit for the problem.

I have seen a lot of problems that I worked my tail off to solve and then someone says "why didn't you just do...." DOH. Hindsight is most always 20/20. I applaud you for getting it regardless of how. I am sure that now that you know there is another way, you will find it fairly easily. (BTW, I did it the hard way first ;-))

Re: Tuesday Puzzle

Consider the circular wall of the city and denote by

N: the north gate

S: the south gate

G: the house of Gary, 3 miles north of N

T: the house of Thomas, 9 miles east of S

Since they can barely see each other, the line GT connecting the houses must be a tangent to the circle (wall).

Let C be the point where this tangent GT touches the circle.

Then TC=TS=9. Let GC=x, and NS=d ( the diameter of the circle).

I will use the notation a^b to mean a to the power b.

From the orthogonal triangle GST we get GT^2 = GS^2 + ST^2 or

(9+x)^2 = (d+3)^2 + 9^2 (eq. 1).

Consider now the power of the point G with respect to the circle.

This gives GN*GS = GC^2 or 3*(d+3) = x^2, from which we get

d+3 = (x^2)/3. Substitute this into (eq. 1) to get:

(9+x)^2 = (x^4)/9 + 9^2.

After some trivial algebra we get x^4 - 9x^2 -9*18x = 0, and dividing by x we further get x^3 - 9x - 9*18 = 0.

This is a third degree equation with integer coefficients. If it happens to have an integer root, this root will be a divisor of the constant term 9*18.

So, with a calculator, we can start experimenting and we may be lucky! And indeed in this case x=6 is a root.

Now we can go back to 3*(d+3) = x^2 and easily find d = 9, and the radius of the wall will be 4.5 miles.

read the post below. It has basically the same

The post by Nikos below is basically it. It resolves to a third order that is fairly easy to solve.

;-)

Thanks for playing, hope you enjoyed it.

and one other way

Let a be the angle between the line from Gary's house to the north gate and the line from the Gary's house to Thomas' house.

sin(a) = r/(r+3)

tan(a) = 9/(2r+3)

cos(a) = sqr(1-sin2(a)) = sqr(1-sin2(r/(r+3))) = sqr(6r+9)/(r+3)

tan(a) = sin(a)/cos(a) = (r/(r+3)) * ((r+3)/sqr(3*(2r+3))) = r/sqr(3*(2r+3))

Equating the two expressions for tan(a):

9/(2r+3) = r/sqr(3*(2r+3))

9 * sqr(3) * sqr(2r+3) = r*(2r+3)

9 * sqr(3) = r * sqr(2r+3)

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

2r^3 + 3r^2 - 243 = 0 factors to (2r-9)*(r^2 + 6r + 27) = 0.

The real solution to this is 4.5

DOH!

This is the line from your explanation that brought a dope slap to my forehead:

"Then TC=TS=9"

Using the (obvious once you see it) fact that the distance from Thomas' house to the tangent point is 9, I got an equivalent 3rd degree polynomial, which I would factor by guessing at the answer, then confirming that it is a root of the equation.

;-)

Would have helped if I read the problem correctly

I just reread the problem and I would have gotten it with my approach, but I drew the initial picture wrong. Never going to get the right answer with the wrong picture...

I did fire up some log dormant neurons yesterday, though.

Thanks for posting.

Steve