Outline · Standard · [ Linear+ ] Math Challenge

[lappy512]

Hey, I have this cool math challenge. Look at the attached image (not to scale).

You have a 100 ft pole, and a 10' by 10' box right next to it. Now, we need to find a place to cut the pole so it can fold and then make a triangle, with one point touching the box, and one point touching the ground. Assume that the box has all 90 degree angles, and the flagpole is perpendicular to the ground.

At what height should we cut the flagpole so it can fold to form the triangle? Show your work, guess and check/graphing not allowed.

It should be possible without guessing and checking, or graphing an equation. You should be able to solve this algebraically.

There should be two answers.

Attached thumbnail(s)

[Senor Halo Blue 9]

Wow I have tried this and it is very interesting...very clever...

[Mynck]

Got the answer! That was easy.

I attached an encrypted zip file. PM me if you want the password.

Attached File(s)
 untitled.zip ( 3.82k ) Number of downloads: 107

[lappy512]

Heh, looked over it, it's definitely incorrect.
(I talked to him, he said it's incorrect after I showed him why)

[Mynck]

Bleh. I see where I messed up. In the second-to-last step, when using the quadratic formula, I should've added instead of subtracting.

But wait. I still get a negative number.

Oh, I see now. I missed one of the x's from the 7th step in the 8th step.

I'll correct and reupload.


-----Edit-----

Either I did it wrong, or this'll take hours to solve.

Attached File(s)
 untitled2.zip ( 3.39k ) Number of downloads: 116

[anthonytc22]

Below is my solution to the problem.

Here is my reference image:

» Click to show Spoiler - click again to hide... «

"a" is how high the pole was cut, and "c" + "d" is the remainder of the pole. (image not to scale)
__________________________________________________________

given: a+c+d=100 feet
given: angle between lines "a" and "b" = 45 degrees (line "b" bisects a 90 degreee angle)

Part 1
the diagonal of the 10 x 10 squre is equivielent to the hypotenuse of a triangle with two sides equal to length 10:

therefore, using the Pythagorean Theorm we get 10² + 10² = b²
and after a couple of steps, we get b = 10 * sqrt(2)


Part 2

Angle P is the same as Angle Q, as they are corresponding interior angles, and are therefore the same:
angle P = angle Q

Part 3

Now I introduce the Law Of Cosines, which states that given a triangle with sides a,b,c and an angle X (that is opposite side c):
c²=a² + b² - 2*a*b*Cos(angle X)

in the above reference image the triangle is abc, where b= 10*sqrt(2) and angle X (opposite side c) = 45 degrees while a=a and c=c

plugging into the law of cosines gives: c²=a² + (10*sqrt(2))² - 2*a*(10*sqrt(2))*Cos(45 degrees)

this simplifies to: c²=a² + 200 - 2*a*10*sqrt(2)*(1/sqrt(2))

this further simplifies after a couple of steps to: c²=a² + 200 - 20*a

Part 4
The trigonomic function Sine gives the ratio between two sides of a triangle. This ratio is of a side opposite an angle and the hypotenuse. Sin(angle)= (side of triangle opposite angle)/(hypotenuse).

I write the equations from the reference image above:
Sin(angle P) = (a-10)/c
Sin(angle Q) = 10/d

since in Part 2 it was shown that angle P = angle Q, it can be said that Sin(angle P)= 10/d

I can then set the two equations togeather: 10/d=(a-10)/c

This can be solved for d, which looks like this: d=(10*c)/(a-10) after several algabraic steps

Part 5
Here are three of the main equations we have:
1) a + c + d = 100 ft
2) c²=a² + 200 - 20*a
3) d=(10*c)/(a-10)

we can now plug in equation 3 into equation 1: a + c + (10*c)/(a-10) = 100

for equation 2 we can solve for c, which then makes it: c = sqrt(a² + 200 - 20*a)

Before we plug this into a + c + (10*c)/(a-10) = 100 , I want to rewrite a + c + (10*c)/(a-10) = 100 so that there are no denominators. So, we rewrite it as follows:
a + c + (10*c)/(a-10) = 100
c + (10*c)/(a-10) = 100 - a
[c + (10*c)/(a - 10) = 100 - a] * (a - 10)
c*(a - 10) + (10 * c) = 100*(a - 10) - a*(a - 10)
c*a - c*10 + c*10 = 100*a - 1000 - a^2 + 10*a
c*a = a^2 + 110*a -1000

at this point we can now substitute for c, and we get:
a*sqrt(a² + 200 - 20*a) = a^2 + 110*a -1000


Now the sqare root needs to go, so we have to square both sides:
(a*sqrt(a² + 200 - 20*a) = a^2 + 110*a -1000)^2

after this there is some VERY nasty algabra. After FOILing and simplifying it becomes:
a^4 - 20*a^3 + 200*a^2 = a^4 - 220*a^3 + 14100*a^2 - 220000*a + 1000000

further simplifying gets to this cubic polynomial:
0 = -200*a^3+13900*a^2-220000*a+1000000

To solve a cubic like this would take many hours to do by hand if its even possible, so I used the aid of a computer to solve it:

1) 9.00544849
2) 11.2820903
3) 49.2124611

These three answers correspond to possible heights for cutting the pole. But as you can see, the first answer is about 9, which obviously doesn't work and is an extraneous solution. However the last two work as they are above 10. So, the two possible answers for cutting the pole are at

1) 11.28 ft
2) 42.21 ft

This post has been edited by lappy512: Mar 13 2006, 07:30 AM

[whimbrel]

Below is my solution to the problem.

I did something similar, but a little cleaner. First, though, I divide everything by 10 to simplify the math. The square has sides of length 1, and the big pole is now length 10.

0) Once you fold the pole, you get a triangle. Get equations for each leg. The side going up is "x", the hypotenuse is (10 -x), and the bottom leg is sqrt(100 - 20x), by solving from the given two legs.

1) You now have a figure with three triangles: the big triangle we care about, and the two triangles above and to the right of the inscribed square. Realize that all three triangles are similar triangles (inside angles, blah blah blah).

2) Because of (1), the trig functions for all three triangles will be identical. Let's choose the easiest to work with, which is the tan function (opposite/adjacent)

3) The upper small triangle has adjacent side = 1, and opposite side = (x - 1). The whole triangle has adjacent side = sqrt(100 - 20x) and opposite side = x. (from step 0)

4) Set these two tangents equal to each other: (x-1)/1 = x/sqrt(100-20x). Solve for x.

5) This results in the cubic: -20x^3 + 139x^2 - 220x + 100= 0.
6) Solve this, giving x = 4.92, .90, or 1.12.
7) Multiply back by 10 to get the real answers of 49.2 and 11.2

Is there a way to do this that doesn't involve solving a cubic?

[Spaceman3750]

My answer below

» Click to show Spoiler - click again to hide... «

Bah! I give up. I probably don't know how to solve it anyway, but I'll kick myself when I find out because it will be easy.

[Mynck]

My password was "manyana".

I also ended up in a cubic.

[GameClaw_268]

Meh, I can't figure it out without trigonometry... I loose...