[bonanova]

In my closet stand two brooms; one is 4 feet long, the other 6 feet long.

One leans against the left wall, the other leans against the right wall, forming a kind of an X pattern.

The base of each broom is directly below the top of the other broom, so it's an X with unequal lengths.

Someone asked how wide my closet is.

I told him that my brooms cross exactly two feet above the floor.

That's not what I want to know, he replied.

I told him to drop by BrainDen and watch this thread.

[Guest]

I know the answer, because I'm sitting at a computer with solid modeling software so I sketched it in and measured. Still working on the proof.

Spoiler for are you sure you want to see it?:

2.46 feet

Edited November 21, 2008 by lazboy

[Guest]

brooms_puzzle.bmp

This is the screenshot of the sketch I drew, with the answer shown. If you construct two triangles with the brooms as hypotenuses, you have cos(A)=x/6 and cos(B)=x/4

You know the height of the triangle formed by the two brooms and the floor of the closet to be 2 feet, so therefore the area of that triangle is 1/2(x)(2) or just x. The area can also be calculated by the formula Area=x²sin(B)sin(A)/2sin(C ), where C is the angle opposite x. Using (180-A-B) for C and x for Area and doing a bunch more substitution, we can solve for A, which is 65.77° (there are actually 3 solutions, but two of them don't make sense (90° and -51.36°)), and therefore x is 6cos(65.77°)=2.46 feet, or 2 feet, 5.55 inches

edit: (C ) was ©

Edited November 21, 2008 by lazboy

[Guest]

I drew it by hand (with the use of a ruler), and I came up with approximately 2 1/2 feet as well. So, I will have to go with the computer software on this one.

[bonanova]

Both right.

My friend has his answer.

[Guest]

I tried writing out the proof, but I got too deep into multiple trig functions to make it work...

w/e time for bed, lol

[bonanova]

I tried writing out the proof, but I got too deep into multiple trig functions to make it work...

w/e time for bed, lol

Fair enough.

It's more of a math problem than a puzzle.

[Prof. Templeton]

I had attempted to prove this one using similar triangles and the resulting equations, since Lazboy had beaten everyone with the answer, but all I ended up with were a couple of equations with too many variables. Bonanova, can this problem be solved that way? If it can I'll go back to the drawing board. If I knew the heights of the brooms where they touched the wall, I could make it work, but only knowing the brooms' lengths I couldn't get there.

Edited November 22, 2008 by Prof. Templeton

[bonanova]

I had attempted to prove this one using similar triangles and the resulting equations, since Lazboy had beaten everyone with the answer, but all I ended up with were a couple of equations with too many variables. Bonanova, can this problem be solved that way? If it can I'll go back to the drawing board. If I knew the heights of the brooms where they touched the wall, I could make it work, but only knowing the brooms' lengths I couldn't get there.

Try to relate the height of the ends of the two brooms and the height of their crossing point.

Go from there.

Draw x-y coordinate axes with origin at the base of the shorter broom.

You can assume it leans from left to right to keep all the x values positive.

[bonanova]

I had attempted to prove this one using similar triangles and the resulting equations, since Lazboy had beaten everyone with the answer, but all I ended up with were a couple of equations with too many variables. Bonanova, can this problem be solved that way? If it can I'll go back to the drawing board. If I knew the heights of the brooms where they touched the wall, I could make it work, but only knowing the brooms' lengths I couldn't get there.

Using a different approach, you can obtain a quartic equation.

Any progress?

[Prof. Templeton]

Using a different approach, you can obtain a quartic equation.

Any progress?

I had come up with a quadratic equation previous, but I discarded my sketch and notes. I'll redraw and see what I can come up with.

[Prof. Templeton]

Let x be the height from the floor to where the shorter broom contacts the wall

Let y be the height from the floor to where the longer broom contacts the wall

Let w be the width of the closet

Let h be the distance of the perpendicular line from the floor to the intersection of the two brooms

Where line h meets the floor it will split w into two parts lets call them a and b

We know h=2

The length of the short broom is 4

The length of the long broom is 6

Because of similar triangle we can establish some ratios

x/w = 2/b

y/w = 2/a

b = 2w/x

a = 2w/y

and

a+b = w

so

2x + 2y = x + y (left a step out)

and then

y = 2x/x-2

On to Pathagarus

w = sqrt(36-y²) and

w = sqrt(16-x²)

36-y² = 16-x²

y²-x²=20

We know that y = 2x/x-2 from above plug that in

(2x/x-2)²-x²=20

Now we need to form the quadratic. After some trial and error (calculus was many years ago) I ended up with

x⁴-4x³+20x²-80x+80=0

A little too difficult for me the solve, but plugging in the known value of x (3.154) I know it does work.

[Prof. Templeton]

Should say quartic above, not quadratic.

I used a computer and got the four values of x, two of which were negative and the other two

3.1522574219524246

1.4017964176594925

Edited November 26, 2008 by Prof. Templeton

[bonanova]

This is the classic crossed ladder problem, popularized by Martin Gardner.

This Wikipedia article outlines a cute approach to taming the unwieldy quartic.

Briefly, if A and B are the heights on the walls of the tops of the ladders

and a and b are the ladder lengths and h is the height of the crossing:

Establish that

[1] A B = h (A + B) h is the harmonic mean of A and B [1/h = 1/A + 1/B]

[2] (a² - b²) = (A + B) (A - B) using Pythagorean thm.

[3] (a² - b²)² = (A+B)³(A+B-4h) square [2] and combine with [1]

let x = (A + B)/ sqrt(a² - b²) and c = 4 h / sqrt(a² - b²)

then [3] becomes

x³(x - c) - 1 = 0

Solve for x; calculate alley width. easy once you know A or B.

A similar problem involves a ladder that contacts the ground, a wall and a box placed under the ladder.

A variation of the box problem was posted here and solved here in this forum nine months ago.