Tuesday Puzzle

At teh top....:)

Or....

Lee Schierer McKean, PA

PI +.565307346

Re: Tuesday Puzzle

Gary Smyth

I get 3.77 combining two right triangles. It may not be perfect placement, but hopefully in Warner's guidelines.

There's got to be a way

Alex Y

that doesn't involve the solution of a 4th-degree polynomial. Letting a spreadsheet program do the solve, I get the same answer (at least to two significant digits) as Gary.

But surely there is a more elegant solution that I am missing?

How about this?

Dan Donaldson

Let a be the length of line segment AD, and let b be the length of line segment CF.

Because of the similarity of the triangles ADE and EFC, the following holds:

a : 1 = 1 : b

so

ab = 1.

According to the Pythagoraen theorem, the following holds:

(AB)2 + (BC)2 = (AC)2

so

(a + 1)2 + (1 + b)2 = 42

which can be rewritten to

a2 + 2 + b2 + 2(a + b) = 16.

Now we use the fact that ab=1, so 2=2ab, and we get:

a2 + 2ab + b2 + 2(a + b) = 16

which can be rewritten to

(a + b)2 + 2�(a + b) - 16 = 0.

Because we know that a+b is greater than 0, using the abc formula we find that

a + b = sqrt(17) - 1.

Because of the similarity of the triangles ADE and EFC, the following holds:

a : 1 = 1 : b

so

b = 1 / a.

Now we know that

a + 1/a = sqrt(17) - 1

so

a2 + ( 1 - sqrt(17) )�a + 1 = 0.

We know that a is greater than 0, and using the abc formula we find that

a = 1/2�( sqrt(17) - 1 + sqrt( ( 1 - sqrt(17) )2 - 4 ) ).

The ladder touches the wall 1 meter higher, which is at about 3.76 meters.

Re: How about this?

Gary Smyth

Quickly. Not elegant or as precise but since the hypotenuse is 4 and that line is the longest measurement in a right triangle we know that the tall (wall) leg must be under 4.

We are given that the lower horizonal leg is over 1, and as it happens the image appears to be in scale. By eye the distance between the end of 1 and the hypotenuse appears to be a little less than 1/3 more than 1.

Given our friend Pythagoras and his theorem A squared + B squared = C squared. Since we know that C squared is 16 and the bottom leg is about 1.3 squared, by adding the squares and taking the square root we can guess at the wall leg. By simple math we get close--3.5 to 3.8. A little test of inserting numbers over and under 1.33 gets to 3.7 with some left. Of course it only works if the drawing was close to scale when making an early assumption. Otherwise it has to be done long hand..

Very nice

Alex Y

I had the a=1/b, but missed the idea of "completing the square" by substituting ab for 1. And completing the square to solve with the quadratic formula twice should have suggested itself to me with a 4th-power polynomial.

Thanks for the problem and the detailed solution.