Ladder

[cinelli]

A barrel of circular cross section rests against a vertical wall, as shown. A ladder rests on the barrel and touches the wall. The barrel has diameter of 1 yard and the ladder is 4 yards and 6 inches long. How far is the foot of the ladder from the wall?

Cinelli

[GoSeigen]

Simple Euclidean geometry... but you've got me over a barrel with this one!

GS

[Boots]

What are these "yards" to which you refer?

I struggled with this, and made a nasty assumption in my calculation.

Based on these things, I can predict that 4yds and 1/4 inch from the base of the wall to the foot of the ladder is wrong!

[GoSeigen]

What are these "yards" to which you refer?

I struggled with this, and made a nasty assumption in my calculation.

Based on these things, I can predict that 4yds and 1/4 inch from the base of the wall to the foot of the ladder is wrong!

I also didn't like the yards and inches so converted to a 6cm diameter cylindrical glass balancing a 25cm ruler. Should work out the same but you have to convert back to yards/feet/inches at the end!

GS

[modellingman]

Couldn't hide the diagram but the rest of the spoiler is ..


The starting point is to draw 3 radius lines from the centre of the circle to the points at which the circle touches the ground, wall and ladder. Four right angled triangles each similar to that formed by the ground, wall and ladder can be constructed. These triangles are labelled A, B, C and D. The sum of the lengths of the hypotenuses of triangles A, B and C equate to the length of the ladder ie 25/6 yards.

Defining a as the angle between the ground and the ladder and using the terms "Opposite" and "Adjacent" to denote the sides of triangles A, B, C and D which are, respectively, opposite and adjacent to this corresponding angle in these triangles, the lengths of all triangle sides can be defined in terms of sin(a) and cos(a) using SOHCAHTOA. The starting point for this is to note that: the adjacent side of triangle A and the hypotenuse of triangle D are both equal to the radius of the circle, ie 1/2; and the adjacent side of triangle B is the same as the opposite side of triangle D.

The required values for the sides of each triangle necessary to derive the hypotenuses of triangles A, B and C are shown below.

.
           |A           |B                |C                    |D           |
Adjacent   |1/2         |(1/2)*sin(a)     |                     |(1/2)*cos(a)|
Opposite   |            |                 |(1+cos(a))/2         |(1/2)*sin(a)|
Hypotenuse |1/(2*cos(a))|sin(a)/(2*cos(a))|(1+cos(a))/(2*sin(a))|1/2         |

The requirement, therefore, is that

(1+sin(a))/(2*cos(a)) + (1+cos(a))/(2*sin(a)) = 25/6

This being a puzzle, it is reasonable to assume that the solution triangle (the triangle formed by the ground, wall and ladder) will be based on a Pythagorean triple. Let the triangle representing this triple have sides P, Q, R in ascending order meaning that sin(a) = P/R and cos(a) = Q/R. Substituting for sin(a) and cos(a) in the requirements equation yields:

(1+P/R)/(2*Q/R) + (1+Q/R)/(2*P/R) = 25/6

or

(R+P)/(2*Q) + (R+Q)/(2*P) = 25/6

or

(P*R + P*P + Q*R + Q*Q)/(2*P*Q) = 25/6

For the numerator of the LHS, note that (by assumption) P*P + Q*Q = R*R, so the numerator can be rewritten as R*(P+Q+R). For the denominator we note that

2*P*Q = (P+Q)*(P+Q) - (P*P + Q*Q) = (P+Q)*(P+Q) - R*R (again by assumption)

and factorising the difference between the two squares yields the denominator as

2*P*Q = (P+Q+R)*(P+Q-R)

The factor (P+Q+R) is common to both the numerator and denominator meaning the requirements equation can be simplified to

R/(P+Q-R) = 25/6

The values 25 and 6 on the RHS are co-prime suggesting that the Pythagorean triple has a value of R which is a multiple of 25. Considering the simplest multiple (25 itself) there are two triples in which 25 is the largest value. These are 7, 24 and 25 and 15,20, 25. Only the former satisfies the requirement that R/(P+Q-R) = 25/6 and therefore this triple provides a basis for the solution triangle.

Since the ladder has length 25/6 yards, it follows that the solution triangle is the triple of 7, 24, 25 scaled to one-sixth size. This means the ladder rests on the ground at 24/6 = 4 yards from the wall and is resting against the wall at 7/6 = 1 yard 6 inches above the ground.

Nothing (except perhaps the hint in Cinelli's diagram) prevents a symmetric solution in which the ladder rests against the wall 4 yards above the ground and on the ground at 1 yard 6 inches from the wall.



modellingman

[modellingman]

I have realised my earlier solution is quite a bit more cumbersome than necessary. Here's a simpler one...



Reflecting on my earlier solution, it did cross my mind that there was probably a relationship between a triangle and its inscribed circle. My maths curriculum at school (well over 50 years ago) was based on the then new-fangled Schools Mathematics Project and included practically no geometry. As I found (courtesy Bing), there is a very simple standard formula relating the sides of a right-angled triangle to the radius of its inscribed circle.

This standard formula leads to the expression for Cinelli's puzzle that

P+Q = 31/6

(where P and Q are the vertical and horizontal sides of the right angled triangle formed by the wall, ground and ladder).

Squaring this expression and manipulating it a bit leads to

PQ = 28/6

Using the first equation to write P in terms of Q and substituting this expression in the second equation yields a quadratic in Q which has two real solutions. Whichever of these solutions is chosen for Q the equations above yield the other solution as the solution for P. These two solutions are the same as those previously derived from the earlier approach: ie 4 and 7/6.



modellingman

[cinelli]

Reply to solutions:

I applaud modellingman for his solutions and for continuing to work in yards, and for a correct solution. Using yards was a bit of a tease and converting into a more common unit was part of the puzzle. But in reply to Boots, yards are not too uncommon. After all we still use miles and yards as a measure for road travel.

But I wonder if anyone has an alternative solution.

Cinelli

[9873210]

Reply to solutions:

But I wonder if anyone has an alternative solution.

Cinelli

Not substantially different but starting as modellingman and getting
(1+sin(a))/(2*cos(a)) + (1+cos(a))/(2*sin(a)) = 25/6
I proceeded algebraically.
Writing s = sin(a) and c = cos(a)

(1+s)/2c + (1+c)/2s =25/6
3(1+s)s + 3(1+c)c =25cs

/// c^2+s^2 = 1 so
3s+3c+3 = 25cs
3(s+1) = c(25s-3)

//// Square both sides and set c^2 = 1-s^2 = (1+s)(1-s)
9(s+1)^2 = c^2 (625s^2-150s+9)
9(s+1)^2= (1-s)(1+s)(625s^2-150s+9)

//// Divide by (s+1). s=-1 is a solution to the quartic but not to the original problem. Induced by multiplying by 6sc above.
9(s+1) = (1-s)(625s^2-150s+9)
9s+9 = (625s^2-150s+9) -s(625s^2-150s+9)
9s = (625s^2-150s ) -s(625s^2-150s+9)

//// Divide by s. solution s=0 is an artifact as above.
9 = (625s-150) -(625s^2-150s+9)
625s^2 -150s +9+9 -625s+150 = 0
625s^2 -775s +168 = 0

//// solve the quadratic. Solutions are rational because discriminant is a perfect square.
//// sqrt(b2-4qc) = 425

s = sin(a) = {1200/1250, 350/1250} = {24/25, 7/25}

This gives the same Pythagorean triple as modellingman so continue as he did.

[GoSeigen]

I'll have another go later today if I get a moment.

GS

[jfgw]

You should't need trigonometry for something like this.

Spoiler...

I would convert the length of the ladder to 150" first. Hold it in place with some bricks at the foot and drag out the barrel.

Now get a shorter ladder and put it underneath the 150" one. I don't know what length it needs to be but it needs to lay parallel to to the longer one and pass through the point 18" from the wall and 18" from the ground (where the exact centre of the barrel was). It will also be 18" from the long ladder (measured perpendicularly).

Now get a try-square and some chalk. Place the square at one end of the short ladder and mark off where the square crosses the long ladder. Do the same the other end. Also mark the short ladder where it is 18" from the wall and ground.

Label the sections of ladder like this:
Mark the top of the long ladder to the top mark "A" .
Mark the top section of the short ladder "B" .
Mark the bottom section of the short ladder "C" .
Mark the bit between the bottom mark and the foot of the long ladder "D" .

If you look at you ladders side-on, you should see that the length of the long ladder is A+B+C+D.

Now fix an eyelet 18" up the wall and attatch a piece of string to it.
Tie a nut onto the other end, the string and nut need to be about a yard long but no longer. Knock a nail into the short ladder where the chalk mark is and hook the string over it so that the wall, the ground and the string make an 18" square.

You will see that the top part of the short ladder, the horizontal bit of string and a bit of the wall make a triangle. If you hold your square against the top of the short ladder again like you did before, you will see that this makes another triangle the same size and shape, just twisted around (compare the angles and the 18" sides). You should see that the height of the top of the long ladder is 18+A+B. Likewise, you should see that the foot of the long ladder is 18+C+D from the wall.

Let's say that the distance of the foot of the long ladder to the wall is x.

Let's also say that the height of the top of the long ladder is y.

x+y=18+A+B+18+C+D

We know that A+B+C+D = the length of the long ladder = 150 so x+y=150+36=186.

So y=186-x

From pythagoras, x^2 + y^2 = 150^2

Substituting for y we get x^2 + (186-x)^2 = 22500

This gives us 2x^2 - 372x + 12096 = 0

Using the quadratic formula gives x=144 or x=42

So the foot of the ladder is either 144" (= 4 yards) from the wall or 42" (= 1 yard 6") from the wall. Since the original drawing shows the ladder relatively horizontal, the answer is 4 yards.


Julian F. G. W.

[GoSeigen]

Using Pythagoras's theorem:


1. I converted the lengths to metres, so the ladder is 25m long(!), the barrel 6m diameter. Think firefighters with a very large supply of beer
2. The centre of the barrel is 3m above the ground and away from the wall.
3. Draw a radius from where the ladder touches the barrel to its centre (forming a right angle with the tangential ladder). Label the upper part of the ladder a, the lower part b. Then a + b = 25m (Eq. 1).
4. Draw a radius parallel to the wall from the centre of the barrel to the ground. The distance between the line and the wall is obviously 3m.
5. Draw a line from the centre of the barrel to where the ladder touches the ground, creating two congruent triangles above and below (the two radiii and shared edge are equal lengths and both are right-angled). Therefore the distance on the ground from the base of the barrel to the base of the ladder is also b and the ladder touches the ground (b+3m) from the wall.
6. Do a similar construction where the barrel and ladder touch the wall, drawing a radius to the wall and a line from the centre to where the ladder touches the wall. We have two more congruent triangles, so the height of the ladder above the centre of the barrel is also a and the total height above the ground is (a+3).
7. So our ladder with the wall and ground make a right-angled triangle with hypotenuse (a+b) and other sides (a+3) and (b+3) = (28-a) by Eq. 1.
8. By Pythagoras (a+b)^2 = (a+3)^2 + (28-a)^2
9. Simplifying and solving the quadratic we get a=4 or a=21, obviously a=4 is the root we need.
10. Therefore b=21 and the foot of the ladder is B+3 = 24m from the wall.

Converting back to yards I make that 4 yards.

[GoSeigen]

Just to clarify, I didn't literally convert yards to metres, I just scaled the lengths and called the resulting values metres (so a scaling factor of 6m for one yard).


GS

[GoSeigen]

My version could have been solved entirely geometrically by using Euclid's construction of a square from a rectangle (application of areas), of which the problem is very reminiscent.

GS

[cinelli]

GoSeigen’s trig-less solution is the one I had in mind. I am glad everyone had a bit of fun with this problem. I also like Julian’s geometric constructions using chalk and a piece of string.

Cinelli

[jfgw]

GoSeigen’s solution certainly looks a lot simpler than mine. We have started by using different methods of showing that the distance from the foot of the ladder to where the ladder touches the barrel, and from the foot to where the ground touches the barrel, are the same. Likewise with the top of the ladder and the wall. We end up with the same quadratic equation, which is easily solved.

Having drawn these out, I think that we could simplify the solution further by saying that, by symmetry, these lengths must me the same.

Here is my solution,


And here is GoSeigen's simpler solution,


An interesting puzzle, thanks for posting it. It is interesting how one can go from not knowing where to start to finding a very simple solution.


Julian F. G. W.

[GoSeigen]

Having drawn these out, I think that we could simplify the solution further by saying that, by symmetry, these lengths must me the same.

Here is my solution,


And here is GoSeigen's simpler solution,

Thanks for the pretty drawings, you've certainly got mine right -- marking the right angles might complete it.

Re: the symmetry comment, how would one prove the symmetry if not by pointing out the congruence? It's no good just saying the diagram looks symmetrical!


GS

[jfgw]

Re: the symmetry comment, how would one prove the symmetry if not by pointing out the congruence? It's no good just saying the diagram looks symmetrical!

I could have made my thinking clearer. The two lines which form tangents are reflections about a bisector of the circle,



Julian F. G. W.

[GoSeigen]

Re: the symmetry comment, how would one prove the symmetry if not by pointing out the congruence? It's no good just saying the diagram looks symmetrical!

I could have made my thinking clearer. The two lines which form tangents are reflections about a bisector of the circle,

D'oh! Sorry we're a bit slow here in the Joe Biden Team..

GS