[Guest]

An old one but a favourite of mine.

A 10 ft ladder is leaning on a wall. The base is touching the floor and the top is touching the wall. Under the ladder there is a box which is 3ft x 3ft x 3ft. and it is pushed hard against the wall on which the ladder is leaning. The ladder is pushed in at the base so that the ladder just touches the corner of the box.

How high does the ladder reach up the wall ?

[EventHorizon]

You've heard of 3-4-5 triangles right?

The 3x3 block cuts the 10' ladder in half (2*5). So you are left with an open side for both triangles... The missing 4'.

So the length up the side of the wall is a total of 3' (the box) plus the 4' so you get a total height of 7'.

what if you want to use your ladder afterwards?

[EventHorizon]

An old one but a favourite of mine.

A 10 ft ladder is leaning on a wall. The base is touching the floor and the top is touching the wall. Under the ladder there is a box which is 3ft x 3ft x 3ft. and it is pushed hard against the wall on which the ladder is leaning. The ladder is pushed in at the base so that the ladder just touches the corner of the box.

How high does the ladder reach up the wall ?

Starting with the two equations...

x/3 = 3/y (where x is the distance away from the box where the bottom of the latter starts, and y is the distance from the top of the box to the top of the ladder)

(x+3)^2 + (y+3)^2 = 100

results eventually in the following quartic equation.

x^4 + 6x^3 - 82x^2 + 54x + 81 = 0

From there you look up how to solve the quartic equation...

(http://planetmath.org/encyclopedia/GaloisTheoreticDerivationOfTheQuarticFormula.html, scroll down for the summary)

I(x) = (x-t1)(x-t2)(x-t3)

= x^3 + 164x^2 + 6724x + 32400

I noticed (by using a calculator.....so sue me) that -100 is a root. Using synthetic division results in

(x-t2)(x-t3) = x^2 + 64x + 324

Noticing also that the only values the quartic needs from t2 and t3 are their sum....that is obviously -64

(because x^2-(t2+t3)x +t2*t3 = x^2 + 64x + 324, so -(t2+t3)x = 64x)

t1 = -100

t2+t3 = -64

continuing through the derivation...

r1+r2 = (-6+sqrt(6^2-4(-100)))/2 = -3 + sqrt(109)

r3+r4 = (-6-sqrt(6^2-4(-100)))/2 = -3 - sqrt(109)

r1*r2 = (36+sqrt(36^2-16*81))/4 = 9

r3*r4 = (36-sqrt(36^2-16*81))/4 = 9

r1 = (-3 + sqrt(109) + sqrt((-3 + sqrt(109))^2-4*9))/2

r2 = (-3 + sqrt(109) - sqrt((-3 + sqrt(109))^2-4*9))/2

r3 = (-3 - sqrt(109) + sqrt((-3 - sqrt(109))^2-4*9))/2

r4 = (-3 - sqrt(109) + sqrt((-3 - sqrt(109))^2-4*9))/2

r3 and r4 turn out negative....not useful...

so x = (-3+ sqrt(109) +/- sqrt( 82 - 6 * sqrt(109)))/2

which is about 5.92 and 1.52.

Since these were the distances from the top of the box....

The height the ladder reaches is either 8.92 or 4.52.

You can double check by noticing that 8.92^2 + 4.52^2 = 100

Aren't you happy they made calculators....

[Guest]

It can't be 8'. This would make the bottom side 6', and with a 3'x3' box you would have the lower similar triangle with two sides equal to 3', thus creating a 45º angle between the ladder and the ground. This can not be so with a triangle of two sides with different lengths.

I've managed to create 4 equations with 4 unknowns, it's just a matter of whittling it down...

Keep it to just two unknowns, from the top of the box to the top of the ladder, and from the base of the box to the base of the ladder.

[Guest]

[spoiler='And the answer is :-

']8.9200 or 4.5203 feet.

[Guest]

1. Pythag

(x+3)^2 + (y+3)^2 = 10^2

2. Similar triangles

(x/3) = (3/y)

[spoiler='And the answer is :-

']8.9200 or 4.5203 feet.

[Guest]

From Pythag

(x+3)^2 + (y+3)^2 = 10^2

(x^2+6x+9 + y^2+6y+9)=100

rearrange

x^2+y^2+6(x+y)+18=100 ..#1

From similar triangles :-

(x/3) = (3/y)

rearrange

xy=9 ...#2

2xy=18 .... #3

sub #3 into #1

x^2+y^2+6(x+y)+2xy=100

rearrange

(x+y)^2 +6(x+y)-100 = 0

solve quadratic

(x+y)= 7.4403 ..#4

From #2 y=(9/x) ...#5

Sub #5 into #4

x+(9/x)=7.4403

mult by x and rearrange

x^2 -7.4403x + 9=0

Solve for x

x=5.9200 or 1.5203

So height is

8.9200 or 4.5203