MATH questions

Sort:
shepi13

I used slope to find it Laughing.

Assuming the intersection of the H side and the ground is the origin, the equation of the ladder is: y-60 = m(x+60)

Then: y = mx + 60m + 60.

It follows that the y intercept (H) is 60m + 60, and the x intercept is -60-60/m.

So (60m+60)^2 + (-60-60/m)^2 = 220^2 by pythagorean theorum.

My calculator solved this for 4 different values, but because the slope has to clearly be greater than 1, the only one that worked was m = 2.3805.

Plugging this into the y intercept of H = 60m+60, I got 202.83.

Didn't use any similar triangles. I'm not really a geometry person though, as I stated earlier.

LoekBergman

Hmm, I used a different method and overlooking my previous answer have to change the number. I made a mistake in my previous calculation. Hope I got it right this time. Math for amateurs, wasn't it? :-)

I used the diagonal of the square 60/60. The length of that diagonal is 60√2.

 

Then I made use of the sinus-rule which says that:

 

A/sin α = B/sin β = C/ sin γ.

 



 

Then I reasoned that there were equal angles in both triangles divided by this diagonal.

 

The angle in the lower right corner beneath the diagonal is equal to the top angle of the big triangle and the angle in the right corner above the diagonal is equal to the left below angle of the big triangle. All those angles must be 45°, otherwise will you not get the square of 60 by 60.

 

Then can you apply the above mentioned sinus rule again and do you get the next calculation:

 

x/ sin 45° = 60√2 /45° and (220-x)/ sin 45° = 60√2 /45°. It is clear that x equals 60√2.

 

With the help of Pythagoras does that imply that the lower bottom of the triangle is 120 wide.

 

The standing height is therefor:

 

(220² – 120²) = 184.

 

That is different then my first answer, because I made an error in that calculation. I took wrong numbers. I had made several attempts and calculated using wrong numbers (110 as the value for x).

 



RomyGer

This question was asked by me in various chess club magazines before, so I know the kind of calculations and answers.

It is too early for me to give more info, but up to and including post 170, nobody gave the correct answer so far.      Sorry, Nobody, But that makes it worthwile to try to find it  --  I hope more posts will follow !

For the Dutch readers : this made me in 2008   "de Bolleboos van de Maand" in the Dutch site of Neurocampus  ( and I am NO high-flyer in maths...)

steve_bute

Two solutions: h=202.8, h=85.2

Assuming some are still working on the problem, I won't post my method.

Thank you, RomyGer, a fun little problem.

TheGrobe

85.2 is the base, assuming that the general proportion is represented in the diagram.

steve_bute
TheGrobe wrote:

85.2 is the base, assuming that the general proportion is represented in the diagram.

The "not drawn to scale" annotation was not enough of a hint, apparently :). It can be the base or the wall height.

TheGrobe

Yeah, so there are two solutions, the 85.2 solution looks like the diagram reflected across the diagonal that runs from the top left to the bottom right.

In the 202.8 solution, the base is 85.2 and in the 85.2 solution the base is 202.8.

LoekBergman

steve_bute is correct that there can be two results. If mine would be good, then would it not only be 184, but 120 too.

steve_bute

I remember a similar problem about having to carry a ladder around a corner in a hallway, where you were given the ladder length and hallway widths on each side of the corner, the question being "Can you get the ladder around the corner?" The most elegant solution I've seen for it was a calculus formulation, where you drew a straight line through the inside corner and minimized the length of the line from wall to wall.

Pre_VizsIa

LoekBergman wrote: 

Then I made use of the sinus-rule which says that:

 

Then can you apply the above mentioned sinus rule again and do you get the next calculation:

 What do noses have to do with this?

Pre_VizsIa

jk of course

TheGrobe
steve_bute wrote:

I remember a similar problem about having to carry a ladder around a corner in a hallway, where you were given the ladder length and hallway widths on each side of the corner, the question being "Can you get the ladder around the corner?" The most elegant solution I've seen for it was a calculus formulation, where you drew a straight line through the inside corner and minimized the length of the line from wall to wall.

How high a ceiling...?

RomyGer

I just read up to and incl. post 179 and still no correct answer.

Loek : draw it on a piece of paper, A4 size and you will notice that 184 and 120 is way out..., sorry...

And, yes, Steve : it is fun and the correct answer is so o o o nice !  ( and I am sure your "method" is wrong, oops...)

SetUCheckM8

What a derp, but 7 + 7 = 17 can be correct :D.

The one who can give the explaination for that will get 3 trophies from me :D

SetUCheckM8

Please quote me when you think you know the solution 

steve_bute

If h=202.8 and h=85.2 are not correct, there must be something about the problem that I have misunderstood:

TheGrobe
RomyGer wrote:

I just read up to and incl. post 179 and still no correct answer.

Are you sure?  202.83036 and 85.20473 check out thoroughly.

202.83036² + 85.20473² = 220²

√((202.83036 - 60)² + 60²) = 154.921 (Hypotenuse of top triangle)

√((85.20473 - 60)² + 60²) = 65.079 (Hypotenuse of bottom triangle)

154.921 + 65.079 = 220 (The sum of the two smaller hypotenuses is the length of the large hypotenuse)

steve_bute

Unless the answer is 220 because he is holding his ladder fully against the side of his house, the box being a decoy.

Remellion

I second steve_bute's answer. Well, if you insist on exact form, I'll give:

H = 30 + 10sqrt130 + 10sqrt(130 - 6sqrt130) or
H = 30 + 10sqrt130 -  10sqrt(130 - 6sqrt130).

And for humour's sake, the ladder is 220 units long.

TheGrobe
SetUCheckM8 wrote:

What a derp, but 7 + 7 = 17 can be correct :D.

The one who can give the explaination for that will get 3 trophies from me :D

I can get 16, but not 17.