 mersenneforum.org The Ladder Against The Wall
 Register FAQ Search Today's Posts Mark Forums Read  2005-06-18, 21:29 #1 Numbers   Jun 2005 Near Beetlegeuse 6048 Posts The Ladder Against The Wall A rectangular shed of dimension h is built against the perpendicular wall of a house. A man stands a ladder on the ground and leans it against the shed so that its top touches the wall of the house. Express the length of the ladder in terms of h. I was given this problem some time ago and found the following partial solution: Let the points P, Q, and R form a right-angled triangle with PQ the hypotenuse. Let there be a point N on PQ such that: 1) A perpendicular line drawn from the point S on PR meets PQ at N. 2) A horizontal line drawn from the point T on QR meets PQ at N. 3) RS = SN = NT = TR = h. Note: Three letters thus NPS means the angle at P formed by NP & PS whilst three letters thus (NPS) means the triangle NP, PS, SN. TNS = 90°, tf (means "therefore") QNT + SNP = 90° NSP = 90°, tf NPS + SNP = 90° tf NPS = QNT, PNS = NQT So that if PS = h, NS = PS = NT = QT and (PNS) = (NQT) Since QT = TR = h, then QR = 2h Since PS = RS = h, then PR = 2h by Pythagoras we get PQ^2 = QR^2 + PR^2 = (2h)^2 + (2h)^2 tf PQ = Sqrt((2h)^2 + (2h)^2) by division we get PQ/h = Sqrt(8) tf PQ = Sqrt(8)h However, If PS h or if PS >h then QT Sqrt(8)h So, my final answer on the subject (for now) is that PQ >= Sqrt(8)h However, my questioner, a Professor of Mathematics, insists that there is a general solution for the question as asked. Any takers?   2005-06-18, 22:34   #2
Wacky

Jun 2003
The Texas Hill Country

44116 Posts Quote:
 Originally Posted by Numbers A rectangular shed of dimension h is built against the perpendicular wall of a house. A man stands a ladder on the ground and leans it against the shed so that its top touches the wall of the house. Express the length of the ladder in terms of h. Let the points P, Q, and R form a right-angled triangle with PQ the hypotenuse. Let there be a point N on PQ such that: 1) A perpendicular line drawn from the point S on PR meets PQ at N. 2) A horizontal line drawn from the point T on QR meets PQ at N.
OK, so far.

Quote:
 3) RS = SN = NT = TR = h.
We may fairly assume that by choosing "h", it is a reference to the height of the shed.
SN = TR = h

But why do you assume that the shed has the same depth as its height?

A rectilinear solid has 3 dimensions. We don't care about one of them (the width), but that does not eliminate the other two.

Last fiddled with by Wacky on 2005-06-18 at 22:37   2005-06-19, 04:31 #3 Numbers   Jun 2005 Near Beetlegeuse 6048 Posts h is the height From the fact that the shed is described as being a rectangle of dimension h, it is safe to assume that it has a single measurement. If it were wider than it is tall it would be described as having dimensions (plural). The shed is therefore a cube, represented in Plane Geometry by a square, but I wrote the question exactly as it was put to me. Keith (my questioner) did not say that my partial solution was wrong, merely that it was incomplete. He also said that the problem could be solved using nothing more advanced than 'O' Level Maths. For those of you not familiar with the British educational system, or those who are but are too young to remember, 'O' level is an exam typically sat by a 16 year-old. Last fiddled with by Numbers on 2005-06-19 at 04:44   2005-06-19, 14:43   #4
Wacky

Jun 2003
The Texas Hill Country

32×112 Posts Quote:
 Originally Posted by Numbers From the fact that the shed is described as being a rectangle of dimension h, it is safe to assume that it has a single measurement. If it were wider than it is tall it would be described as having dimensions (plural). The shed is therefore a cube…
I don't accept this argument. It is equally plausable to assume that omitted dimensions do not matter. (That is standard procedure in typical "puzzles". In fact, you can sometimes use the omission of a fact to help figure out the answer) This is clearly the case in this example that the width of the shed does not matter.

But, even assuming that the shed has square cross section of dimension "h", the ladder can be of any length not less than h*sqrt(8). A very long ladder would lie nearly horizontally (or vertically). As the length decreases, the pitch increases until we reach the minimum length which must sit at 45 degrees.

I don't see how one can be any more specific than you were.

Quote:
 … too young to remember, 'O' level is an exam typically sat by a 16 year-old.
Even that does not really help. When I was in school, differential calculus wasn't taught until the second year of University. Now it is taught in secondary school. I'm assuming that the 'O' exam includes plane geometry, but not trig, analytic geometry, or calculus.

Last fiddled with by Wacky on 2005-06-19 at 14:47   2005-06-19, 15:24 #5 Numbers   Jun 2005 Near Beetlegeuse 22·97 Posts Dimension v Measurement Not being a real mathematician like yourself (I am not due to start University until September) when I read the word "dimension" I did not understand it to have the meaning used in "a three-dimensional object". In English (as opposed to mathematics), dimension can also mean "measurement", as in "The dimensions of my swimming pool are 25ft x 10ft". So when I read that it was a rectangular shed of dimension h, I took that to mean that in any meaningful direction (excluding diagonals and so forth) it measured h. I also have the advantage of knowing Keith, the poser of the original question, which means that I am much less likely than otherwise to suspect him of subterfuge or deception; he just isn't like that. I think that he would have honed his presentation very carefully before posing the question to ensure that it had only the meaning he intended. I am encouraged by your comment that you don't see how one can be any more specific than I was (I have not yet discovered how to do a quote), because that would imply that if there is indeed a more complete solution then perhaps it is not quite as straighforward as Keith supposes. Your assumption about what is (or rather was, it no longer exists) included in an 'O' Level maths course is correct. I passed the thing but had to teach myself calculus and know no more about analytical geometry than how to spell it. Last fiddled with by Numbers on 2005-06-19 at 15:26   2005-06-19, 17:39   #6
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

205210 Posts Quote:
 Originally Posted by Wacky Even that does not really help. When I was in school, differential calculus wasn't taught until the second year of University. Now it is taught in secondary school. I'm assuming that the 'O' exam includes plane geometry, but not trig, analytic geometry, or calculus. A little clarification here. I think I was in my final year(1954) of 'O' levels before you wacky. At the time the Senior Cambridge, in what was the equivalent or higher than the G.C.E Level, also offered 'A' level subjects to students specialising in particular subjects. I did mine in Additional maths 'A' level.
So though the other subjects were at O level, including O level math , the 'A' level done at the same time consisted of Pure and Applied mathematics and embraced the subjects you mention plus applied mechanics at a higher level than in most engineering courses at University. Thats how I was able to breeze thru University! Mally    2005-06-27, 04:52 #7 Ken_g6   Jan 2005 Caught in a sieve 1100010112 Posts Using your points, the length of the ladder l is l=sqrt((PS)^2+h^2)+sqrt((QT)^2+h^2) Call the height of the wall w (which I'm considering to be PR, but it's symmetrical). Let's derive a relation between PS and QT, so that l will be a function of two variables (w and h). Triangles PNS and NQT are similar. They are both right triangles, and NS is parallel to QT. So PS/NS = NT/QT. NS=NT=h, so PS/h = h/QT. Cross-multiply to get PS*QT = h^2 Here I'll say that PS = w-h (since w=PR). So QT = h^2/(w-h) So, l=sqrt((w-h)^2+h^2)+sqrt((h^2/(w-h))^2+h^2) It's simpler as PS: l=sqrt((PS)^2+h^2)+sqrt((h^2/(PS))^2+h^2) l=sqrt((PS)^2+h^2)+sqrt(h^4/(PS)^2+h^2) Let's see Keith get the PS out of this one! (In other words, I think you're right and Keith is wrong.)   2005-06-27, 14:11 #8 tom11784   Aug 2003 Upstate NY, USA 2×163 Posts I think I *may* have your generalization What's not specified and makes a huge difference on the length is the angle with respect to the ground with which the ladder is placed. We have that the size of the shed is h, and the distance from the shed to the base of the ladder is PS. Then we have tan(QPR)=h/PS, meaning PS=h*tan(QPR). Using the fact proven in above posts that PS*QT=h2 this gives that QT=h/tan(QPR). Now we know that the length of the ladder is: L = sqrt( (PS+h)2 + (QT+h)2 ) = sqrt( (h*tan(QRP)+h)2 + (h/tan(QRP)+h)2 ) Now we pull out the h terms as follows L = sqrt ( h2(tan(QRP)+1)2 + h2(1/tan(QRP)+1)2 ) L = h * sqrt( (tan(QRP)+1)2 + (1/tan(QRP)+1)2 ) Exapnding the inside terms, placing in terms of sin and cos (very messy in this text box - but I'll try to help using the bold, italics, and underlining) and noticing a neat regrouping: L = h * sqrt( (tan2(QRP) + 2*tan(QRP) + 1) + (1/tan2(QRP) + 2/tan(QRP) + 1) ) L = h * sqrt( (sin2(QRP)/cos2(QRP) + 2sin(QRP)/cos(QRP) + 1) + (cos2(QRP)/sin2(QRP) + 2cos(QRP)/sin(QRP) + 1) ) L = h * sqrt( (sin2(QRP)/cos2(QRP) + 1) + (2sin(QRP)/cos(QRP) + 2cos(QRP)/sin(QRP)) + (cos2(QRP)/sin2(QRP) + 1) ) L = h * sqrt( (sin2(QRP)+cos2(QRP))/cos2(QRP) + (2sin2(QRP)+2cos2(QRP))/(sin(QRP)cos(QRP)) + (sin2(QRP)+cos2(QRP))/sin2(QRP) ) Now we use our favorite trig identity sin2 + cos2 = 1 to reduce and then substitute with sec(QRP) and csc(QRP) to finally get rid of those nasty denominators: L = h * sqrt( 1/cos2(QRP) + 2/(sin(QRP)cos(QRP)) + 1/sin2(QRP) ) L = h * sqrt( sec2(QRP) + 2sec(QRP)csc(QRP) + csc2(QRP) ) And now we notice that this is in the proper format for a squared sum... L = h * sqrt ( (sec(QRP) + csc(QRP))2 ) L = h * (sec(QRP) + csc(QRP)) for 0 2005-06-27, 15:28 #9 Ken_g6   Jan 2005 Caught in a sieve 39510 Posts Yes, but now you have L(h,QRP). I got L(h,PS) and L(h,w). But what he wants is L(h). I think I can prove that's impossible. Assume there exists L(h). Then w must be a function of h. Now assume there exists an inverse of w at at least two points. Then the functions h(w) and thus L(w) are defined at at least two points. One point we know: for w=0, h=0, and L=0. But this states that there exists a non-zero value for w which defines the value of L. This is absurd! Pick two different h's, and you'll get two different L's. There's one other possibility: maybe the inverse of w exists at only one point. But then w(h) must be a constant, which is clearly false. Pick two h's; for L to remain the same, w(h) must vary. Thus there is no function w(h), so L cannot be a function of only h.   2005-06-27, 17:19   #10
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

80416 Posts Quote:
 Originally Posted by Ken_g6 Yes, but now you have L(h,QRP). I got L(h,PS) and L(h,w). But what he wants is L(h). I think I can prove that's impossible. Assume there exists L(h). Then w must be a function of h. Now assume there exists an inverse of w at at least two points. Then the functions h(w) and thus L(w) are defined at at least two points. One point we know: for w=0, h=0, and L=0. But this states that there exists a non-zero value for w which defines the value of L. This is absurd! Pick two different h's, and you'll get two different L's. There's one other possibility: maybe the inverse of w exists at only one point. But then w(h) must be a constant, which is clearly false. Pick two h's; for L to remain the same, w(h) must vary. Thus there is no function w(h), so L cannot be a function of only h
: I think you are quite right Ken-G6. How about Keith enlightening us?
I have not really taken a shot at this problem but posed it to my math mentor, a real genius. He straight away said 'insufficient data'. Mally    2005-06-27, 19:28 #11 Numbers   Jun 2005 Near Beetlegeuse 38810 Posts Keng6, I’m not sure I follow your logic. You say that: “Assume there exists L(h). Then w must be a function of h.” According to my rather primitive understanding of function, w can only be a function of h if the value of w changes as the value of h changes. Since in the above problem h is a constant (the height and width of the shed) then neither w nor anything else can be a function of h. It is therefore not surprising that you should conclude “there is no function w(h),”. However, I had not thought of approaching the problem from the point of trying to prove that no solution exists; thank you for bringing this interesting new dimension to the thread. Last fiddled with by Numbers on 2005-06-27 at 19:38   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post Gandolf Math 49 2017-04-13 20:00 Dubslow Soap Box 17 2012-05-14 08:51 ewmayer Soap Box 25 2009-06-17 23:07 fivemack Factoring 13 2007-04-13 23:26

All times are UTC. The time now is 21:12.

Thu Oct 28 21:12:01 UTC 2021 up 97 days, 15:41, 0 users, load averages: 1.52, 1.47, 1.42