Date: 03/16/2001 at 14:38:56 From: Craig M Chamberlain Subject: Triangle ladder puzzle I found a very interesting math puzzle back when I was in high school, and every so often I return to it to see if I can solve it myself - but no such luck. It was in Popular Science magazine circa 1980-1985, and I am not sure if they ever gave the answer so I am going to type the exact question and see what help you can provide. There was also a diagram which I have duplicated here very simply. "Cussed Ladders" "The January Calcu-letter introduced an ancient and tough ladder puzzle with a comparatively mild warm-up. You were told of two ladders leaning opposite ways between two buildings, one touching eight feet above the ground, the other twelve. At what height above the ground would they cross? |\ /| | \ / | | \ / | | \ / | | \ / | 8'| /|\ | 12' | / | \ | | / |x \ | | / | \ | |/ | \| ----------------- a b "The problem could have been solved using geometry, algebra, trigonometry, analytic geometry - or by selecting an arbitrary value for the distance between the buildings, since the height must be the same no matter what distance is used. "Maybe the simplest way is to label the distance on the ground from each building to line x as a and b, which allows you to set up two simple ratios based on similar triangles: 12 x 8 x ---- = --- and --- = --- a+b a a+b b "Multiplying both equations through and substituting for the common terms yeilds what may have been intuitively obvious: b = 3/2a. Now you can rewrite the above equations using 3/2a in place of b and 5/2a in place of a+b. Solving both equations for x (the a's cancel) gives the height above the ground at which the ladders intersect: 4.8 feet. "Now you are ready to try the second variation of the cussed ladders. This time, ladder lengths are given as 40 and 30 feet, and they cross at a point 10 feet from the ground. How far apart are the buildings against which the ladders are leaning? "Because in this puzzle you have three dimensions given instead of two, it may at first glance look to be the easier problem. It is not. This one will require that you use your calculator - as well as your head" So that is the problem as it was presented with no answer. My thought was again to use similar triangles, this time using 10' instead of x and again using a and b for the base. Now we have two similar triangles: /|\ / | \ 10 b 10 30 / | \ ---- = ---- and ---- = ---- / | \ a 10 b a+b / |10' \ / | \ --------------- a b Once I follow this through (I hope you can see my reasoning here), as in the first problem I substitute in b for the other ratio and proceed to solve for the other part of the base, yet when I use these to check my answer by using Pythagoris to get back 30 and 40 for the hypotenuse, it doesn't work. I am stuck! Craig.
Date: 03/17/2001 at 19:00:13 From: Doctor Floor Subject: Re: Triangle ladder puzzle Hi, Craig, Thanks for writing. Let's say that the walls are the lines x = 0 and x = D in a rectangular grid, and that the x-axis is the floor. Let's say that the shorter, 30-foot-long ladder is placed at O(0,0). This ladder reaches x = D at point A(D, sqrt(900-D^2)) (where we used the Pythagorean theorem to calculate the y-coordinate). [Note: of course 900-D^2 must be greater than 10, which shows that D must be smaller than sqrt(800).] The point where this ladder reaches the 10-foot-high mark is P( 10D/sqrt(900-D^2) , 10 ). The line through P and B(D,0) has gradient 10 - 0 10sqrt(900-D^2) ---------------------- = ------------------- 10D/sqrt(900 - D^2)-D D(10-sqrt(900-D^2)) so the equation is 10sqrt(900-D^2) 10sqrt(900-D^2) y = ------------------- x - --------------- D(10-sqrt(900-D^2)) 10-sqrt(900-D^2) (Point B now fits fine). This means that the y-intercept is: 10sqrt(900-D^2) ---------------- sqrt(900-D^2)-10 Since the second ladder is 40 feet long, this must be equal to sqrt(1600-D^2). So we can solve D from the equation: 10sqrt(900-D^2) ---------------- = sqrt(1600-D^2) sqrt(900-D^2)-10 so 10sqrt(900-D^2) = sqrt[(900-D^2)(1600-D^2)] - 10 sqrt(1600-D^2) and after regrouping 10[sqrt(900-D^2) + sqrt(1600-D^2)] = sqrt[(900-D^2)(1600-D^2)] After that, squaring gives 100[2500-2D^2 + 2sqrt[(900-D^2)(1600-D^2)]] = (900-D^2)(1600-D^2) and simplifying yields 200sqrt[1,440,000 - 2500D^2 + D^4] = 1,190,000 - 2300D^2 + D^4 Squaring this again, we find an equation that is quartic in D^2. From there it is theoretically possible to solve it, but the computations are tedious. For some information on how to solve quartic equations, see the Dr. Math FAQ: Cubic and quartic equations http://mathforum.org/dr.math/faq/faq.cubic.equations.html If you need more help, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
Date: 03/14/2001 at 09:55:59 From: David Sabath Subject: Geometry This is a problem for which I have never seen a satisfactory answer. I have worked on it on and off for quite a while, and have presented it to several people who should know how to do it. I searched your archives and found a similar problem. The similar problem was the problem of the week for June 6-10, 1994. Your problem of the week was as follows: Two ladders, one 20 feet and one 30 feet, are leaning on opposite walls of an alley 10 feet wide. A basketball hoop hangs at the intersection of the ladders and the question is how high the hoop is. This is a fairly straightforward problem and the answer is rather easily computed. My problem is this: Still two ladders, one 20 and one 30 feet long, leaning on opposite walls of an alley. However, I want to know how wide the alley has to be in order for the intersection point of the ladders to be exactly 10 feet high. This problem appears simple at first glance, but I can assure it is not. I would appreciate any help you can give me. Thanks, David Sabath Math Enthusiast
Date: 03/22/2001 at 15:21:11 From: Doctor Jackpo Subject: Re: Geometry Hi David, Let's look at a diagram so we can talk more easily about the geometry problem. I'll prove in a moment that 1/AC + 1/DB = 1/FG. If this is true, then it must be the case that 1/AC + 1/DB = 1/10 and also that AC^2 + AB^2 = 900 and BD^2 + AB^2 = 400. Solving these 3 simultaneous equations, you would get AC = 27.3572325237.. BD = 15/7612871098... and finally AB = 12.3118572378... So, why is it true that 1/AC + 1/DB = 1/FG ? Begin by looking at triangles ABC and ABD. We know that triangle FGB is similar to CAB and triangle DBA is similar to FGA. Therefore, GB / (GB + GA) = FG / AC and GA / (GB + GA) = FG / DB Adding these two equations gives us FG ( 1/DB + 1/AC ) = GA + GB / (GB + GA) which simplifies to FG (1/DB + 1/AC) = 1 (1/DB + 1/AC) = 1/FG I hope this helps. If you have more questions, write back. - Doctor Jackpo, The Math Forum http://mathforum.org/dr.math/
Date: 03/23/2001 at 03:02:38 From: Doctor Floor Subject: Re: Geometry Hi, David, In addition to the letter Dr. Jackpo wrote as an answer to your question, let me write you a way to (theoretically) find the exact solution. I quote from Dr. Jackpo: so we know that 1/AC + 1/DB = 1/10  and we also know that AC^2 + AB^2 = 900  and BD^2 + AB^2 = 400.  Now we can go on: - gives AC^2 - DB^2 = 500   rewrites into 1/AC + 1/DB = 1/10 1/AC = 1/10 - 1/DB 1/AC = (DB - 10)/(10DB) AC = 10DB / (10-DB) And when we substitute this into  we find: 100DB^2/(10-DB)^2 -DB^2 = 500 100DB^2 - DB^2(10-DB)^2 = 500(10-DB)^2 This reduces to a quartic equation. A quartic equation can be solved in exact roots. It would be too tedious to do this now here. For a description of the method, see from the Dr. Math FAQ: http://mathforum.org/dr.math/faq/faq.cubic.equations.html I hope this addition is helpful to you. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
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