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Using the Pythagorean theorem and rules about similar triangles X=116.32
There is probably a simpler way because after substituting the the reduced similar triangle equation into the Pythagorean theorem equation I ended up with a fourth order polynomial equation to solve (used a root finder on the web to solve it). Errol Groff wrote: Ladder_Problem.PDF is an interesting math problem I found in the book Machine Shop Trade Secrets by James A Harvey. The solution I used to find the answer will be posted as Ladder_Problem_Answer.PDF http://metalworking.com/DropBox/ Errol Groff Instructor, Machine Tool Department H.H. Ellis Technical High School 643 Upper Maple Street Dantieson, CT 06239 New England Model Engineering Society www.neme-s.org |
Interesting Math Problem
Ladder_Problem.PDF is an interesting math problem I found in the book
Machine Shop Trade Secrets by James A Harvey. The solution I used to find the answer will be posted as Ladder_Problem_Answer.PDF http://metalworking.com/DropBox/ Errol Groff Instructor, Machine Tool Department H.H. Ellis Technical High School 643 Upper Maple Street Dantieson, CT 06239 New England Model Engineering Society www.neme-s.org |
Errol Groff wrote:
Ladder_Problem.PDF is an interesting math problem I found in the book Machine Shop Trade Secrets by James A Harvey. The solution I used to find the answer will be posted as Ladder_Problem_Answer.PDF http://metalworking.com/DropBox/ Errol Groff Instructor, Machine Tool Department H.H. Ellis Technical High School 643 Upper Maple Street Dantieson, CT 06239 New England Model Engineering Society www.neme-s.org I was able to do it on my cad program and that took some nudging around. I still can't think of a trig answer for it? |
"Errol Groff" wrote in message
... Ladder_Problem.PDF is an interesting math problem I found in the book Machine Shop Trade Secrets by James A Harvey. Simple similar triangles problem. Well, not perfectly simple, but not impossible. Let the top point be A, lower left-hand corner C, far right corner E, and the middle left point B, middle bottom point D and middle hypotenuse, F. A |\ B-F | |\ C-D-E Line segments BC, CD, DF and BF are given as 24.000 inches (I had no idea ladders could be positioned so exactly). Segment AE is given as 120". Let AC = X. Given the straight lines, parallel lines and right angles, find X. There are three basic right triangles defined, ACE, ABF and FDE. We know the base of ABF is 24", the left side of FDE is 24" and the hypotenuse of ACE is 120". BF = 24 DF = 24 X = 24 + AB AB = X - 24 DE = CE - 24 CE = 24 + DE AF = 120 - EF EF = 120 - AF AE = 120 BC = BF = CD = DF = 24 They are similar triangles because of the equal angles, thus: AB/X = AF/AE = CD/CE (top over whole) DF/X = EF/AE = DE/CE (bottom-right over whole) and AB/DF = AF/EF = CD/DE (top over bottom-right) Now substitute like an M.F.... (X-24)/X = (120-EF)/120 = 24/CE 24/X = (120-AF)/120 = (CE-24)/CE (X-24)/24 = (120-EF)/EF = 24/DE Split the parenthesis: 1 - 24/X = 1 - EF/120 = 24/CE 24/X = 1 - AF/120 = 1 - 24/CE X/24 - 1 = 120/EF - 1 = 24/DE (Bottom left relation) X/24 = 120/EF = X = 120/24*EF = X = 120/24*(120 - AF) (Center by center right relation) -1\\ + AF/120 = -1\\ + 24/CE = AF = 24*120/CE X = 120\\\/24*(120\\\ - 24*120\\\/CE) = 1/(24 - 24*24/CE) = 1/[24 - 24*24/(24+DE)] = [1/1] / [24*(24+DE) - 24*24]/(24+DE) = (24 + DE) / 24*24\\\\\ + 24*DE - 24*24\\\\\ = 24/24*DE + DE/24*DE = X = 1/DE + 1/24 Um, and so on... it seems I'm too tired to finish this tonight. Tim -- "California is the breakfast state: fruits, nuts and flakes." Website: http://webpages.charter.net/dawill/tmoranwms |
almost any height you want
you didnt specify how far awat the bottom is. Doug "Errol Groff" wrote in message ... Ladder_Problem.PDF is an interesting math problem I found in the book Machine Shop Trade Secrets by James A Harvey. The solution I used to find the answer will be posted as Ladder_Problem_Answer.PDF http://metalworking.com/DropBox/ Errol Groff Instructor, Machine Tool Department H.H. Ellis Technical High School 643 Upper Maple Street Dantieson, CT 06239 New England Model Engineering Society www.neme-s.org |
Tim Williams wrote:
Simple similar triangles problem. ... Um, and so on... it seems I'm too tired to finish this tonight. HEH! I did the same thing and did finish, coming up with the amazing answer: X = ... .. .. .. .. .. .. .. .. .. X !!! I was too annoyed and bored to try again. Bob |
Doug Schultz wrote:
almost any height you want you didnt specify how far awat the bottom is. Well, no. The bottom can only be a fixed distance away. Closer and the top won't touch the building because the cube is holding it away. Further away and it won't touch the cube. Try it with the building and cube drawn on a piece of paper and use a ruler as the ladder. Bob |
On Sun, 29 May 2005 01:05:13 GMT, Errol Groff wrote:
Ladder_Problem.PDF is an interesting math problem I found in the book Machine Shop Trade Secrets by James A Harvey. The solution I used to find the answer will be posted as Ladder_Problem_Answer.PDF http://metalworking.com/DropBox/ Errol Groff Instructor, Machine Tool Department H.H. Ellis Technical High School 643 Upper Maple Street Dantieson, CT 06239 New England Model Engineering Society www.neme-s.org This isn't a trig problem. Rather it's an algebra problem that, surprisingly, requires the solution to a quartic equation. However, by some clever selection of variables, the quartic can be separated into two quadratics. b = side of box l = length of ladder y = height of smallest triangle x = base of medium sized triangle Then the base of the large triangle is x+b and its height is y+b. From similar triangles, we have: y/b = b/x = x*y = b^2 = y = b^2/x (1) Applying Pythagoras to the large triangle: (x+b)^2 + (y+b)^2 = l^2 or: x^2 + 2*b*x + b^2 + y^2 + 2*b*y + b^2 = l^2 Substituting y = b^2/x from (1) yields: x^2 + 2*b*x + b^2 + b^2/x^2 + 2*b*b^2/x + b^2 = l^2 (2) Now define: a = x + b^2/x (3) so: a^2 = x^2 + 2*b^2 + b^4/x^2 Then (2) becomes: a^2 + 2*b*a - l^2 = 0 Solve this quadratic for a. Now substitute a into (3) and solve the resulting quadratic to find x. a = x + b^2/x (3) x^2 - a*x + b^2 = 0 Grinding through the numbers (I used a program I wrote): Length of ladder [25] ? 120 Side of box [6] ? 24 Solutions to: +1.0000 * x^2 +48.0000 * x^1 -14400.0000 = 0 a real: 98.376468 imaginary: 0.000000 real: -146.376468 imaginary: 0.000000 a selected = 98.3765 Solutions to: +1.0000 * x^2 -98.3765 * x^1 +576.0000 = 0 a real: 92.124028 imaginary: 0.000000 real: 6.252440 imaginary: 0.000000 Base and height of large triangle a 116.1240, 30.2524 or 30.2524, 116.1240 So the value labeled X in the PDF is 116.124 Regards, Marv Home Shop Freeware - Tools for People Who Build Things http://www.geocities.com/mklotz.geo |
"Bob Engelhardt" wrote in message
... Doug Schultz wrote: almost any height you want you didnt specify how far awat the bottom is. Well, no. The bottom can only be a fixed distance away. Closer and the top won't touch the building because the cube is holding it away. Further away and it won't touch the cube. Try it with the building and cube drawn on a piece of paper and use a ruler as the ladder. Bob Whoops didnt see the length of the ladder. thanks for pointing that out. Doug |
That jives with my cad drawing answer. Imaginary numbers give me a
headache :-) Marvin W. Klotz wrote: On Sun, 29 May 2005 01:05:13 GMT, Errol Groff wrote: Ladder_Problem.PDF is an interesting math problem I found in the book Machine Shop Trade Secrets by James A Harvey. The solution I used to find the answer will be posted as Ladder_Problem_Answer.PDF http://metalworking.com/DropBox/ Errol Groff Instructor, Machine Tool Department H.H. Ellis Technical High School 643 Upper Maple Street Dantieson, CT 06239 New England Model Engineering Society www.neme-s.org This isn't a trig problem. Rather it's an algebra problem that, surprisingly, requires the solution to a quartic equation. However, by some clever selection of variables, the quartic can be separated into two quadratics. b = side of box l = length of ladder y = height of smallest triangle x = base of medium sized triangle Then the base of the large triangle is x+b and its height is y+b. From similar triangles, we have: y/b = b/x = x*y = b^2 = y = b^2/x (1) Applying Pythagoras to the large triangle: (x+b)^2 + (y+b)^2 = l^2 or: x^2 + 2*b*x + b^2 + y^2 + 2*b*y + b^2 = l^2 Substituting y = b^2/x from (1) yields: x^2 + 2*b*x + b^2 + b^2/x^2 + 2*b*b^2/x + b^2 = l^2 (2) Now define: a = x + b^2/x (3) so: a^2 = x^2 + 2*b^2 + b^4/x^2 Then (2) becomes: a^2 + 2*b*a - l^2 = 0 Solve this quadratic for a. Now substitute a into (3) and solve the resulting quadratic to find x. a = x + b^2/x (3) x^2 - a*x + b^2 = 0 Grinding through the numbers (I used a program I wrote): Length of ladder [25] ? 120 Side of box [6] ? 24 Solutions to: +1.0000 * x^2 +48.0000 * x^1 -14400.0000 = 0 a real: 98.376468 imaginary: 0.000000 real: -146.376468 imaginary: 0.000000 a selected = 98.3765 Solutions to: +1.0000 * x^2 -98.3765 * x^1 +576.0000 = 0 a real: 92.124028 imaginary: 0.000000 real: 6.252440 imaginary: 0.000000 Base and height of large triangle a 116.1240, 30.2524 or 30.2524, 116.1240 So the value labeled X in the PDF is 116.124 Regards, Marv Home Shop Freeware - Tools for People Who Build Things http://www.geocities.com/mklotz.geo |
On Sun, 29 May 2005 01:05:13 GMT, Errol Groff
wrote: Ladder_Problem.PDF is an interesting math problem I found in the book Machine Shop Trade Secrets by James A Harvey. The solution I used to find the answer will be posted as Ladder_Problem_Answer.PDF http://metalworking.com/DropBox/ One approach is use of the Pythagoras theorem and ratios of sides of similar triangles. No trig is necessary, just simultaneous equations. X = 116.124 |
On Sun, 29 May 2005 17:22:04 GMT, machineman
wrote: That jives with my cad drawing answer. Imaginary numbers give me a headache :-) Marvin W. Klotz wrote: On Sun, 29 May 2005 01:05:13 GMT, Errol Groff wrote: Ladder_Problem.PDF is an interesting math problem I found in the book Machine Shop Trade Secrets by James A Harvey. The solution I used to find the answer will be posted as Ladder_Problem_Answer.PDF http://metalworking.com/DropBox/ Errol Groff Instructor, Machine Tool Department H.H. Ellis Technical High School 643 Upper Maple Street Dantieson, CT 06239 New England Model Engineering Society www.neme-s.org This isn't a trig problem. Rather it's an algebra problem that, surprisingly, requires the solution to a quartic equation. However, by some clever selection of variables, the quartic can be separated into two quadratics. b = side of box l = length of ladder y = height of smallest triangle x = base of medium sized triangle Then the base of the large triangle is x+b and its height is y+b. From similar triangles, we have: y/b = b/x = x*y = b^2 = y = b^2/x (1) Applying Pythagoras to the large triangle: (x+b)^2 + (y+b)^2 = l^2 or: x^2 + 2*b*x + b^2 + y^2 + 2*b*y + b^2 = l^2 Substituting y = b^2/x from (1) yields: x^2 + 2*b*x + b^2 + b^2/x^2 + 2*b*b^2/x + b^2 = l^2 (2) Now define: a = x + b^2/x (3) so: a^2 = x^2 + 2*b^2 + b^4/x^2 Then (2) becomes: a^2 + 2*b*a - l^2 = 0 Solve this quadratic for a. Now substitute a into (3) and solve the resulting quadratic to find x. a = x + b^2/x (3) x^2 - a*x + b^2 = 0 Grinding through the numbers (I used a program I wrote): Length of ladder [25] ? 120 Side of box [6] ? 24 Solutions to: +1.0000 * x^2 +48.0000 * x^1 -14400.0000 = 0 a real: 98.376468 imaginary: 0.000000 real: -146.376468 imaginary: 0.000000 a selected = 98.3765 Solutions to: +1.0000 * x^2 -98.3765 * x^1 +576.0000 = 0 a real: 92.124028 imaginary: 0.000000 real: 6.252440 imaginary: 0.000000 Base and height of large triangle a 116.1240, 30.2524 or 30.2524, 116.1240 So the value labeled X in the PDF is 116.124 Regards, Marv Home Shop Freeware - Tools for People Who Build Things http://www.geocities.com/mklotz.geo True, CAD will get you the answer but it's just so... inelegant. There's a big difference between getting the answer and solving the problem. Kind of like the difference between buying and making a needed part. Regards, Marv |
Marvin W. Klotz wrote:
On Sun, 29 May 2005 01:05:13 GMT, Errol Groff wrote: Ladder_Problem.PDF is an interesting math problem I found in the book Machine Shop Trade Secrets by James A Harvey. The solution I used to find the answer will be posted as Ladder_Problem_Answer.PDF http://metalworking.com/DropBox/ Errol Groff Instructor, Machine Tool Department H.H. Ellis Technical High School 643 Upper Maple Street Dantieson, CT 06239 New England Model Engineering Society www.neme-s.org This isn't a trig problem. Rather it's an algebra problem that, surprisingly, requires the solution to a quartic equation. However, by some clever selection of variables, the quartic can be separated into two quadratics. b = side of box l = length of ladder y = height of smallest triangle x = base of medium sized triangle Then the base of the large triangle is x+b and its height is y+b. From similar triangles, we have: y/b = b/x = x*y = b^2 = y = b^2/x (1) Applying Pythagoras to the large triangle: (x+b)^2 + (y+b)^2 = l^2 or: x^2 + 2*b*x + b^2 + y^2 + 2*b*y + b^2 = l^2 Substituting y = b^2/x from (1) yields: x^2 + 2*b*x + b^2 + b^2/x^2 + 2*b*b^2/x + b^2 = l^2 (2) Now define: a = x + b^2/x (3) so: a^2 = x^2 + 2*b^2 + b^4/x^2 Then (2) becomes: a^2 + 2*b*a - l^2 = 0 Solve this quadratic for a. Now substitute a into (3) and solve the resulting quadratic to find x. a = x + b^2/x (3) x^2 - a*x + b^2 = 0 Grinding through the numbers (I used a program I wrote): Length of ladder [25] ? 120 Side of box [6] ? 24 Solutions to: +1.0000 * x^2 +48.0000 * x^1 -14400.0000 = 0 a real: 98.376468 imaginary: 0.000000 real: -146.376468 imaginary: 0.000000 a selected = 98.3765 Solutions to: +1.0000 * x^2 -98.3765 * x^1 +576.0000 = 0 a real: 92.124028 imaginary: 0.000000 real: 6.252440 imaginary: 0.000000 Base and height of large triangle a 116.1240, 30.2524 or 30.2524, 116.1240 So the value labeled X in the PDF is 116.124 Regards, Marv Home Shop Freeware - Tools for People Who Build Things http://www.geocities.com/mklotz.geo I just wrote equations and combined until I got: x^2 + 576/(1-24/x)^2 = 14400 and I solved that numerically to get x = 116.124" but I didn't see any elegant way to get this algebraically. GWE |
Errol Groff wrote:
Ladder_Problem.PDF is an interesting math problem I found in the book Machine Shop Trade Secrets by James A Harvey. The solution I used to find the answer will be posted as Ladder_Problem_Answer.PDF http://metalworking.com/DropBox/ Errol Groff Instructor, Machine Tool Department H.H. Ellis Technical High School 643 Upper Maple Street Dantieson, CT 06239 New England Model Engineering Society www.neme-s.org I did a back of the napkin approximation: Since 24" is only 1/5 of 120", I figured that I could ignore the difference between the actual radius and the vertical height to get the horizontal length of the large triangle. Since the horizontal distance of the triangle that is 4/5 the size of the real one, the real one would be: 5/4 x 24" = 30" Solving the python theorem: 120^2 = H^2 + 30^2 H = 116.19 which is not a bad approximation. Bob |
Solve for n.
| |\ | \ | \5+n y| \ | \ |__2__\ | |\ 2| 2| \5-n |_____|__\___ 2 x Using the Pythagorean theorem: x^2 = (5-n)^2 - 4 = n^2 - 10n + 21 y^2 = (5+n)^2 - 4 = n^2 + 10n + 21 Using rule of similar triangles: y/2 = 2/x yx = 4 y^2 * x^2 = 16 (n^2 - 10n + 21) * (n^2 + 10n + 21) = 16 n^4 - 58n^2 + 425 = 0 Solving quadratic equation: ax^2 + bx + c = 0 x = (-b + Sqrt(b^2 - 4ac))/2a = 49.396 x = (-b - Sqrt(b^2 - 4ac))/2a = 8.604 x = n^2 n = srqt(49.396) or n = sqrt(8.604) n = 7.0282287953651594252678749478686 (Damn, this can't be right!) or n = 2.9332575747792760110661745961244 (Hmmm, maybe) y^2 = (5+n)^2 - 2^2 y^2 = 7.9332575747792760110661745961244^2 -4 y^2 = 58.936575747792760110661745961237 y = 7.6770160705701769496450099398666 y+2 = 9.6770160705701769496450099398666 ft. y+2 = 116.1241928468421233957401192784 in. What a headache. -- Steve Walker (remove wallet to reply) |
I used Mathcad:
Given (which is a keyword in Mathcad) sqrt(y^2 + 24^2) + sqrt(x^2 + 24^2) = 120 sqrt( (y+24)^2 + (x+24)^2 ) = 120 Find (y,x)= 92.125, 6.251 which means the ladder touches the wall at 116.125 inches height (and comes out 30.251 inches from the wall). and also the interchanged solution which you can tell by the drawing is wrong, and two negative solutions which are just theoretical, since Mathcad wouldn't accept x0, y0 as constraints for some reason. with y being the distance from the top edge of the cube to where the ladder touches, and x being the distance from the right edge of the right edge to the ladder. Doug |
Interesting solution.
However, using your 2 values of n and substituting into the hypotenuse of the upper small triangle one gets: 5+n = either 12 or 8 (approximately) which is longer than the total ladder. Using the negative roots for n gives the lower triangle a longer hypotenuse that the total ladder. How is this possible? Art "Steve Walker" wrote in message news:I4tme.6402$m%3.2273@trnddc02... Solve for n. | |\ | \ | \5+n y| \ | \ |__2__\ | |\ 2| 2| \5-n |_____|__\___ 2 x snip n = 7.0282287953651594252678749478686 (Damn, this can't be right!) or n = 2.9332575747792760110661745961244 (Hmmm, maybe) |
Wood Butcher wrote:
Interesting solution. However, using your 2 values of n and substituting into the hypotenuse of the upper small triangle one gets: 5+n = either 12 or 8 (approximately) which is longer than the total ladder. Using the negative roots for n gives the lower triangle a longer hypotenuse that the total ladder. How is this possible? Art "Steve Walker" wrote in message news:I4tme.6402$m%3.2273@trnddc02... Solve for n. | |\ | \ | \5+n y| \ | \ |__2__\ | |\ 2| 2| \5-n |_____|__\___ 2 x snip n = 7.0282287953651594252678749478686 (Damn, this can't be right!) or n = 2.9332575747792760110661745961244 (Hmmm, maybe) I dunno. G I remember the formulas. Theory sometimes escapes me if I don't use it often enough. -- Steve Walker (remove wallet to reply) |
Many thanks to all who participated in this discussion. I learned some new math, and a new way to use Excel and, all in all, very much enjoyed the conversation. We even managed to stay pretty much on topic!! I am back to school tomorrow and will create packets of the interesting methods for solving this problem and distribute them to the math teachers. I have no doubt that any of them have the math skills to solve this thing but with the number of students they have to deal with, and since they must get so focused on what they are teaching at the given moment, I have not had much feedback from them. I would not want to be a classroom teacher. Whatever the pay it is not enough. OTOH when I express admiration for those who have to face a new group of dis-interested students every 60 minutes they usually reply that they know that in 60 minutes those students will have moved on where as I have my students ALL day for six to ten days in a row. I suppose it is all in what you get used to. Thanks again, Errol Groff Instructor, Machine Tool Department H.H. Ellis Technical High School 643 Upper Maple Street Danielson, CT 06239 New England Model Engineering Society www.neme-s.org |
I think the easy way to approach this problem would be to make a scale
drawing, and then measure the distance X. This would give you a close approximation. Now you could use the brute force trial and error method to get to the result. Looking at the algebra that Marv provided, kinda gives me a headache. I am keeping this to have one of my learned friends explain the algebra to me in terms I can understand. -- Roger Shoaf About the time I had mastered getting the toothpaste back in the tube, then they come up with this striped stuff. "Errol Groff" wrote in message ... Ladder_Problem.PDF is an interesting math problem I found in the book Machine Shop Trade Secrets by James A Harvey. The solution I used to find the answer will be posted as Ladder_Problem_Answer.PDF http://metalworking.com/DropBox/ Errol Groff Instructor, Machine Tool Department H.H. Ellis Technical High School 643 Upper Maple Street Dantieson, CT 06239 New England Model Engineering Society www.neme-s.org |
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