reverse Posted May 10, 2005 Posted May 10, 2005 Oh , you may be right, I just get some idea that this is the way to do it. when I get a spare second I will get a pencil and paper and nut it out. I was thinking maybe the fact that both sets of angles must equal 180 degrees…maybe some sort of simultaneous equation. Forget it until I try a few drawings. I'm sort of thinking something like x+w+20=180, and X+Y=140... and substitutions and that sort of thing?.
Callipygous Posted May 10, 2005 Posted May 10, 2005 yeah, but the problem is all these equations are based on the properties of triangles and therefore all include the number 180 in one way or another. they all represent the same relationship which means when you plug them in with the other equations your variables cancel out and leave you with 140=140 or something similar.
dan19_83 Posted May 10, 2005 Posted May 10, 2005 your right callipygous. took me half an hour of going around in circles with simultaneous equations to figure that one out!!
reverse Posted May 11, 2005 Posted May 11, 2005 yet my instinct tells me that: the person who devised this problem wanted us to use (the fact that three angles of a triangle must add up to 180) and also make us do (a paradigm shift in the way the problem is being viewed). Hmmmm. is there any way to slide one of the triangles over to get a 90 degree total somewhere??
reverse Posted May 11, 2005 Posted May 11, 2005 Ok….. I have had a quick play with this on paper. Nothing normal works. I’m gonna start extending out the lines past the main ABC triangle till I can see another triangle that I can work with. (this is good brain exercise.)
dan19_83 Posted May 12, 2005 Posted May 12, 2005 i think we need thud&blunder, the creator of the thread, to give us a clue cause i for one am stumped!
Callipygous Posted May 12, 2005 Posted May 12, 2005 im not willing to put in the effort, but one method i find helpful for BSing tests is to plug in some values for things. if you cant find it by the general rules, maybe a specific one will work : P
reverse Posted May 13, 2005 Posted May 13, 2005 At the moment I'm waiting to get a spare half hour to play around with the idea of an isosceles triangle projected out the side of our starting triangle. I need to do it with a ruler and protractor to see if there are any obvious relationships. I think this is going to be the approach the designer of the question had in mind. or not...I'm just guessing at the moment.
reverse Posted May 14, 2005 Posted May 14, 2005 Oh sure. But “we” don’t want to kill a fly with a hammer. We want the most efficient and intelligent solution.
ski_power Posted May 14, 2005 Posted May 14, 2005 oh, then mathematically it'd take us a long time, BUT we'd arrive at an answer. I'll try it out and then post it here.
Flareon Posted May 14, 2005 Posted May 14, 2005 I pretty much got to where everyone else seems to be...Good to see that I'm not the only one stuck on this. Strangely, our dear riddler hasn't been back for over a week. I was hoping he could give some light clues or at least some feedback.
dan19_83 Posted May 15, 2005 Posted May 15, 2005 maybe he made an error with the puzzle and it is impossible to solve and now he's too ashamed to come back!!
Aeternus Posted May 15, 2005 Posted May 15, 2005 That's what i get the triangle out to be from reading - The isosceles triangle ABC has AB = AC and angle BAC = 20 degrees. The point D lies on AC such that angle DBC = 60 degrees. The point E lies on AB such that angle ECB = 50 degrees. Find angle EDB. The diagrams drawn by everyone else seem to have DBC and ECB the other way round which seems strange but it doesnt really matter as its just B and C reversed so whatever i come out with can simply be swapped round. Thats the same basically (ignoring the difference in B and C) as everyone else has. So i figured perhaps sin and cosine rules could help by working out the lengths of the lines in blue and purple and taking away the lines in red and green respectively to get the lengths of X and Y. Then using the cosine rule to get the yellow line and then using that and the angle along with the sine rules to get the angles. I could have made a mistake but i get XA to be 80 and YA to be 30 using a little python script i knocked together (rather than typing it all into the calculator or trying to figure out how different things canceled to give more sensible numbers etc). Python script is Here. I may have made a mistake or something, if so sorry, please correct me.
jdurg Posted May 16, 2005 Posted May 16, 2005 Well there is definitely an error in the initial poster's diagram as it appears as if he has the 30 degrees and 20 degrees switched on his diagram.
reverse Posted May 17, 2005 Posted May 17, 2005 I’m trying to solve it using three rules alone. 1 three angles of a triangle add to 180. 2 Isosceles triangles are symmetrical. 3 If you skew a square, two pairs of sides remain parallel. I’m stubborn like that.
Callipygous Posted May 17, 2005 Posted May 17, 2005 thats all well and good, but there doesnt seem to be a square in the situation. the closest thing is a quadrilateral which doesnt even appear to be a rhombus.
reverse Posted May 17, 2005 Posted May 17, 2005 Isn't this a totally cool puzzle! ps. Those diagrams are really wonky.
Aeternus Posted May 17, 2005 Posted May 17, 2005 Can anyone confirm whether i came out with the right answer? or did i make a mistake somewhere? Also , as reverse suggests are things like the sine and cosine rules off bounds?
reverse Posted May 17, 2005 Posted May 17, 2005 Hey man, if it works, it works. Can’t argue with results! I figure EDB = 30? I’m just trying to show that triangle ADE is the same as the 50 20 110 triangle via parallelism.
reverse Posted May 17, 2005 Posted May 17, 2005 OK try this. From the last colour diagram, Draw a line parallel to EC…but shift it up so it intersects point D. Now think like a mirror.
dan19_83 Posted May 17, 2005 Posted May 17, 2005 mmmmm.... think like a mirror. aaagggghhhh i can see myself!!!!!! (really bad humour, i know) seriously though, i don't see this what angle you are refering to after you draw the parallel line. i.e what new angle is at that side of the triangle that will help us solve the puzzle?
uncool Posted May 17, 2005 Posted May 17, 2005 Hey guys...I posted the answer. anyone wanna check it? -Uncool-
dan19_83 Posted May 17, 2005 Posted May 17, 2005 well that's just great!! all these wasted posts and the answer is in post nine. i only saw it now!! just gave a quick glance through it. are you sure it's right?
reverse Posted May 18, 2005 Posted May 18, 2005 seriously though' date=' i don't see this what angle you are refering to ?[/quote'] oh yea. because the drawings so far are not that good BAC is drawn badly making it NOT look like 20 degrees. it looks more like 30 degrees in the rough drawings. once you actually draw it as 20 degrees you will see three nested symmetrical triangles. that will enable you to use only two common sense laws to solve the puzzle. (I think)
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