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Topic: Length of hypotenuse (Read 1286 times) |
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NickH
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Length of hypotenuse
« on: Oct 17th, 2002, 4:12pm » |
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ABC is a right triangle with angle ABC = 90°. D is a point on AB such that angle BCD = angle DCA. E is a point on BC such that angle BAE = angle EAC. If AE = 3 inches and CD = 4 inches, find AC.
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Eric Yeh
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Re: Length of hypotenuse
« Reply #1 on: Oct 18th, 2002, 4:38pm » |
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Yeesh. Geometry and trig were never my strong points, and now it's been far too long. I tried a couple attacks: in the first I use the radius of the incircle together with one of the half-angles, and get two equations in two unknowns that is solveable but messy. Better, I just use the law of sines to get one equation in the [full] angle -- the cotangent half-angle formula beautifully drops out the half angles involved so it is in sines and cosines of alpha. Unfortunately, to get it into just sin alpha requires squaring the equation so that it becomes a quartic. Well, solving that would not be tough w a computer (and I may even be able to do it by hand, but I didn't want to think too hard this time of the week), but I guess there should be a more elegant solution. Is that true?
Best,
Eric
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icon
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Re: Length of hypotenuse
« Reply #2 on: Oct 21st, 2002, 6:29am » |
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on Oct 17th, 2002, 4:12pm, NickH wrote:| ABC is a right triangle with angle ABC = 90°. D is a point on AB such that angle BCD = angle DCA. E is a point on BC such that angle BAE = angle EAC. If AE = 3 inches and CD = 4 inches, find AC. |
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highlight to read:>
nick lets say you draw triangle letters a on left, c on top b on right 90 angle with c being 30 and a 60. as we draw line e from a and d from c and intersecting point of lines d and e we can name x, can we assume that cx=3 and dx=1?
i think am wrong but after 12 hr night shift it looks plausible
since axc=135 degress and axd 45 degrees
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| « Last Edit: Oct 21st, 2002, 6:30am by icon » |
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Eric Yeh
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Re: Length of hypotenuse
« Reply #3 on: Oct 21st, 2002, 6:52am » |
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Icon,
I could easily have made a mistake but I don't believe what you say is true. Drop a perp from your X to AB and call it F; similarly to AC and call it G. Now DXF is 15 degrees, so with your assumption that DX = 1, XF = cos 15. Then we also have XG = cos 15 since XG = XF = r, the radius of the incircle. But then XCG = 15 degrees also implies that XG/XC = sin XCG = sin 15, but then XC = XG/sin 15 = cos 15/sin 15 != 3 as you claimed.
Best,
Eric
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icon
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Re: Length of hypotenuse
« Reply #4 on: Oct 21st, 2002, 2:20pm » |
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hi
am not really sure as well cause my backround isnt math at all i just like riddles hehe, what also curious that the cd and ae, when they are dropped so they break ab and cb into 2 equal sides with ad=bd and ce=be? 
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Eric Yeh
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Re: Length of hypotenuse
« Reply #5 on: Oct 21st, 2002, 2:28pm » |
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Icon,
Not sure exactly what you're saying, but as the problem is defined AD != BD and CE != BE. Or are you suggesting a new problem??
Eric
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SWF
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Re: Length of hypotenuse
« Reply #6 on: Oct 22nd, 2002, 5:50pm » |
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Since this is a right triangle BCD+BAE=45o and cos(2*BCD)=sin(2*BAE).
There are two ways to project a hypotenuse to get length BC:
BC = AC*cos(2*BCD) = 4*cos(BCD)
Similarly for AB:
AB = AC*cos(2*BAE) = 3*cos(BAE)
Combine those two to eliminate AC:
3*cos(BAE)/cos(2*BAE) = 4*cos(BCD)/cos(2*BCD) = 4*cos(45o-BAE)/sin(2*BAE)
where BCD was eliminated by using the relations between BCD and BAE given on first line above. Rearranging the two expressions containing only BAE and letting t=tan(BAE) gives
t3+t2+(3*sqrt(2)/2-1)*t -1 =0
Solving for t gives
t=-1/3+cuberoot(h+g)-cuberoot(h-g)
where h=sqrt((50*sqrt(2)-39)/72) and g=(32+27*sqrt(2))/108.
The expression for AB above can be rearranged for AC in terms of t:
AC=3*sqrt(1+t2)/(1-t2)
Numerical values are t=0.521983281 and AC=4.6514833758.
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NickH
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Re: Length of hypotenuse
« Reply #7 on: Oct 23rd, 2002, 1:11pm » |
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SWF,
That's a tour de force! Had I known the solution when I posted the problem, I'd have set AE and CD so the cubic could be solved by inspection!
Nick
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Eric Yeh
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Re: Length of hypotenuse
« Reply #8 on: Oct 23rd, 2002, 3:24pm » |
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Yes, very impressive SWF -- kudos! I guess the key that I missed was the substitution of tan x, dropping the otherwise sixth degree equation to a cubic -- very clever. (And I was already happy with myself for dropping it to cuartic  Unfortunately, using the tan x innovation doesn't work for bringing the problem to a quadratic, though it's only off by one term!).
Nick: If I may make a suggestion, I strongly believe you should explicitly note in your first post if you do not know the solution to a problem. The reason is that a good thinker for any kind of question (not just riddles) automatically uses the full context as an input to the problem-solving process; why not, it is an available information source! My hope for this forum -- hopefully shared by others, which is true from what I've seen thus far -- is that there are only two kinds of questions presented: 1) (the main class) interesting problems with elegant solutions, and 2) (the great minority but happily accepted) unsolved problems that the proposer either believes would be interesting or just can't sleep without knowing the answer to. These are really two different kinds of problems that require two different kinds of thinking.
Example: Here I dropped my development of any solution as soon as it got over quadratic, since such a result seemed to me inelegant -- something I assumed it would not be based on its posting. Had it been for a contest for $$, I surely would have simply solved the equation that day of my first post. And if it had just been posted with a comment that it was unsolved (therefore not necessarily elegant)? Well then, somewhere in between; I may or may not have finished the problem off.
So, is it still solveable without further contextual information? Of course. So why specify? It is not because the problem is "too hard otherwise" or any such -- I for one do not shirk from difficulty, only from uninterestingness. (A tough challenge is always welcome. And as I mentioned, in this case it would have actually encouraged to work on it more!) To me, it would be because I would enjoy the total set of problems immensely more if I knew for each whether it was in the first or second camp. The information would suggest very roughly what kind of strategy is required for each problem, and thereby allows you to choose which kind of problem you wish to tackle at the time. I think this is invaluable.
Anyhow, sorry this became such a long discourse. It was not my intention to be pedantic, only to bring up an interesting topic, and to promote what I believe is in the best interest of the site. But I would love to hear alternate opinions; what do you think? Please forgive me if this was in any way offensive, as it was not meant to be.
Best,
Eric
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NickH
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Re: Length of hypotenuse
« Reply #9 on: Oct 23rd, 2002, 4:10pm » |
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Eric,
Point taken. I understand what you're saying.
I might add this is the first problem I have posted here that I haven't had a solution for. Elegance of solution should be a primary consideration for any puzzle, I agree. My only defence can be that this puzzle is simple and elegant _in statement_.
I guessed the solution would also be elegant; so it is, in fact, as long as the resulting cubic is easily solvable! (As it stands, it's horrendous.) Substitute 9 for 3, and 8*sqrt(2) for 4, and (IMHO) you still have a neat puzzle.
Nick
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Eric Yeh
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Re: Length of hypotenuse
« Reply #10 on: Oct 26th, 2002, 5:28pm » |
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Glad you agree, Nick -- didn't want to offend you. Keep up the good work!
BTW, I meant to say a long time ago that your icon is really cool. But I didn't have anything more relevant to post at the time, so I decided to skip it.
Best,
Eric
P.S. Not to be further argumentative, but purely for my two cents: I think there are quite a few elegantly phrased questions out there that don't make good puzzles -- their solutions are very often inelegant. And as for this question, I do think it's reasonably nice, but (IMHO) anything that requires solving even a factorable cubic is not what I'd consider particularly elegant. It's very straightforward to get a sixth-power equation here, which is only a little harder to solve than a cubic in general (both you'd basically want a computer for). There are many other "harder" geometry questions out there that require more intuition, and less number crunching.
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