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Posted: 11/25/2014 11:38:47 AM EDT
[#1]
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You do realize it has nothing to do with the math problem. Quote HistoryQuoted:
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Quoted:Right over the head it went.....
Apparently. AF = BD but depends on how you label it. I have so many labels lying around, at this point AYQ may be a triangle somewhere in there  You do realize there is no math involved other than geometry. |
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Posted: 11/25/2014 11:53:19 AM EDT
[#2]
Quote HistoryQuoted: You do realize there is no math involved other than geometry. Quote HistoryQuoted: Quoted: Quoted: Quoted:Right over the head it went..... Apparently. AF = BD but depends on how you label it. I have so many labels lying around, at this point AYQ may be a triangle somewhere in there You do realize there is no math involved other than geometry. |
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Posted: 11/25/2014 12:44:03 PM EDT
[#3]
Did anyone look at my link?
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Posted: 11/25/2014 12:53:56 PM EDT
[#4]
Is this a trick question? I got X into terms of the 3 other unknown angles and came up with 0=0, which is true, but doesn't that make the problem indeterminate?
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Posted: 11/25/2014 1:11:32 PM EDT
[#5]
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https://www.duckware.com/tech/worldshardesteasygeometryproblem.html My head hurts. I lost this proof at the jump to FG = FD. In his second solution he made a mistake calling angle ADB angle AEB for the known angles. ADB is 40 and AEB is still 30. |
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Posted: 11/25/2014 1:27:39 PM EDT
[#6]
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Is this a trick question? I got X into terms of the 3 other unknown angles and came up with 0=0, which is true, but doesn't that make the problem indeterminate? http://i61.tinypic.com/2ln7i4y.jpg The problem is that the four equations are not linearly independent. See my solution on page 2. Right now the only solution that I can conceive requires analytic geometry, i.e., write equations for the lines, then calculate the angle. |
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Posted: 11/25/2014 1:53:08 PM EDT
[#7]
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The problem is that the four equations are not linearly independent. See my solution on page 2. Right now the only solution that I can conceive requires analytic geometry, i.e., write equations for the lines, then calculate the angle. Quote HistoryQuoted:
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Is this a trick question? I got X into terms of the 3 other unknown angles and came up with 0=0, which is true, but doesn't that make the problem indeterminate? http://i61.tinypic.com/2ln7i4y.jpg The problem is that the four equations are not linearly independent. See my solution on page 2. Right now the only solution that I can conceive requires analytic geometry, i.e., write equations for the lines, then calculate the angle. In the website, the guy throws lines on there and assumes they are those angles. With no linear dimensions and the drawing not being scaled, he can't really do that, can he? |
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Posted: 11/25/2014 1:56:56 PM EDT
[#8]
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In the website, the guy throws lines on there and assumes they are those angles. With no linear dimensions and the drawing not being scaled, he can't really do that, can he? Quote HistoryQuoted:
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Is this a trick question? I got X into terms of the 3 other unknown angles and came up with 0=0, which is true, but doesn't that make the problem indeterminate? http://i61.tinypic.com/2ln7i4y.jpg The problem is that the four equations are not linearly independent. See my solution on page 2. Right now the only solution that I can conceive requires analytic geometry, i.e., write equations for the lines, then calculate the angle. In the website, the guy throws lines on there and assumes they are those angles. With no linear dimensions and the drawing not being scaled, he can't really do that, can he? Which line in particular do you have a problem with? |
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Posted: 11/25/2014 1:57:33 PM EDT
[#9]
Quote HistoryQuoted: The problem is that the four equations are not linearly independent. See my solution on page 2. Right now the only solution that I can conceive requires analytic geometry, i.e., write equations for the lines, then calculate the angle. Quote HistoryQuoted: Quoted: Is this a trick question? I got X into terms of the 3 other unknown angles and came up with 0=0, which is true, but doesn't that make the problem indeterminate? http://i61.tinypic.com/2ln7i4y.jpg The problem is that the four equations are not linearly independent. See my solution on page 2. Right now the only solution that I can conceive requires analytic geometry, i.e., write equations for the lines, then calculate the angle. i feel less stupid now.
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Posted: 11/25/2014 1:59:15 PM EDT
[#10]
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Which line in particular do you have a problem with? Quote HistoryQuoted:
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Is this a trick question? I got X into terms of the 3 other unknown angles and came up with 0=0, which is true, but doesn't that make the problem indeterminate? http://i61.tinypic.com/2ln7i4y.jpg The problem is that the four equations are not linearly independent. See my solution on page 2. Right now the only solution that I can conceive requires analytic geometry, i.e., write equations for the lines, then calculate the angle. In the website, the guy throws lines on there and assumes they are those angles. With no linear dimensions and the drawing not being scaled, he can't really do that, can he? Which line in particular do you have a problem with? The blue lines on the pictures, more specifically the second solution. He is assuming that the triangle is symmetrical about the center vertical axis. Can he do that if he doesn't know the scale or the dimensions of the lines? |
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Posted: 11/25/2014 2:07:57 PM EDT
[#11]
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We can all get the other angles but those four are the most important and you're not showing how outside of guessing. Well, I started over and came up with 20. 
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Posted: 11/25/2014 2:08:50 PM EDT
[#12]
Quote HistoryQuoted: Is this a trick question? I got X into terms of the 3 other unknown angles and came up with 0=0, which is true, but doesn't that make the problem indeterminate? http://i61.tinypic.com/2ln7i4y.jpg This is what I did, and the same result. I was able to substitute variables given their other values in terms of a different variable, but ultimately cannot solve for X with the existing angle relationships. An additional triangle has to be drawn to create additional relationships.
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Posted: 11/25/2014 2:13:59 PM EDT
[#13]
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The blue lines on the pictures, more specifically the second solution. He is assuming that the triangle is symmetrical about the center vertical axis. Can he do that if he doesn't know the scale or the dimensions of the lines? Quote HistoryQuoted:
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Is this a trick question? I got X into terms of the 3 other unknown angles and came up with 0=0, which is true, but doesn't that make the problem indeterminate? http://i61.tinypic.com/2ln7i4y.jpg The problem is that the four equations are not linearly independent. See my solution on page 2. Right now the only solution that I can conceive requires analytic geometry, i.e., write equations for the lines, then calculate the angle. In the website, the guy throws lines on there and assumes they are those angles. With no linear dimensions and the drawing not being scaled, he can't really do that, can he? Which line in particular do you have a problem with? The blue lines on the pictures, more specifically the second solution. He is assuming that the triangle is symmetrical about the center vertical axis. Can he do that if he doesn't know the scale or the dimensions of the lines? Step 4, then? Click To View Spoiler |
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Posted: 11/25/2014 2:27:23 PM EDT
[#14]
I am sorry, I have to go back, I thought the triangle was the same, solved two different ways on the website.
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Posted: 11/25/2014 3:21:43 PM EDT
[#15]
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The problem is that the four equations are not linearly independent. See my solution on page 2. Right now the only solution that I can conceive requires analytic geometry, i.e., write equations for the lines, then calculate the angle. Quote HistoryQuoted:
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Is this a trick question? I got X into terms of the 3 other unknown angles and came up with 0=0, which is true, but doesn't that make the problem indeterminate? http://i61.tinypic.com/2ln7i4y.jpg The problem is that the four equations are not linearly independent. See my solution on page 2. Right now the only solution that I can conceive requires analytic geometry, i.e., write equations for the lines, then calculate the angle. I used your drawing with my overlays to show how it can be calculated, no equations involved. End of page 2 Here I'll paste: 
Note triangles AGC and AEC are the same. I outline one in blue, the other mostly in red. I didn't break down the angles on the equilateral triangles, those seem obvious. I did mark as many congruent sides as possible. I'll list the congruent sides: AB = AG = GB = CE (marked in blue,) AC = BC, AE = CG, DG = GF = FD = EF |
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Posted: 11/25/2014 3:23:27 PM EDT
[#16]
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We can all get the other angles but those four are the most important and you're not showing how outside of guessing. Well, I started over and came up with 20. http://s30.postimg.org/t7htwvp41/image.jpg There ya go.  We were on the same page, all the way to the first incorrect answer. I see you went my route too.
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Posted: 11/25/2014 3:24:39 PM EDT
[#17]
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The blue lines on the pictures, more specifically the second solution. He is assuming that the triangle is symmetrical about the center vertical axis. Can he do that if he doesn't know the scale or the dimensions of the lines? Quote HistoryQuoted:
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Is this a trick question? I got X into terms of the 3 other unknown angles and came up with 0=0, which is true, but doesn't that make the problem indeterminate? http://i61.tinypic.com/2ln7i4y.jpg The problem is that the four equations are not linearly independent. See my solution on page 2. Right now the only solution that I can conceive requires analytic geometry, i.e., write equations for the lines, then calculate the angle. In the website, the guy throws lines on there and assumes they are those angles. With no linear dimensions and the drawing not being scaled, he can't really do that, can he? Which line in particular do you have a problem with? The blue lines on the pictures, more specifically the second solution. He is assuming that the triangle is symmetrical about the center vertical axis. Can he do that if he doesn't know the scale or the dimensions of the lines? Yes, definition of isosceles triangle (unless I'm misunderstanding your question) |
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Posted: 11/25/2014 3:26:03 PM EDT
[#18]
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I used your drawing with my overlays to show how it can be calculated, no equations involved. End of page 2 Quote HistoryQuoted:
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Is this a trick question? I got X into terms of the 3 other unknown angles and came up with 0=0, which is true, but doesn't that make the problem indeterminate? http://i61.tinypic.com/2ln7i4y.jpg The problem is that the four equations are not linearly independent. See my solution on page 2. Right now the only solution that I can conceive requires analytic geometry, i.e., write equations for the lines, then calculate the angle. I used your drawing with my overlays to show how it can be calculated, no equations involved. End of page 2 I started making a few sketches this morning in a hunt for similar triangles. Considered a bisector, but didn't add it to the sketch. |
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Posted: 11/25/2014 4:24:03 PM EDT
[#19]
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Doesn't that still leave us with 4 equations and x is indeterminent? Quote HistoryQuoted:
Doesn't that still leave us with 4 equations and x is indeterminent? No. We know that: EFG is 60 and FGE is 60. Therefore, FE and FG are equivalent lengths. DFG is 40 and DGF is 40. Therefore, DF and DG are equivalent lengths. If you draw that out, you see a kite. And because it's a kite, we know that we can draw a line from FDG to FEG and bisect the shape, producing two equivalent triangles. |
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Posted: 11/25/2014 5:35:49 PM EDT
[#21]
Quote HistoryQuoted: Don't ask about geometry in GD. These motherfuckers think a tesseract has infinite sides because a movie director said so. |
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Posted: 11/25/2014 7:53:20 PM EDT
[#25]
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I promise to use only my imaginary compass from now on. http://s11.postimg.org/pj25annzn/imaginary_compass.jpg Quote HistoryQuoted:
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Sorry, I made a typo in the notes. DEG is an isosceles. Anyhow, the right angle in DFE is made by dividing line BC in half which can be done using a compass. "Not to scale" means you can't rely on any physical representation.  I promise to use only my imaginary compass from now on. http://s11.postimg.org/pj25annzn/imaginary_compass.jpg That's better. |
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Posted: 11/26/2014 9:47:33 PM EDT
[#26]
Any more thoughts on this?
I've come to the conclusion that this is a problem with insufficient constraints. I can define every angle in the picture, either with an actual value or a derived value based on X. If you call "F" the point where AE and BD cross, then: DEF = X EDF = 130 - X CDE = X + 10 CEF = 150 - X All the other angles have numeric values that are easily calculated as others have shown. Based on the derivations above, X can be any value < 130 degrees (to keep all the values positive). There is nothing else to constrain it. I think what's lacking is the length of at least one of the line segments. Or maybe I'm missing something else here. |
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Posted: 11/27/2014 2:25:37 AM EDT
[#27]
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Did anyone look at my link? Yup. I love seeing all the bullshit answers after the solution was provided. |
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Posted: 11/27/2014 9:57:04 PM EDT
[#28]
Quote HistoryQuoted: Sorry, I made a typo in the notes. DEG is an isosceles. Anyhow, the right angle in DFE is made by dividing line BC in half which can be done using a compass like so: http://s2.postimg.org/uhf8ad7e1/bisectual.jpg Quote HistoryQuoted: Quoted: Quoted: Quoted: We can all get the other angles but those four are the most important and you're not showing how outside of guessing. Well, I started over and came up with 20. I don't understand a couple things you did. 1, how did you construct right angle DFE? And 2, how is Triangle DFE isosceles? Sorry, I made a typo in the notes. DEG is an isosceles. Anyhow, the right angle in DFE is made by dividing line BC in half which can be done using a compass like so: http://s2.postimg.org/uhf8ad7e1/bisectual.jpg Understood on the typo. Regarding DFE, I don't think you can draw a right angle to FE, and logically expect that it touches point D. It LOOKS like it touches point D, but there is no way to prove or construct it that way. |
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Posted: 11/28/2014 12:02:18 AM EDT
[#29]
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Understood on the typo. Regarding DFE, I don't think you can draw a right angle to FE, and logically expect that it touches point D. It LOOKS like it touches point D, but there is no way to prove or construct it that way. Quote HistoryQuoted:
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We can all get the other angles but those four are the most important and you're not showing how outside of guessing. Well, I started over and came up with 20. I don't understand a couple things you did. 1, how did you construct right angle DFE? And 2, how is Triangle DFE isosceles? Sorry, I made a typo in the notes. DEG is an isosceles. Anyhow, the right angle in DFE is made by dividing line BC in half which can be done using a compass like so: http://s2.postimg.org/uhf8ad7e1/bisectual.jpg Understood on the typo. Regarding DFE, I don't think you can draw a right angle to FE, and logically expect that it touches point D. It LOOKS like it touches point D, but there is no way to prove or construct it that way. Since BCD has 20 degree angles on two corners, it is an isosceles triangle and by bisecting BC, a right angle is created. This would not only split the side (BC) in half, but also the triangle BCD. The line DF also splits the angle at D in half and intersects precisely at D. That right angle (F) remains a right algle in DEF. |
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Posted: 11/28/2014 12:03:31 PM EDT
[#30]
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We can all get the other angles but those four are the most important and you're not showing how outside of guessing. Well, I started over and came up with 20. http://s29.postimg.org/63aijh287/image.jpg I don't understand how anyone believes a solution using a graphics engine thinks it satisfies the original question. I constructed a solution in about a minute, but that doesn't mean I answered the question with a general geometry solution. |
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Posted: 11/28/2014 12:13:26 PM EDT
[#31]
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Any more thoughts on this? I've come to the conclusion that this is a problem with insufficient constraints. I can define every angle in the picture, either with an actual value or a derived value based on X. If you call "F" the point where AE and BD cross, then: DEF = X EDF = 130 - X CDE = X + 10 CEF = 150 - X All the other angles have numeric values that are easily calculated as others have shown. Based on the derivations above, X can be any value < 130 degrees (to keep all the values positive). There is nothing else to constrain it. I think what's lacking is the length of at least one of the line segments. Or maybe I'm missing something else here. Refer to my last post with diagram. You have to add lines (that conform to standard geometric theorems/laws) When you do, you can surmise the rest of the angles, legitimately. The red lines in my diagram were the ones I added. I bisected the main triangle ABC, also created a parallel line to AB that I labeled DF I believe. Also, made a reflected but equal line AF that is the same as BD. Once you add those lines, you can surmise the rest from there. I even marked congruent sides based upon the angles opposite to show you how I got those angles. |
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Posted: 11/28/2014 12:15:10 PM EDT
[#32]
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I don't understand how anyone believes a solution using a graphics engine thinks it satisfies the original question. I constructed a solution in about a minute, but that doesn't mean I answered the question with a general geometry solution. Quote HistoryQuoted:
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We can all get the other angles but those four are the most important and you're not showing how outside of guessing. Well, I started over and came up with 20. http://s29.postimg.org/63aijh287/image.jpg I don't understand how anyone believes a solution using a graphics engine thinks it satisfies the original question. I constructed a solution in about a minute, but that doesn't mean I answered the question with a general geometry solution. You were on the right track. I used a pen and paper, no ruler so it looks all crooked. Doesn't matter as it isn't to scale. Just add the red lines I did above and you'll see you can create congruent lines and triangles and equilateral triangles which are all crucial to solving for X. 
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Posted: 11/28/2014 12:23:44 PM EDT
[#33]
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Refer to my last post with diagram. You have to add lines (that conform to standard geometric theorems/laws) When you do, you can surmise the rest of the angles, legitimately. The red lines in my diagram were the ones I added. I bisected the main triangle ABC, also created a parallel line to AB that I labeled DF I believe. Also, made a reflected but equal line AF that is the same as BD. Once you add those lines, you can surmise the rest from there. I even marked congruent sides based upon the angles opposite to show you how I got those angles. Quote HistoryQuoted:
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Any more thoughts on this? I've come to the conclusion that this is a problem with insufficient constraints. I can define every angle in the picture, either with an actual value or a derived value based on X. If you call "F" the point where AE and BD cross, then: DEF = X EDF = 130 - X CDE = X + 10 CEF = 150 - X All the other angles have numeric values that are easily calculated as others have shown. Based on the derivations above, X can be any value < 130 degrees (to keep all the values positive). There is nothing else to constrain it. I think what's lacking is the length of at least one of the line segments. Or maybe I'm missing something else here. Refer to my last post with diagram. You have to add lines (that conform to standard geometric theorems/laws) When you do, you can surmise the rest of the angles, legitimately. The red lines in my diagram were the ones I added. I bisected the main triangle ABC, also created a parallel line to AB that I labeled DF I believe. Also, made a reflected but equal line AF that is the same as BD. Once you add those lines, you can surmise the rest from there. I even marked congruent sides based upon the angles opposite to show you how I got those angles. Why do I have to add lines? Like I said above, every angle in the diagram is either easily found or can be derived as a function of X. Therefore there is no single solution. If you say X=20, I agree that is a valid solution. But I argue that X=21, 55, or 87 (the ARFCOM preferred value) are also all valid. I don't have means to draw pictures here. Please humor me.... pick a value for X between 1-129 degrees, and label the original picture (without any added lines) with the resulting values, showing me why it wouldn't work. . |
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Posted: 11/28/2014 12:43:57 PM EDT
[#34]
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Why do I have to add lines? Like I said above, every angle in the diagram is either easily found or can be derived as a function of X. Therefore there is no single solution. If you say X=20, I agree that is a valid solution. But I argue that X=21, 55, or 87 (the ARFCOM preferred value) are also all valid. I don't have means to draw pictures here. Please humor me.... pick a value for X between 1-129 degrees, and label the original picture (without any added lines) with the resulting values, showing me why it wouldn't work. . Quote HistoryQuoted:
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Any more thoughts on this? I've come to the conclusion that this is a problem with insufficient constraints. I can define every angle in the picture, either with an actual value or a derived value based on X. If you call "F" the point where AE and BD cross, then: DEF = X EDF = 130 - X CDE = X + 10 CEF = 150 - X All the other angles have numeric values that are easily calculated as others have shown. Based on the derivations above, X can be any value < 130 degrees (to keep all the values positive). There is nothing else to constrain it. I think what's lacking is the length of at least one of the line segments. Or maybe I'm missing something else here. Refer to my last post with diagram. You have to add lines (that conform to standard geometric theorems/laws) When you do, you can surmise the rest of the angles, legitimately. The red lines in my diagram were the ones I added. I bisected the main triangle ABC, also created a parallel line to AB that I labeled DF I believe. Also, made a reflected but equal line AF that is the same as BD. Once you add those lines, you can surmise the rest from there. I even marked congruent sides based upon the angles opposite to show you how I got those angles. Why do I have to add lines? Like I said above, every angle in the diagram is either easily found or can be derived as a function of X. Therefore there is no single solution. If you say X=20, I agree that is a valid solution. But I argue that X=21, 55, or 87 (the ARFCOM preferred value) are also all valid. I don't have means to draw pictures here. Please humor me.... pick a value for X between 1-129 degrees, and label the original picture (without any added lines) with the resulting values, showing me why it wouldn't work. . Angle x has a finite value determined by the 60 degree and 70 degree lines, and the line that connects their intersections with the sides of the outer triangle. There is one solution. |
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Posted: 11/28/2014 12:46:41 PM EDT
[#35]
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Angle x has a finite value determined by the 60 degree and 70 degree lines, and the line that connects their intersections with the sides of the outer triangle. There is one solution. Quote HistoryQuoted:
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Any more thoughts on this? I've come to the conclusion that this is a problem with insufficient constraints. I can define every angle in the picture, either with an actual value or a derived value based on X. If you call "F" the point where AE and BD cross, then: DEF = X EDF = 130 - X CDE = X + 10 CEF = 150 - X All the other angles have numeric values that are easily calculated as others have shown. Based on the derivations above, X can be any value < 130 degrees (to keep all the values positive). There is nothing else to constrain it. I think what's lacking is the length of at least one of the line segments. Or maybe I'm missing something else here. Refer to my last post with diagram. You have to add lines (that conform to standard geometric theorems/laws) When you do, you can surmise the rest of the angles, legitimately. The red lines in my diagram were the ones I added. I bisected the main triangle ABC, also created a parallel line to AB that I labeled DF I believe. Also, made a reflected but equal line AF that is the same as BD. Once you add those lines, you can surmise the rest from there. I even marked congruent sides based upon the angles opposite to show you how I got those angles. Why do I have to add lines? Like I said above, every angle in the diagram is either easily found or can be derived as a function of X. Therefore there is no single solution. If you say X=20, I agree that is a valid solution. But I argue that X=21, 55, or 87 (the ARFCOM preferred value) are also all valid. I don't have means to draw pictures here. Please humor me.... pick a value for X between 1-129 degrees, and label the original picture (without any added lines) with the resulting values, showing me why it wouldn't work. . Angle x has a finite value determined by the 60 degree and 70 degree lines, and the line that connects their intersections with the sides of the outer triangle. There is one solution. I agree X is finite, between 1-129 degrees as I showed in my simple equations above. And yes, X is related to the 60 and 70 degree lines. But I see nothing that limits it to one value. Try setting it to a random value and show me why it fails. ETA: There are four unknown values: DEF = X EDF = 130 - X CDE = X + 10 CEF = 150 - X I can substitute any X between 1-129 degrees and produce a valid solution. |
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Posted: 11/28/2014 12:51:07 PM EDT
[#36]
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I don't understand how anyone believes a solution using a graphics engine thinks it satisfies the original question. I constructed a solution in about a minute, but that doesn't mean I answered the question with a general geometry solution. Quote HistoryQuoted:
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We can all get the other angles but those four are the most important and you're not showing how outside of guessing. Well, I started over and came up with 20. http://s29.postimg.org/63aijh287/image.jpg I don't understand how anyone believes a solution using a graphics engine thinks it satisfies the original question. I constructed a solution in about a minute, but that doesn't mean I answered the question with a general geometry solution. The only tool of computer wizardry I used was Photoshop and that was after I believed I'd thought it through in my head to arrive at the answer of 30. When asked how I got that, I tried again and came up with another answer (20), then checked it against other angles so I made another drawing. There aren't any fancy math engines in this basement, but occasionally there's an overabundance of free time.
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Posted: 11/28/2014 3:45:12 PM EDT
[#37]
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I agree X is finite, between 1-129 degrees as I showed in my simple equations above. And yes, X is related to the 60 and 70 degree lines. But I see nothing that limits it to one value. Try setting it to a random value and show me why it fails. ETA: There are four unknown values: DEF = X EDF = 130 - X CDE = X + 10 CEF = 150 - X I can substitute any X between 1-129 degrees and produce a valid solution. Quote HistoryQuoted:
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Any more thoughts on this? I've come to the conclusion that this is a problem with insufficient constraints. I can define every angle in the picture, either with an actual value or a derived value based on X. If you call "F" the point where AE and BD cross, then: DEF = X EDF = 130 - X CDE = X + 10 CEF = 150 - X All the other angles have numeric values that are easily calculated as others have shown. Based on the derivations above, X can be any value < 130 degrees (to keep all the values positive). There is nothing else to constrain it. I think what's lacking is the length of at least one of the line segments. Or maybe I'm missing something else here. Refer to my last post with diagram. You have to add lines (that conform to standard geometric theorems/laws) When you do, you can surmise the rest of the angles, legitimately. The red lines in my diagram were the ones I added. I bisected the main triangle ABC, also created a parallel line to AB that I labeled DF I believe. Also, made a reflected but equal line AF that is the same as BD. Once you add those lines, you can surmise the rest from there. I even marked congruent sides based upon the angles opposite to show you how I got those angles. Why do I have to add lines? Like I said above, every angle in the diagram is either easily found or can be derived as a function of X. Therefore there is no single solution. If you say X=20, I agree that is a valid solution. But I argue that X=21, 55, or 87 (the ARFCOM preferred value) are also all valid. I don't have means to draw pictures here. Please humor me.... pick a value for X between 1-129 degrees, and label the original picture (without any added lines) with the resulting values, showing me why it wouldn't work. . Angle x has a finite value determined by the 60 degree and 70 degree lines, and the line that connects their intersections with the sides of the outer triangle. There is one solution. I agree X is finite, between 1-129 degrees as I showed in my simple equations above. And yes, X is related to the 60 and 70 degree lines. But I see nothing that limits it to one value. Try setting it to a random value and show me why it fails. ETA: There are four unknown values: DEF = X EDF = 130 - X CDE = X + 10 CEF = 150 - X I can substitute any X between 1-129 degrees and produce a valid solution. Those equations are not linearly independent. Write the equations from my unreduced form on page 2 in matrix format, then test the determinant. The geometry in the problem provides a set of boundary conditions that determine every angle and line length, any solution that does not satisfy those conditions is not a solution of the problem. |
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Posted: 12/15/2014 5:58:06 AM EDT
[#38]
For the next time this shows up.
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