Metasolution Considered Unbecoming
July 22, 2009 8:52 AM Subscribe
How was I supposed to have solved this math problem?
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In case it isn't clear, the 7 is a chord length and we are looking for X, the remainder of that line. Obviously, finding the entire line length is also acceptable. The 12 line is tangent to the circle.
The source of this problem is a problem-a-day calendar, so I don't think we should need to use much advanced analysis. Then again, the depth of the problem has a lot of variation. Also, I know the right answer (from the calendar) and I got that answer, but my solution wouldn't work "in the wild" so I'm thinking it couldn't be the right way.
I'm going to leave my "cheat" answer out of the question to avoid poisoning your minds. I'll post it later if necessary. Partial answers or helpful observations are welcome as they may stimulate my or others' thinking.
Image
In case it isn't clear, the 7 is a chord length and we are looking for X, the remainder of that line. Obviously, finding the entire line length is also acceptable. The 12 line is tangent to the circle.
The source of this problem is a problem-a-day calendar, so I don't think we should need to use much advanced analysis. Then again, the depth of the problem has a lot of variation. Also, I know the right answer (from the calendar) and I got that answer, but my solution wouldn't work "in the wild" so I'm thinking it couldn't be the right way.
I'm going to leave my "cheat" answer out of the question to avoid poisoning your minds. I'll post it later if necessary. Partial answers or helpful observations are welcome as they may stimulate my or others' thinking.
I remember nothing from high school geometry but still curious, what is the name of the calendar?
posted by caelumluna at 9:08 AM on July 22, 2009
posted by caelumluna at 9:08 AM on July 22, 2009
Is the start of the x+7 line directly opposite the start of the 12 line? (ie a line drawn between the two is the diameter of the circle?
posted by latentflip at 9:11 AM on July 22, 2009
posted by latentflip at 9:11 AM on July 22, 2009
Response by poster: The calendar is this.
a line drawn between the two is the diameter of the circle
The original diagram doesn't show it that way and there's no construction of it, so I guess not.
posted by DU at 9:16 AM on July 22, 2009
a line drawn between the two is the diameter of the circle
The original diagram doesn't show it that way and there's no construction of it, so I guess not.
posted by DU at 9:16 AM on July 22, 2009
*Throws current scribbles away*
posted by latentflip at 9:18 AM on July 22, 2009
posted by latentflip at 9:18 AM on July 22, 2009
I can't get my head around how you'd even start that without some sort of angle or additional dimension. It just looks one number away from anything you'd need to start with, to me.
Will they be looking for an actual answer? Or an equation giving the answer? It wouldn't be the first question to leave out pertinent information, of course, but it's been a long, long time since I've done any of this.
posted by Brockles at 9:18 AM on July 22, 2009
Will they be looking for an actual answer? Or an equation giving the answer? It wouldn't be the first question to leave out pertinent information, of course, but it's been a long, long time since I've done any of this.
posted by Brockles at 9:18 AM on July 22, 2009
Response by poster: Will they be looking for an actual answer? Or an equation giving the answer?
The answer is 9.
It wouldn't be the first question to leave out pertinent information, of course...
It would also not be the first problem in this calendar that suffers from ill-formedness. So it's possible that there's no direct way to calculate the answer and my metasolution is the only way.
(I say "my" but it was actually a collaboration with my officemate.)
posted by DU at 9:23 AM on July 22, 2009
The answer is 9.
It wouldn't be the first question to leave out pertinent information, of course...
It would also not be the first problem in this calendar that suffers from ill-formedness. So it's possible that there's no direct way to calculate the answer and my metasolution is the only way.
(I say "my" but it was actually a collaboration with my officemate.)
posted by DU at 9:23 AM on July 22, 2009
If you draw the diagonal to the circle and then from the tangent to where the chord intersects the circumference you have two right angle triangles.
Let's call the diagonal distance "d".
This gives one right angle triangle with hypotenuse "d" and edge 7, intersecting angle "a" (the other edge is not labelled, nor interesting).
The other right angle triangle has edge "d" and hypotenuse "7 +x" with the same intersecting angle "a". Likewise it has edge 12 and hypotenuse "7 +x" with intersecting angle "90 - a".
From here, I'm spinning my wheels a bit trying to remember my trig to calculate "d".
(This all assumes that the line from the tangent to the start of the chord goes through the diameter. If it doesn't do that, then all bets are off).
posted by devbrain at 9:24 AM on July 22, 2009
Let's call the diagonal distance "d".
This gives one right angle triangle with hypotenuse "d" and edge 7, intersecting angle "a" (the other edge is not labelled, nor interesting).
The other right angle triangle has edge "d" and hypotenuse "7 +x" with the same intersecting angle "a". Likewise it has edge 12 and hypotenuse "7 +x" with intersecting angle "90 - a".
From here, I'm spinning my wheels a bit trying to remember my trig to calculate "d".
(This all assumes that the line from the tangent to the start of the chord goes through the diameter. If it doesn't do that, then all bets are off).
posted by devbrain at 9:24 AM on July 22, 2009
Best answer: In the Wikipedia entry for circle, under Theorems it mentions a "Tangent-Secant Theorem" that applies here, regardless of whether the 12-tangent and the 7+x secants intersections mark a diameter.
posted by reegmo at 9:26 AM on July 22, 2009
posted by reegmo at 9:26 AM on July 22, 2009
It doesn't have to go through the diameter, and I get the answer as 9.
It's just a simple chord rule (apparently, I had to google it).
I shall not ruin it yet by giving the answer here, but if you go to this link and look at Theorem 3, you should be able to get the answer with a little further algebra.
posted by latentflip at 9:28 AM on July 22, 2009
It's just a simple chord rule (apparently, I had to google it).
I shall not ruin it yet by giving the answer here, but if you go to this link and look at Theorem 3, you should be able to get the answer with a little further algebra.
posted by latentflip at 9:28 AM on July 22, 2009
Too late :(, good work reegmo.
posted by latentflip at 9:28 AM on July 22, 2009
posted by latentflip at 9:28 AM on July 22, 2009
The theorem you need tells you that 12 squared is (7 +x)*x and you get that from similar triangles. To get the similar triangles you need to know theorems about angles formed by chords and tangents. X^2+7x-144=0 factors as (x-9)(x+16) =0 which means x=9.
posted by Obscure Reference at 9:28 AM on July 22, 2009
posted by Obscure Reference at 9:28 AM on July 22, 2009
Response by poster: Never heard of the tangent-secant theorem. That's pretty awesome. Thanks!
posted by DU at 9:35 AM on July 22, 2009
posted by DU at 9:35 AM on July 22, 2009
I'm a little late to the party, but I'm curious - what was the metasolution?
posted by pemberkins at 9:41 AM on July 22, 2009
posted by pemberkins at 9:41 AM on July 22, 2009
Response by poster: More generally, it seems to refer to the power of a point which I have also never heard of.
posted by DU at 9:41 AM on July 22, 2009
posted by DU at 9:41 AM on July 22, 2009
Response by poster: The metasolution: Imagine a new problem that's identical to this one only the circle is slightly larger. I can do this because the angle and diameter aren't given. I can still have a tangent line of length 12. I can still have a chord of length 7. I can still join these two lines. The only thing that might change is x. However, the assumption behind the question is that x is unique. So x must be a constant as I change the size of the circle.
Therefore I can transform this problem into one where the circle is exactly the right size such that the chord IS the diameter. From here the solution is easy.
posted by DU at 9:46 AM on July 22, 2009
Therefore I can transform this problem into one where the circle is exactly the right size such that the chord IS the diameter. From here the solution is easy.
posted by DU at 9:46 AM on July 22, 2009
you can't assume that X is fixed to prove that it is a constant.
posted by mary8nne at 10:00 AM on July 22, 2009
posted by mary8nne at 10:00 AM on July 22, 2009
Response by poster: Sure it is. I've just added a given to the ones shown on the diagram. The displeasing part is the source of the given, not the soundness of the proof using the given. (I could have stated the proof more rigorously, though.)
posted by DU at 10:01 AM on July 22, 2009
posted by DU at 10:01 AM on July 22, 2009
I just drew this up in Solidworks (it's more fun than editing schematics).
DU's comment at 10:46 is correct, you can make the circle any size greater than or equal to 7. If you make it 7 then the answer is (7/2)^2 + 12^2 = (x+(7/2))^2.
So, the metasolution is to simply substitute variables in for 7 and 12. ie: (C/2)^2 + T^2 = (x+(C/2))^2
I think the real key is to realize the the circle can change sizes, this was not immediately obvious.
posted by Confess, Fletch at 10:31 AM on July 22, 2009
DU's comment at 10:46 is correct, you can make the circle any size greater than or equal to 7. If you make it 7 then the answer is (7/2)^2 + 12^2 = (x+(7/2))^2.
So, the metasolution is to simply substitute variables in for 7 and 12. ie: (C/2)^2 + T^2 = (x+(C/2))^2
I think the real key is to realize the the circle can change sizes, this was not immediately obvious.
posted by Confess, Fletch at 10:31 AM on July 22, 2009
Actually DU, your metasolution is perfectly valid (and is very impressive).
First imagine the circle shrunk down into a point; the problem makes no sense that way since the length labeled 7 is zero. As you grow the circle, obviously the smallest circle for which the problem is valid has diameter 7.
Now imagine that the circle continues to grow. Clearly, by pivoting the longer line around the point that the two lines intersect, you can always cut off a chord of length seven without changing the length of the line at all, no matter how big the circle gets. Therefore the solution for the circle of diameter 7 always works.
This is no more than a variation on what Confess, Fletch already said.
posted by jamjam at 1:13 PM on July 22, 2009
First imagine the circle shrunk down into a point; the problem makes no sense that way since the length labeled 7 is zero. As you grow the circle, obviously the smallest circle for which the problem is valid has diameter 7.
Now imagine that the circle continues to grow. Clearly, by pivoting the longer line around the point that the two lines intersect, you can always cut off a chord of length seven without changing the length of the line at all, no matter how big the circle gets. Therefore the solution for the circle of diameter 7 always works.
This is no more than a variation on what Confess, Fletch already said.
posted by jamjam at 1:13 PM on July 22, 2009
I guess I should say that the circle can change sizes without changing x.
If anybody wants my Solidworks file memail me.
posted by Confess, Fletch at 1:42 PM on July 22, 2009
If anybody wants my Solidworks file memail me.
posted by Confess, Fletch at 1:42 PM on July 22, 2009
Your solution boils down to the fact that you assume all circles give the same answer, because the problem does not state otherwise, hence you can use an easy one to get it. If you really want to be thorough, you have to prove this. Of course, the easiest way to prove it would be to calculate x in some more orthodox manner.
When I was preparing for university entrance exams, we had a math teacher that would stop us from exploting loopholes like this by giving us problems with an unknown (possibly zero) number of solutions.
By the way, you are not an engineer, are you? I'm a physicist working with engineers and I see them do this kind of thing all the time.
posted by Dr Dracator at 10:57 PM on July 22, 2009
When I was preparing for university entrance exams, we had a math teacher that would stop us from exploting loopholes like this by giving us problems with an unknown (possibly zero) number of solutions.
By the way, you are not an engineer, are you? I'm a physicist working with engineers and I see them do this kind of thing all the time.
posted by Dr Dracator at 10:57 PM on July 22, 2009
Response by poster: I am a software engineer. But you'll note that my officemate (also working as a software engineer, but I think with a BS in math) and I both didn't like this solution. It was just the only thing we could think of.
This isn't the first problem we've come across that involved learning (or rediscovering) a theorem we didn't already know. But I wasn't able to google the "arcane chord-related theorem" my first comment referred to.
posted by DU at 4:23 AM on July 23, 2009
This isn't the first problem we've come across that involved learning (or rediscovering) a theorem we didn't already know. But I wasn't able to google the "arcane chord-related theorem" my first comment referred to.
posted by DU at 4:23 AM on July 23, 2009
Response by poster: Ahh, I just proved the tangent-secant theorem, which is the basis for this whole thing. I never knew or suspected this. I'm going to be using this all the time. All kinds of chord problem just got a whole lot easier!
posted by DU at 6:25 AM on July 23, 2009
posted by DU at 6:25 AM on July 23, 2009
Another way to look at this is to ignore the circle.
If you just draw the 'tangent' and 'secant' without the circle it become obvious that you can change the angle between them without changing their lengths.
Now it's just a matter of choosing an angle that makes for convient solving.
posted by Confess, Fletch at 7:47 AM on July 23, 2009
If you just draw the 'tangent' and 'secant' without the circle it become obvious that you can change the angle between them without changing their lengths.
Now it's just a matter of choosing an angle that makes for convient solving.
posted by Confess, Fletch at 7:47 AM on July 23, 2009
Response by poster: Hey, there we go. You have two lines. One is 12 units long, the other is X units long. X is greater than 7, because there's also a point marked on that line 7 units from the end.
You can make the angle between the two lines anything you want. Furthermore, as long as the angle is greater than 0 and less than 180, you can always draw a circle connecting the ends of the first and second lines plus that 7 unit mark.
Now just choose a useful circle.
posted by DU at 8:59 AM on July 23, 2009
You can make the angle between the two lines anything you want. Furthermore, as long as the angle is greater than 0 and less than 180, you can always draw a circle connecting the ends of the first and second lines plus that 7 unit mark.
Now just choose a useful circle.
posted by DU at 8:59 AM on July 23, 2009
Response by poster: Whoops, no, that doesn't work. I can draw a circle connecting those three points, but I need the circle to ALSO be tangent at the 12 line.
posted by DU at 9:03 AM on July 23, 2009
posted by DU at 9:03 AM on July 23, 2009
The Problem with DUs solution is that it assumes THERE IS a solution before solving. Now there is nothing in the original problem that explicitly states THERE IS a solution. ie if this were an exam question the answer could just as easily be that there exist multiple solutions for X or none.
and DU you did not prove anything.
posted by mary8nne at 2:15 AM on July 24, 2009 [1 favorite]
and DU you did not prove anything.
posted by mary8nne at 2:15 AM on July 24, 2009 [1 favorite]
Response by poster: The Problem with DUs solution is that it assumes THERE IS a solution before solving.
Yes, exactly.
Now there is nothing in the original problem that explicitly states THERE IS a solution.
Well, there kind of is. The way the calendar is set up, the answer to each day is the day's date. So every problem has a single, numerical answer.
and DU you did not prove anything.
GIVEN that there is a single solution, my proof works. And we do in fact know that there is a single solution, from the above. Our discomfort, and the reason I posted this AskMe, arises from the fact that our knowledge of that uniqueness wasn't explicitly a "given". It's meta-information from outside the problem.
Anyway, I'm very happy to have a cleaner proof now. Particularly because it involves a previously unknown-to-me theorem that's going to be pretty useful.
posted by DU at 4:45 AM on July 24, 2009
Yes, exactly.
Now there is nothing in the original problem that explicitly states THERE IS a solution.
Well, there kind of is. The way the calendar is set up, the answer to each day is the day's date. So every problem has a single, numerical answer.
and DU you did not prove anything.
GIVEN that there is a single solution, my proof works. And we do in fact know that there is a single solution, from the above. Our discomfort, and the reason I posted this AskMe, arises from the fact that our knowledge of that uniqueness wasn't explicitly a "given". It's meta-information from outside the problem.
Anyway, I'm very happy to have a cleaner proof now. Particularly because it involves a previously unknown-to-me theorem that's going to be pretty useful.
posted by DU at 4:45 AM on July 24, 2009
This thread is closed to new comments.
Boy was that unclear. What I mean is, the variation among problems in this calendar is high. One problem will be like 20 and then the next one is like this, where it seems to require some arcane chord-related theorem.
posted by DU at 9:04 AM on July 22, 2009