Chief Delphi - Math Quiz 4

Chief Delphi (http://www.chiefdelphi.com/forums/index.php)

- Math and Science (http://www.chiefdelphi.com/forums/forumdisplay.php?f=70)

- - Math Quiz 4 (http://www.chiefdelphi.com/forums/showthread.php?t=130175)

Quote:

Originally Posted by Ether (Post 1395583)

No takers?

Advanced math not required.

But it does help to think outside the box...

I'm going to post another hint after school starts and CD traffic patterns change.

Maybe this thread will catch the eye of a resourceful student.

Here’s the solution I came up with. It uses a coordinate system instead of trigonometry. It seems like it should work, but I have not gotten it to work out correctly yet.

I set all of the given lengths and ratios to variables for covenience.
DR=a
DL=b
OL=c
DE/DF=d

If the location of point D is (0,0), then it is easy to figure out the coordinates for points O, R, and L as well as the equations for lines OL, OR, DL, and OD.

D= (0,0)
R= (0,a)
L= (0,b)
O=(-c,b)
OD: y=(-b/c)*x
OR: y=((b-a)/-c)*x+a
OL: y=b

If the slope of line ED= s, then it is possible to write equations for the rest of the lines.

ED: y=sx
OC: y=-x/s+b-c/s

Point F is the intersection of ED and OR.

F=(a*c/(c*s+b-a), s*(a*c/(c*s+b-a)))

Because DH is on the y axis, the length of DC is the y intercept of OC.

C= (0, b-c/s)

C is the midpoint of HD, so it is easy to find point H.

H= (0, 2(b-c/s))

It is then possible to find the equation of line OH. After that, it is possible to find the coordinates of point E because it is the intersection of lines OH and ED.

OH: y=(b-2*c/s)*x/c+2*b-2*c/s
E= ((2*b*c*s-2*c^2)/(c*s^2-b*s+2*c), (2*b*c*s-2*c^2)*s/(c*s^2-b*s+2*c))

At this point if we find the distance between point E and point D and divide that by the distance between point F and point D and set that ratio equal to the ratio of DE to DF, it is possible to solve for s (the slope of line ED). By plugging that back into the equation representing the y coordinate of point H, it will be the equation representing the length of line HD.

Distance between E and D: sqrt(((2*b*c*s-2*c^2)/(c*s^2-b*s+2*c))^2+((2*b*c*s-2*c^2)*s/(c*s^2-b*s+2*c))^2)

Distance between F and D:
sqrt((a*c/(c*s+b-a))^2+(s*(a*c/(c*s+b-a)))^2)

At this point, I think it should be possible to set the ratio of the length of ED to the length of FD equal to the given ratio, then solve for s, however when I plugged this into an online equation solver, it said that it was impossible to solve for s (the slope of line ED). Is there a flaw in my logic, or did I just make a stupid mistake somewhere?

Quote:

Originally Posted by echin (Post 1397821)

Is there a flaw in my logic, or did I just make a stupid mistake somewhere?

It's a bit circuitous, but I see no flaw or stupid mistake. Well done. Reps to you.

Quote:

when I plugged this into an online equation solver, it said that it was impossible to solve for s

In math jargon, what is it called when this happens? ...and how do you find a solution?

Quote:

Originally Posted by echin (Post 1397821)

At this point, I think it should be possible to set the ratio of the length of ED to the length of FD equal to the given ratio, then solve for s, however when I plugged this into an online equation solver, it said that it was impossible to solve for s

Eric, are you still following this thread?

Yes, I am.

If it is impossible to solve for s, wouldn't that mean that the equation would simplify to something like s=s or 1=2? My other thought was that the equation solver I used might not be capable of solving an equation of that length. This might be the case because I later plugged in the same equation that I tried to solve the first time and attempted to solve for all of the other variables. The only one that it could solve was d(ratio of DE to DF) and that was given.

Quote:

Originally Posted by echin (Post 1398055)

wouldn't that mean that the equation would simplify to something like s=s or 1=2?

No. It just means that your equation solver was unable to find a function f such that s = f(DR,DL,OL,DEoverDF).

That may be because your equation solver isn't clever enough¹ to find it, or because there is no such function (that your solver considers to be an acceptable closed-form solution). What solver did you use?

Can you think of a simple relation g(x,y)=0 for which you cannot find a closed-form solution y=f(x)? And yet, if given a numerical value for x, how would you go about finding a numerical solution for y?

¹ Sometimes these solvers just need a little "help". For example, instead of trying to solve DE/DF = DEoverDF, try solving DE^2 = DF^2*DEoverDF^2. This gets rid of the square roots

Eric seems to have lost interest.

Should I post the solution, or is someone out there still working on it?

Mentors, teachers, college students, professors, engineers, math mavens of all stripes are invited to chime in here.

Quote:

Originally Posted by Ether (Post 1394687)

For the mathematically inclined, looking for a ~~summer~~ back-to-school challenge

There's still a lot of meat left on the bone here:

1) Using only half a dozen or so simple trig and geometry rules, set up a relation T(DR,DL,OL,DEoverDF,DH)=0 which can be numerically or graphically solved for DH, if given values for the other variables as stated in the original post1 problem statement.

2) Using only high-school-freshman math and the 4 basic arithmetic operations (+ - * /), set up a relation A(DR,DL,OL,DEoverDF,DH)=0 which can be numerically or graphically solved for DH, if given values for the other variables as stated in the original post1 problem statement.

3) Find an explicit solution DH=f(DR,DL,OL,DEoverDF) which uses only (a finite number of) add, subtract, multiply, divide, and square root operations.

Mentors, teachers, college students, professors, engineers, and math mavens of all stripes are invited to post.

Quote:

Originally Posted by Ether (Post 1398755)

1) 2) 3)

This thread continues to accumulate views, so I assume there is still some quiet interest lurking out there.

Monday evening I will post a solution for 1, 2, or 3. Vote for your choice.

Quote:

Originally Posted by Jim Wilks (Post 1399228)

My vote is for 2)

http://www.chiefdelphi.com/media/papers/3053

Quote:

Originally Posted by Ether (Post 1398755)

There is a 4th way to solve this that doesn't even require high school math. Tools :-)
I plugged the scenario and constraints into Solidworks and set constants for the givens, so the driven variables (DF which is tied to the DE/DF ratio) and DH which is the actual solution.
Not sure if you consider this cheating...I consider it to be solving a problem "outside the box".

Of course, this method does have its benefits. I'm going to throw this on the 3D printer shortly:

Couldn't do that if it was kept purely as a mathematical exercise!

Quote:

Originally Posted by seg9585 (Post 1402818)

There is a 4th way to solve this that doesn't even require high school math. Tools :-)

All three solutions I posted involved the use of computer tools for plotting and root finding (Maxima and Excel).

Quote:

Originally Posted by seg9585 (Post 1402818)

Not sure if you consider this cheating...I consider it to be solving a problem "outside the box".

Not cheating at all. If your personal skillset includes how to use CAD tools in this manner, then that is an option. Nice job, reps to you!

I gave it a go, but I am stuck :(. my trig is rusty and I when learned it was in different language.

I went as far as figuring out ROD and ORD angles, but after that I suspect I am missing some pertaining formulas to move forward.

OD=SQRT(DL*DL+OL*OL)
OR=SQRT((DL-DR)*(DL-DR)+OL*OL
ROD = ARCOS((OD*OD+OR*OR-DR*DR)/(2*OD*OR))

I thought I retained a lot of my math, but I was humbled by this experience ;)
Nevertheless, it was a good brain excersize