Math Quiz 4
 

For the mathematically inclined, looking for a summer challenge.

Re: Math Quiz 4
 

No takers?

Advanced math not required.

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

Re: Math Quiz 4
 

Edit: Incorrect, please disregard
~~~~~~~~~~~~~~~~~~~~~~~~
I would have responded earlier, but I was at an off-season.

Also, I will not have any pretty MS-Paint pictures due to limited computer access. Please bear with me.

The 3 things we know are:
1. OL is perpendicular to DH.
2. DE is perpendicular to OC
3. DE = CH.

Based on 1 and 3, I can safely state that triangles OLC and OLD are similar right triangles reflected along line OL. This is known because both triangles share a leg, a right angle, a hypotenuse, and a terminating point O.

I am given DR, DL, OL, and DE/DF.

From triangles OLC and OLD's similarity, I know LC and DL are the same length. I can also compute OD (which is equal to OC) from the Pythagorean theorem using OL and CL

I know the angle PCL = OCL and that angle can be found by taking arctan(OL/LC)

I can now find PD based on the law of sines. For all of you who have forgotten geometry, the length of a side in a triangle devided by the sine of the angle opposing it js a constant for the triangle. All three side/and pairs share this property. CD/sin(CPD) = PD/sin(PCD). We know both angles and length CD (angle PCD is known from arctan(OL/CL).

Now that I have established PD, I will try and find ED. I know the total angle of COR is arctan(CL/OL) + arctan(RL/OL). OP can be found by using the Pythagorean theorem on the triangle OPD as both PD and OD are known and angle OPD is right by 2.

Angle COR is also and POF. Using triangle POF is right by 2, PF = OP×tan(POF). As we know know PF and PD, DF = PD-PF. DE = DE/DF (given) × DF.

EP = DE-PD. We know the angle HOL is equal to arctan(EP/OP)+arctan(PD/OP)-arctan(DL/OL). We also know triangle HOL is right by 1.

DH =tan(HOL)×OL.

If I have made a mistake, please let me know. I think my solution is correct, but I may be wrong.

Re: Math Quiz 4
 

3 reads DC=CH in the image... I believe that correction to your post messes up the solution pretty badly...

Re: Math Quiz 4
 

It does, my apologies.

Re: Math Quiz 4
 

I've thought about this a little, but haven't had time to sit down and work out the math yet. With what's provided, you can easily figure out lengths and angles for everything about triangles OLR and OLD, but that alone doesn't help you.

I suspect the answer lies in using the DC=CH relationship, along with the DE/DF ratio to infer some other relationships. As point C (note Point H moves in relation to point C!) approaches point L, the line DE approaches line DH. As such, the length of DE is important - that length has to change in a very specific way as you move point C along the DH line. This seems to imply that the length of DF is critically important, as it tells you the length of DE... Can we find DF using what we know of triangles OLR and OLD and the relationship the angle FDR has with the length of DC and DH?

Anyways, those are my lunchtime musings... They may lead you towards a solution, or down a blind alley from which there is no return! Either way, it's the journey that matters, not the destination :)

Re: Math Quiz 4
 

Yup. Sometimes the simplest way to solve a problem is to draw your own lines...

Re: Math Quiz 4
 

I am at an off-season event right now (showme-showdown), but I am working on this throughout the day. I suggest those who are trying it to look into Stewart's theorem.

Re: Math Quiz 4
 

I'm really surprised no student has posted a solution yet... I know I stumped one of our students with it at our meeting Saturday - She's a Dean's List Winner and couldn't give me an answer after 3 hours of staring at the whiteboard!

Re: Math Quiz 4
 

Here's a golfed version of the solution i came up with, ill neaten it up in a second

(2*sqrt(z^2+a^2)*sin(90-arcsin(a/sqrt(z^2+a^2))))/sin(180-(90-arcsin(a/sqrt(a^2+(z-y)^2))+180-(180-(180-arcsin(a/sqrt(a^2+(z-y)^2))+arcsin(a/sqrt(z^2+a^2)))+arcsin((bsin(180-(180-arcsin(a/sqrt(a^2+(z-y)^2))+arcsin(a/sqrt(z^2+a^2)))))/(y)))))

E/ i just noticed that I accidentally wrote o instead of sin(o) when i golfed this code.... woooooooooooooooops

2e/ pay no attention to the golfed version above

3e/ fixed the golfed version. ignore the 180-180-180-...

where


y=DR
z=DL
a=OL
b=DE/DF

e/ explanation (i know i accidentally used "b" twice, in the expalantion "b" will be marked as "#")

b (OD) = sqrt(z^2+a^2)
c (<ODL)= arcsin(a/b)
d (<DOL)= 90-c
e (OR) = sqrt(a^2+(z-y)^2)
f (<ORL)= arcsin(a/e)
g (<ROL) = 90-f
h (<DRO)= 180-f
i (<DOR) = 180-(h+c)
//temporary stepping out of alphabetical order
l (DF) = b^2 + e^2 - 2*b*e*cos(i)
//back again
j (<DFO)= arcsin((bsini)/(l))
k (<ODF)= 180-(i+j)
m (DE) = #L
n (<DOC)= 90-C
o (<DCO) = 180-(g+k)
sin(n)/x = sin(o)/b :: 1/x = sin(o)/bsin(n) :: x = bsin(n)/sin(o) ::

solution = 2 * DO * sin(<DOC)/sin(<DCO)

Dean's List winner doesn't necessarily mean a math whiz... most of our deans list entries aren't mathletes (except for our 2012 entry, who won at MSC)

Re: Math Quiz 4
 

j = arcsin((bsini)/(y)) <- error here

Re: Math Quiz 4
 

Woops, my bad, that should be lowercase L.

e/ L is after J in the alphabet. Right. I would need to solve for triangle DRF somehow. I'll post a different solution in a bit

When I get home I'll see if i can get a neater final solution. (though I only just went through trig (starting BC) so I don't know many tricks)

2E/ fixed that error. working on a latex solution

3E/ i feel like i should be doing something with all of those cos(arcsin() - arcsin())

Re: Math Quiz 4
 

I see at least 3 errors in your derivation. Check your work carefully.

Or better yet...

Think outside the box. Consider other approaches.

Re: Math Quiz 4
 

Hmmph, I suppose I'm a bit rusty, haven't had to do anything with my brain since last build season. Although, I don't see how using cartesian coordinates will help, since most of the numbers given are variables.

by "errors in [my] derivation", are you referring to issues I have with the fundamental concepts of trigonometry or just slips like the first error I posted? the only thing I know i purposely avoided was the ambiguous law of sines case.

Re: Math Quiz 4
 

You seem to understand trigonometry but there are obvious (careless I assume) mistakes in your derivation. Check each step carefully.

Or better yet, think a bit more about other (simpler) approaches.

Re: Math Quiz 4
 

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.

Re: Math Quiz 4
 

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?

Re: Math Quiz 4
 

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

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

Re: Math Quiz 4
 

Eric, are you still following this thread?

Re: Math Quiz 4
 

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.

Re: Math Quiz 4
 

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

Re: Math Quiz 4
 

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.

Re: Math Quiz 4
 

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.

Re: Math Quiz 4
 

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.

Re: Math Quiz 4
 

My vote is for 2)

Re: Math Quiz 4
 

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

Re: Math Quiz 4
 

Nice work!

Re: Math Quiz 4
 

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!

Re: Math Quiz 4
 

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

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!

Re: Math Quiz 4
 

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

Re: Math Quiz 4
 

I posted the solutions here:

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

Re: Math Quiz 4
 

Can you elaborate on following statement:
They are equal because they are similar, seem like a shaky foundation.

Re: Math Quiz 4
 

Please re-read the statement more carefully. Triangles CDP and LOC are similar, therefore angles CDP and LOC are equal.

If you are questioning how to show that the two triangles are similar, please rephrase your question.

Re: Math Quiz 4
 

I was not aware of the special meaning of "similar triangles" term (thanks for the link).
However, it is still not clear to me how process arrived to a logical conclusion that "triangles CDP and LOC are similar".

Re: Math Quiz 4
 

Triangles CDP and LOC are both right triangles. And they share a common angle PCL. Therefore they are similar.

Re: Math Quiz 4
 

Thanks, that's the step I was missing
Now I know why there were so many lines and triangles on it: to confuse people. :)

Re: Math Quiz 4
 

Actually, no. The problem arose as a result of a real-world optics problem I was working on. It occurred to me it would make an interesting quiz for FRC students.

Re: Math Quiz 4
 

In this case it would be nice to know "real life" details. It would be a good example of math relevancy to real life.