Math Quiz 4

[Ether]

For the mathematically inclined, looking for a summer challenge.

[Ether]

Originally Posted by Ether

For the mathematically inclined, looking for a summer challenge.

No takers?

Advanced math not required.

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

[Aaron.Graeve]

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.

[Jon Stratis]

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

[Aaron.Graeve]

It does, my apologies.

[Jon Stratis]

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

[Karthik]

Originally Posted by Ether

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

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

[faust1706]

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.

[Jon Stratis]

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!

[Arhowk]

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)

She's a Dean's List Winner and couldn't give me an answer after 3 hours of staring at the whiteboard!

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)

[Ether]

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

[Arhowk]

Originally Posted by Ether

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

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())

\frac{2 * \sqrt{z^2 + a^2}*cos(arcsin(a/b))}{cos(arcsin(a/\sqrt{a^2+(z-y)^2})-arcsin(\frac{bsinc}{\sqrt{a^2+(z-y)^2}})-arcsin(\frac{a}{b})-arcsin(\frac{bsin(arcsin(\frac{bsinc}{\sqrt{a^2+(z-y)^2}})-arcsin(\frac{a}{b}))}{\sqrt{2*b*\sqrt{a^2+(z-y)^2}*cos(arcsin(\frac{bsinc}{\sqrt{a^2+(z-y)^2}})-arcsin(\frac{a}{b}))}))}}

[Ether]

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

Or better yet...

Think outside the box. Consider other approaches.

[Arhowk]

Originally Posted by Ether

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

Or better yet...

Think outside the box. Consider other approaches.

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.

[Ether]

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?

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.