*For the mathematically inclined, looking for a summer challenge.
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No takers?
Advanced math not required.
But it does help to think outside the box…
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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.3 reads DC=CH in the image… I believe that correction to your post messes up the solution pretty badly…
It does, my apologies.
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 
Yup. Sometimes the simplest way to solve a problem is to draw your own lines…
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.
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!
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 - 2be*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)
*j = arcsin((bsini)/(y)) <- error here
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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}))}))}}
```<br><br><br><br><br><br>*I see at least 3 errors in your derivation. Check your work carefully.
Or better yet…
Think outside the box. Consider other approaches.
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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.
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.
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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=(ac/(cs+b-a), s*(ac/(cs+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-2c/s)x/c+2b-2c/s
E= ((2bcs-2c^2)/(cs^2-bs+2c), (2bcs-2c^2)s/(cs^2-bs+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(((2bcs-2c^2)/(cs^2-bs+2c))^2+((2bcs-2c^2)s/(cs^2-bs+2*c))^2)
Distance between F and D:
sqrt((ac/(cs+b-a))^2+(s*(ac/(cs+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?
It’s a bit circuitous, but I see no flaw or stupid mistake. Well done. Reps to you.
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?
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.