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There is a mistake, but thats not it.
Base = sqr(R2^2+S2^2-2R2S2Cos(180 - Theta1))
But since Cos(180 - Theta1) = - Cos(Theta1)
Then
Base = sqr(R2^2+S2^2+2R2S2Cos(Theta1))
No Typo!
:)
Now, I checked it out 2 more times, by hand, and found a mistake
Towards the end. I'll fix the graphic, and then I'll do a test in a scratch progie.
BTW, Now I think the answer is 22.
:)
I'm now 99 and 44/100 Percent Positive it is 22.
And, its just standard Word 97.
Whats So puzzling?
:)
i looked through your solution, i did not check ur algebra but everything else seems right. that is a pretty slick way, nicejob. i could use it to solve any problem of this type!
Ooh, now I don't know who's right.:confused: Oh well, I'll fiddle around with both your methods and see which one takes less to write down;)
I'd be interested to know where you got your 99 and 44/100 from...:D
I have found that 66/100ths Times that You KNOW something, that it is invariably wrong.
Hence, My Rule of Thumb:
Always Doubt Reality until Proven Otherwise!
Which means, Even though You've Triple checked for oncoming buses already, and none are coming, Keep Checking as you cross the street. If you survive, then I guess you were right!
:p
Anyways, I've attached an image, anticipating a question on how
I came up with a cetain equation in the above image. Basically, I
derived an equation for expressing the area of a triangle in terms
of its sides, then I used it on the Sub triangle Stradling the Base
of the overall Triangle.
http://www.vbforums.com/attachment.p...postid=1152492
found another equation! problem solved! if everything is right i will get what NotLKH got.
the ratio of the triangle with area 10 to the ratio of the whole triangle is:
hp/(p+1)*hq/(q+1)/(hp/(p+1)+hq/(q+1)):h
so we have equation
(x+8+10+5)*p/(p+1)*q/(q+1)/ (p/(p+1)+q/(q+1))=10
and i know:
p=18/(x+5)
q=15/(x+8)
hmmm it doesn't work i got 15*18/23=10 wierd
first rule about geometry, always name your points so you know what you are talking about. http://www.vbforums.com/attachment.p...postid=1152523
first a clarification, there is 2 Ls in the diagram. the bottom one will be L'.
let the ratio of ML:LK=p:1 (redefining p here, its changed from original posts) MH:HI=q:1
let the area of the whole triangle be A, and A (ABC) denote the area of triangle ABC, so we can obtain:
A (MLI):A (LIK) = Ap/(1+p):A/(1+p)=(x+8)/15
simplifying yields:
p=(x+8)/15
similarly:
A (MKH):A (HKI) = Aq/(1+q):A/(1+q)=(x+5)/18
simplifying yields:
q=(x+5)/18
let the height of the triangle be h (MJ) so by similar triangles the height LL'=h/(p+1)
and the height HM=h/(q+1)
ok?
now assume GJ is the height of triangle KGI (i forgot to name another point, so J will do since it doesn't matter), let LL'=a, HM=b,JL=c,ML=d, GJ be z, then MJ=ML-JL=d-c
i want to find z in terms of a and b
by similar triangles:
GJ:LL'=MJ:ML
z/a=(d-c)/d=1-c/d
similarly
GJ:HM=JL:ML
z/b=c/d
combining the 2 equations:
z/a=1-z/b
z/a+z/b=1
z(a+b)/ab=1
z=ab/(a+b)
the basically the height of the GKI: z=ab/(a+b)
where a=h/(p+1)
b=h/(q+1)
and
p=(x+8)/15
q=(x+5)/18
1/(p+1)=15/(x+23)
1/(q+1)=18/(x+23)
so we have:
z=h^2(15*18/(x+23)^2)/(h(23/(x+23))
simiplifying yields:
z=270h/(23(x+23))
so A (GKI): A (MKI)=z:h
A (MKI)=23+x
so:
10/(23+x)=270h/(23(x+23)h)
simplifying yields: 10=270/23
or 230=270
so whats wrong with this?
shouldn't a + b = 33/(x+23)Quote:
z=ab/(a+b)
......
p=(x+8)/15
q=(x+5)/18
1/(p+1)=15/(x+23)
1/(q+1)=18/(x+23)
so we have:
z=h^2(15*18/(x+23)^2)/(h(23/(x+23))
and not
23/(x+23)
???
:)
yes. but the answer still don't make sense! i am very screwed up these days :( :(
But you've now proved 11 = 9!
:D
Perhaps I could help if you could explain how you get:
A (MLI):A (LIK) = Ap/(1+p)
their base are in the ratio of 1:p. they share the same height. since A=bh/2, so their area is porportional to their base.
Just while someone's still on the subject of probability, I conducted a random survey in GTA3. I found that if you're driving quite fast and crash due to anything at all, the probability you end up facing the way you came from when you finally get control back is almost 3/4!
Come on, I was bored...
Yer, GTA3 rox!
WOW!
10 sheets of Paper {2 Sets of Screwups, 1 set good!},
and I showed that the Unknown area is indeed 22.
Bugz, You might want to check it out, since I used proportions
Throughout.
Its a Zipped PDF, and I only Outlined the steps, but included the
result of each step.
I'll DblChk it, make sure My Typed version has no mis memos,
BUT, The Theory is Good, and I reccommend Bandit should work it
out.
God, My Hand Hurts!
:p
{I like my first method best, You only have to Crunch Two
Massive Equations. This Newer way is just One Long Series of
Crunching One Massive Equation Right After Another, Over And
Over!}
i don't know what impresses me more, ur solution or ur ability to use tools to create solutions lol
My sources suggest that the copy of Adobe Acrobat you were using was pirated :D
They have been wrong before...
Hey, What can I say, I worked for a Digital Printing company for 9 years, I got good! {And, I should be back doing it in the next month, I hope!}
BTW, I rechecked, and It was 100% Good!
:D
Except....
:p
In Step 1, I had Two Subtraction Signs, but they were supposed to be Addition.
I fixed it, and Now the attachment is Perfect.
I assume You're the 1 download, so you might want to redownload the new one, and toss the old.
Looks Like Bandit just arrived!
What Do Y'All Think? Am I right?
:)
Now, I don't think you can say its pirated, since the company IQuote:
Originally posted by A$$Bandit
My sources suggest that the copy of Adobe Acrobat you were using was pirated :D
They have been wrong before...
worked for was shut down, and the owner of the Acrobat
Package gave it to me as a parting gift. {Actually, it was tossed
by another coworker, along with about 20 other apps, and, when
I pointed that out to My boss, the actual Procurer of those apps,
he told me I could have anything there.}
So, who's your source?
:)
And, BTW, I'll delete my attachments, and never bother to help you out again if thats the way your going to be!
:p
That would be telling;)
But yeh it is wrong like half the time!
Thanks loads for helping me with this problem, I fully get it now (I used your first right method), hehe.
Pffftttt .. all this fancy maths and he still gets 100 - 44 = 66 :D :eek:Quote:
Originally posted by NotLKH
I have found that 66/100ths Times that You KNOW something, that it is invariably wrong.
Well, It's unlucky to be right All the Time!Quote:
Originally posted by beachbum
Pffftttt .. all this fancy maths and he still gets 100 - 44 = 66 :D :eek:
;)
I don't know if this is any help at all, but I can prove that q<5.
Using Bugz's original equations:
p=18/(x+5) and q=15/(x+8),
the following can be derived:
x+5=18/p
x=(18-5*p)/p (1)
and
x+8=15/q
x=(15-8*q)/q (2)
Equating (1) and (2), we get:
(18-5*p)/p=(15-8*q)/q
18*q-5*p*q=15*p-8*p*q
3*p*q-15*p+18*q=0
p*q-5*p+6*q=0
p*q-5*p+6*q-30=-30
(p+6)*(q-5)=-30
Since p and q are both positive, then the only way that p+6 and q-5 will multiply to a negative number is if q<5.
Hope this helps!
But if q<5, and if q=15/(x+8) and x=(15/q)-8, then x is negative.
Since x cannot be negative, this question has no solution.
However, many of the previous solutions seem to be correct.
Something is very wrong...
this is the ORIGINAL VALUES for p and q, but nevertheless they are workable.
p=18/(x+5)
q=15/(x+8)
x=(18-5p)/p
x=(15-8q)/q
from here, since p>0,q>0,x>0
18-5p>0
18>5p
p<18/5=3.6
q<15/8=1.875
now if you substitute these values in, [hopefully] you won't have a problem.
true 0<p<5, but actually 0<p<3.6
Ok, I see it now. Thx, Bugz!:D
Great!Quote:
Originally posted by A$$Bandit
(I used your first right method), hehe.
So, do you understand it?
You worked everything out alright?
If you're not careful, the Teacher will think you Pirated the answer if you just did something out of thin air, with nothing to back it up.
Such as useing the area formula without deriving it, unless you've
used it in class already.
-Lou
So, since you're using my Meth #1, did you notice that:
A0*A2*A4 = A1*A3*A5
Or, in other words, a more impressive rearrangement:
1 = { A0*A2*A4 }/{ A1*A3*A5 }
No, in fact, I happened NOT to notice that :D
I understand it all completely, I was a bit stuck with Bugz method because I'm not much good without diagrams. Yours was great! I just needed pointing in the right direction and as soon as I realized you'd added that extra line in I knew the rest.
And no, I don't think I've pirated the formula... I'll just check...
OK. I've been working on this for a bit, and I've come down to a set of 2 equations, with 1 unknown. Unfortunately, they both amount to 5th order polynomials. So, I'm brainstorming other directions, but its a good challange for all.
Suppose you know A1, A3, and A5. Find any or all A2, A4, A0.
Think you can do it?
:)
So, A$$, How'd your class go? What was the teachers solution?
Guess its too much for you!Quote:
Originally posted by NotLKH
Suppose you know A1, A3, and A5. Find any or all A2, A4, A0.
Think you can do it?
:)
:p
A$$Bandit, we helped you to the best of our abilities, so post ur Math God's solution here --- you own it to us this much.
Bugz,
I found an even simpler method, Using Shared Heights.
Check it Out:
http://www.vbforums.com/attachment.p...postid=1158919
Yeh I'm real sorry, but I only got it today, hehe!
OK it's coming up and believe it or not it's pretty much what NotLKH had, although it's much simpler (would you believe...)
I'll get right on it...
Well I looked again and it's pretty much the most minimalistic version of NotLKH's method. In order to make it look even smaller than it already is, I've left out comments because you shouldn't need them...
(A + B + 8)/15 = A/5
2A - B = 8
(A + B + 5)/18 = B/8
5B - 4A = 20
A = 12
B = 10
X = 22...
That's it. Essentially it's the same as your method only this one's more basic. Thanks all for helping, let's just hope it pays off!
Is it simpler than the method I posted just above this post of yours?Quote:
Originally posted by A$$Bandit
Yeh I'm real sorry, but I only got it today, hehe!
OK it's coming up and believe it or not it's pretty much what NotLKH had, although it's much simpler (would you believe...)
I'll get right on it...
:D
Blatantly!:D :D :cool: