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Euclid contest, question 7a, 7b, and 9a

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Euclid contest, question 7a, 7b, and 9a

Post  karooomph on Thu Apr 09, 2009 8:01 pm

believe it or not, I've actually given up on drawing diagrams.... Smile

So please see attached pic, should contain questions and diagram.

Please do 7a, 7b, and 9a.

Thanks!



btw, 7a, hole punched out word is down, if you're ever wondering


Last edited by karooomph on Thu Apr 09, 2009 9:28 pm; edited 4 times in total (Reason for editing : the host image wont work -_-)
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Re: Euclid contest, question 7a, 7b, and 9a

Post  karooomph on Thu Apr 09, 2009 8:13 pm

bounce
it says my scanned pic's too big or corrupted... ideas? confused


edit: YOU would not believe how long that took

and srry, but the jpeg kept on deteriorating, tried tiff, but too big.

see if you can decipher what it says, any questions, feel free to ask

Thanks
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Re: Euclid contest, question 7a, 7b, and 9a

Post  bfrsoccer on Fri Apr 10, 2009 12:08 am

It's midnight and my mom's trying to kick me off of the computer, so here are some concise explanations to help you get the basic idea (I can elaborate more later if you want):

7a. y=x^2-2x+4
y=x^2-2x+1-1+4; complete the square to turn into friendlier form
y=(x-1)^2+4
y=(x-1-p)^2+4-q; moved p units to the right and q units down
Now you can get two equations by plugging 0 for y and each of the x intercepts for each x, and I'm guessing that this could lead to equations for finding p or q

7b. area ΔABC is 4
1/2*b*h=4
1/2*b*4=4 (the height is 4 because B is at (0,-4) and the top is on the x-axis)
base=2
which means that A is two units to the left of C at (2,0) and D is at (3,?)

Since we know the coordinates of A and B, which lie on the x-axis, we can set up the equation: y=?*(x-2)(x-4), where ? is some unknown constant. Plug in B=(0,-4) into the equation to find ?

Then the base of ΔDBC equals 4 because of the distance between B and C is known, and the height of ΔDBC is ?-(-4). The area can be found from there using 1/2b*h

9a. For the last once, you can covert everything to log base 8:

EDIT: Last part taken out, as it was full of typos and didn't make sense. See the new explanation a few posts below


Last edited by bfrsoccer on Sun Apr 12, 2009 12:30 pm; edited 2 times in total
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Re: Euclid contest, question 7a, 7b, and 9a

Post  karooomph on Fri Apr 10, 2009 12:27 pm

I'm not done reading, but in case i forget:

y=x^2-2x+1-1+4; complete the square to turn into friendlier form
y=(x-1)^2+ 3

i think you forgot to subtract the 1.
I'll continue reading now : )

(posted at 1:08 am, bravo!)

EDIT: in case if you're wondering how you do 7a:
y = x2 - 2x + 4
change to vertex form
y = (x-1)2 + 3
p right, q down
y= (x-1-p)2 + 3 - q (-p moves right, +p goes left) (+q goes up, -q goes down)
so we get two lines when substituting in intercepts:
0 = (3-1-p)2 + 3-q ---> 0 = (2-p)2 + 3-q
0 = (5-1-p)2 + 3-q ---> 0 = (4-p)2 + 3-q
if p and q are the same in both equations, (2-p)2 must equal (4-p)2
(2-p)2 = (4-p)2
4 - 4p + p2 = 16 - 8p + p2
-12 = -4p
3 = p now we sub that into original sentences and find q

0 = (2-3)2 + 3-q
0 = (4-3)2 + 3-q

0 = (-1)2 + 3-q
0 = (1)2 + 3-q

0 = 1 + 3 - q
0 = 1 + 3 - q

q = 4

therefore p=3, and q=4

EDIT2: please check if im right, i asked someone and they came up with something different. thx

*now i still gotta do 9a Surprised

EDIT3: I got 9a now too.

log2x , (1+log4x) , log84x
instead of converting to base8, I converted to base2. -
log2x , log22 + (log2x / log24) , log24x / log28
= log2x , log22 + (log2x / 2) , log24x / 3
= log2x , 2log22 / 2 + log2x / 2 , log24x / 3
= log2x , (log222 + log2x) / 2 , log24x / 3
= log2x , log24x / 2 , log24x / 3
now divide 2nd and 3rd terms to get common ratio of geometric sequence:
(log24x / 3) / (log24x / 2)
= (log24x / 3) * (2 / log24x)
log24x cancels
= 2 / 3, the common ratio
so now we look at the first term. The first term * 2/3 should give the equivalent of the second term.
log2x * 2/3 = 2log2x / 3
= log2x2 / 3
the second term as already known, is log24x / 2
Therefore, log2x2 / 3 must equal log24x / 2
log2x2 / 3 = log24x / 2
multiply 6 BS
2log2x2 = 3log24x
log2(x2)2 = log2(4x)3
log2x4 = log2(4x)3
x4 = (4x)3
cube-root BS
(x3 * x)1/3 = 4x
x * (x)1/3 = 4x
x1/3 = 4x / x
x1/3 = 4
64 = x

so now only 5b and 7b left now! (5b's the question posted as "triangle trig")

EDIT27:
done 7b, too lazy to post thou Neutral


Last edited by karooomph on Mon Apr 13, 2009 11:35 am; edited 5 times in total
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Re: Euclid contest, question 7a, 7b, and 9a

Post  karooomph on Fri Apr 10, 2009 12:47 pm

i'm done reading!

7b. area ΔABC is 4
1/2*b*h=4
1/2*b*4=4 (the height is 4 because B is at (0,-4) and the top is on the x-axis)
base=2
which means that A is two units to the left of C at (2,0) and D is at (3,?)

i get the above. but...

Since we know the coordinates of A and B, which lie on the x-axis, we can set up the equation: y=?*(x-2)(x-4), where ? is some unknown constant. Plug in B=(0,-4) into the equation to find ?

Then the base of ΔDBC equals 4 because of the distance between B and C is known, and the height of ΔDBC is ?-(-4). The area can be found from there using 1/2b*h

...i dont get.

especially this:
y=?*(x-2)(x-4), where ? is some unknown constant. Plug in B=(0,-4) into the equation to find ?

I have 7a and 9a done and good, would it be possible for you to elaborate/solve 7b for me though?

Thanks cyclops
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Re: Euclid contest, question 7a, 7b, and 9a

Post  bfrsoccer on Sun Apr 12, 2009 12:17 am

For 7b:

Sorry, using "?" for two different things, an unknown constant in the equation for the parabola and the y-coordinate of D (which I didn't intend to do, but oh well), was misleading.

y=?*(x-2)(x-4) is the equation for the graph shown because it has zeroes at x=2 and x=4, and the general equation for a parabola with x-intercepts a and b is y=?*(x-a)(x-b) (not that this is a standard form that you'd use in school, but if you think about a parabola having to pass through two points on the x-axis, you have (x-a)(x-b), and all you can do from there is scale the parabola, which is where the ? comes in).

Besides the zeroes of this function, we know that it passes through B at (0,-4). So:

-4=?*(0-2)(0-4)
?=-1/2

Now, from here, we can find the coordinate of D by plugging in x-coordinate of D, 3, into -1/2*(x-2)(x-4), which gives 1/2. So, D is at (3,1/2).

There are a few ways to proceed from here. You can use a number of formulas, especially this one.
Additionally, one way to use 1/2*a*b would be:

- Treat BC as the base of ΔDBC. It's length is sqrt( (4-0)2 + (0-(-4))2 ), or sqrt(32)=sqrt(2*2*2*2*2)=4sqrt(2).

- Now we have to find the height, or the distance from D to BC. There are a few ways to do this, including a nice trick with vectors, but going with a traditional algebra method, you know the slope of BC is 4/4=1, so the slope of the line segment from D to BC (whose length is the distance between the two) is -1.

- Using point-slope form with the coordinate of D (3,1/2), we have equation y-1/2=-(x-3). The equation for the line through BC is y=x-4. Finding when these two is a matter of plugging in y=x-4 into y-1/2=-(x-3) for y, so:

x-4-1/2=-(x-3)
x-4.5=-x+3
2x=7.5
x=3.75
y=x-4=-.25.

- The distance between D and BC = h = sqrt( (3-3.75)2 + (1/2-(-.25))2 )

Now we know b, 4sqrt(2), and h is the result of the above calculation, and we get the area of ΔDBA with 1/2*b*h.

There are a few other ways using 1/2*b*h you could have found the area once you knew the coordinates of D, such calling the point of intersection of segments AC and BD point E, finding the coordinates of E, and then using that information to add the areas of ΔEDC and ΔECB.
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Re: Euclid contest, question 7a, 7b, and 9a

Post  karooomph on Sun Apr 12, 2009 9:51 am

thank you.

Something came up for 9a though, the log question. I got 64 as the only answer, but if you try 1/4 for x, you'll see that it's technically a geometric sequence, as you get {-2, 0, 0}, (0 is the common ratio then).

So where did I go wrong then?

Thanks
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Re: Euclid contest, question 7a, 7b, and 9a

Post  bfrsoccer on Sun Apr 12, 2009 1:24 pm

OK, the last paragraph of my 9a solution didn't make much sense.

Anyway, "now divide 2nd and 3rd terms to get common ratio of geometric sequence:"

r * log2(4x) / 2 = r^2 log24x / 3
r^2 log24x / 3 - r * log2(4x) / 2 = 0
r*(r*log24x / 3 - log2(4x) / 2) = 0
r=0 or r*log24x / 3 - log2(4x) / 2; r=0 or r=2/3
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Re: Euclid contest, question 7a, 7b, and 9a

Post  karooomph on Mon Apr 13, 2009 11:41 am

I'm going to assume all those exponents are bases instead. Very Happy


Anyway, "now divide 2nd and 3rd terms to get common ratio of geometric sequence:"

r * log2(4x) / 2 = r^2 log24x / 3
r^2 log24x / 3 - r * log2(4x) / 2 = 0
r*(r*log24x / 3 - log2(4x) / 2) = 0
r=0 or r*log24x / 3 - log2(4x) / 2; r=0 or r=2/3

a general geometric sequence goes as follows:
a, ar, ar2, ar3, ... arn-1
(where r is the common ratio)
so we're compared the 2nd and 3rd terms, therefore it's ar and ar2
so shouldn't it be :
r * log2x = r2 * log2x
? But then that doesn't lead you to anywhere
or i dunno, please tell me what you did
thx
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Re: Euclid contest, question 7a, 7b, and 9a

Post  bfrsoccer on Mon Apr 13, 2009 7:07 pm

I got mixed up with the terms, but regardless, some weird dividing issues still do occur in this problem.

log2x , log24x / 2 , log24x / 3
r2 * log2x = r*log24x / 2 = log24x / 3
r2 * log2x = r*log24x / 2
r*(r*log2x - log24x / 2)=0
r=0 or r*log2x - log24x / 2=0
How do we solve r*log2x - log24x / 2=0? To get the stuff involving "x" to cancel, we use the relationship r2 * log2x=log24x / 3 -> log2x=log24x / 3r and substitute log24x / 3r for log2x
r*log24x / 3r - log24x / 2=0
r=2/3
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Re: Euclid contest, question 7a, 7b, and 9a

Post  karooomph on Mon Apr 13, 2009 7:13 pm

no, but what im saying is, in a, ar, ar^2, shouldnt "a" remain constant?
instead of what your doing,
r2 * log2x = r*log24x / 2 = log24x / 3

Thanks

EDIT, oh, wait, i c where you're coming from
disregard ALL OF ABOVE

Thanks
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Re: Euclid contest, question 7a, 7b, and 9a

Post  karooomph on Mon Apr 13, 2009 8:08 pm

bfrsoccer wrote:
How do we solve r*log2x - log24x / 2=0? To get the stuff involving "x" to cancel, we use the relationship r2 * log2x=log24x / 3 -> log2x=log24x / 3r and substitute log24x / 3r for log2x
r*log24x / 3r - log24x / 2=0
r=2/3

shouldnt r2 * log2x=log24x / 3 -> log2x=log24x / 3r2???

wouldn't that change he answer you got?
THanks

EDIT-
*oh wait, the extra r cancels?
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Re: Euclid contest, question 7a, 7b, and 9a

Post  bfrsoccer on Mon Apr 13, 2009 9:11 pm

Yeah, we use log24x / 3r because we are substituting for r*log2x, not log2x (so we only divide r2 * log2x=log24x / 3 by r and not r2).

I guess I did accidentally say "substitute log24x / 3r for log2x" instead of "substitute log24x / 3r for r*log2x", but I think all of the math manipulation shown is correct.
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and after all these posts...

Post  karooomph on Tue Apr 14, 2009 5:35 pm

i get it now

thank you
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Re: Euclid contest, question 7a, 7b, and 9a

Post  caseycolin on Mon Aug 02, 2010 5:19 am

i got it also...
thanks for the help.




Study in UK|Education|Study in USA


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Re: Euclid contest, question 7a, 7b, and 9a

Post  mafiafran on Thu Aug 12, 2010 10:37 pm

thank for the help

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Re: Euclid contest, question 7a, 7b, and 9a

Post  LeeRain on Fri Sep 10, 2010 1:05 am

YOU would not believe how long that took




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