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

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

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 -_-)

**karooomph**- Posts : 74

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

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

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

**karooomph**- Posts : 74

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

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:

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

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+

i think you forgot to subtract the 1.

I'll continue reading now : )

(posted at 1:08 am, bravo!)

y = x

change to vertex form

y = (x-1)

p right, q down

y= (x-1-p)

so we get two lines when substituting in intercepts:

0 = (3-1-p)

0 = (5-1-p)

if p and q are the same in both equations, (2-p)

(2-p)

4 - 4p + p

-12 = -4p

3 = p now we sub that into original sentences and find q

0 = (2-3)

0 = (4-3)

0 = (-1)

0 = (1)

0 = 1 + 3 - q

0 = 1 + 3 - q

q = 4

therefore p=3, and q=4

*now i still gotta do 9a

log

instead of converting to base8, I converted to base2. -

log

= log

= log

= log

= log

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

(log

= (log

log

= 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.

log

= log

the second term as already known, is log

Therefore, log

log

multiply 6 BS

2log

log

log

x

cube-root BS

(x

x * (x)

x

x

64 = x

so now only 5b and 7b left now!

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 = x

^{2}- 2x + 4change to vertex form

y = (x-1)

^{2}+ 3p 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-q0 = (5-1-p)

^{2}+ 3-q ---> 0 = (4-p)^{2}+ 3-qif 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 + p

^{2}= 16 - 8p + p^{2}-12 = -4p

3 = p now we sub that into original sentences and find q

0 = (2-3)

^{2}+ 3-q0 = (4-3)

^{2}+ 3-q0 = (-1)

^{2}+ 3-q0 = (1)

^{2}+ 3-q0 = 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

**EDIT3:**I got 9a now too.log

_{2}x , (1+log_{4}x) , log_{8}4xinstead of converting to base8, I converted to base2. -

log

_{2}x , log_{2}2 + (log_{2}x / log_{2}4) , log_{2}4x / log_{2}8= log

_{2}x , log_{2}2 + (log_{2}x / 2) , log_{2}4x / 3= log

_{2}x , 2log_{2}2 / 2 + log_{2}x / 2 , log_{2}4x / 3= log

_{2}x , (log_{2}2^{2}+ log_{2}x) / 2 , log_{2}4x / 3= log

_{2}x , log_{2}4x / 2 , log_{2}4x / 3now divide 2nd and 3rd terms to get common ratio of geometric sequence:

(log

_{2}4x / 3) / (log_{2}4x / 2)= (log

_{2}4x / 3) * (2 / log_{2}4x)log

_{2}4x 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.

log

_{2}x * 2/3 = 2log_{2}x / 3= log

_{2}x^{2}/ 3the second term as already known, is log

_{2}4x / 2Therefore, log

_{2}x^{2}/ 3 must equal log_{2}4x / 2log

_{2}x^{2}/ 3 = log_{2}4x / 2multiply 6 BS

2log

_{2}x^{2}= 3log_{2}4xlog

_{2}(x^{2})^{2}= log_{2}(4x)^{3}log

_{2}x^{4}= log_{2}(4x)^{3}x

^{4}= (4x)^{3}cube-root BS

(x

^{3}* x)^{1/3}= 4xx * (x)

^{1/3}= 4xx

^{1/3}= 4x / xx

^{1/3}= 464 = x

so now only 5b and 7b left now!

*(5b's the question posted as "triangle trig")***done 7b, too lazy to post thou**

EDIT27:EDIT27:

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

i'm done reading!

i get the above. but...

...i dont get.

especially this:

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

Thanks

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

**karooomph**- Posts : 74

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

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)

- 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)

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.

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

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

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

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 * log

r^2 log

r*(r*log

r=0 or r*log

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

r * log

^{2}(4x) / 2 = r^2 log^{2}4x / 3r^2 log

^{2}4x / 3 - r * log^{2}(4x) / 2 = 0r*(r*log

^{2}4x / 3 - log^{2}(4x) / 2) = 0r=0 or r*log

^{2}4x / 3 - log^{2}(4x) / 2; r=0 or r=2/3**bfrsoccer**- Administrator
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## Re: Euclid contest, question 7a, 7b, and 9a

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

a general geometric sequence goes as follows:

a, ar, ar

(where r is the common ratio)

so we're compared the 2nd and 3rd terms, therefore it's ar and ar

so shouldn't it be :

r * log

? But then that doesn't lead you to anywhere

or i dunno, please tell me what you did

thx

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

r * log_{2}(4x) / 2 = r^2 log_{2}4x / 3

r^2 log_{2}4x / 3 - r * log_{2}(4x) / 2 = 0

r*(r*log_{2}4x / 3 - log_{2}(4x) / 2) = 0

r=0 or r*log_{2}4x / 3 - log_{2}(4x) / 2; r=0 or r=2/3

a general geometric sequence goes as follows:

a, ar, ar

^{2}, ar^{3}, ... ar^{n-1}(where r is the common ratio)

so we're compared the 2nd and 3rd terms, therefore it's ar and ar

^{2}so shouldn't it be :

r * log

_{2}x = r^{2}* log_{2}x? 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

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

log

r

r

r*(r*log

r=0 or r*log

How do we solve r*log

r*log

r=2/3

log

_{2}x , log_{2}4x / 2 , log_{2}4x / 3r

^{2}* log_{2}x = r*log_{2}4x / 2 = log_{2}4x / 3r

^{2}* log_{2}x = r*log_{2}4x / 2r*(r*log

_{2}x - log_{2}4x / 2)=0r=0 or r*log

_{2}x - log_{2}4x / 2=0How do we solve r*log

_{2}x - log_{2}4x / 2=0? To get the stuff involving "x" to cancel, we use the relationship r^{2}* log_{2}x=log_{2}4x / 3 -> log_{2}x=log_{2}4x / 3r and substitute log_{2}4x / 3r for log_{2}xr*log

_{2}4x / 3r - log_{2}4x / 2=0r=2/3

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

no, but what im saying is, in a, ar, ar^2, shouldnt "a" remain constant?

instead of what your doing,

r

Thanks

disregard ALL OF ABOVE

Thanks

instead of what your doing,

r

^{2}* log_{2}x = r*log_{2}4x / 2 = log_{2}4x / 3Thanks

**EDIT,**oh, wait, i c where you're coming fromdisregard ALL OF ABOVE

Thanks

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

bfrsoccer wrote:

How do we solve r*log_{2}x - log_{2}4x / 2=0? To get the stuff involving "x" to cancel, we use the relationship r^{2}* log_{2}x=log_{2}4x / 3 -> log_{2}x=log_{2}4x / 3r and substitute log_{2}4x / 3r for log_{2}x

r*log_{2}4x / 3r - log_{2}4x / 2=0

r=2/3

shouldnt r

^{2}* log

_{2}x=log

_{2}4x / 3 -> log

_{2}x=log

_{2}4x /

**3r**???

^{2}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

Yeah, we use log

I guess I did accidentally say "substitute log

_{2}4x / 3r because we are substituting for r*log_{2}x, not log_{2}x (so we only divide r^{2}* log_{2}x=log_{2}4x / 3 by r and not r^{2}).I guess I did accidentally say "substitute log

_{2}4x / 3r for log_{2}x" instead of "substitute log_{2}4x / 3r for r*log_{2}x", but I think all of the math manipulation shown is correct.**bfrsoccer**- Administrator
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## and after all these posts...

i get it now

thank you

thank you

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

Last edited by caseycolin on Wed May 04, 2011 1:42 am; edited 1 time in total

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