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

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

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

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

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

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

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
x^1/3 = 4x / x
x^1/3 = 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

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

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.

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

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

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

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, ... ar^n-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

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

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

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?

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.

i get it now

thank you