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[MaximumZero]

AGH, MY EYES!

[Ultimuh]

AGH, MY EYES!

More like.. brain.. unless you have an unnatural fear of math.
At any rate, my brain hurt just by casually looking at that.

[da_nang]

Spoiler (click to show/hide)

Q2:
f(x) = 3x-2
f(g(h(x))) = 3g(x/2) - 2 = f(g(x/2))
h(h(8 )) = ?

g(h(x)) = g(x/2)

h(x) = x/2

h(h(8 )) = h(4) = 2

Spoiler (click to show/hide)

Q3:
y = +-sqrt(81-x^2), x € R | -9<= x <= 9

d = sqrt((x-8)^2 + (y - 6)^2) € N

(x-8)^2 + (y - 6)^2 must be a perfect square i.e. (x-8)^2 + (y - 6)^2 = n^2 for an n € N

1. y = sqrt(81-x^2)

(x-8)^2 + (sqrt(81-x^2) - 6)^2 = x^2 - 16x + 64 + 81-x^2 - 12sqrt(81-x^2) + 36

= 181 - 16x - 12sqrt(81-x^2) = f(x)

f'(x) = -16 - 12*(-2x)/(2sqrt(81-x^2) = 12x/sqrt(81-x^2) - 16
Search for roots:
12x/sqrt(81-x^2) - 16 = 0
3x = 4sqrt(81-x^2)
9x^2 = 16(81-x^2)
25x^2  = 16*81
x^2 = 16*81/25
x = +- 4*9/5 = +- 36/5

By Fermat's theorem, global maximum and global minimum is found at -9 and 7.2 respectively. Insertion tells us f(x) is rising between 7.2 and 9, falling between -9 and 7.2.
Number of points where the distance is an integer is thusly the number of squares between (inclusive ends) between the global minimum and the global maximum plus those that are between the global minimum and local maximum at x=9.

f(-9) = 325
f(7.2) = 1
f(9) = 37

N₁ = floor(sqrt(f(-9))) - ceiling(sqrt(f(7.2))) + 1 + floor(sqrt(f(9))) - ceiling(sqrt(f(7.2))) + 1 = 18 - 1 +1 + 6 -1 +1 = 24

2. y = -sqrt(81 - x^2)

(x-8)^2 + (-sqrt(81 - x^2) - 6)^2 = (x-8)^2 + (sqrt(81 - x^2) + 6)^2 = x^2 - 16x + 64 + 81-x^2 + 12sqrt(81-x^2) + 36

= 12sqrt(81-x^2) - 16x + 181 = g(x)
g'(x) = -12x/sqrt(81-x^2) - 16 = 0
3x = -4sqrt(81-x^2)
25x^2 = 16*81
x = +-36/5

Fermat's theorem gives global maximum and global minimum at -7.2 and 9 respectively. Insertion tells us g(x) is rising between -9 and -7.2 and falling between -7.2 and 9. Number of points where the distance is an integer is thusly the number of squares between (inclusive ends) between the global minimum and the global maximum plus those that are between the global maximum and local minimum at x=-9.

g(-9) = 325
g(-7.2) = 361
g(9) = 37

N₂ = floor(sqrt(g(-7.2))) - ceiling(sqrt(g(9))) + 1 + floor(sqrt(g(-7.2))) - ceiling(sqrt(g(-9))) + 1 = 19 - 19 +1  + 19 - 7 +1 = 1 + 13 = 14

Total number of points: N₁ + N₂ = 24 + 14 = 38.

[Skyrunner]

Answer to Q3 is incorrect. Q2 is right. :D

Also, xD So much advanced math! >.>

[da_nang]

Answer to Q3 is incorrect. Q2 is right.

Also, xD So much advanced math! >.>

Dammit, I double-counted! Since f(7.2) and g(-7.2) are perfect squares means that both the left and right path share the same point.
Total number of points = 38 - 2 = 36
Right?

[Skyrunner]

Still nope.

As a hint, it was multiple choice.

• 8
• 11
• 12
• 13
• 14

[Darvi]

I get three. Welp.

Did some reworking and now I'm pretty sure the answer's eight.


Holy shit I thought of an easier way.

Spoiler (click to show/hide)

Twelve.

Absolutely. I'll write up how I figured it out once I'm in uni:V


Actually, I'll do it now.

Spoiler (click to show/hide)

A has a distance of 10 to O(0,0). Therefore it has a distance of 7 to the circle at its closest point and 13 at its furthest. That's 7 distances accounted for. But the ones from 8 to 12 count twice, once clockwise and once counterclockwise. So add another 5 and you get twelve.

Holy shit that was easy once I thought of is. But then again so is most of maths.

[da_nang]

Still nope.

As a hint, it was multiple choice.

• 8
• 11
• 12
• 13
• 14

I misread 9 as 9^2... Well that explains the madness. Yes, the answer is 12 if you count the intersections of constant functions on this relation.
Still, if the circle had a radius of 9 the number of points would be 36.

[Skyrunner]

Darvi got it right the math book way xD
da_nang gets the Advance College Mathz award, but a few points off becase calculator

I see the lines and equations problem is left untouched.
So... if you all are interested, maybe I'll post more/create a simple maths thread for the slightly bored.

On the other hand, if you are thinking something along the lines of "stop posting maths and clogging ip the terrified thread" or "screw math" do let me know xD

Anyways, my math test is in 12 hours. Terrified I am. I just need to not make mistakes, and half my problem's gone :<

[da_nang]

Darvi got it right the math book way xD
da_nang gets the Advance College Mathz award, but a few points off becase calculator

I see the lines and equations problem is left untouched.
So... if you all are interested, maybe I'll post more/create a simple maths thread for the slightly bored.

On the other hand, if you are thinking something along the lines of "stop posting maths and clogging ip the terrified thread" or "screw math" do let me know xD

Anyways, my math test is in 12 hours. Terrified I am. I just need to not make mistakes, and half my problem's gone :<

Hey, that previous method still works as long as you adjust for a radius of 3.

As for Q1, it's either a general solution (infinite amount of y-intercepts or no intercept at all) or I'm terrible at translating English math questions. Stupid language barrier. 

[Darvi]

I don't know what an intercept is so no idea about the solution.

[da_nang]

I don't know what an intercept is so no idea about the solution.

It's f(0) for a function f(x). Basically, the point at which the function intersects the y-axis.

[Darvi]

In that case, I guess it's the point (0,9)?

[Skyrunner]

I was creating a vector graphic, then I clicked "preview" and Darvi solved it xD

Eh, I suppose they weren't as hard as I thought :<

[Darvi]

I don't even know how I did it. My brain's weird like that.

Also, I think it's actually (0,17).

No. Shut up. Shut up brain. I'm trying to think and you're not helping. Stfu.