Serious Daily math problem thread

mNFwTJ3wz9

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grondilu

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Holy fuck that is complex as shit
Yeah I may have made it look more complex than necessary.

We just have to consider the angles a and b of respectively the minutes hand and the hours hand, as separating them from the midnight position. To satisfy the problem requirement, they must be opposite modulo 2 pi.

a = -b + k 2 pi

By definition we also have :

a = 2 pi t/3600 and b = 2 pi t/(12 * 3600)

so :

2 pi t/3600 = - 2 pi t/ (12 * 3600) + k 2 pi

t /3600 = - t /(12 * 3600) + k

t/3600 (1 + 1/12) = k

t = 43200/13 k

Notice that this cycle lasts precisely twelve thirtheenth of an hour. About 55 minutes.
 
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WorldYamataizer

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Problem #0006

In the diagram, GOER is a rectangle. Circles of radii 3,4,5 are drawn in as shown. What are the dimensions of the rectangle?
Edit: For clarity, each circle is tangent to anything it touches.
Caution! I think this is a challenging problem. Have fun!
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mNFwTJ3wz9

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dimensions of rect are GO = x and GR = y

lines tangent to a circle from a point are equal length
Circle 1 eqns :
x-3 + OW - 3 = GW
Circle 3 eqns :
x - 5 + y - 5 = GE (GE = (x^2 + y^2)^0.5)

Heron formula on triangle enclosing circle 2: (area of this 2 should be are of GOE - GOW)
(s *( s- a) *(s - b) *( s - c )) ^ 0.5
[some equation in terms of GW,WE,GE]
(WE = y - OW)

Triangle 1 eqn
x ^ 2 + OW ^2 = GW ^2

That gives 4 eqn, 4 variables, => solvable

OW = a
GW = b


x - 3 + a - 3 = b
(x-5 + y-5)^2 = x^2 + y^2

s = (b + y-a + (x^2 + y^2)^0.5) / 2
(s * (s-b) * (s-y+a) * (s - (x^2 + y^2)^0.5) ) = (x*y/2 - x*a/2)^2

(s * (s-b) * (s-y+a) * (s - (x^2 + y^2)^0.5) ) = x^2 * (y - a)^2 * (1/4)

x^2 + a^2 = b^2
 
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WorldYamataizer

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dimensions of rect are GO = x and GR = y

lines tangent to a circle from a point are equal length
Circle 1 eqns :
x-3 + OW - 3 = GW
Circle 3 eqns :
x - 5 + y - 5 = GE (GE = (x^2 + y^2)^0.5)

Heron formula on triangle enclosing circle 2:
(s *( s- a) *(s - b) *( s - c )) ^ 0.5
[some equation in terms of GW,WE,GE]
(WE = y - OW)

Triangle 1 eqn
x ^ 2 + OW ^2 = GW ^2

That gives 4 eqn, 4 variables, => solvable
Have fun with that :feelsmega:. At that point, i would use Wolfram Alphas system equation solver.
 
mNFwTJ3wz9

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Have fun with that :feelsmega:. At that point, i would use Wolfram Alphas system equation solver.
That was just me trying to "check" how solvable it is tbh
I'm moderately certain it will simplify later
dimensions of rect are GO = x and GR = y

lines tangent to a circle from a point are equal length
Circle 1 eqns :
x-3 + OW - 3 = GW
Circle 3 eqns :
x - 5 + y - 5 = GE (GE = (x^2 + y^2)^0.5)

Heron formula on triangle enclosing circle 2: (area of this 2 should be are of GOE - GOW)
(s *( s- a) *(s - b) *( s - c )) ^ 0.5
[some equation in terms of GW,WE,GE]
(WE = y - OW)

Triangle 1 eqn
x ^ 2 + OW ^2 = GW ^2

That gives 4 eqn, 4 variables, => solvable

OW = a
GW = b


x - 3 + a - 3 = b
(x-5 + y-5)^2 = x^2 + y^2

s = (b + y-a + (x^2 + y^2)^0.5) / 2
(s * (s-b) * (s-y+a) * (s - (x^2 + y^2)^0.5) ) = (x*y/2 - x*a/2)^2

(s * (s-b) * (s-y+a) * (s - (x^2 + y^2)^0.5) ) = x^2 * (y - a)^2 * (1/4)

x^2 + a^2 = b^2

x + a - 6 = b
let's remove b from everywhere tbh

x^2 + a^2 = (x + a - 6)^2
(x + y - 10)^2 = x^2 + y^2
(x + y - 8) * ( y - a - 2) * (x + a - 8) * (8) = x^2 * (y - a)^2 (refer below)
s = ((x + a - 6) + (y - a) + (x + y - 10))/2
s = (x + a - 6 + y - a + x + y - 10)/2
s = (2x + 2y - 16)/2
s = x + y - 8

(x + y - 8) * (x + y - 8 - x - a + 6) * (x + y - 8 - y + a) * (x + y - 8 - x - y + 10) = x^2 * (y - a)^2 * (1/4)
(x + y - 8) * ( y - a - 2) * (x + a - 8) * (8) = x^2 * (y - a)^2
eq 1
x^2 + a^2 = x^2 + a^2 + 2ax -12a -12x + 36
12a + 12x - 36 = 2ax
x + a = ax/6 - 3

eq 2
x^2 + y^2 = x^2 + y^2 + 2xy -20x -20y +100
20x + 20y - 100 = 2xy
x + y = xy/10 - 5

eq 3
(x + y - 8) * ( y - a - 2) * (x + a - 8) * (8) = x^2 * (y - a)^2
(xy/10 - 13) * (y - a - 2) * (ax/6 - 11) * 8 = x^2 * (y-a)^2
a,b,x,y = 8,17,15,20 (this gives wrong inradius for circle 2)
 
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grondilu

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prob 0006:

I got a rectangle of 12x35 using Mathematica and the formula for the inradius of a triangle (mentioned in a previous problem):

Solve[{
3^2 == ((s1 - GO) (s1 - OW) (s1 - GW))/s1,
s1 == (GO + OW + GW)/2, GW^2 == GO^2 + OW^2,
4^2 == ((s2 - GW) (s2 - EW) (s2 - EG))/s2,
s2 == (GW + EW + EG)/2, EG^2 == GO^2 + (OW + EW)^2,
5^2 == ((s3 - ER) (s3 - GR) (s3 - EG))/s3,
s3 == (ER + GR + EG)/2,
ER == GO,
GR == OW + EW
}, {s1, s2, s3, GO, OW, GW, EW, EG, GR, ER}]
 
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WorldYamataizer

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prob 0006:

I got a rectangle of 12x35 using Mathematica and the formula for the inradius of a triangle (mentioned in a previous problem):

Solve[{
3^2 == ((s1 - GO) (s1 - OW) (s1 - GW))/s1,
s1 == (GO + OW + GW)/2, GW^2 == GO^2 + OW^2,
4^2 == ((s2 - GW) (s2 - EW) (s2 - EG))/s2,
s2 == (GW + EW + EG)/2, EG^2 == GO^2 + (OW + EW)^2,
5^2 == ((s3 - ER) (s3 - GR) (s3 - EG))/s3,
s3 == (ER + GR + EG)/2,
ER == GO,
GR == OW + EW
}, {s1, s2, s3, GO, OW, GW, EW, EG, GR, ER}]
What problems can Mathematica NOT solve? I want to find a problem where you cant cheat with Mathematica. Its so broken tbh
 
mNFwTJ3wz9

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That was just me trying to "check" how solvable it is tbh
I'm moderately certain it will simplify later


x + a - 6 = b
let's remove b from everywhere tbh

x^2 + a^2 = (x + a - 6)^2
(x + y - 10)^2 = x^2 + y^2
(x + y - 8) * ( y - a - 2) * (x + a - 8) * (8) = x^2 * (y - a)^2 (refer below)
s = ((x + a - 6) + (y - a) + (x + y - 10))/2
s = (x + a - 6 + y - a + x + y - 10)/2
s = (2x + 2y - 16)/2
s = x + y - 8

(x + y - 8) * (x + y - 8 - x - a + 6) * (x + y - 8 - y + a) * (x + y - 8 - x - y + 10) = x^2 * (y - a)^2 * (1/4)
(x + y - 8) * ( y - a - 2) * (x + a - 8) * (8) = x^2 * (y - a)^2
eq 1
x^2 + a^2 = x^2 + a^2 + 2ax -12a -12x + 36
12a + 12x - 36 = 2ax
x + a = ax/6 - 3

eq 2
x^2 + y^2 = x^2 + y^2 + 2xy -20x -20y +100
20x + 20y - 100 = 2xy
x + y = xy/10 - 5

eq 3
(x + y - 8) * ( y - a - 2) * (x + a - 8) * (8) = x^2 * (y - a)^2
(xy/10 - 13) * (y - a - 2) * (ax/6 - 11) * 8 = x^2 * (y-a)^2
a,b,x,y = 8,17,15,20 (this gives wrong inradius for circle 2)
the
a,b,x,y = 9,15,12,35 got removed while editing
 
Divergent_Integral

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What problems can Mathematica NOT solve? I want to find a problem where you cant cheat with Mathematica. Its so broken tbh
I might be able to help with that, jfl.

Btw, @grondilu has provided an entirely correct solution to #0005 in posts #100 and #105. Well done!

Now we get

Problem #0007

In order to make everyone as happy as possible, I've decided to give three subproblems, each of different degree of difficulty. You are strongly encouraged to give the answer only for the most difficult problem that you can solve.

a. How many positive integers are there (in base 10) that are less than 1000 and that have no two or three digits the same.

b. I have a defective pocket calculator of which only the number keys, the parentheses keys, the minus key, and the reciprocal key ( x^(-1) ) still work. Can I somehow still perform the four basic arithmetic operations with this calculator? If yes, how?

c. A finite set S of more than three non-negative real numbers has the following properties: the sum of its members is 3, and the sum of the squares of its members is > 1. Prove that there exists a subset of S with exactly three elements and whose sum is > 1.

@WorldYamataizer @SelfCrucified @grondilu @mNFwTJ3wz9 @Diocel @mental_out @your personality @Irredeemable @Caesercel @nihility
 
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grondilu

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What problems can Mathematica NOT solve? I want to find a problem where you cant cheat with Mathematica. Its so broken tbh
People sometimes say Mathematics is a bicycle for the mind.

If so, Mathematica is a motorcycle.

You don't have to put any effort, but you still have to know where you're going.
 
WorldYamataizer

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Here is my solution to problem #0006.
@grondilu and @mNFwTJ3wz9 , good work on the painful system of equations to obtain the correct answer. My work may not contain too many words of explanation, but it should be easy to follow


People sometimes say Mathematics is a bicycle for the mind.

If so, Mathematica is a motorcycle.

You don't have to put any effort, but you still have to know where you're going.
It seems like a useful tool. I’m just amazed since ive never seen it in use before. I’m probably the “worst” at mathematics on this thread tbh, so In trying to learn from much more experienced mathcels.
 
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WorldYamataizer

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Here's a solution to problem #0007-a
Case#1: The number has one digit. Well, we have 1,2,3,4,5,6,7,8,9. --->9 integers.

Case#2: The number has two digits. The number is of the form ab, For the first position, a, we have 9 choices of non-zero digits: namely 1-9. Now for the second position, b, we have 10 choices of any digits 0-9 except a, so we have 9 choices. --->9*9=81 integers.

Case#3: The number has three digits. The number is of the form abc. For the first position, we have 9 choices of non-zero digits: namely 1-9.
Now for the second position, b, we have 10 choices of any digits 0-9 except a, so we have 9 choices. For the final digit, c, we have 10 choices of any digits 0-9 except what we chose to a and b, so we have 8 choices. --->9*9*8=
648 integers.

In total we have
9+81+648=738 positive integers which satisfy the given conditions. All results follow from the multiplicative counting principle.
 
Divergent_Integral

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Here's a solution to problem #0007-a
Case#1: The number has one digit. Well, we have 1,2,3,4,5,6,7,8,9. --->9 integers.

Case#2: The number has two digits. The number is of the form ab, For the first position, a, we have 9 choices of non-zero digits: namely 1-9. Now for the second position, b, we have 10 choices of any digits 0-9 except a, so we have 9 choices. --->9*9=81 integers.

Case#3: The number has three digits. The number is of the form abc. For the first position, we have 9 choices of non-zero digits: namely 1-9.
Now for the second position, b, we have 10 choices of any digits 0-9 except a, so we have 9 choices. For the final digit, c, we have 10 choices of any digits 0-9 except what we chose to a and b, so we have 8 choices. --->9*9*8=
648 integers.

In total we have
9+81+648=738 positive integers which satisfy the given conditions. All results follow from the multiplicative counting principle.
Entirely correct, sir!
 
mNFwTJ3wz9

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Here is my solution to problem #0006.
@grondilu and @mNFwTJ3wz9 , good work on the painful system of equations to obtain the correct answer. My work may not contain too many words of explanation, but it should be easy to follow

View attachment 387885View attachment 387886

It seems like a useful tool. I’m just amazed since ive never seen it in use before. I’m probably the “worst” at mathematics on this thread tbh, so In trying to learn from much more experienced mathcels.
My system of equations was wrong AF now that I look at it JFL.
Heron's formula should have never been used, as it does not add ANY useful information. In fact I'm puzzled how wolfram even gave an answer. (I think this could be a nice problem #0006.5 tbh, how the hell did my equations spit out any answers)

I should have used
perimiter * inradius * 0.5 = Area ((base *height) not (area1 - area2))
which makes it hand solvable

lines tangent to a circle from a point are equal length
Circle 1 eqns :
x-3 + OW - 3 = GW
Circle 3 eqns :
x - 5 + y - 5 = GE (GE = (x^2 + y^2)^0.5)

Area equation:
(GW + WE + GE) * 4 * 1/2 = WE * x * 1/2

Triangle 1 eqn:
x ^ 2 + OW ^2 = GW ^2

That gives 4 eqn, 4 variables, => solvable
OW = a
GW = b

x - 3 + a - 3 = b
(x-5 + y-5)^2 = x^2 + y^2
(y-a +b + (x^2 + y^2)^0.5) * 4 * 1/2 = (y-a)*x*1/2
x^2 + a^2 = b^2
x + a - 6 = b
let's remove b from everywhere

(x + y - 10)^2 = x^2 + y^2

(y - a + x + a - 6 + (x^2 + y^2)^0.5) * 4 * 1/2 = (y-a)*x*1/2

x^2 + a^2 = (x + a - 6)^2
20x + 20y - 100 = 2xy

(y + x - 6 + x + y - 10) * 4 * 1/2 = (y-a)*x*1/2

12a + 12x - 36 = 2ax
10x + 10y - 50 = xy

8x + 8y - 64 = xy-ax

6a + 6x - 18 = ax
10x + 10y - 50 = xy

8x + 8y - 64 = 10x + 10y - 50 - 6a - 6x + 18

6x - 18 = a*(x - 6)
10*(x - 5) = (x-10)*y
32 = 4x - 2y + 36*(x - 3)/(x-6)
32 = 4x + 36*(x - 3)/(x-6) - 20 * (x - 5)/(x-10)
-------------------------------------------------------------
4*(8-x)*(x-6)*(x-10) = 36*(x-3)(x-10) - 20*(x-5)*(x-6)
-------------------------------------------------------------
-4x^3 + 80x^2 -504x + 1440 = 0
-------------------------------------------------------------
[Solvable manually tbh, but using wolfram anyways]
x = 12
x = 4 + i*(14^0.5)
x = 4 - i*(14^0.5)
 
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Yeah this has gotten too complicated for me to follow tbh
 
WorldYamataizer

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My system of equations was wrong AF now that I look at it JFL.
Heron's formula should have never been used, as it does not add ANY useful information. In fact I'm puzzled how wolfram even gave an answer. (I think this could be a nice problem #0006.5 tbh, how the hell did my equations spit out any answers)

I should have used
perimiter * inradius * 0.5 = Area ((base *height) not (area1 - area2))
which makes it hand solvable

lines tangent to a circle from a point are equal length
Circle 1 eqns :
x-3 + OW - 3 = GW
Circle 3 eqns :
x - 5 + y - 5 = GE (GE = (x^2 + y^2)^0.5)

Area equation:
(GW + WE + GE) * 4 * 1/2 = WE * x * 1/2

Triangle 1 eqn:
x ^ 2 + OW ^2 = GW ^2

That gives 4 eqn, 4 variables, => solvable
OW = a
GW = b

x - 3 + a - 3 = b
(x-5 + y-5)^2 = x^2 + y^2
(y-a +b + (x^2 + y^2)^0.5) * 4 * 1/2 = (y-a)*x*1/2
x^2 + a^2 = b^2
x + a - 6 = b
let's remove b from everywhere

(x + y - 10)^2 = x^2 + y^2

(y - a + x + a - 6 + (x^2 + y^2)^0.5) * 4 * 1/2 = (y-a)*x*1/2

x^2 + a^2 = (x + a - 6)^2
20x + 20y - 100 = 2xy

(y + x - 6 + x + y - 10) * 4 * 1/2 = (y-a)*x*1/2

12a + 12x - 36 = 2ax
10x + 10y - 50 = xy

8x + 8y - 64 = xy-ax

6a + 6x - 18 = ax
10x + 10y - 50 = xy

8x + 8y - 64 = 10x + 10y - 50 - 6a - 6x + 18

6x - 18 = a*(x - 6)
10*(x - 5) = (x-10)*y
32 = 4x - 2y + 36*(x - 3)/(x-6)
32 = 4x + 36*(x - 3)/(x-6) - 20 * (x - 5)/(x-10)
-------------------------------------------------------------
4*(8-x)*(x-6)*(x-10) = 36*(x-3)(x-10) - 20*(x-5)*(x-6)
-------------------------------------------------------------
-4x^3 + 80x^2 -504x + 1440 = 0
-------------------------------------------------------------
[Solvable manually tbh, but using wolfram anyways]
x = 12
x = 4 + i*(14^0.5)
x = 4 - i*(14^0.5)
This is somewhat of a brute force technique, but I’m pleased that it actually works out:feelsokman:
I often resort to brute force techniques to get an answer, and then look for a more elegant, aesthetic approach.

In the problem, the circles of radii 3,4, and 5 suggests that a 3-4-5 right triangle may be hidden somewhere. Once you find it, the problem falls apart:feelsthink:
 
mNFwTJ3wz9

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This is somewhat of a brute force technique, but I’m pleased that it actually works out:feelsokman:
I often resort to brute force techniques to get an answer, and then look for a more elegant, aesthetic approach.

In the problem, the circles of radii 3,4, and 5 suggests that a 3-4-5 right triangle may be hidden somewhere. Once you find it, the problem falls apart:feelsthink:
Makes sense, in general I don't always get good numbers or even solvable problems. So I'm more interested if the number of variables = number of equations, and if it can be solved.

I'm still trying to figure out why my initial solution gave me ANY answers tbh.
 
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engineercel, still everyone iq mogs me here
 
mNFwTJ3wz9

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b is pretty easy tbh
c. A finite set S of more than three non-negative real numbers has the following properties: the sum of its members is 3, and the sum of the squares of its members is > 1. Prove that there exists a subset of S with exactly three elements and whose sum is > 1.
Sort in decending ->
x1 + x2 + x3 = a
x4 + ... + xn = b
a + b = 3 or b = 3 - a

x4,x5...,xn < a/3
b < (a/3) * (n-3)
3 - a < (a/3) * (n-3)
9 - 3a < (a) * (n-3)
9 < a * n
a > 9/n

so a > 1 is definitely true for n<=9

let's see what happens if we do squares
sq(x1) + sq(x2) + sq(x3) = c
sq(x4) + ...... + sq(xn) = d
c + d > 1 or d > 1 - c

d < (c/3) * (n-3)
1 - c < (c/3) * (n-3)
3 - 3c < nc - 3c
3 < nc
c > 3/n

fuck man why do I always do these in the night and fuck up

Lets try some other logic. We want the sum of the top 3 to be minimum while maintaining minimum sum of squares as 1 and total sum as 3.

lets say all elements are equal at initial at 3/n, and then we change every element by yi, such that sum of y is 0.
We also sort y as ascending.
The sum of squares of x increases by sum of squares of y.
lets reuse c and d as variables

sq(y1) + sq(y2) + sq(y3) = c
sq(y4) + ...... + sq(yn) = d
c + d > 1 - 9/n or d > 1 - 9/n - c

wait I think this solves it


d < (c/3) * (n-3)
1 - c - 9/n < (c/3) * (n-3)
3 - 3c - 27/n < nc - 3c
3 - 27/n < nc
c > 3/n - 27/(n^2)

lol no that failed.

we need to prove y1 + y2 + y3 > 1 - 9/n, given c + d > 1 - 9/n

fuck man sets were never my strong point
It's over for you on a quantum level if you think math is fun ngl
tbhngldedsrsgoodnight
 
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Reformulated problem.
Alright, so we have numbers y1, y2, y3 ......yn in descending order.
sum of these is 0
sum of squares is > 1 - 9/n
we need to prove that sum of y1 + y2 + y3 > 1-9/n
and n>9

y1 + y2 + y3 = a
y4 + ..... + yn = -a
sq(y1) + sq(y2) + sq(y3) = b
sq(y4) + ...... + sq(yn) = c
-3/n<yi<1-3/n
b + c > 1 - 9/n


So we want to minimize a, given these constraints.

given some value of x3, in the last values we should use only x3 and -3/n and maybe 1 other number to for compensating for a to maximize c


fuck this does not make any sense. I'll just do part b
 
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For part b : -
i = sqrt(-1)
so let us assume a case where a = i and b = i
in the complex plane, an inverse is basically a 180 degree rotation and a magnitude flip.
let us assume we have all possible multiples of i available. None of these can be added to give a number not on the imaginary axis, and inverting the magnitude with a 180 rotation does not give us any new numbers to work with.
Therefore there is no formula, involving only a,b,inverse and subtraction that yields a+b
@Divergent_Integral is this correct?
 
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For part b : -
i = sqrt(-1)
so let us assume a case where a = i and b = i
in the complex plane, an inverse is basically a 180 degree rotation and a magnitude flip.
let us assume we have all possible multiples of i available. None of these can be added to give a number not on the imaginary axis, and inverting the magnitude with a 180 rotation does not give us any new numbers to work with.
Therefore there is no formula, involving only a,b,inverse and subtraction that yields a+b
@Divergent_Integral is this correct?

I was assuming that it's a calculator with only the conventional number keys. So real numbers only. (I should have mentioned that.)
 
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I was assuming that it's a calculator with only the conventional number keys. So real numbers only. (I should have mentioned that.)
No I'm saying that it doesn't matter if a and b are real or complex. I just proved that there is no possible formula that involves only a, b, subtraction and inverse that result in a*b

I also typo'd that pretty badly, it should be:
Therefore there is no formula, involving only a,b,inverse and subtraction that yields a*b
 
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Solutions to Problem #0007

a.
See post #114 by @WorldYamataizer

b. Addition is easy to derive: x + y = x - (0 - y). Multiplication is a bit harder. We start by observing that 1 / (x - x^2) = 1 / (1 - x) - 1 / (0 - x). By taking the multiplicative inverse we get x - x^2, from which we can subtract x to get -x^2. Next we obtain x^2 = 0 - (-x^2). Now that we have found a way to square a number, the next step is to observe that -2xy = (x - y)^2 - x^2 - y^2. Then, in order to get xy, we only need to divide out the factor -2, which can be done as follows: u / (-2) = 1 / ((-2) / u) = 1 / ( (0 - 1/u) - 1/u).

c. Suppose that S contains n > 3 numbers. Label these numbers with indices 1, 2, 3, ... , n, in such a way that x_1 >= x_2 >= x_3 >= .... >= x_n. We're going to prove by reductio ad absurdum that x_1 + x_2 + x_3 > 1. For suppose to the contrary that x_1 + x_2 + x_3 <= 1. Then

x_1 + x_2 + x_3 - (x_1 - x_3)(1 - x_1) - (x_2 - x_3)(1 - x_2) <= 1,

because the two products have only non-negative factors. But we also see that

x_1 + x_2 + x_3 - (x_1 - x_3)(1 - x_1) - (x_2 - x_3)(1 - x_2) =
x_1^2 + x_2^2 + (3 - x_1 - x_2)x_3.

So we have x_1^2 + x_2^2 + (3 - x_1 - x_2)x_3 <= 1, whence

x_1^2 + x_2^2 + x_3(x^3 + x_4 + ... + x_n) <= 1, which would imply

x_1^2 + x_2^2 + x_3^2 + x_4^2 + ... x_n^2 <= 1, contradicting the assumption that the sum of the squares of the members of S is > 1.

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Addition is easy to derive: x + y = x - (0 - y). Multiplication is a bit harder. We start by observing that 1 / (x - x^2) = 1 / (1 - x) - 1 / (0 - x). By taking the multiplicative inverse we get x - x^2, from which we can subtract x to get -x^2. Next we obtain x^2 = 0 - (-x^2). Now that we have found a way to square a number, the next step is to observe that -2xy = (x - y)^2 - x^2 - y^2. Then, in order to get xy, we only need to divide out the factor -2, which can be done as follows: u / (-2) = 1 / ((-2) / u) = 1 / ( (0 - 1/u) - 1/u).
fuck I did not think about using numbers like 1 and 2. Good problem tbh.
 
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Good problems good problems. I like that both the remaining problems for me only required insightful, clever moves.
 
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I like that both the remaining problems for me only required insightful, clever moves.
I'm not going to accept that tbh
problem c solution
fuck I think I was a bit close with my initial solution

Sort in decending ->
x1 + x2 + x3 = a
x4 + ... + xn = b
a + b = 3 or b = 3 - a
sq(x1) + sq(x2) + sq(x3) = c
sq(x4) + ...... + sq(xn) = d
c + d > 1

d < x3*b
x3 < a/3
d < a*b/3

if a < 1
c < a^2

c + d > 1
a^2 + (a*b)/3 > 1
a^2 + (a * (3 - a))/3 > 1
3*a^2 + 3a - a^2 > 3
2a^2 + 3a > 3
a^2 + 1.5a - 1.5 > 0
a > 1/4 ( sqrt(33) - 3 ) = 0.686
FUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUU JUST LOL

wait no.
We just need a better bound on c.
I'm going to cheat a little bit and use wolfram alpha ngl
So after some checking we need to bound c with a^2 - 2/3*a
we have to prove given x+y+z < 1

x^2 + y^2 + z^2 < (x + y + z)^2 - 2/3*(x + y + z)
or
2xy + 2yz + 2xz < 2/3*(x + y + z)
or
3xy + 3yz + 3xz < (x + y + z)
or
x(z + y - 1) + y(x + z - 1) + z(x + y - 1) + xy + yz + xz < 0
we know (z + y - 1) < -x , etc.
so finally :
xy + yz + xz < x^2 + y^2 +z^2
THIS IS TRUE [consider (x-y)^2 + (y-z)^2 + (z-x)^2 > 0]
FUCK YEAH, YOU DON'T NEED INSIGHT.
solved 3 without insight @WorldYamataizer
 
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Problem #0008

A square is drawn inside a rectangle of length x and width y, with one vertex of the square on the diagonal of the rectangle, as shown. If the square has side z, prove that
(1/z)=(1/x)+(1/y)

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Problem #0008
View attachment 389336
A square is drawn inside a rectangle of length x and width y, with one vertex of the square on the diagonal of the rectangle, as shown. If the square has side z, prove that
(1/z)=(1/x)+(1/y)

@Divergent_Integral @nihility @grondilu @mNFwTJ3wz9 @Diocel @mental_out @your personality @Irredeemable @Caesercel @SelfCrucified
Couldn't solve it tbh
 
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Problem #0008
View attachment 389336
A square is drawn inside a rectangle of length x and width y, with one vertex of the square on the diagonal of the rectangle, as shown. If the square has side z, prove that
(1/z)=(1/x)+(1/y)

@Divergent_Integral @nihility @grondilu @mNFwTJ3wz9 @Diocel @mental_out @your personality @Irredeemable @Caesercel @SelfCrucified

If we consider the other rectangle, the one on the other side of the diagonal, it's not difficult to see it's necessarily of the same area than the square. You just have to shift triangles around. Or if you prefer, just notice that this common area measure is xy/2 - (x-z)z/2 - (y-z)z/2.

So, by decomposing the area of the rectangle, xy, into its different subparts :

xy = (x-z)z + (y-z)z + 2z²

and thus :

xy = xz -z² + yz - z² + 2z²
xy = xz + yz
xy = (x+y)z
z = xy/(x+y)
(1/z) = (x+y)/(xy) = (1/x) + (1/y)
 
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If we consider the other rectangle, the one on the other side of the diagonal, it's not difficult to see it's necessarily of the same area than the square. You just have to shift triangles around. Or if you prefer, just notice that this common area measure is xy/2 - (x-z)z/2 - (y-z)z/2.

So, by decomposing the area of the rectangle, xy, into its different subparts :

xy = (x-z)z + (y-z)z + 2z²

and thus :

xy = xz -z² + yz - z² + 2z²
xy = xz + yz
xy = (x+y)z
z = xy/(x+y)
(1/z) = (x+y)/(xy) = (1/x) + (1/y)
Nicely done, grondilu. I liked that you used areas.

Anyone else that wants to give the problem a try, but wants to use a different method, try using similar triangles
 
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If we consider the other rectangle, the one on the other side of the diagonal, it's not difficult to see it's necessarily of the same area than the square. You just have to shift triangles around. Or if you prefer, just notice that this common area measure is xy/2 - (x-z)z/2 - (y-z)z/2.

So, by decomposing the area of the rectangle, xy, into its different subparts :

xy = (x-z)z + (y-z)z + 2z²

and thus :

xy = xz -z² + yz - z² + 2z²
xy = xz + yz
xy = (x+y)z
z = xy/(x+y)
(1/z) = (x+y)/(xy) = (1/x) + (1/y)
I did it like this: choose a Cartesian coordinate system OXY with origin O in the lower left-hand corner of the rectangle and the X- and Y-axes oriented in the usual way. Then the equation of the diagonal is Y = x/y X, and the lower right-hand corner of the square is at (z, xz/y). Hence we have z + xz/y = x, from which follows z = xy / (x + y), etc.
 
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Anyone else that wants to give the problem a try, but wants to use a different method, try using similar triangles
Ok. I made this figure:

First, to show that ABC and ADE are similar triangles, it is easy to see that they have the same measurements for their angles. E is a right angle by the definition of a rectangle, and B is a right angle as it is supplementary to the right angle of the red square. The diagonal of the rectangle makes equal angles BAC and ADE. Thus, angles B = E, BAC = ADE, and lastly ACB = DAE.

The corresponding sides of similar triangles are proportional; triangles ABC and DEA are proportional:
(y - z) * k = y and
k * z = x, by dividing both we have:
(y - z) / z = y / x
==> y / z - 1 = y / x
divide both sides by y and rearrange terms to get
1 / z = 1 / x + 1 / y.

btw if i made any errors in my language (eg using equal instead of congruent) I would appreciate any criticism.
 
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The corresponding sides of similar triangles are proportional; triangles ABC and DEA are proportional:
(y - z) * k = y and
k * z = x, by dividing both we have:
(y - z) / z = y / x

It's a bit messy. I think we can just apply Thales' theorem in the upper left triangle. Amazingly, that's all we need.

z/x = (y-z)/y

zy = x(y-z)
zy + xz = xy

etc.
 
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Ok. I made this figure:
View attachment 389368
First, to show that ABC and ADE are similar triangles, it is easy to see that they have the same measurements for their angles. E is a right angle by the definition of a rectangle, and B is a right angle as it is supplementary to the right angle of the red square. The diagonal of the rectangle makes equal angles BAC and ADE. Thus, angles B = E, BAC = ADE, and lastly ACB = DAE.

The corresponding sides of similar triangles are proportional; triangles ABC and DEA are proportional:
(y - z) * k = y and
k * z = x, by dividing both we have:
(y - z) / z = y / x
==> y / z - 1 = y / x
divide both sides by y and rearrange terms to get
1 / z = 1 / x + 1 / y.

btw if i made any errors in my language (eg using equal instead of congruent) I would appreciate any criticism.
Easiest solution to follow so far.
I was getting stuck with square root terms because of using the Pythagorean Theoreom on all the triangles and trying to solve for x, y and z in relation to each other that way.
 
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Foid * dreams = Chad's cock
 
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Problem #0009

a (easy).
Assuming the addition formulae for the sine and cosine, prove that tan(x/2) = (1 - cos x) / sin x.

b (medium). Using some or all of the four basic arithmetic operations, construct a binary operation x * y such that all of the four basic arithmetic operations on x and y may be expressed in terms of just x, y, * and as many parentheses as needed. (Hint: you may use Problem #0006(b), see posts #111 and #128.)

c (hard). Let S be a countably infinite set of points in the Euclidean plane, and suppose that the distance between any two of these points is always an integer. Prove that S is a subset of a straight line.
 
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c (hard). Let S be a countably infinite set of points in the Euclidean plane, and suppose that the distance between any two of these points is always an integer. Prove that S is a subset of a straight line.
Going for this
Let's just shift all the points such that one of them is on 0,0
And then rotate so that one of the points is on the x axis
so far, the set contains (0,0) , (0,x0)
let us add (y1,x1) to the set and see what kind of equations we get
(y1^2 + x1^2)^0.5 = int
no wait that goes nowhere
no actually this is legit
so if we add some point not on the x axis, then the number of points on the x axis is limited
this can be proved I think
so consider a very far off point on the x axis
coordinate (0,x2) x2 is an integer obviously
so the distance between the far off point and the point not on the x axis has to be atleast abs(x1 - x2)
Fuck I'm sleepy, goodnight.
 
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Problem #0009 a
Assuming the addition formulae for the sine and cosine, prove that tan(x/2) = (1 - cos x) / sin x.
Solution

As a consequence of the addition formulae for sin and cos, sin(x)=2sin(x/2)cos(x/2) and cos(x)=(cos(x/2))^2-(sin(x/2))^2

Therefore, (1-cos(x))/sin(x)=(1-((cos(x/2))^2-(sin(x/2))^2))/2sin(x/2)cos(x/2)=(2(sin(x/2))^2)/(2sin(x/2)cos(x/2))=sin(x/2)/cos(x/2)=tan(x/2)

As required.
 
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I wis i knew that a ibnarty operarin is.
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Going for this
Let's just shift all the points such that one of them is on 0,0
And then rotate so that one of the points is on the x axis
so far, the set contains (0,0) , (0,x0)
let us add (y1,x1) to the set and see what kind of equations we get
(y1^2 + x1^2)^0.5 = int
no wait that goes nowhere
no actually this is legit
so if we add some point not on the x axis, then the number of points on the x axis is limited
this can be proved I think
so consider a very far off point on the x axis
coordinate (0,x2) x2 is an integer obviously
so the distance between the far off point and the point not on the x axis has to be atleast abs(x1 - x2)
Fuck I'm sleepy, goodnight.
continuing from here
so given that minimum distance between the far off point and our y point has to be larger than abs(x1 - x2), but also has to be integer.
int(abs(x1-x2)) + 1 is therefore the new minimum distance, between our point and the y point.
 
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I wis i knew that a ibnarty operarin is.
I believe binary refers to the number of 'arguments' or 'parameters' being operated on.

I will bring together in this post the appropriate problem/solution that @Divergent_Integral referred to earlier:
b. I have a defective pocket calculator of which only the number keys, the parentheses keys, the minus key, and the reciprocal key ( x^(-1) ) still work. Can I somehow still perform the four basic arithmetic operations with this calculator? If yes, how?
b. Addition is easy to derive: x + y = x - (0 - y). Multiplication is a bit harder. We start by observing that 1 / (x - x^2) = 1 / (1 - x) - 1 / (0 - x). By taking the multiplicative inverse we get x - x^2, from which we can subtract x to get -x^2. Next we obtain x^2 = 0 - (-x^2). Now that we have found a way to square a number, the next step is to observe that -2xy = (x - y)^2 - x^2 - y^2. Then, in order to get xy, we only need to divide out the factor -2, which can be done as follows: u / (-2) = 1 / ((-2) / u) = 1 / ( (0 - 1/u) - 1/u).
 
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continuing from here
so given that minimum distance between the far off point and our y point has to be larger than abs(x1 - x2), but also has to be integer.
int(abs(x1-x2)) + 1 is therefore the new minimum distance, between our point and the y point.
however at some point, the y coordinate will not be able to satisfy the Pythagoras theorem.
(int (abs(x1 - x2)) + 1 )**2 > y1**2 + (x1 - x2)**2
let's say abs(x1 - x2) = a + b
a = integer part, b = fractional part or 0, never 1
(a + 1)^2 < y1^2 + (a + b)^2
a^2 + 1 + 2a < y1^2 + a^2 + b^2 + 2ab
1 + 2a - b^2 - 2ab < y1^2
(1 + b)*( 1 - b) + 2a * (1 - b) < y1^2
(1 + 2a + b) * (1 - b) < y1^2

abs(x1 - x2) < ((y^2)/(1 - b)) - 1 + b
putting a finite bound on how far the point can be on the x axis.

Ok but how the fuck does this help
Ok no wait
So Let us assume an arbitrary triangle(with integer sides and non zero angles)
so we have (0,0), (0,x0) and (y1,x1)

and a point (y2, x2), such that it's distance from (0,0) is large
Ok I'll get back to this in some time.
 
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Alright, I've had some food and ready to take a shot at this
so y2,x2 is far away, and we need to prove that integer distances fail
 

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