Serious Daily math problem thread

[mNFwTJ3wz9]

Fuck I'm going to start using proper coordinate systems, I don't know why the fucked it so badly.

So Let us assume an arbitrary triangle(with integer sides and non zero angles)
so we have (0,0), (x0,0) and (x1,y1)

x0 > 0 for convenience

and a point (x2,y2), such that it's distance from (0,0) is large

Case 1 -
lets say x2 > x0 WLOG

so both (x2^2 + y2^2)^0.5 and ((x2-x0)^2 + y2^2)^0.5 are integers

but this puts a bound on how high (x2^2 + y2^2)^0.5 can be, because the minimum difference between these distances has to be 1.
(x2^2 + y2^2)^0.5 >= ((x2-x0)^2 + y2^2)^0.5 + 1

squaring both sides :
(x2^2 + y2^2) >= ( (x2-x0)^2 + y2^2) + 1 + 2 * ( (x2-x0)^2 + y2^2)^0.5
x2^2 + y2^2 >= x2^2 - 2*x2*x0 + x0^2 + y2^2 +1 + 2 * ( (x2-x0)^2 + y2^2)^0.5
2*x2*x0 - x0^2 - 1 >= 2 * ( (x2-x0)^2 + y2^2)^0.5
squaring both sides again
1 + 2*x0^2 + x0^4 - 4*x0*x2 - 4 x0^3*x2 + 4*x0^2*x2^2 >= 4 * (y2^2 + x2^2 + x0^2 - 2*x0*x2)
oh fuck this sucks
x0^4 - 4*x0^3*x2 + 4*x0^2*x2^2 - 2*x0^2 + 4*x0*x2 - 4*x2^2 - 4*y2^2 + 1 >= 0
No wait, x0^2 - 1 can be taken common
(x0^2 - 1)*(x0 - 2*x2 - 1)*(x0 - 2*x2 + 1) >= 4 y2^2
FUUUUUCK YEAAHHH
We can now bound y2^2

 

[mNFwTJ3wz9]

Oh man I don't think this is good
So given 2 points we have a fucked up bound on hieght
and then we have the same bound ROTATED and TWICE
Oh man brute forcing this is super bad.
Unless
Oh wait
I'll just tamper with the bounds a little,
(x0^2 - 1)^0.5 * x2 > y2
obviously y > 0 WLOG
So we can exclude these regions for any 2 points

 

[Divergent_Integral]

I wis i knew that a ibnarty operarin is.
@Divergent_Integral
U are my idol.
HWhn i make wonderlan, you are invited

Thanks, brocel. A binary operation is a function with two arguments, such as addition.

 

[WorldYamataizer]

Problem #0010

While Lino was simplifying the fraction (A^3+B^3)/(A^3+C^3) he cancelled the threes (A^3+B^3)/(A^3+C^3)
to obtain the fraction (A+B)/(A+C). If B ≠ C, determine a necessary and sufficient condition on A,B and C for Lino's
method to actually yield the correct answer, i.e.. for (A^3+B^3)/(A^3+C^3)=(A+B)/(A+C).

@Divergent_Integral @SelfCrucified @grondilu @mNFwTJ3wz9 @Diocel @mental_out @Irredeemable @Caesercel @nihility @Liszt

 

[grondilu]

prob #0010 can be solved by Mathematica, ngl :

Spoiler: Mathematica solution

$ wolframscript
Wolfram Language 12.1.1 Engine for Linux x86 (64-bit)
Copyright 1988-2020 Wolfram Research, Inc.

In[1]:= Solve[(A^3+B^3)/(A^3+C^3) == (A+B)/(A+C)]

Out[1]= {{B -> -A}, {C -> A - B}, {C -> B}}

 

[mNFwTJ3wz9]

Problem #0010

While Lino was simplifying the fraction (A^3+B^3)/(A^3+C^3) he cancelled the threes (A^3+B^3)/(A^3+C^3)
to obtain the fraction (A+B)/(A+C). If B ≠ C, determine a necessary and sufficient condition on A,B and C for Lino's
method to actually yield the correct answer, i.e.. for (A^3+B^3)/(A^3+C^3)=(A+B)/(A+C).

@Divergent_Integral @SelfCrucified @grondilu @mNFwTJ3wz9 @Diocel @mental_out @Irredeemable @Caesercel @nihility @Liszt

(A^3+B^3)/(A^3+C^3)=(A+B)/(A+C)
case 1 : numerator = 0 or Denominator = 0 or both
A = -B or A = -C or both

Otherwise :
(A^3+B^3) * (A + C) = (A^3+C^3) * (A + B)
A^4 + A^3 * C + B^3*A + B^3*C = A^4 + A^3 * B + C^3*A + C^3*B
A^3 * (C-B) + A * (B^3 - C^3)+ B^3*C = B*C( C^2 - B^2)

A^3 * (C-B) + A * (B^3 - C^3)+ B^3*C = B*C( C^2 - B^2)

dividing by B-C everywhere and B=C is a solution

-A^3 + A * (B^2 + BC + C^2) + B*C(B+C) = 0
A*(B^2 + C^2 - A^2) + B*C*(A + B + C) = 0
Now Apply Quadratic formula on B or C to get some really fucked Values I guess

Wolfram Language 12.1.1 Engine for Linux x86 (64-bit)
Copyright 1988-2020 Wolfram Research, Inc.

In[1]:= Solve[(A^3+B^3)/(A^3+C^3) == (A+B)/(A+C)]

Out[1]= {{B -> -A}, {C -> A - B}, {C -> B}}

Probably will get these
OK yeah it does simplify to :
-(A + B)*(A + C)*(A - B - C) = 0

 

[WorldYamataizer]

Useful Fact for Problem #0010

a^3+b^3=(a+b)*(a^2-ab+b^2) for all real numbers a and b.

Probably will get these
OK yeah it does simplify to :
-(A + B)*(A + C)*(A - B - C) = 0

Nice algebraic persistence.
EDIT: Your brute force methods never fail to impress me

 

[mNFwTJ3wz9]

EDIT: Your brute force methods never fail to impress me

Thanks man.

I think brute force is really nice because even the proof for Riemann hypothesis can be brute forced in finite time (Even if it is BB(700 something))

Sadly I don't have enough persistence to continue with 9(c) The equations after rotation are just too bad, and the way of forcing integer values on the resultant equations are not obvious.

 

[mNFwTJ3wz9]

Sadly I don't have enough persistence to continue with 9(c) The equations after rotation are just too bad, and the way of forcing integer values on the resultant equations are not obvious.

Incel trait : Too busy having sex to bruteforce math equations.

@Diocel @mental_out @Irredeemable @Caesercel @nihility @Liszt

Truecel trait : Can't even take a 2 min break from seducing jb's to look at the math problems.

@Divergent_Integral @WorldYamataizer
Chad trait : Can't have sex, so cope with making math problems to solve with incels.

 

[Divergent_Integral]

Solutions to Problem #0009

a. See the solution by @WorldYamataizer, post #144.

b. Using the result of Problem #0006(b), we only need to find a binary operation that expresses "minus" and "reciprocal", because we can express the four basic arithmetic operations in terms of those. In a sense both minus and reciprocal are self-inverse (conveniently disregarding the fact that reciprocal is unary and minus is binary). It seems reasonable then to combine them into one operation. One of the simplest such combinations is 1 / (x - y). And indeed x * y = 1 / (x - y) is immediately seen to work, since x * 0 = 1 / x and (x * y) * 0 = x - y.

c. Let A and B be two points in the set S. Then, with d denoting Euclidean distance, for all points C that belong S, we have d(A,C) - d(B,C) = an integer <= d(A,B), because of the triangle inequality and the assumption that all mutual distances be integers. Using the definition of a hyperbola as a locus of points such that the difference of distances from two foci is a constant, we thus see that C must lie on one element of a finite set of hyperbolas with one of their axes along line AB.

We will now argue by reductio ad absurdum. Suppose there is a point D of S that does not lie on the line AB. Then, by similar reasoning as before, all points C of S must lie on a finite set of hyperbolas with one of their axes along line BD. Thus we see that all points C of S must lie on the intersections of finitely many distinct hyperbolas. But that would make S a finite set, contrary to assumption.

 

[mNFwTJ3wz9]

c. Let A and B be two points in the set S. Then, with d denoting Euclidean distance, for all points C that belong S, we have d(A,C) - d(B,C) = an integer <= d(A,B), because of the triangle inequality and the assumption that all mutual distances be integers. Using the definition of a hyperbola as a locus of points such that the difference of distances from two foci is a constant, we thus see that C must lie on one element of a finite set of hyperbolas with one of their axes along line AB.

We will now argue by reductio ad absurdum. Suppose there is a point D of S that does not lie on the line AB. Then, by similar reasoning as before, all points C of S must lie on a finite set of hyperbolas with one of their axes along line BD. Thus we see that all points C of S must lie on the intersections of finitely many distinct hyperbolas. But that would make S a finite set, contrary to assumption.

Ok I'm going to need some time to process that

 

[Babica Yaga]

Do some statistics problems

 

[WorldYamataizer]

Solutions to Problem #0010

Additionally, see post #156 by @mNFwTJ3wz9 for his solution.

 

[JosefMengelecel]

a = 0.2
b = 0.3
c = 0.5

 

[mNFwTJ3wz9]

Solutions to Problem #0010

High IQ

 

[RainerNLM]

 

[WorldYamataizer]

High IQ

Thx!!!

 

[hikicel]

stem is for faggots

 

[WorldYamataizer]

Problem #0011

Welcome to .co Math Club! Here's a problem to work on for today.

A diameter AB of a circle intersects a chord CD at the point E. If CE=7, DE=1 and ∠BED=pi/4 (45 degrees), then determine the radius of the circle.

@Divergent_Integral
@SelfCrucified
@grondilu
@mNFwTJ3wz9
@Diocel
@mental_out
@Irredeemable
@Caesercel
@nihility
@Liszt

 

[mNFwTJ3wz9]

A diameter AB of a circle intersects a chord CD at the point E. If CE=7, DE=1 and ∠BED=pi/4 (45 degrees), then determine the radius of the circle.

isnt there some theorem about the multiplication of the sections of the chords being some fixed value?

 

[WorldYamataizer]

View attachment 392257

Yes, that diagram works. What program do u use to make these pics? I use desmos tbh

 

[mNFwTJ3wz9]

Fuck it, let's do it the old way

Yes, that diagram works. What program do u use to make these pics? I use desmos tbh

MS paint lmao

eqn 1
x1^2 + y1^2 = x2^2 + y2^2 = (x3 + y3)^2
now let's do the areas of triangles

 

[WorldYamataizer]

isnt there some theorem about the multiplication of the sections of the chords being some fixed value?

Yes
Maybe this wikipedia article has the one youre thinking of

Intersecting chords theorem - Wikipedia

en.m.wikipedia.org

 

[mNFwTJ3wz9]

Yes
Maybe this wikipedia article has the one youre thinking of

Intersecting chords theorem - Wikipedia

en.m.wikipedia.org

I'll look it up after tbh
Let's just try to brute force for now, since it seems peaceful enough
1/2 * sin(135) * y3 * 7 + 1/2 * sin(45) * x3 * 7 = x1* y1 * 1/2
1/2 * sin(45) * y3 * 1 + 1/2 * sin(135) * x3 * 1 = x2* y2 * 1/2
lets simplify these 2 for now
since sin 135 and sin 45 are 1/sqrt(2)
x1*y1*sqrt(2) = 7y3 + 7x3
x2*y2*sqrt(2) = y3 + x3
lol I think this just derived the chord theorem
x1*y1 = 7*x2*y2

adding some Pythagoras stuff with the "height"
x1^2 = (x3 - 7/sqrt2)^2 + ( 7/sqrt2)^2
y1^2 = (y3 + 7/sqrt2)^2 + ( 7/sqrt2)^2


x2^2 = (x3 + 1/sqrt2)^2 + ( 1/sqrt2)^2
y2^2 = (y3 - 1/sqrt2)^2 + ( 1/sqrt2)^2
So the complete list of equations so far is :

x1^2 + y1^2 = x2^2 + y2^2 = (x3 + y3)^2
x1*y1 = 7*x2*y2 = (y3 + x3) * 7/sqrt2
x1^2 = (x3 - 7/sqrt2)^2 + ( 7/sqrt2)^2
y1^2 = (y3 + 7/sqrt2)^2 + ( 7/sqrt2)^2
x2^2 = (x3 + 1/sqrt2)^2 + ( 1/sqrt2)^2
y2^2 = (y3 - 1/sqrt2)^2 + ( 1/sqrt2)^2

which looks like it should be enough...
Lets start with
x1^2 + y1^2 = (x3 + y3)^2
(x3 - 7/sqrt2)^2 + ( 7/sqrt2)^2 + (y3 + 7/sqrt2)^2 + ( 7/sqrt2)^2 = (x3 + y3)^2
- 7 * sqrt2 * x3 + 49/2 + 49/2 + 7 * sqrt2 * y3 + 49/2 + 49/2 = 2 * x3 * y3
An nice, I think this somewhat solves things

 

[grondilu]

Prob 0011. We can take E as an origin and fix C and D in a reference frame where AB will be on the horizontal axis (so we can use the 45° constraint). Let's note r2 the square root of 2.

C : (-r2/2*7, -r2/2*7)
E : (0, 0)
D : (r2/2, r2/2)

We just need to search the center of the circle, and it's the intersection of the horizontal axis with the bisector of CD. That should not be hard to put in equations. The diagram below is exact. I used Kig.



Let's name I = (w, z) the middle of CD.
A point M(x,y) is on that bisector if and only if (x-w) + (y-z) = 0. We want the point for which y = 0, so x = w + z

Thus x = -3r2/2 + -3r2/2 = -3r2

So the radius is sqrt( (-3r2-r2/2)² + (r2/2)² ) = sqrt( (7r2/2)² + 1/2 ) = sqrt( 49/2 + 1/2 ) = sqrt(50/2) = sqrt(25) = 5

The fact that I get an integer gives me confidence I didn't make a mistake, but that's no guarantee ofc.

 

[mNFwTJ3wz9]

which looks like it should be enough...
Lets start with
x1^2 + y1^2 = (x3 + y3)^2
(x3 - 7/sqrt2)^2 + ( 7/sqrt2)^2 + (y3 + 7/sqrt2)^2 + ( 7/sqrt2)^2 = (x3 + y3)^2
- 7 * sqrt2 * x3 + 49/2 + 49/2 + 7 * sqrt2 * y3 + 49/2 + 49/2 = 2 * x3 * y3
An nice, I think this somewhat solves things

x2^2 + y2^2 = (x3 + y3)^2
(x3 + 1/sqrt2)^2 + ( 1/sqrt2)^2 + (y3 - 1/sqrt2)^2 + ( 1/sqrt2)^2 = (x3 + y3)^2
x3 * sqrt2 + 1/2 + 1/2 - y3 * sqrt2 + 1/2 + 1/2 = 2 * x3*y3
Finally:
x3 * sqrt2 + 1/2 + 1/2 - y3 * sqrt2 + 1/2 + 1/2 = - 7 * sqrt2 * x3 + 49/2 + 49/2 + 7 * sqrt2 * y3 + 49/2 + 49/2
4 * sqrt2 * x3 + 1 = 4 * sqrt2 * y3 + 49
x3 = y3 + 6 * (2^0.5)
y3 + 6 * (2^0.5) + 2^0.5 - y3 = 2^0.5 * (y3 + 6 * (2^0.5)) *y3

y3 = 5 - 3 * (2^0.5)
x3 = 5 + 3 * (2^0.5)

 

[Appetite4Destrction]

jfl @ solving math problems on an incel forum
you niggas.. inceldom is a hell of a drug..

 

[mNFwTJ3wz9]

radius = (x3 + y3 ) / 2 = 5
@WorldYamataizer

Yes
Maybe this wikipedia article has the one youre thinking of

Intersecting chords theorem - Wikipedia

en.m.wikipedia.org

huh, this just gives x3 * y3 = 7
would have been somewhat useful for solving manually, because quadratics would not be involved in the last step

 

[WorldYamataizer]

@grondilu
@mNFwTJ3wz9

Yes, good job, the both of you. It turns out that the radius is exactly 5.

Here is my solution

Edit: not AP, i meant the centre of the circle to P. Anyways, I hope my solution is clear

 

[grondilu]

@grondilu
@mNFwTJ3wz9

Yes, good job, the both of you. It turns out that the radius is exactly 5.

Here is my solution
View attachment 392347
Edit: not AP, i meant the centre of the circle to P. Anyways, I hope my solution is clear

Nicely done, probably the simplest way to solve it.

 

[dreadtheblackpill]

high iqcels only

 

[IncelCream]

Shut up GrAYboi

At least he is making us think

 

[waste matter]

no math for your face

 

[Animecel2D]

High IQ thread but didn’t read

 

[Jungle]

high iq

 

[cutthroat]

maths me

 

[grondilu]

Prob 0011 : I wanted to see how Mathematica could "solve" it :

$ wolframscript
Wolfram Language 12.1.1 Engine for Linux x86 (64-bit)
Copyright 1988-2020 Wolfram Research, Inc.

In[1]:= EuclideanDistance[#[a], #[i]] &@Association[                                                    
          RandomInstance[GeometricScene[                                                                
             {a, b, c, d, e, i},                                                                        
             {                                                                                          
              EuclideanDistance[c, e] == 7,                                                             
              EuclideanDistance[e, d] == 1,                                                             
              PlanarAngle[e -> {b, d}] == \[Pi]/4,                                                      
              EuclideanDistance[a, i] == EuclideanDistance[i, b] ==                                     
               EuclideanDistance[i, c] == EuclideanDistance[i, d],                                      
              GeometricAssertion[{a, i, b}, "Collinear"],                                               
              GeometricAssertion[{Line[{a, b}], Line[{c, d}]}, {"Concurrent", e}]                                                                                     
              }                                                                                         
             ]                                                                                          
            ]["Points"]                                                                                 
          ]                                                                                             

Out[1]= 5.

 

[WorldYamataizer]

Problem #0012
Hello, .co Math Club.
Due to increased responsibilities from my classes, I’ll take a few days off from posting interesting math problems, before resuming.

If you have any suggestions on specific types of problems you would like to make/share then let me know (e.g. more calculus, geometry, algebra etc.)

Triangle ABC is isosceles with AB=AC and BC=65. P is a point on BC such that the perpendicular distances from P to AB and AC are 24 and 36, respectively. Determine the area of triangle ABC, correct to one decimal.
@Divergent_Integral
@SelfCrucified
@grondilu
@mNFwTJ3wz9
@Diocel
@mental_out
@Irredeemable
@Caesercel
@nihility
@Liszt

 

[grondilu]

Prob 0012.

Spoiler: Mathematica solution

I know, it's cheating, but at the same time I get to learn to use Mathematica.

Notice that Mathematica informs me that there is an other solution, with a different area, if we don't impose the points P, R, and S to be on the triangle (which I believe means at least one angle is obtuse).

 

[WorldYamataizer]

Prob 0012.

Spoiler: Mathematica solution

View attachment 392768

I know, it's cheating, but at the same time I get to learn to use Mathematica.

Good. Daily math problems for incels. Nice

 

[grondilu]

Good. Daily math problems for incels. Nice

I also added a subtitle for good measure :

 

[mNFwTJ3wz9]

Problem #0012
Hello, .co Math Club.
Due to increased responsibilities from my classes, I’ll take a few days off from posting interesting math problems, before resuming.

If you have any suggestions on specific types of problems you would like to make/share then let me know (e.g. more calculus, geometry, algebra etc.)

View attachment 392734
Triangle ABC is isosceles with AB=AC and BC=65. P is a point on BC such that the perpendicular distances from P to AB and AC are 24 and 36, respectively. Determine the area of triangle ABC, correct to one decimal.
@Divergent_Integral
@SelfCrucified
@grondilu
@mNFwTJ3wz9
@Diocel
@mental_out
@Irredeemable
@Caesercel
@nihility
@Liszt

So lets try to not brute force this one
by similar triangles : RP/SP = BP/BC
also area(ABC) = area(ACP) + area(ABP)
= 0.5 * (AC*SP + AB* RP)
lets say AC = AB = x
(x - 10)^2 + 24^2 = (x - 15)^2 + 36^2
x = 169/2
area = x * (24 + 36) * 1/2
area = 2535

 

[WorldYamataizer]

So lets try to not brute force this one
by similar triangles : RP/SP = BP/BC
also area(ABC) = area(ACP) + area(ABP)
= 0.5 * (AC*SP + AB* RP)
lets say AC = AB = x
(x - 10)^2 + 24^2 = (x - 15)^2 + 36^2
x = 169/2
area = x * (24 + 36) * 1/2
area = 2535

Slick. I like it!

The key to this problem was realizing the similar triangles given to you. Here's the way I did it.

Since ABC is isosceles, then angle ACB= angle ABC.
Therefore, triangles BRP and CSP are similar. (Angle, angle)
Suppose BP=x, then PC=65-x
By similarity, 24/36 = x/(65-x). Solving this equation gives x=26. So BP=26 and PC=39.
By the Pythagorean Theorem, BR=10 and CS=15

Dropping a perpendicular A onto BC, call the foot of the altitude M.
Then triangle AMC is similar to triangles BRP and CSP. (shared angle, right angle)
By similarity, AM/36 = (65/2)/15
AM=78.

The area ABC is (1/2)(65)(78)=2535

 

[Incellectual]

ER == GO

based

 

[Snowstormhigh]

3+4

 

[wei#3959]

3+4

7

you all iq mog me to oblivion btw

 

[WorldYamataizer]

Problem #0013

Hello .co Math Club. Today, we will try to find side lengths of a triangle that satisfy a certain condition. Enjoy.

Three side lengths of a triangle are consecutive integers. The largest angle of the triangle is two times the smallest angle. What are the side lengths of the triangle?

It may be helpful to draw a useful diagram.

@Divergent_Integral
@SelfCrucified
@grondilu
@mNFwTJ3wz9
@Diocel
@mental_out
@Irredeemable
@Caesercel
@nihility
@Liszt

 

[grondilu]

Problem #0013

Hello .co Math Club. Today, we will try to find side lengths of a triangle that satisfy a certain condition. Enjoy.

Three side lengths of a triangle are consecutive integers. The largest angle of the triangle is two times the smallest angle. What are the side lengths of the triangle?

It may be helpful to draw a useful diagram.

@Divergent_Integral
@SelfCrucified
@grondilu
@mNFwTJ3wz9
@Diocel
@mental_out
@Irredeemable
@Caesercel
@nihility
@Liszt

Spoiler

As the triangle gets larger, it looks more and more like an equilateral triangle, and such triangle is obviously not the solution, so one can safely assume the solution is small. So we can try with small integer values, and I think {2,3,4} fits. I used the cosine rule to calculate the angles :

a² = b² + c² - 2bc cos(alpha)

The cosines of the two relevant angles are -7/8 and 1/4 (straightforward application of the above formula). Let's name the smallest one x and the largest one y.

cos(2x)=cos(x)² - sin(x)² = cos(x)² - (1 - cos(x)²) = 2 cos(x)² - 1

So :
cos(2x) = 2 (1/4)² - 1 = 2/16 - 1 = 1/8 - 1 = - 7/8,

So 2x has the same cosine as y. Since we're in a triangle the angles are positive and inferior to pi, on which domain the cosine function is injective, thus y = 2x, and therefore the triangle with side lengths 2,3,4 does fit the requirement.

 

[mNFwTJ3wz9]

Problem #0013

Hello .co Math Club. Today, we will try to find side lengths of a triangle that satisfy a certain condition. Enjoy.

Three side lengths of a triangle are consecutive integers. The largest angle of the triangle is two times the smallest angle. What are the side lengths of the triangle?

It may be helpful to draw a useful diagram.

@Divergent_Integral
@SelfCrucified
@grondilu
@mNFwTJ3wz9
@Diocel
@mental_out
@Irredeemable
@Caesercel
@nihility
@Liszt

triangle abc
angles ABC opposite to corresponding sides
a = x,b = x+1,c = x+2 WLOG
C = 2*A

using the area sin rule -
1/2 * a * b * sin(C) = 1/2 * b * c * sin(A)
a * sin(C) = c * sin(A)
x * sin(2*A) = (x + 2) * sin(A)
x * 2 * sin(A) * cos(A) = (x+2) * sin(A)
cos(A) = (x+2)/ (2*x)

Now another formula -
area = sqrt(s*(s-a)*(s-b)*(s-c))
1/2 * b * c * sin(A) = sqrt( 1.5 * (x+1) *( 0.5*(x + 3) )*( 0.5*(x + 1) )*( 0.5*(x-1) ) )
sin(A) * 0.5 * (x+1) * (x+2) = 0.5 * (x+1) * sqrt( 1.5 *(x + 3) * 0.5 * (x - 1) )
sin(A) = sqrt( 0.75 * (x - 1) * (x + 3) ) / (x + 2)

Finally
sin(A)^2 + cos(A)^2 = 1
((x+2)/(2*x))^2 + 0.75 * (x - 1) * (x + 3) / ((x + 2)^2) = 1
Oh man this is kinda bad, but lets do it...
(x+2)^4 + 0.75 * (x - 1) * ( x + 3 ) * (2x)^2 = (x+2)^2 * (2x)^2
(x^2 + 4x + 4) * (x+2)^2 + 3 * (x^2 + 2x - 3) * x^2 = 4 * x^2 * (x+2)^2
(4x + 4) * (x+2)^2 + 3 * (x^2 + 4x + 4 - 2x - 7) * x^2 = 3 * x^2 * (x+2)^2
(4x + 4) * (x+2)^2 + (- 2x - 7) * x^2 = 0
-2*x^3 - x^2 + 32*x + 16 = 0
2 x^3 + x^2 - 32 x - 16 = 0
Ok I'm not doing a cubic by hand.
x = 4 or -4 or -1/2.
Going with 4
triangle is 4,5,6

2,3,4

You sure?

 

[grondilu]

You sure?

I mean, yeah. I'm pretty sure, unless I missed something about the problem statement.

    cosineRule[a_, b_, c_] := (a^2 - b^2 - c^2)/(2 b c)
    ArcCos /@ Simplify@Table[cosineRule @@ RotateLeft[{n, n + 1, n + 2}, i], {i, 0, 2}]
    % /. n -> 2
    {Max[%],Min[%]}

does return {ArcCos[-7/8], ArcCos[1/4]}

Mathematica then fails to see that the ratio between those two values is exactly 2, but the numerical value suggests it strongly, and there is the demonstration I gave about it which I think is fairly simple.