 #### Linesnap99

##### Officer
★★★★
meanwhile chads measuring foids ass size with their pp #### mNFwTJ3wz9

##### Everyone except me is a Fed ﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽
★★★★★
I mean, yeah. I'm pretty sure, unless I missed something about the problem statement.

Code:
``````    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. cosineRule[a_, b_, c_] := (a^2 - b^2 - c^2)/(2 b c)
I think you missed a negative sign.

Last edited: #### grondilu

##### Overlord
★★★★★
View attachment 395444

I think you missed a negative sign. Oh fuck you're right, in the cosine rule formula.

Indeed, n = 4 works, not n = 2 : Last edited: #### mNFwTJ3wz9

##### Everyone except me is a Fed ﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽
★★★★★
meanwhile chads measuring foids ass size with their pp
I mean, what else am I supposed to do... Over for mathematicacels #### grondilu

##### Overlord
★★★★★
Over for mathematicacels
I just don't know Mathematica well enough yet. I should have remembered there is a function to define a triangle from its side lengths, and then there is a function to get the angles. There's a function for almost everything in Mathematica.

I'm slightly disappointed that Mathematica can't pick up on remarkable ratios of ArcCosines, though. It should have simplified ArcCos(1/8)/ArcCos(3/4) into 2.

My earlier demonstration still works, though :

2*(3/4)² - 1 = 2*9/16 - 1 = 9/8 - 1 = 1/8

Last edited: #### WorldYamataizer

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

Code:
``````    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.
There must be a mistake!
If the largest angle is twice the smallest angle, then the interior angles of the triangle are x , 180-3x , and 2x.
But x < 180-3x < 2x
This gives that 36<x<45

arccos(1/4) is approximately 76 degrees #### grondilu

##### Overlord
★★★★★
There must be a mistake!
Yes there was, as I wrote. You were right about a sign mistake. In the cosine rule I divided by 2 b c instead of -2 b c #### WorldYamataizer

##### Banned
-
Yes there was, as I wrote. You were right about a sign mistake. In the cosine rule I divided by 2 b c instead of -2 b c
Sorry, i left this tab open for like half an hour, so I couldn't catch up on new posts to catch up #### mNFwTJ3wz9

##### Everyone except me is a Fed ﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽
★★★★★
arccos(1/4) is approximately 76 degrees
did you do that in your head?  #### WorldYamataizer

##### Banned
-
did you do that in your head? No, I used desmos and punched it in.  #### mNFwTJ3wz9

##### Everyone except me is a Fed ﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽﷽
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No, I used desmos and punched it in. The "approximate" made me think that lol #### WorldYamataizer

##### Banned
-
Problem #0014

A general point P(c,d) is selected on the circle x^2+y^2=a^2+b^2 ; from P, tangents are drawn to the ellipse

(b^2)(x^2)+(a^2)(y^2)-(a^2)(b^2)=0​

Show that, no matter where P is selected on the circle, the two tangents from P to the ellipse are always perpendicular.

Here is my solution to Problem #0013
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

##### Overlord
★★★★★
A general point P(c,d) is selected on the circle x^2+y^2=a^2+b^2 ; from P, tangents are drawn to the ellipse

(b^2)(x^2)+(a^2)(y^2)-(a^2)(b^2)=0​
Show that, no matter where P is selected on the circle, the two tangents from P to the ellipse are always perpendicular.
This problem sounds poorly formulated. #### WorldYamataizer

##### Banned
- #### LDARbeforeROPE

##### Omega Male
★★★
Pre cal is getting hard to follow  #### Caesercel

##### Baldcel wristcel framecel oldcel ethnicel
★★★★★ #### LDARbeforeROPE

##### Omega Male
★★★
Lol wtf is this childhood level shit.
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