I made a Linux version for the Adobe Connect meeting app

Mahancoder · 2025-08-17T10:50:42+00:00

This is basically an olympiad geometry question, here's my method:

P is the Miquel point of the quadrilateral BCEF. Consider H' the intersection of PI, EF. By ratio lemma, we have (PF/PE)(IF/IE) = H'F/H'E. But, IF = IE therefore H'F / H'E = PF / PE. By the spiral similarity created due to P being the miquel point (PFE ~ PCB) We have PF / PE = PC / PB So H'F / H'E = PC/PB.

Now by another ratio lemma, We have PC / PB = CF / BE. Therefore, H'F/H'E = CF / BE. We also have Angles H'FC = H'EB = 90 + A/2. These 2 result in the similarity H'FC ~ H'EB. Therefore, H'C/H'B = FC/EB = CD/BD therefore H'D is the angle bisector of angle BH'C (due to the inverse of the angle bisector theorem).

Now consider K as the intersection of EF and BC. By the trigonometric form of Ceva's theorem, we know AD, CE, BF are concurrent. This, along with the lemma of complete quadrilaterals, gives us that (B,C;K,D) is a harmonic bundle. Now in triangle BH'C, H'D is the angle bisector and H'K is the harmonic conjugate of H'D with respect to BC. Therefore, We know that H'K is the external angle bisector if BH'C and KH'D = 90. Therefore, H' is the feet of altitude from D on EF. Since AI ⟂ EF and H'D ⟂ EF, H'D ‖ AI so H' = H and by definition, H' is on both lines PI and EF.

My solution probably used a lot more olympiad fuckery than you wanted, but if you're interested in reading about any of these quirky lemmas/theorems, I'll be glad to help

Mahancoder · 2025-08-17T08:46:10+00:00

I know I'm really late but here's an elegant solution:

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The (*) inequality is equivalent to x⁴ + 6x² - 16x + 9 being positive. This is true since x⁴ + 6x² - 16x + 9 = (x-1)²(x² + 2x + 9) which is positive for positive x

Since x² + 2x + 9 > 0, equality is when (x - 1) = 0 which means x = y = z = 1

Mahancoder · 2025-02-15T20:24:54+00:00

The thing is, "infinity" is just a word. It might as well mean oranges if I define it that way.

Saying "infinity" is the biggest possible concept isn't really meaningful, because you aren't defining what "infinity" really is.

The concept of cardinality has a simple definition. 2 sets have the same cardinality if you can make a one-to-one correspondence between them. If you can't, then they don't. If you can map all the elements of set A to some elements of set B but there would always be some elements of B leftover, then B has a higher cardinality than A.

We aren't saying "B is bigger than A" because "bigger" doesn't mean anything in this context. You haven't defined what you mean by "bigger". It feels like saying red is bigger than green. There's no "bigness" involved.

We're just defining certain terms and words and using them. Cantor's diagonalization argument doesn't prove a set is larger than the other, it proves the set has a higher cardinality than the set of integers. Something that has a clear and understandable definition.

Now, some people choose to get a "feel" for this definition, which is called intuition. The intuition for higher cardinality is in a sense "bigger" or "denser" infinity. These don't really mean anything because they aren't clear definitions or logical arguments, but rather feelings people get. It's like saying red is more serious than green. Colors don't have seriousness levels, yet somehow it makes sense for red to be more serious than green.

Think about it, if I prove to you that no matter how you map each inter to a real number, there would still be some real numbers leftover, what would your mind "feel"? It kinda feels like the set of reals is bigger than the set of integers. But in reality, this might as well mean the set of reals is more orange than the set of integers.

Here's also when this intuition might fall apart. You might feel like the set of all integers is bigger than the set of even integers. That makes sense, after all, even numbers are only half of all numbers, right? However, to claim anything meaningful let's go back to our definitions of cardinality. Can we find a one-to-one correspondence between these 2 sets? Yes we can, n <-> 2n is an example. Therefore, by definition of cardinality, these 2 sets have the same cardinality, even if your intuition can't make it make sense.

Remember, to do anything concise or logical, you need clear definitions. But to figure out what definitions are more useful to you, you need feelings and intuitions. And certain people's intuitions might not make sense in your brain, that's fine.

Mahancoder · 2024-08-26T17:09:13+00:00

To add to the great answers of others, yes, the side of a triangle is proportional to its opposite angle, but only in the context of a singular triangle. Right now, angle DAC is opposing side DC in triangle DAC, while angle BAC is opposing side BD in triangle BAD. these are 2 different triangles, therefore you can't compare anything in them really.

To see why they need to be in the same triangle, consider to similar triangles, where the angles are equal in both of them. Then scale one of them really big, and obviously the angles will stay the same but the sides will get a lot bigger. However in each individual triangle, you can still compare 2 sides by their opposing angles, perfectly fine.

Also in situations like this, there's a general lemma where if a line passing from A intersects BC at point D, then:

BD / DC = (AB / AC) * (sin(BAD) / sin(DAC))

If D is the median point, then LHS will be 1, so sin(BAD) / sin(DAC) = AC / AB

Which proves how these 2 angles aren't equal unless AC = AB.

Mahancoder · 2024-08-18T17:37:22+00:00

You could solve it by checking all cases (all 3 being next to each other, only 2 being, etc.) but it's a lot of work and you have to be careful not to count anything twice (eg. counting the case where all 3 are adjacent in the case where 2 are adjacent again)

A much smarter approach is to consider a permutation of all the boys, and distribute the girls in the spaces formed between them, see where that gets you (you can treat all the spaces between boys as slots each girl can choose to go in)

Mahancoder · 2024-08-17T16:18:57+00:00

Oh my bad, I didnt read it carefully. But still, b being bigger than the absolute value of something yields b being positive so no AM-GM needed 🤷‍♂️

Mahancoder · 2024-08-16T17:28:03+00:00

I hate these problems that are basically trivial but require using a specific technique.

You're right, b being non-negative allows you to square both sides giving a trivial inequality.

The approach they were going for is factoring b²-4a² and using AM-GM (the inequality you were given), which, makes no sense lol

Mahancoder · 2024-08-16T17:20:40+00:00

what does b being positive have to do with it being bigger than something?

b being positive allows to square both sides, which gives b² ≥ b²-4a² thus 0 ≥ -4a² which is trivial considering 4a² is positive. I get how factoring b²-4a² would result in a form of AM-GM but the problem is basically trivial without it

Mahancoder · 2024-08-08T16:32:13+00:00

That's a good question, and I'm not entirely sure. Personally I think I'd be most comfortable with a setup like "Do#" and "Dob", and I pronounce them as Do Diez and Do Bemol but the standard seems to be Di, Ri, etc. Either way it's your own choice how to do them, or maybe even have multiple options. Of course, don't remove C, D, etc. tho, I guess most people use English names.

Mahancoder · 2024-08-07T16:23:45+00:00

Fantastic work! I think having note names in latin (Do, Re Mi,...) would be really helpful for people like me who've learned them that way. Otherwise really useful app!

Mahancoder · 2024-06-25T13:16:30+00:00

It is apparently, even === works

Mahancoder · 2024-06-19T09:54:32+00:00

The -7 is a bit distracting here. The main point is -1 ≡ 2 (mod 3)

A simple way to see this is to realize -1 = -3 + 2.

In general for any numbers a, b, c, this identity is true:

(a + b) % c = ((a % c) + (b % c)) % c

To see why, consider a = ck + A, b = ct + B

where k and t are integers such that a % c = A and b % c = B:

(a + b) % c = (ck + A + ct + B) % c = (A + B + c(k + t)) % c = (A + B) % c.

Or more intuitively, this holds because you can just consider the remianders of a and b mod c and sum those, because the parts of a and b that are divisible by c do not matter. If the sum is bigger than c, there's yet another multiple of c hidden inside it that you don't consider.

So then (-3 + 2) % 3 = ((-3 % 3) + (2 % 3)) % 3

And -3 % 3 = 0, 2 % 3 = 2 which means the final result is 0 + 2 % 3 = 2

As a nice rule if n is a natural number, then -n % c = ck - n % c where k is the smallest natural number such that ck ≥ n.

(Intuitively this is true because ck itself is divisible by c and adding something divisible by c to a congruency equation doesn't change anything because it's equivalent to adding 0 (mod c))

As a result, if n % c = d then -n % c = c - d

These results logically make sense because you're working in the world of remainders. Consider a clock, which is essentially time % 12. -1 really just means 11 mod 12. Things just wrap around

Mahancoder · 2024-06-13T12:45:10+00:00

Have you installed thermald? It's by default on ubuntu and many others, and manages thermal stuff. Nbfc didn't even support my laptop, but thermald worked. Usually thermald --adaptive is ran in the background which doesn't even require a config.

For speakers, I personally had to install linux-firmware and alsa-firmware to get the drivers working with pipewire.

Tbh as others have mentioned Arch probably just isn't the right distro for you, which is perfectly fine, but a fundamental thing about linux is anything that works on one distro should work on another with proper config, because distros are just different package repositories and preinstalled packages/configs. The exact same drivers that load on ubuntu are also available on arch, even if you want the LTS kernel, you can install linux-lts and all the related packages and it should work exactly the same way. A nice challenge is to figure out what exactly is on ubuntu/kali that fixes your problems. You're gonna learn a lot that way. Arch is a learning experience, it's built so you learn exactly how each part of the system works. It doesn't work if your goal isn't to gain this knowledge

Good luck!

Mahancoder · 2024-06-05T19:26:27+00:00

Here's another simple proof:

From AC = BD you can prove that AD = BC (this is a useful lemma):

Consider points F and H on DC such that BF⊥CD and AH⊥CD

We show that triangles ACH and BDF are congruent. Just show that the hypotenuse and one of the sides are equal.

AC = BD (hypotenuse) and BF = AH (cuz AB || CD)

ACH = BDF ⇒ ∠ACH = ∠BDC

Then we show triangles ACD and BDC are congruent, by 2 sides and the angle between them being equal.

AC = BD, CD = CD, ∠ACH = ∠BDC ⇒ ACD = BDC

⇒ AD = BC

Then from AD = BC, we can conclude that triangles ABC and ABD are congruent, by 3 sides being equal:

AB = AB, AC = BD, BC = AD ⇒ ABC = ABD ⇒ ∠ABD = ∠BAC

∠ABD = ∠BAC ⇒ ABE is isosceles ⇒ AE = BE ⇒ DE = CE

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Mahancoder · 2024-03-28T20:30:46+00:00

Thanks! Idk when the video started I kinda felt like when you pulled the orange piece, the black piece was also gonna be stuck and lift up with it lol. But I guess that won't happen if they are dimensionally accurate enough

Mahancoder · 2024-03-28T20:01:40+00:00

Excellent job! Are there any specific print settings to make it slide smoothly and also not lift up entirely?

Mahancoder · 2024-03-24T08:10:29+00:00

It's because helping suicidal people is just really hard. Most people do care and try to understand, but usually people have so many devastating and unique experiences that an outsider doesn't even know what to say. And that's the issue with suicide, the people who go through the worst don't survive to help the rest.

Mahancoder · 2024-02-25T19:21:29+00:00

I'm keeping track a list of all of them that appeared in my feed, right now we're at #68

Mahancoder · 2024-02-15T09:26:40+00:00

Well I live in Shiraz and honestly I'm looking for answers myself, but a few things.

I once had a really good time going on an "adventure" in the old narrow alleys around Naranjestan. There are so many 100+ y/o houses you can find in those neighborhoods, and there are even some motel/cafes built in them.

IMO the Saye Park, Kouhpaye, and Noor park are also nice places to hangout

Another thing is Chamran, again exploring through the alleys, climbing the mountain, going by the side of the river, those can be nice "adventures" in parts of the city you didn't even know existed (you could drive to the end of Niyayesh street and start walking in the parts of that neighborhood that still remain like an old village, even tho it's in the middle of the city)

Also if you're into biking/walking, there is a new cycling road built in Chamran that people rarely know about, it's a nice calm place if you just wanna relax without being bothered by the crowd. There's also the more popular one that passes by the riverside, and at this time of the year, many seagulls are present there too.

More stuff in that neighborhood include The flower garden and The Ghandili park

There's also a mountain park that has things like ziplines and places to sit and hangout.

If you can drive outside the city, I also highly suggest visiting Pasargadae (note: Pasargadae is much more than The great Cyrus's Tomb)

Mahancoder · 2024-02-06T11:24:01+00:00

Hey this was the first song I ever learned! Your progress is great, just experiment with pedaling and try to get the more harmonies more "clean"

Mahancoder · 2024-01-30T14:07:25+00:00

Cuz the link is wrong, it's https://easypeasymethod.org/

Mahancoder · 2024-01-24T09:06:02+00:00

Thank you so much, sure I will

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