IMO LIVESOLVE DAY 1

[00:00:05]Okay. So, I think we're live right now.

[00:00:07]Let me check.

[00:00:21]Okay. Yeah, I think I'm live now.

[00:00:27]Okay.

[00:00:30]So, let me uh Okay.

[00:01:16]Okay, I'm going to wait for a bit until I get a few more viewers and then I'm going to start Okay. So, I think I can get started now.

[00:03:31]Okay. So, what I'm going to do is I'm going to go ahead and put a timer up here. So, let me search up timer.

[00:03:41]see if Okay.

[00:03:47]All right. So, I have my timer set here.

[00:03:50]So, what I'm going to do now is I'm going to go ahead and open the international math website.

[00:03:58]And if I click on the IMO competitions or probably yeah, if I click on this problems tab, then I should see the problems immediately.

[00:04:10]Okay. So, I guess I'm going to get started right now. And I what I'm going to be doing is I'm going to go through each problem one at a time or maybe go back and forth between them and I'm going to be solving them live on this whiteboard right here.

[00:04:26]So, I'm going to like Yeah, I'll be doing all my work up here so you can see how I'm solving them. All right, without further ado, let's go ahead and get started. Let me go ahead and click the problems and open them and I'll start the timer. Okay, let me Okay, yeah, here are the problems. And the timer starts now.

[00:04:53]Okay, so there are 20 26 integers greater than one written on this blackboard and they're not necessarily different.

[00:05:01]And Confucius chooses two different integers which are greater than one or they're just two integers I guess and replaces them with GCD and LCM over GCD and regardless of the choices on of this person after finally Benny moves we want to prove that one in exactly one integer is greater than n.

[00:05:28]So, okay. Oh, there are 20 26 integers integers here. So, oh, I think I think I see what's going on. Let me go ahead and uh screenshot this problem and then put it on the whiteboard.

[00:05:42]Okay, there we go.

[00:05:45]So what what we're trying to do now is so we have 20 26 integers greater than one. And so we're replacing any two integers.

[00:05:58]We're choosing two of them and replacing them with these guys here.

[00:06:03]And we want to prove that at some point the the board will just stay the same.

[00:06:11]it will only have one one one all the way to m and then this value of m is independent of the choices that he makes h so we're basically replacing suppose suppose that we have dx and dy are two numbers so that's where d is the gcd then we're basically basically replacing dx and dy with d and then xy.

[00:06:47]So what exactly stays the same here?

[00:07:13]Okay, let's try an example.

[00:07:16]So suppose that I have suppose let's start with two integers on the board. Then we're going to get suppose we start with like I don't know suppose we have uh 14 and 34 and yeah you lost the game then the GCD is equal to 2 and the LCM is equal to 14 * 34 all over 2 and then we have to divide that by two again and then this becomes 19 or yeah 119 and then when we take the GCD we end up with one.

[00:08:03]Okay, so the two integers case was like pretty simple. Let's try the three integers case.

[00:08:11]I I lowkey think that something like 6 10 15 might be interesting because all the GCDs are pair wise different but they're complete GCDs. one.

[00:08:26]Okay.

[00:08:28]So, okay. So, let's first try doing it on 610. Then the GCD is 2. The LCM is 30 I believe. And this becomes 15. Then 15.

[00:08:53]And let's try moving on 1515. Now if we end up doing this, we end up getting uh two uh okay so GCD of these two is 15. Oh, and then this LCM over GCD is equal to one.

[00:09:14]So GCD equals 1 is not the only way to get one. We could have that the two numbers are equal making the GCD equal the LCM and yeah so we can't do anything with this number now wait what if we use induction because if we can show that one number will always become one if if we can show that there will always be a number that's equal to one then that means that we're done automatically because if we have a bunch of numbers here and then we make one of them one by induction we can only apply on these rest and therefore we will have one one one all the way to one and then an M here.

[00:10:05]Okay, so that would solve part A if we can show that there will always be a one. But then part B is interesting.

[00:10:13]Let's wait let me not get ahead of myself and let me calculate the number here. So it seems like the number here is one going to be like one. So it seems like the magic number here is 30.

[00:10:27]Okay then we have uh 131.

[00:10:33]So now let's go ahead and try uh wait.

[00:10:37]Yeah. Okay. So let's try uh dude. Dude, I hate Microsoft.

[00:10:43]Okay, so let's try 10 and 15 first. So now if we do 10 and 15, it becomes 65.

[00:10:53]The GCD is again equal to 30. So then it's 30 / 5, which is 6.

[00:10:59]And then if we move on five and six, then it becomes 6130, I believe.

[00:11:05]And then we have the GCD of these two is equal to six.

[00:11:12]And then we have the LCM is 30. So it becomes 615.

[00:11:20]And finally we get 1 130. Again, it definitely feels like I don't know.

[00:11:29]It seems like the magic number here is just equal to the greatest common is equal to sorry the least common multiple here. But I'm actually I actually don't know because it could be something else entirely.

[00:11:49]Let's try out like eight and 20 because that might be interesting. So the greatest common divisor of these is four and the LCM is equal to I believe it's 40 and we divide it by four.

[00:12:06]So that's equal to 10.

[00:12:11]So now we have four and 10. Then the greatest common divisor of these is two.

[00:12:18]And then the LCM is 20 this time. So we get 20 divided by two is 10.

[00:12:27]And then we have 2 and 10. Then the greatest common divisor is two. Then we get two and five.

[00:12:37]And finally that gives 1 10. So 10 is our magic number now. Huh.

[00:12:45]That's interesting. Am I doing everything correctly?

[00:12:54]So we have the greatest common divisor is four and the LCM is 40. So LCM over GCD is just four.

[00:13:09]And then the GCD is two, the LCM is 20.

[00:13:13]So that's 2 and 20 over two is 10.

[00:13:18]So yeah, it's not the greatest common divisor.

[00:13:27]Wait, because we have these GCDs and LCMs and we have that this number here is kind of similar to this one and this one. I guess we could try using uh just we could just try solving the problem prime by prime, but I don't know if this will work.

[00:13:44]Yeah, this may or may not work.

[00:14:03]Okay. So suppose that we so if I guess if we rephrase it in terms of the primes it might become something different. So suppose we have a bunch of numbers like a1 a2 all the way to a 2026.

[00:14:22]Then if we operate on a1 a2 we replace them with minimum of a1 a2 because that's the vp of the the gcd.

[00:14:37]And then this one becomes like max minus min.

[00:14:43]So if we're assuming that a1 is less than a2 then we just replace them with a1 a2 minus a1.

[00:14:53]So we want to end up proving I guess that all of these numbers end up collapsing down into some number then 0000 0.

[00:15:06]The problem is for the primes that might not line up.

[00:15:11]This could probably be be solved if we solve it for each prime and then we have bunch of other things and then all the if we solve it for each prime then then all the numbers on the board are relatively prime and then all of them just become one immediately.

[00:15:33]So we want to prove that if we operate on this list here and these are non- negative integers and yeah if if we prove that they're all if we end up getting something 0000 then we're set. Hold on just a second.

[00:15:58]Okay, let me re chat.

[00:16:06]Oh my goodness. Two immo team members are inside my chat.

[00:16:17]Okay. Anyways, so maybe some sort of invariant. I don't know.

[00:17:19]Let's start off simple and try out like probably one two three.

[00:17:26]Then this one becomes one one three and then these and then if we apply on these two then we get one one two.

[00:17:40]Wait, this really really feels like the uklidian algorithm.

[00:17:45]Yeah. Yeah. Wait a second. This just is this just the uklidian algorithm?

[00:17:53]Cuz if it is, it would be hilarious.

[00:17:59]Yeah, like the the V2 of this is three and the V2 of this is two and the greatest common divisor of three and two is one and that's one here.

[00:18:12]And then if you have then for the V5 you have this is zero and this is one and the greatest common divisor is one.

[00:18:23]Wait, this might be it. This might be it.

[00:18:28]Okay. So, because of the sum just keeps decreasing and decreasing, we're going to get to a point where only like we're going to be stuck at some point because all the numbers we have on the board are greater than zero and the sum just keeps decreasing.

[00:18:50]So, but yeah, the sum is monovariant. So that means that the sum on the sum of the numbers on the board is just yeah it just keeps decreasing. Why did I say that twice anyways? So what we end up with is because we end up stuck with 0 0 by uklidian algorithm. This number has to be gcd of a1 all the way to a 2026.

[00:19:22]This might be it.

[00:19:33]Okay. And now that we have all the primes are in different spots, we have that all the numbers are relatively prime.

[00:19:44]So suppose we have like relatively number relatively prime numbers A1 all the way to like A 2026. Now I'll call them B so that I don't get confused.

[00:19:56]Then we just apply this move all the every single time.

[00:20:02]And what we end up with is B is B1 all the way to B 2026.

[00:20:09]So I think this solves a problem because I think that the fixed number on the board is if we have the numbers like a1 all the way to a 2026 to start with it is like 2 to the power of gcd of new 2 of a1 all the way to new 2 of a 2026.

[00:20:36]And then you do this for all the other primes.

[00:20:43]Okay, wait. Let me write this up to make sure that I actually solved it.

[00:20:48]So yeah.

[00:21:28]Okay.

[00:25:40]I made a writing error here.

[00:32:46]Okay, so I think I finished the red up.

[00:32:50]I honestly think this is a really really nice problem. Like I especially like the idea of using the GCD with all of the VPs because usually the thing that you do with VPs is you have a GCD or you or whenever you have a GCD, you end up eliminating the VPs.

[00:33:11]And by eliminating the VPs here, we end up getting this new problem which actually involves GCDs again. And when we involve GCDs again, that's the part I really really like.

[00:33:26]So it's kind of like you have like a GCD inside of a GCD, which is very very strange.

[00:33:35]Okay, I'm going to take like a five minute break and look at the chat and then I'm going to look at problem two.

[00:33:44]Okay, so I've taken an hour so far. Oh, wait. It's not been an hour, has it not?

[00:34:01]Yeah, it's not been an hour. Wait, what is this timer doing?

[00:34:06]Wait, chat, what time did I start exactly?

[00:34:11]Cuz cuz I think this timer just sped up in time.

[00:34:21]Okay, so I might have set it for 4 hours. Oh, I did set it for 4 hours. 30 minutes.

[00:34:29]Does anyone know what happened here?

[00:34:33]Hi, Eric. Hi, Zong Shu.

[00:34:54]The two ways to interpret GCD mean in a lattice and generator generator of an ideal. This is true.

[00:35:01]But I'm talking about like in the in when you're looking at specifically like Olympia number theory.

[00:35:16]Wait, do you know what time exactly I started?

[00:35:20]Oh, I I I've been live for like 35 minutes. So, I'm going to assume I've been I this has been going on for 30 minutes. So, I'm going to go ahead and set up a timer. uh on my on my system and I'm going to set it for four hours.

[00:35:40]Okay. So, you can't see the timer right now, but it's a 4 hour timer up here.

[00:35:46]And yeah, you can probably base it off of the live stream time. If the live stream time hits uh around 4 hours and 35 minutes, then that's approximately when the time is up. So yeah, the live stream time is basically the Yeah. Yeah, the live stream timer is just basically the my timer for this test.

[00:36:12]This is literally what the problem is.

[00:36:14]Alo, wait. Actually, wait, what?

[00:36:56]All right, I'm going to go ahead and start the timer and then start looking at problem two.

[00:37:02]Okay, so I'm going to go way down here.

[00:37:05]So that's behind. Okay, so problem two.

[00:37:10]Oh my goodness, it's a geo.

[00:37:15]Okay, let me go ahead and screenshot it.

[00:37:17]Oh my goodness, this is a geo problem.

[00:37:20]Two geos are so back.

[00:37:31]OM equals O.

[00:37:39]Okay, let me get Okay. So, so we have our triangle ABC here and M and N are the midpoints of the sides. So that's M that's N.

[00:38:03]Then points K and L are chosen inside the B MC and BNC. So maybe K could be here and then K is inside ABL and L is inside AKC.

[00:38:26]So maybe something like this.

[00:38:32]Okay. So suppose that KBA is equal to ACL.

[00:38:37]LBK equals LNC and LCK equals BMK.

[00:38:50]And if we have that O is a circumc center up here, then we want to prove that OM equals O.

[00:39:14]So, ABK equals ACL.

[00:39:20]I guess this could mean that if you uh wait yeah that if you extend the sides here then this quadrilateral here is cyclic.

[00:39:34]That's the only thing I can really think of.

[00:39:39]LBK equals LNC.

[00:39:48]Oh, LNC.

[00:39:54]So, it's equal to it's equal to Oh, I meant to put this one here. So it's LBK is equal to ln c a l and that's ncl and then ln c. So that means that's this angle here. So angle CBL is equal to B minus the sum of those two angles.

[00:40:36]So that's like B minus 180 minus I think that would be angle NLC which is B - 180 plus angle NLC.

[00:41:02]I'm yeah I don't know if this is useful or not.

[00:41:08]I think I'm just saying random stuff right now.

[00:41:45]Okay, let's try something different.

[00:42:34]You know what? I'm going to go ahead and uh draw in the angles just to be sure.

[00:42:40]So erase all this the line and okay so the line BN is not needed at all neither is this line here I can just clear that okay so first we know that I'm going to mark this in this color these two angles are equal I'm going to redraw this diagram.

[00:43:29]So, first we know that these two I need to change the color.

[00:43:36]So, we know that We draw this in and maybe L should be over here.

[00:43:51]Then we know that these angles here are equal and then LBK is equal to L and C. Okay, I drew the diagram slightly wrong.

[00:44:06]Um, so what if L is further down here?

[00:44:38]I think L being here is reasonable.

[00:44:42]So we now have that these angles are equal.

[00:44:46]Then LBK.

[00:44:52]So suppose that's screen angle is equal to LNC. So if I equal this angle here and likewise if I construct this line here then I should probably construct this one as well.

[00:45:19]Then these blue angles are equal.

[00:45:28]And the end goal is to prove that if O is a circumc center of AKL, then OM equals O N.

[00:45:40]I've seen this type of thing before where you can prove that the power with respect to M of the circumcircle of AKL is the same as the power of N. This is like that 2009 IMO2. So maybe we could try using that. Especially since this circumstance seems very out of the blue, so to say.

[00:46:08]So what do we do?

[00:46:27]We also haven't used the fact that they are midpoints.

[00:46:52]Yeah. So I want to go that back to this idea earlier where I extend these two lines here and suppose you made a X and Y How do we connect this angle here to this angle?

[00:49:03]I was considering taking a homophy to A that sends M to B and N to C, but that just makes it a bit more complicated because even though M becomes B and N becomes C, we get those two points out there.

[00:51:19]Okay, this might be a complete uh this might just be out out of nowhere, but what if uh CLK is similar to MKB.

[00:51:31]Obviously, it doesn't look like it in the diagram, but if it w it is, then I'd be happy because it definitely appears that LKB is similar to CLN.

[00:52:24]So we want to prove that this is like 180 - red minus green.

[00:52:31]So that's 180 - red minus green. And this is B minus R - G where B is the the angle B I don't think this will work.

[00:54:15]This is 180 minus green.

[00:58:58]I mean I think I might know where this could be going. So if I intersect a k and a l with the sides, this will make it a bit messy. Then it's equivalent to prove that the perpendicular to L and the perpendicular to K meet on the perpendicular bis sector of BC by taking the homoid scale to a but again this doesn't seem really helpful because we still have those mid those relations with the midpoints.

[00:59:45]So bas basically any point on this line here we want to show that the line here is just or actually we just want to show the antipode is this the on the perpendicular bis sector Hold on.

[01:00:26]If I intersect K with this side here at this point X, then this one here with point Y, then wait, I think I see where I think I like where this is going cuz we let this be X and let this be Y.

[01:00:45]Then because these green angles here are equal, we know that uh B X and L is cyclic.

[01:00:53]Because the blue angles are equal, we know that C Y M K is cyclic.

[01:00:59]And we because the red angles are equal, we know that BXY C is cyclic.

[01:01:16]Yeah, I was considering using radical center on these two circles, but the problem is that yeah, the obvious problem is that these two circles don't really have a nice radical center or that these two circles don't have a nice radical axis.

[01:01:50]And we also know that by reusing theorem that X Y MN is cyclic.

[01:01:59]Oh, hold on. The radical axis becomes a lot nicer, I think.

[01:02:24]Okay, wait. If you apply radical axis on these three circles here, then you get that uh xn and ym me and the radical axis of the two circles end up meeting at a point on the radical. Wait, so a lies on the radical axis of the two circles?

[01:02:52]Wow, what a surprise. So a lies on the radical axis of the two circles, but and so does this other point here.

[01:03:07]I'm going to go ahead and highlight the quadilaterals in purple so I don't forget what they are.

[01:03:13]So So the line through A and this point here is just the radical axis.

[01:09:08]Just a second.

[01:09:31]there. We're going back.

[01:09:40]Can we see anything else with these quadrilaterals?

[01:11:46]Wait.

[01:11:48]Angle XL LB is equal to angle X NB, which is angle A and B. And this angle's fixed.

[01:12:31]draw that in.

[01:16:33]Wait a second.

[01:16:35]So, hear me out. So, if I extend this line here, oh, let me mark it in orange.

[01:16:42]So, if I extend this line here and end up letting this circumcircle intersect this line C Y again at this point L prime then L prime also satisfies like the problem condition. Technically it does cuz I mean like the points L and R O L O L O L O L O L O L O L O L O L O L prime are kind of equivalent which would mean that I think yeah I think this would mean that A prime KL is cyclic which seems completely random.

[01:17:38]Oh yeah, I think I know that. Okay. So if I extend this line here and then construct the other circumcircle.

[01:18:00]So this humongous one and they intersect they intersect somewhere like over here.

[01:18:11]Then this is yeah this is a very very bad diagram. My bad.

[01:18:17]But like if I intersect it at this point K prime then K L K prime L and A should be all conc.

[01:18:26]And I know that KL, K prime, and L are conc.

[01:18:32]I actually don't know how I never labeled this point before.

[01:18:36]But because if we label this intersection point Q, then KQ * Q prime by that one cyclic quadrilater by that one cyclic quadrillateral is equal to it's equal to YQ * QC, which is XQ * QB. And that is K Q prime and that is yeah L Y times L Q * QY.

[01:19:06]Okay. So K, L, K prime and L primer.

[01:19:11]The obvious question is to show that A also lies on this circle.

[01:19:34]Yeah, I think that my motivation for like this keep this working out. I have no idea if this will work out, but like my motivation for doing this is like the point L and L prime are equivalent. So the problem statement would just still hold if I replace L with L prime or something like that. I actually don't know at all.

[01:21:23]I wonder if it helps that A lies on the radical axis.

[01:26:38]Okay. So, one thing I still haven't used yet is the fact that M and N are the midpoints of the sides.

[01:28:53]Oh wait a second.

[01:28:56]So we know that power of a with respect to this circle here. So power with respect to omega 2 of a is equal to That would be if we let this intersection point be Z then a Z * A C.

[01:29:17]But the power with respect to omega 1 is equal to um ax * a n.

[01:29:28]So that means that this point z is actually the midpoint of ax which is really surprising.

[01:29:36]So in other words, if we construct a midpoint Z of the side AX, then this entire circle here is then then this and Z is on this orange circle.

[01:29:50]And yeah, it's the diagram is becoming worse by the day.

[01:29:55]Like now our circle has like a vertical side.

[01:30:01]Actually, let me try redrawing the circle.

[01:30:14]That might make it easier.

[01:30:38]then it doesn't pass through K. I think this diagram is just cooked.

[01:30:45]I'm just going to draw it like very very weirdly like that.

[01:31:10]You know what? Screw it. I'm going to draw a new diagram.

[01:31:17]Hopefully, it will be easier to draw this new diagram given that I know a lot more information.

[01:31:28]So then m and n are on the sides and we have the points x and y up here. I know such that this is cyclic and I constructed x and y with that in mind.

[01:31:52]So if I draw in those two lines, let them intersect at a point Q and then A and B is this length here. So X and B then.

[01:32:17]So if I draw that in Then this is the point L.

[01:32:28]Actually, that's accurate. It's kind of lopsided.

[01:32:33]So, if I can draw this in correctly, this should be good. In which case, this is L.

[01:32:48]And then then I have to draw in uh so Bxn then C YM.

[01:33:00]So that's C YM then YM and then it intersects this side here.

[01:33:32]Okay, surely this is a bit better.

[01:33:38]Actually made made a big mix big mistake. I need to draw these circles in in a different color so that like I can look at them more easily.

[01:34:16]And this does not pass through the points.

[01:34:44]And I erased line. See why in the process This is L. This is K.

[01:35:10]That's a circumc center. We want to prove that it's equidistant from M&N.

[01:35:15]Okay, cool.

[01:35:21]Okay, now I can go ahead and rehighlight these quadrilaterals.

[01:35:26]There, there, there, there, then there, there, there, there. And I constructed them in mind with them being cyclic. So hopefully this is better now.

[01:36:02]These two circles seem very wrong because A is not on the radical axis.

[01:36:08]H I'll see if I can redraw the circles so that they actually are on the radical axis.

[01:37:16]Okay, this should be a bit better.

[01:37:20]So that AQ is a radical axis.

[01:37:29]Then I draw in these guys and they intersect at L prime and K prime.

[01:38:58]Oh yeah, I also know that I guess I also do know that it intersects the midpoint. So Make these circles.

[01:39:27]Slightly flatter top.

[01:39:32]Slightly flatter.

[01:40:59]Okay, let's try restating the problem because like I might be going too far in a wrong direction or trying to over complicate it. It feels like I have all the main constructions.

[01:41:11]Okay. So let ABC be a triangle and let X and Y be points on AC and AB such that B C XY is cyclic and let the two segments BX and C Y intersecting Q.

[01:41:26]Let the circumcircle of Bxn intersect C Y at point L and let this circumcircle of CMY intersect segment BX at point K. We want to show that M and N are equidistant from the circumc center of AKL.

[01:42:18]Now, if we can prove that A lies on the KL KL prime circle, then we're set or we might be set because then we eliminated the point A from the problem entirely perhaps.

[01:43:51]Okay, if I let this point be uh suppose m prime and n prime then because axy is similar to ABC we know that when we do this opposite similarity m prime gets sent to m and n prime gets sent to n.

[01:44:09]So this means that like let me switch to a different color now that angle y m prime x is equal to angle b mc and y prime x is just y prime c which is oh yeah that's just going backwards.

[01:44:39]It's my bad.

[01:45:57]No, MN prime is parallel to BQ.

[01:46:06]Oh, so MM prime K prime is an isosles trapezoid.

[01:46:31]Yeah. So L prime N prime NL is an isosles trapezoid.

[01:46:41]Oh wait be okay so because ln n prime l prime is an isosles trapezoid the perpendicular bis sector of ll prime is just an l so that's the perpendicular bis sector and then the perpendicular bis sector of the other one is mm prime and because m n prime m prime n is cyclic that means that the circumcenter of kl K prime L prime does in fact lie on the perpendicular bis sector of MN. So yeah, if we prove that A lies on the circle, we are set.

[01:47:19]Oh, what a finding.

[01:47:31]Yeah, this does seem like a really really nice finding, but again, I think I'm over complicating things way too much.

[01:47:41]Okay, please let me prove that AL prime KL is cyclic and let me just end the problem here.

[01:47:50]Oh my goodness.

[01:48:56]Please let me finish the problem.

[01:52:44]Oh yeah. I I guess that because of this uh perpendicular bis sectors fact, the fact that the center here is just a circumcenter of MN, Mr. n prime.

[01:52:54]We have that if like this point O prime here is a circumc center of B CXY then O should be the midpoint of that of that Let me put this little point.

[01:56:52]So I wanted to add in the midpoint of B Y here and show that A is conclic with that point that point and that point.

[01:57:00]Again very bad diagram. So, Okay.

[02:00:27]Okay. So, approximately 2 and 1/2 hours remaining and I've reduced it to this one thing.

[02:02:26]Let's constructing this midpoint here again.

[02:04:19]Okay, let's try focusing on BX BCXY for a second.

[02:04:38]Then L lies on the circumcircle of Bxn.

[02:04:52]Okay, let's go ahead and look at this like sub diagram here.

[02:04:56]So, so we have like a cigar quadrilateral here.

[02:05:12]You got a that's y z.

[02:05:20]Then you construct the midpoint of cx.

[02:05:28]And then you look at the circum this is n or yeah that's yeah that's uh actually you look at the midpoint of AC so that's N and then you consider the circumcircle of B and X and also if you let O be the center here you also look at the circum center of you also look at the circle with diameter A.

[02:06:01]And the question here is must the radical axis of these two circles be the line C Y.

[02:06:20]C lies on that radical axis because if you let like as I said earlier Z be this midpoint then you know that the circle of with diameter AO is just A O C A O Z so the power is CZ * CA and that's equal to CX * CN.

[02:06:41]So C lies on the radical axis.

[02:06:45]The grand question is does Y Oh, it also passes through this other midpoint here.

[02:07:16]So the power with respective to the circle diameter ao of y is half of a. So 12 of a y * y b then you have the circum circle of bn x.

[02:07:50]How do we find the power of y with respect to that?

[02:07:55]Like if we could show that these midpoints here are also colar midpoint of a y midpoint of acx.

[02:08:06]Oh wait this just does it. Oh, I think I solved it because Okay, so basically so we have that this point mid point a y here we have that if we like we have this sub diagram here and when we have the sub diagram we can prove that that the that l that a o prime lz is cyclic Because by we can do this by proving that c y is a radical axis of the circle with diameter a o prime which is a circumcircle of a o prime z and also the circumcircle of b and x.

[02:08:53]So we and we want to find the power and the power of point C with respect to each circle is just CZ Z * ZA which is C CX * CN and then the power of Y with respect to both circles is if we add in this midpoint here then we have that then the power with respect to circumc circle of bxn of y is if we let this circle intersect here then it's just the midpoint.

[02:09:30]So that's the midpoint by power of a point. So that means that the power with respect to that other that one circle of y is half of this quantity here. And same thing goes for the other circle because the foot from O to this side here is just the midpoint of B Y.

[02:09:51]Okay. So therefore you prove that A O prime LZ is cyclic.

[02:09:58]And this also means that okay so this means that A L O prime is equal to 90° and then similarly a ko prime is equal to 90° as well.

[02:10:12]And therefore O prime is O prime is actually the antipode of A with respect to the circumcircle of AKL.

[02:10:22]And because O prime is on the perpendicular bis sector of BC by by construction because it's just the circumc center of that circle.

[02:10:32]We know that O prime B equals O prime C.

[02:10:36]And then we take a homophy scale 1/2 to get that OM equals O N.

[02:10:42]Okay, but I just need to make sure that we're not going backwards anywhere cuz going backwards is definitely a huge pitfall in this proof. And I don't want to say that I solved it before I confirmed that I'm not going backwards.

[02:11:26]Okay. So the first thing we did is construct the points X and Y and note that BX YC is cyclic. Then we also noted that X and LB is cyclic as is uh as is YMKC and then we constructed the circum center of BYXC which is O prime and we end up proving that a prime lz is cyclic by proving that it lies on the radical axis. Um, oh yeah, I constructed this midpoint over here in this proof here. My bad.

[02:12:40]So we can actually just construct this other point W. I forgot to construct that because it was very off diagram. So if I do that then that. So that's W.

[02:12:51]Then we note that the circumcircle of a z prime w and the circumcircle of b and x has radical axis c y because c has equal power with respect to both as does y because of this midpoint observation.

[02:13:14]And this midpoint here lies on the circumcircle by a power of a point and then we have anyways because we have the they're on the radical axis we therefore know that L lies on the circumcircle of A prime Z which means that AO prime is 90°.

[02:13:44]as is angle a ko prime and this just solves a problem. Okay, I'll go ahead and write it up. I'm very convinced that this works, but I'm going to take around uh 30 minutes to write it up cuz I want to make sure that nothing goes wrong.

[02:14:11]Dude, this is like I have no clue how I didn't see this until now. How I didn't see that circumcenter observation barley.

[02:14:29]Oh. Oh. Oh. Uh, YouTube auto translated it. That's silly.

[02:14:42]Wait, how many?

[02:14:50]Cool.

[02:14:59]Okay, let me go ahead and on my wiper back up and we'll write it up.

[02:15:07]Dude, this solution is so strange.

[02:15:14]And keep this diagram here in the corner just for my reference.

[02:16:05]Oh yeah, we never even use the points K prime and L prime. Yeah, but I guess they helped in like finding the circumc center observation.

[02:16:20]In a way, this feels a lot like 2024 you move problem five where constructing where there were two basically two possible points and those two possible points gave the main construction.

[02:16:35]In this case, the two possible points K prime K and K prime and L and L prime ended up giving me the circum center that I needed.

[02:16:47]And then what's with the other ones? So B, X and L cyclic and then we have Y M KC.

[02:17:11]Oh, and that's why the midpoints are used because we're using mid the circumcenter center introduces the midpoints naturally.

[02:19:07]of a of ax. Yeah.

[02:19:58]And this should be O prime.

[02:20:59]with respect to omega is equal to cz * c a * ca which is equal to by midpoints cx * cn which is power with respect to omega 2 that's omega oh that's omega 1 my Yeah, that's omega 1.

[02:21:28]Um, omega 1 of C.

[02:21:34]And then we also have power with respect to omega of I think That should be y is equal to yw * y a which is equal to again by midpoints y n prime * y or yb * yn prime I should not mix up the order and that's equal to the power with respect to omega one of y.

[02:22:21]So c y is the radical axis of omega and omega 2. So, and because we know that L lies on the radical axis and L lies on the circle omega 2, we have L is on omega.

[02:22:51]And as such, angle A L O prime equals 90°.

[02:22:59]And similarly, angle A KO prime equals 90°.

[02:23:08]So that means that K that O prime All right.

[02:23:59]Okay, we solved it.

[02:24:01]Yay! We solved it.

[02:24:25]Okay, I'm going to take a bit of a break right now and look at the chat.

[02:24:35]Wait, Zongu is still here. What What was the first reduction step that I did that you were pointing out to Okay.

[02:26:19]Okay.

[02:26:22]All right. I think I should look at problem three now. At least give it a try. There's a good chance I may not be able to do it, but hey, I might be able to sweep. Who knows?

[02:26:33]Okay, Blue Bang and Xangu have a stick of length one. Want to divide it between the cells?

[02:26:42]Leo marks at most n points on the stick and then jang marks at most n points on the stick. So n and n mark points are distinct. Then the stick is cut at all marked points creating a number of pieces. Afterwards they take turns claiming any unclaimed piece of the stick with Leo going first.

[02:27:05]Each player's goal is to maximize the total length of their own pieces.

[02:27:12]So, what's the largest C that we may guarantee a total l of regardless of Young's play?

[02:27:38]Oh, I shouldn't look at the day two problems. I have to stay up here.

[02:27:56]I guess I should make some progress on this or the most I can do.

[02:28:03]So, Oh, Leo goes first. Okay.

[02:28:33]Okay. So, what if n equals 1?

[02:28:36]Then we have a stick here.

[02:28:42]stick is of length one. Yeah.

[02:28:47]Suppose that Leo goes here first.

[02:28:51]And if Jang moves inside this chunk here, then Leo just takes that one first and wins.

[02:29:03]Or actually, he doesn't win cuz like or like it's like very suboptimal, I guess. And then if Jung moves here, then Leo ends up taking this one and this one.

[02:29:30]So, it's optimal for him to move at the midpoint here.

[02:29:36]So that if this is X and n equals one case is already kind of non-trivial.

[02:30:18]Okay. So put that that we mark this point here and that's at point X on the stick.

[02:30:36]Then Jang can mark this one here and then we takes that one and that one.

[02:30:44]So then this total score would be x + 1 - y.

[02:30:52]And that's if y is closer to the end of the stick than to x.

[02:30:59]And if y is closer to x than the end of the stick in this chunk, then the total length is equal to y.

[02:31:09]And that's so in any case if y is chosen to be in this chunk then the total length that le can claim is at most 1 + x over two.

[02:31:23]Okay.

[02:31:25]So if I take this here, I should probably put a message in chat saying that uh saying that I'm uh that I saw P1 and P2.

[02:32:36]Okay. So now let's suppose that Jung moves in this chunk here.

[02:32:43]So suppose that's why and then there are once again two cases either y is over here like very close to that in which case the total length is y + 1 - x.

[02:33:07]So if y is greater than or equal to that's like 2x - 1 then it's y + 1 - x all over 2 and then when y is minimized it becomes x the length is x.

[02:33:32]So then we get 1 + x / 2 then x then if y is in this other is further away then jang just takes the and we're going to assume that x is greater than 0.5 for our purposes because it's just symmetric and then in which case Jang then in this case Lou just claims the length of oh wait I mess it up I messed it up uh if y is in this lower half then we claim a length of x regardless because this length is larger but if y is in this other part and we claim those other ones. So if y is less than or equal to 2x - 1 which is less than or equal to let's see oh it would be 1 + x over it would be x over 2.

[02:34:51]Then the total length we claim is equal to y + 1 - x which is then 1 - x / 2.

[02:35:02]So I'm going to do dude 1 - x over 2 x over two then x then 1 - x. So that's 1 - x over 2.

[02:35:21]And then finally if y is in the other chunk where or actually Now finally, but if y is over here such that this chunk here is the largest and then and then it's this chunk then it's that chunk over here and we end up getting a length of okay if we get x then y then okay so we first we claim end up claiming these two chunks assuming that this length here is larger. So in that case that it's 1 - y and the only way that this length can be larger is if it's y = 1 - x.

[02:36:13]So then in this case we get x again.

[02:36:16]And finally if y is on this side here then we end up claiming this one and this one. So that's just x again. So therefore the what Luke claims is going to be the minimum of 1 + x / 2 1 - x / 2 and x.

[02:36:43]We want to maximize this minimum.

[02:36:52]Okay, let's graph it.

[02:36:55]So that's one. That's one half.

[02:37:00]And then this is one. This is one up here.

[02:37:05]And then 1/2 34s all the way up to one. So that's one line.

[02:37:15]1/2 then 34s. Then down there another one. And then finally x.

[02:37:26]So it seems like the 1 + x over two is always a larger one.

[02:37:31]So therefore when we let x be basically as large as possible, we end up getting a length of 1/2.

[02:37:50]And when x is as large as possible.

[02:37:54]Oh, wait. We're looking at the minimum.

[02:37:55]My bad. We're not the maximum.

[02:38:01]So something like this.

[02:38:04]So this is our minimum function. So x is equal to this downward facing y 1 - x / 2 when x is equal to 2 over 3. And therefore the maximum we can obtain is 2 over 3. obtained when taking x= 2 over3.

[02:38:22]Honestly, I kind of expected this like I kind of expected this to happen where you do it at where the optimal strategy for uh Leu is to take the points at the equally spaced parts and then we can claim at least two over three somehow.

[02:40:32]I should probably take the time to reread my problem one and problem two solutions.

[02:41:39]Okay. So we first consider prime by prime.

[02:41:56]and BPM= X then yeah and then we're just using uklidian algorithm at some point should be instead at some.

[02:42:34]Yeah, actually I guess like at some point works.

[02:42:39]So then zero is so that terminates. And finally, because we used we did this process with all the BBs, we get that all the numbers on the board are relatively prime and their product is that and then we're done.

[02:43:22]All right. I guess we can continue looking at it. Let me put the problem up here again.

[02:43:39]Um, Let me see if the Okay, now I'll see if the x= 2/3 work. x= 2/3.

[02:44:19]Oh, so what my math says is that this point here, Leo choosing that when n equals 1 should be the optimal strategy.

[02:44:31]And then with that, Leo can always guarantee 2/3.

[02:44:38]So if Jang moves in this chunk then Lou just claims the big one. Otherwise, and then if it's closer to this one in the chunk, then we again win.

[02:45:14]Suppose this over here.

[02:45:20]Oh, and then this zone here is just zero.

[02:45:24]So that means that in this zone here, we have that this is the largest chunk.

[02:45:29]This is the second largest and this is the smallest.

[02:45:40]Oh, I think I see what happened here.

[02:45:42]So, because when we're moving in this chunk, we have that it's always going to be these two are always going to be the largest and smallest in some order because they form an arithmetic sequence with 1/3.

[02:46:01]So, that's why we're able to guarantee at least 2/3.

[02:46:07]And obviously when we move here then then we just win.

[02:46:13]Okay.

[02:46:18]Now the case uh yeah the case where n is greater than or equal to two is definitely a bit more interesting. So I'm going to put this down for now. n= 1 gives the answer 2/3.

[02:46:36]Now let's look at n= 2.

[02:46:55]And I could honestly just bash this.

[02:47:24]I would expect the answer this time to be uh something like three fifths.

[02:47:49]Suppose that the truth is this point out here.

[02:47:54]And this point out here where this is 1/5 then Jang picks uh two other points and then we choose the largest the middle one and the smallest one.

[02:48:30]What's a good way to distribute these lengths here such that we always have this idea of like the arithmetic sequence cuz it worked out really nicely with the other one.

[02:48:48]Actually, we can guarant we cannot guarantee a naratic sequence with this one because we have two because Jung has two points.

[02:50:00]Okay, let's just take cases.

[02:50:04]This will be a humongous case bash.

[02:50:23]or it may not be a case a huge case bash. We'll we'll see.

[02:50:34]Okay. First if if Jang moves in these here then that means that we can guarantee a length of at least four fifths.

[02:50:46]And similarly if we move here so really we and if Jang moves here and here we can already guarantee the length of three fifths as we need.

[02:51:02]So let's go ahead and assume that there's a point in here.

[02:51:06]So there are two cases we need to consider.

[02:51:09]One is where we have length in here and the other is we have two points in here and the other is that we have one point in here and one point in there.

[02:51:20]So first suppose we have two points in here and these are links X Y and Z.

[02:51:30]Then we need to order X, Y, and Z somehow. And then take the smallest then largest then middle one of the 1/5.

[02:51:40]Yeah.

[02:51:44]Okay. So the largest one of the 1/5 So the largest one of these five is guaranteed to be one of X, Y, or Z.

[02:51:52]Because if all of these were smaller than 1/5 then then this the length of the segment is less than one. So suppose that like Z is the largest one and then and similarly the length of the smallest one must be either X Y or Z because if 1/5 of the smallest is the smallest one then the length is greater. So suppose that x is the smallest then either y or 1/5 is a middle one.

[02:52:24]So if y is a middle one then we guarantee at least uh z + x + y which is 3 fifths and if z is a and if 1/5 is a middle one. So you have something like x y 1/5 and then z.

[02:52:49]Oh and then 1/5 is just greater than y.

[02:52:52]So that means that we again get greater than uh 3 fths.

[02:52:57]So in the case where the two points are inside the middle of that chunk, we get that three fifths is possible.

[02:53:06]So, let me take these away.

[02:53:20]And then suppose that this time we have one point in here. Oh, I got rid of the entire line.

[02:53:28]Can I get rid of this one squiggle here?

[02:53:31]There.

[02:53:34]And suppose now we have one point here.

[02:53:37]Oh, I have my eraser. And one point here. So that's 1/5, 1/5.

[02:53:44]And I'm going to go ahead and mark these up. So this is going to be x, then y, then z, then w. So x + y = 35ths, and z + w = 1/5.

[02:54:00]And we want to do stuff with this.

[02:54:04]So once again, what's the largest one?

[02:54:08]Well, the largest one again has to be one of XYZ or W because if the largest is 1/5, then then you're cooked.

[02:54:18]So that means that suppose without loss of generality, actually we can't assume without loss of generality because X + Y is fixed and so is Z plus W.

[02:54:31]Actually wait because this length is 3 fths one of x over y is greater than 3/10th and 3/10 is greater than a fifth. So that means that it's already going to be larger than z and w combined.

[02:54:46]So we can assume without loss of generality that Y is the largest one.

[02:54:53]And then similarly the smallest one is one of XYZ or W as well.

[02:54:59]And again we can assume with alpha loss of generality that W is the smallest one. So we have W Y and then X5 and Z here somewhere.

[02:55:20]And now we have to figure out which one is the middle one.

[02:55:41]So Z is smaller than 1/5 and yeah Z is smaller than 1/5 by because it's inside this segment here.

[02:55:56]So that means that 1/5 can't be over here. It has to be either Z 1/5.

[02:56:07]Yeah. So Z and 1/5 have to be in that order.

[02:56:44]Wait. Actually, wait. Let me get some water real quick. Okay.

[02:58:51]Okay, I'm Okay, so I can now erase this.

[02:59:06]Okay, so back to this problem here.

[02:59:51]H.

[03:00:01]Okay. So suppose that uh Z is the middle one because then that means that this is yeah if Z is the middle one then you're then you're fixed.

[03:00:18]So Z 1/5 then X then you're looking at W plus then you want to minimize X + 1/5 or at least J wants to because X is a larger one or X or J wants to maximize it my bad.

[03:00:47]So then by setting x= 3/10.

[03:00:57]Wait. Oh yeah. Oh yeah. X has to be less than 1/5.

[03:01:03]Yeah. X has to be less than 1/5 because X is because X is less than Z which is less than 1/5. Oh yeah because X is less than 1/5 on the list here.

[03:01:14]And that means that we want to minimize x plus one maximize x plus 1/5 and that's at two fifths. And that gives again three fifths.

[03:01:27]Okay. So then there's a case where 1/5 is a middle one and then once again z is fixed as is as is x.

[03:01:47]Then we want to maximize the length of x + z where we know that x is less than 3/10 and then z is greater than oh and z is less than 15. So that gives x + z = two 1/2.

[03:02:19]Huh?

[03:02:27]Wait a second.

[03:02:30]This might not work.

[03:02:33]This might not be our lucky construction.

[03:02:37]I erase the line and then I take get rid of that.

[03:02:43]So what we're saying is that wait let me go ahead and enable my pen again.

[03:02:55]So what we're saying is that if we let if Gang marks this point here and this point here basically or any point super super close then this is 1/5 3/10th 3/10th and then that's basically 1/5 minus some quantity epsilon and epsilon.

[03:03:21]What we're saying is that we can only guarantee 3/10 plus 1/5 plus epsilon.

[03:03:34]Okay, so this doesn't really work.

[03:03:39]So, and it seems like this 3/10th just cooked us a lot because 3/10 + 1/5 is greater than or actually wait I'm wait I'm dumb. I am dumb because actually whenever we have our points because 1/2 is less than three fifths. Yeah, actually wait. Never mind. I'm actually not dumb.

[03:04:15]The most that John can confirm in this case. Yeah, I'm not trolling.

[03:04:25]Yeah, sorry about that. Was just was just blinking there a bit for a second.

[03:04:45]And then this means that we have to get 3/10 + 1/5 which is 1/2 which is less than 3 fths and then plus some small pocket change.

[03:05:00]So let's assume that for now we can. So this doesn't work and yeah again the 310 is what cooked us.

[03:05:27]So suppose that we let it be at so that's 1/5 this is two fifths this is two fifths then in this case suppose that it's like that and that that and here.

[03:05:58]Then it's then we clean the two fifths first and then we get the 1/5.

[03:06:10]So that so the strategy we presented earlier with the 1/5 35th one case distribution doesn't exactly work here. So this is very promising.

[03:07:11]So you have two 1/5 two- fifths two- fifths. Now let's do the case bash again.

[03:07:21]Let me erase this.

[03:07:27]that.

[03:07:29]So now the case mesh.

[03:07:36]So if I Okay. So if I choose two points in here, then we again snatch the three- fifths as we need.

[03:07:47]Otherwise we and similarly if we choose to point it here, we again snatch the three-fifths.

[03:07:58]Actually, we don't immediately snatch the three fifths H.

[03:08:17]So this is again non-trivial. So suppose we have that this is X, Y and Z.

[03:08:25]We want to snatch it. And that's a total length two fifths.

[03:08:39]Then we know that two- fifths is the largest.

[03:08:42]So two fifths plus and then one of x, y, or z is the smallest.

[03:08:51]So I'll do plus z. So I'm going to do Z then two fifths. Then what we have left is uh X Y and two fifth and 1/5s.

[03:09:06]And then there are a few cases from here.

[03:09:10]The first case is where we have 1/5 is the largest and then X and Y are this part down there. In which case this is 21 15 plus 25ths plus another 21 15.

[03:09:44]And actually the most we can guarantee with this strategy is we want to maximize the length of the middle segment which is X.

[03:10:00]This length of the middle segment.

[03:10:15]Oh, it can be just half of that. So that's 1/5. And then you get again get three fifths.

[03:10:23]Then there's a case where 1/5 is in the middle.

[03:10:30]In which case you get that is x then y then you already claim three fifths.

[03:10:38]Yeah. And then if 1/5 is on the side here then you again get 3 fths because it's larger than 2 fths plus 1/5 plus c.

[03:10:47]Okay. So in this case where they're on the same side of it or they're on the smaller branch then we solve the problem.

[03:11:02]The more interesting case comes when we look at when both of them are in this cuz obviously first Gang picks this one then no first Leo picks this one then Gang picks this one and then what we're left with is we pick two of the links in there and they have to sum to 1/5 and otherwise you're clipped.

[03:11:36]And this is not always possible sadly.

[03:11:49]Okay.

[03:11:55]So this means that this once again does not work.

[03:11:59]Sad but true.

[03:12:06]What if we try making it like centered?

[03:12:10]This could this could be a good idea. So let me go ahead and erase this entire thing.

[03:12:19]Then we have shoot uh we have this line here.

[03:12:26]This is two fifths and this is three fifths. So now we have two fifths 1/5 two- fifths.

[03:12:35]This could be the one that ends up working.

[03:12:39]And this looks very very symmetric. So yeah, so we don't have to deal with many other random cases.

[03:12:48]Anyways, suppose that we have x and the two points are here first. Then we have x, y, and z where x + y + z = 2. And this is just basically the same as the case we did earlier cuz the 1/5 and two fifths are swapped.

[03:13:10]And then I'll mark the ends of these.

[03:13:13]Now suppose and the same case is that if those they're both at that end. Now what if there's one at each end then we have x y z w and a 1/5 in the middle x + y = 2s and z + w = 2 fths. Can we guarantee that we can that we can always pick two of these at sum to at least three of these at sum to at least three fifths the largest middle and smallest.

[03:13:54]Okay. So once again just like in the other case if 1/5 is a is the largest one then that means that the total length of the stick is less than one.

[03:14:06]So that means that one of XYZ or W is the longest one. So assume without loss of generality that it's W.

[03:14:15]And then we also have a smallest one.

[03:14:18]This time we actually Z let's assume that Z is the largest one.

[03:14:23]Then we also have without loss of generality a smallest one.

[03:14:28]And sadly we can't really take cases on that.

[03:14:48]Oh wait, one of these is also larger than 1/5. So suppose that is zy and then the other two are smaller. So 1/5 then x and w suppose or something like this.

[03:15:07]And this Z and Y can be without loss of reality on and that W X and W cannot. So if it's W down here then it becomes three fifths as we need. And again if it's oh and then if it's oh oh wait if it's X then W then this might be an issue. So, so X is less than W and then Y is less than Z.

[03:15:41]Oh, and that just gives a contradiction.

[03:15:45]So, it has to be the W and the X and that gives three fifths.

[03:15:56]Okay.

[03:15:58]All right. Cool.

[03:16:00]Oh, yeah.

[03:16:02]So now we've solved the case where it's like one on either side.

[03:16:10]So now we can safely assume that there's always going to be one in the middle.

[03:16:34]And this point to be erased here.

[03:16:49]Okay. So now suppose that we have a point here in the middle or we have two points here in the middle.

[03:16:57]Then two- fifths and two-fifths are the two largest ones. So we claim those. So those are claimed. And then we once again run into the issue. Dang it.

[03:17:08]Dang it. We ran into the issue again.

[03:17:34]I think the problem with having two lengths of two- fifths is that Jang can just guarantee to pick one of them up and then you're cooked for the three-fifth strategy.

[03:18:00]So you don't want two lengths of two in your original construction.

[03:18:22]Honestly, let's just go for a more symmetric strategy here.

[03:18:26]So, let's go ahead and erase all this.

[03:18:31]So, and let's just split it into 1/3, 2/3, and 2/3.

[03:18:43]This is just like a last effort strategy before before I stop working on this problem.

[03:18:55]Okay. So 2/3 no 1/3 and 1/3.

[03:19:05]So once again there are two strate there are two cases here.

[03:19:10]One is where they are both on the same segment. So we have x y z and in this case we guarantee a length of at least 1/3 + 1 9th which is 4 9th which is sadly not greater than 3 fths and that means that 3 fifths is probably not the answer. I'm very convinced that 3 fths is not the answer for n equals 2.

[03:19:38]And then we have the case where they're on different segments. So xyz w each of their sums is 1/3.

[03:19:46]So it's going to be y z 1/3 x w something like that. And then that gives a length a sum of 2/3.

[03:20:06]Oh wait, it's not 4 9ths. It's 5 9ths.

[03:20:10]and still that's not greater than three fifths.

[03:20:14]So the answer in this in this configuration is 5 9ths.

[03:20:52]And but like the problem with going doing the equidistant strategy is that this does not always work. For example, when n equals 1 as I worked out earlier, we can do better than this. We can do better than just looking at the 1/2 case because we can always guarantee a length of at least 2/3.

[03:21:26]Weirdly enough, actually, wait, I'm starting to question even that now. Actually, yeah, we can.

[03:21:35]My bad.

[03:21:51]I was going to say 1 one and 1/5 35th as a last effort strategy, but then now I realize that this is equivalent to the 1/5 35th 1/5 case I tried earlier.

[03:22:11]I want there to be a way where I can just snatch away the one bit or Yeah. Or I can just snatch away the one fifth or the snatch away the three fifths. I meant just like we could do in the n equals 1 case.

[03:22:36]That's Like I genuinely don't know what I should do.

[03:24:04]N equals 2 is tricky, but once you get the construction, you should get an idea of how to do this.

[03:24:26]Okay.

[03:24:28]Yeah, I'm very I'm like very very convinced that the answer is 3 fths for n= 2 and in general it's like n + 1 over 2 n + one for general n.

[03:24:42]Like I agree that like if I figure out what the construction is for n= 2 then this should hopefully generalize to higher end as does the bound So, like I'm still trying to figure out how we can like snatch away the three folks that we need.

[03:25:57]And it seems like the the bad case happens when when Jang just just carves into the smallest segments because then you're just plain like you're just trying to take things away from the largest segments and then what's left is you're just battling it all left among the smallest ones.

[03:26:23]So it's like you're just splitting up the smallest segment into many many ones.

[03:26:30]And that's where the all the strategies are failing. If you take if you're looking at the smallest segment then you're kind of cooked.

[03:26:44]So, you want to figure out a way to make sure that even if you carve out the smallest segments, you're still fine.

[03:27:01]And in the 1/5, 35th, 1/5 case, even if you carve out the carving out the smallest segments still gives you the three fifths that you need.

[03:27:15]But what if you could h what if you get exactly the three-fifths bound that you need by doing this the smallest carved out strategy? What if you consider the smallest one for which this possibly works?

[03:27:34]So suppose that we have our points.

[03:27:40]Suppose we have our first point x here and a point y here.

[03:27:48]Then we end up carving out of these two.

[03:27:52]And then we take the largest one of these two and the largest two of these eyes here.

[03:28:22]the largest value of the middle one is 1/2.

[03:28:25]So you're basically taking uh 1/2 of this plus 1/2 of this or something like that or maybe not one half of this. So you're taking like x over 2 plus like okay suppose that this is x y z. So x over 2 plus max of y and z and and we're assuming that y and z are both greater than x.

[03:29:29]So what if you do like 1 - So x over 2 + 1 - 2x 1 - 3x / 2 = 35ths 3 x / 2 = 2s x = 4 over 15.

[03:29:47]4 over 15.

[03:29:53]We have the two and then the so we do four fths, four fths, 41 15s, 41 15 and 7 15. Then we claim the 71 15 immediately and then we look at the so it becomes like this might work doing four fifth 41 15 4 15 and 7 15 16 Okay, so let's go ahead and erase this.

[03:30:34]Yeah, 41 15 and 7 15 might work.

[03:30:40]cuz this in a sense it's the one that like barely works, you know.

[03:30:47]Oh, I'm in eraser mode.

[03:30:53]So suppose we have like 4 15 4 15 4 15 7 Okay. So, if you carve one out of each of these guys, if you carve out two points inside this segment, then as I said earlier, you're done.

[03:31:38]If you carve out two points in this segment, then um then you can do 4 15 um oh wait so 71 15 / 3 is 745ths.

[03:32:01]Actually 415 actually this case is non-trivial because 7 cuz the largest one could be in the 71 15.

[03:32:20]In every single case though, uh, both Leu and Jang are going to claim four 15s.

[03:32:31]So Leo is going to claim two of these and Jang is going to claim one of them.

[03:32:40]So we have 4 15 and 4 15 are in here.

[03:32:44]That's X, Y, and Z.

[03:32:47]then then Leo claims X and Z and similarly if they're on the other side then if it's 4 15 4 15 so basically Jang just claims the outer two in these two in all cases so that just gives 4 15 plus uh 7 15 by 2 which is 830ths + 7 over 30 which is 12 1/2 Oh, cuz then Jon can carve into this one.

[03:33:58]Ooh.

[03:34:23]Oh yeah, you're cooked. If you if you had to select only one of these four 15 Wait, how is this or this? Doesn't even work.

[03:34:59]The 415th one.

[03:35:13]Joey, how does this even work? This is not even worse because like the 415 strategy doesn't even work.

[03:35:35]Let's see.

[03:35:47]Okay, back to the drawing board with the carving out idea. So suppose you have X, Y, Z. Those are our distribution.

[03:35:58]Then we take these out of the smallest segment or actually suppose that we have that those two are carved out then we take the maximum of y and See?

[03:36:40]Okay. Wait. I'm trying to look at this.

[03:36:41]What happened in this case? So 17= 2.

[03:36:47]Okay.

[03:36:53]I'm slowly becoming less and less convinced that the answer is actually three fifths or n= 2 So, it's like either carving out of the small segment or carving out of the big segment are the ones that fail.

[03:37:36]Cuz if you carve out of the small segment, then at some point you take Jang takes the half of the big segments away and you're only left with a pocket change. and you can't and you can't get to the three discs.

[03:37:52]And meanwhile, if you carve out of the large se large segments, then you're left with only small segments to work with.

[03:38:05]So you want to somehow stop both of those from happening.

[03:38:11]So x over 2 plus max y And then if you carve out of the biggest one.

[03:38:37]Then what you're left with is all these very small segments.

[03:39:20]If you carve out of this larger one, then you might get into issues.

[03:39:37]I think this I think this the solution to getting around this is to make the smaller one smaller, the smallest one smaller and the largest one larger. So you So if you carve into the smaller one, then you're guaranteed to get the larger ones first. And if you get And if you carve into the larger one, then you can still win.

[03:40:01]So you want this to be equal to like three fifths.

[03:40:07]somehow.

[03:40:15]So that means that max y z is equal to 35ths - x / 2.

[03:40:23]So we have x uh 2 fths + x / 2 35ths - x / 2 and and because we have this we have x is less than 1/5 And if it is 1/5 then we get 2 fths + 110th which is 12.

[03:41:09]Then 3 fths minus 1/10th is again 1/2. Huh?

[03:41:25]Oh, I can't subtract.

[03:41:28]Oh, I'm dumb.

[03:41:30]It's a minus sign.

[03:41:34]So, we have both of these guys.

[03:41:42]And in the case where those two were equal over here that gave the x= 415 case which sadly failed.

[03:41:55]So what if we make x like and we're assuming that x is the smallest. So yeah x is less than 41 15.

[03:42:06]So what if x is smaller than that?

[03:42:15]Okay, let's figure it out.

[03:42:17]What could be the ones that work? So we have x then 2 fths - x then 3 fths - x construction 1/5 1/5 35th fails for the same reason that you realize for 41 15th 415 71 15th yeah exactly it's because like you either carve in to the large segment or you carve into the life segment and you're stuck with 1/5 plus actually 1/5 plus 1/2 does work but yeah it's 1/5 plus three yeah so so the construction for yeah basically the idea of carving out the large one so how can we get away with this H.

[03:43:20]So I want X to be like sufficiently small.

[03:43:40]Wait, actually something weird is happening. Oh, this is X over two and X over two. Yeah. Yeah. Yeah. Yeah.

[03:43:55]Okay. And now we're left and let's see what other progress I can make with the n equals 2 case.

[03:44:32]Okay. So, let's carve out of the large one.

[03:44:37]So, we have like So, we have like Oh, why is it moving?

[03:44:42]Right. Anyways, so we have that and that.

[03:44:50]So the largest one of these is Yeah. So the largest one of these is going to be uh so we have like 1/5 - x / 6 or something like that.

[03:45:18]Actually, let's set x to be like an epsilon.

[03:45:24]So like instead of like x being somewhere like over here, we have like epsilon 2s - epsilon and then 35ths minus epsilon / 2.

[03:45:44]Then whenever we split this, we're guaranteed that one of these guys is the largest. So if we get X, Y, and Z, we're actually we're still not guaranteed anything.

[03:46:07]So epsilon is going to be claimed last.

[03:46:09]That's what we know.

[03:46:12]Or it might be claimed somewhere like it's either going to be like last or second to Best.

[03:46:59]two fifths and three fifths.

[03:47:02]Then the smallest is going to be suppose that epsilon is like zero as a joke. Then the smallest is x. No, the smallest is zero. Second smallest is x.

[03:47:37]Then we have either y z or two fifths.

[03:47:43]And if two- fifths is is the largest one then we know that by the pigeon hole principle that one of y or z is at least 1/5. So that gives a two- fifth one. If two fifths is here and Z is there, Okay.

[03:48:42]Yeah. And two- fifths has to be here in this case.

[03:48:48]I'm just trying to get a feel for whether this epsilon thing can work out or this x thing in general.

[03:48:57]So y is two- fifths and z and we want to look at y plus or we want to uh oh yeah we want to maximize x and the maximum value of x is just best.

[03:49:51]Oh, wait. If two-fifths is over here, then you might be cooked already. But depending, this could be not so good. Uh I suppose that then can cut it like so that this is two fifths like I don't Okay. So we have epsilon 2 minus epsilon / 2. Then to resolve this case you just like take like that that and that and that's like so this is 3 fifths minus some small factor I'll call it like delta and then this is like delta delta delta So you end up claiming the should just be delta delta. My bad.

[03:51:58]It's like delta delta 35ths minus 2 delta. So you first claim the 35ths minus 2 delta then the two then John claims the two- fifths. Then you're left with one of the deltas and then the epsilon.

[03:52:18]Okay. So, you can't like take it to be too small.

[03:52:22]Okay. This is going to be a bit more complicated than I thought.

[03:52:33]And Joey probably saw this from a mile away. Like, yeah, maybe I should have seen it coming from a mile away, too.

[03:52:45]Yeah.

[03:52:47]So again, let's return to the x then 2x then you have like 2x over 5 - 1. Oh, it's 2 - 12 x Let's go back to the 415 415 715th case up here. Because if we go into here, then we claim one of the 415 and the other 415.

[03:53:34]And then what you're left with is just then this part here.

[03:53:46]So you want these two to be like relatively far apart but not too far apart because when they're too far apart the thing with the that guarantees at most two fifths happens.

[03:54:57]Yeah, like if X is like even reasonably close to 415 then it will again fail.

[03:55:45]Wait, because it seems nice. Let's try out x= 1/5.

[03:55:51]So 1/5 2 - 110th is 310th.

[03:55:59]Then you add 1/5 to that. So that's So then this big guy here is 1/2.

[03:56:09]Does the carving out strategy here work?

[03:56:16]Suppose you have like x y z x + y + z is equal to 12.

[03:56:28]So we first claim the largest one.

[03:56:37]If the largest one has len uh oh wait you can do like 3/10th then one10th then one/10enth suppose then you do 3/10th then 3/10th then 1/5 then oh then this gives three fifths exactly exactly three fifths But what if you do like then you do 3/10 then 3/10 then 1/5 then 210 - epsilon and then go hog This motivates a general strategy. What if I do uh instead of doing so 2 fths - 12 x then 110th 110th then you start off with then you claim one of these guys then you claim the x and then you claim you claim one of the two fifths one of the one10th so you get 12 - 1/2 x and then plus x which is 12 + uh x / 2 which must be less than three which must be greater than 3 fths. My bad.

[03:58:31]1 + x is greater than.

[03:58:50]Okay. Wait. So you two fthsus one and then you have the 110th and then you have the plus x.

[03:59:02]x over two.

[03:59:12]So the most we you can guarantee with the strategy here is uh the formula doesn't match for what I got here. So I think I sied oh this should be two.

[03:59:28]This should be 210 minus epsilon and epsilon.

[03:59:37]Okay. So we first take the two fifths 12x plus the larger of these two values.

[03:59:46]note then you take the larger of plus max of like x and then the 2/10 minus epsilon plus epsilon we want this to be greater than or equal to 3 fths so max of x 210us epsilon I'll reduce that actually 1/5 minus epsilon must be greater than or equal to 3 1/5 plus 12 x and this 1/5 minus epsilon doesn't work at all. So this requires x to be greater than 1/5 + 12 x or x is greater than two fths and then you're cooked because then that's not the smallest segment that you can do.

[04:00:48]So, I don't think three-fifths works.

[04:00:54]Wow.

[04:00:57]Wow. Three-fifths doesn't work.

[04:01:04]Yeah.

[04:01:18]Yeah, I know you can set epsilon equal to zero.

[04:01:21]Yeah, but I'm just saying like in case that taking epsilon not equal to zero, it works.

[04:01:32]But like I very very highly believe the answer is not three fifths now because I just bashed all the cases and this just gave something really weird.

[04:01:57]Okay, I have 35 minutes to hopefully figure out figure out a correct solution.

[04:02:06]Okay, so if the answer is not three fifths, then what is is the answer exactly?

[04:02:25]Okay, I'm going to do a bit of bashing now. So, wish me luck. Suppose that our answer is C as the problem statement suggests.

[04:02:35]Then we have X, Y, and Z. And then you have them in this order. By the cutting out principle, we know that x over 2 plus max of y and z, it must be greater than or equal to c.

[04:03:04]So that means that this has to be like C C - X / 2 and then 1 - C - X / 2.

[04:03:20]And then because X is the smallest one, we know that X is less than uh 23 C.

[04:03:34]We know X is we have and 2/3 C. Yeah.

[04:03:50]I originally had three fifths actually. Wait, hold on. Uh, did I screw something up when doing the the three fest case?

[04:04:14]Cuz then that's that's the max.

[04:04:20]Oh yeah, that's the max. My bad.

[04:04:28]So that means that it's going to be So this is C - X / 2.

[04:04:36]This is 1 - C - X / 2.

[04:04:43]And by minimum assump assumptions this means that 1 minus C is less than C. So C is greater than 12.

[04:05:01]Okay. So now we have this configuration here.

[04:05:06]Now let's see what happens if we Yeah. So now let's see what happens if we carve these out. If we take like if we take like 1 - c - x / 2 then what's left over here is 2 c - one and then this is c minus then we can do like 2 c minus one and zero Then we pick 1 - c - x / 2 first and we also claim that other one and then plus zero plus then either x or c minus one.

[04:06:10]Oh yeah and x is less than equal to 2/3 1 - c.

[04:06:15]So suppose that the largest one is 2 c minus one then it becomes c - x over 2 and then you're cooked. So that means that the largest one has to be x and this is greater than c.

[04:06:40]So 1 - c + x / 2 is greater than or equal to c.

[04:06:47]So x / 2 is greater than or equal to 2 c - 1.

[04:06:55]So 1 - c over 3 is greater than or equal to 2 c - 1. 1 - greater than or equal to 6 c - 3. C is greater than or equal to 47.

[04:07:09]is four se seven. Four sevens might be the answer. Who knows?

[04:07:29]And then that would mean that we have 27th, 27th, and 37th.

[04:07:35]Hold on.

[04:07:48]Okay, so we have 27ths, 27ths, and 37ths. Wait, 47th is correct. Let's go.

[04:07:57]Okay, well, I first need to like prove it. Yeah, like I was like relatively convinced that the answer might be 47s.

[04:08:10]Also, if I do end up solving this on stream, then I'm just going to count it as like solved with chat.

[04:08:17]Okay. Anyways, so we have 27th, 27th, and 37th. This might be our ultimate construction because if we carve out in this one here, then what we're doing is we have Then we take the we first take the 37th and then we can guarantee to take one seventh from here. So that's 47th.

[04:08:46]Then from here we end up carving out.

[04:08:53]So we first snatch the 27th and then the 27th and then we have to play this game in there. And that's 47th again.

[04:09:04]And then if we do this then this then this is xyz w.

[04:09:11]So then we had to find out some combination of xyz and w which give at least Would you give at least two sevenths? Well, I guess X and Y just do that. But so we know that X and Y are on either side of the 27th.

[04:09:59]and 27th might be at the largest one. So 27th is the largest one.

[04:10:11]Yeah, this can be really really weird because Yeah.

[04:10:49]and first let me first let me resolve this case first.

[04:10:53]So we first pick the 37ths that's given and then my pigeon hole one of these is so we have a greater than 17th here and a greater than 17th here and we just pick that one.

[04:11:08]So we solved both in this one both in this one two different ones and we solve both in this one. Now we just have to solve this last one. And then hopefully this will give the construction for sevens.

[04:11:36]Okay. So as a cheese. Okay. At this point I have 25 minutes left. And what I can do is either try finish working this out, but I'm gonna first guess the answer. If and I'm gonna for the jokes, I'm going to just guess the answer and just assume that it's one point. So if n equals 0, then you can guarantee 1, which is 1 over 1, then 2 over 3, then maybe 4 over 7.

[04:12:01]So conjecture the answer is equal to 2 the n / 2 n + 1 - 1.

[04:12:08]This is my conjecture. And if the answer gets a point, then hooray.

[04:12:14]But I'm not going to expect anything. So yeah.

[04:12:30]Oh, two three. 223 is wrong. Wait, wait. Why is it wrong for the same reason?

[04:12:54]Wait, why does two 223 not work? I literally the things that 223 worked for ended up the the things that 113 and 447 didn't work for were the things that 2 two two and that 23 got right.

[04:13:18]Oh.

[04:13:21]Oh, okay.

[04:13:26]So, all xxx 1 - 2x give 1/2.

[04:13:33]Oh, okay. Yeah.

[04:14:32]Okay. So we have this one x 1 1 - c - x / 2 and then c - x / 2 here.

[04:14:39]So if C is equal to 47 as I want then it's X 37 - X / 2 and then 47 - X / 2 Oh, wait. This is true.

[04:15:58]Yeah. The x over 2 plus max xyz principle.

[04:16:11]I'm trying to find a suitable x value of x that works cuz then x is less than or equal to 27th.

[04:16:50]Okay.

[04:17:21]Okay, if I can't get any more progress within like the next 5 or 10 minutes, if I can't get a construction for a value of X within the next 10 minutes, I'm just going to call it GG's.

[04:17:37]Hopefully, it's something of this form.

[04:17:44]But what if I do like 17th 514 and this is 1/2 and this is of the above form so it could work.

[04:18:22]Yeah. If this construction doesn't work, I think I think I'm going to call GG's because I only have 18 minutes left. So Okay, wait. Let me lock in. Uh Okay. 54 + 1/4 1/4 is greater than 314. So if you cut both into five the 51 14 one then we're done.

[04:19:49]And what if we cut into one of each?

[04:19:54]Then we start with the 1/2 and because 54 / two is equal uh no never mind. Uh oh this doesn't work cuz you could carve it like here and here. That's 114th. 114th.

[04:20:26]Then this is like zero. This is 54th.

[04:20:30]This is 1/2. Then you do 1/2 plus you'd get the 54s. Oh, this does work. Wait, what?

[04:20:43]It literally just barely works. What the hell?

[04:21:33]Okay. Uh, what I meant by GG's is like if I don't get the construction for Nol two, if this doesn't work for N= 2, I'm just going to end the timer and end it and And the mock was 770 or 771.

[04:22:11]But this like has a chance of working I guess.

[04:22:18]Oh, if you carve both into here, then you can like can you always guarantee a total length of 17th?

[04:23:27]cuz I literally only need 114th.

[04:23:36]Oh, wait. I can just get the 114th from either this one or this one by pigeon hole.

[04:23:42]Okay. So that means that he have to carve into the 1/2 one.

[04:23:59]Okay. Suppose you carve both into the 1/2 one. It's like that.

[04:24:06]Then you first take the 514s Oh, because 1/2 time over two is greater than seven. My bad. Uh, but in any case, you're only left going to get like 54 plus like 1/4, which is we need a 17th. And this is a Oh, wait. This just does it. Oh, wait.

[04:24:33]No. Yeah. 1/4 is greater than 314s. What am I doing?

[04:24:39]Okay, I'm very convinced that 17th, 514, 1112th works.

[04:24:58]Wait. So suppose you first grab the 51 14ths.

[04:25:02]Then you need need at least uh the one then you need at least uh 34s.

[04:25:16]So if you try taking the 17th then you're only going to be left with like this pocket change here. So you want it.

[04:25:24]So you don't want to take the 17th and you want Jang to take the 17th.

[04:25:31]So then you want to take two of these guys and by take and you can always take two of them that sum to at least 1/4.

[04:25:41]So that's that case. If you do this then this then you end up getting a Okay. Then 514s. Then you need a 314s.

[04:25:59]Oh, and then my pigeon hole. Uh, no, you can't use pigeon hole.

[04:26:06]Oh, this fails.

[04:26:10]Actually, be if we cut one of them here and one of them here, then it's not necessarily the case that you're going to fail. Because what I was thinking is that Jan could just take this cut here, but then you just claim the 1/2 first.

[04:26:33]So the only reason why you'd claim the 514 first is if this leng if this is here greater than 27th and we cut in here as well.

[04:26:48]So we take the 51 14s and then oh and then one of these is at least two two sevenths. So yeah and then the case where it's like that then you take this one first.

[04:27:13]So you first start off with 51 14s.

[04:27:22]No, you first start off with this. Then you take the then the other person takes the that one and then this one. This one might fail and I'm going to be a bit So So if you do like 114th So that's uh then this becomes 37th. So you first take the 37th and suppose that this is like there you first take the 37th then the other guy takes 51 14 then you then you have like a 114th there and that gets you to 1/2. But you can't take the 114th.

[04:28:19]So you have to.

[04:28:37]If you take the 37th, then you need 17th more.

[04:28:47]If you take the 114th first, then you're cooked.

[04:28:50]If you take the So that means that you have to take one of these ones first.

[04:28:58]And then if Jang doesn't take this one, then he's cooked. So he has to take this one.

[04:29:05]So all in all, Jang takes 51 14 plus 14th, which is 47th.

[04:29:12]No. No, 54 is 37th and therefore Leo gets 47ths.

[04:29:19]So yeah, I still haven't found anything wrong with inherently wrong with this construction yet. And I think this still works if you take uh the cut to be closer to here. Now the main culprit that I'm wondering about is if you have two cuts in here like that.

[04:29:40]Then I actually wonder what you're going to do in here.

[04:29:46]And this may mean that the construction ends up not working.

[04:29:51]Yeah, I'm just being very very pessimistic here because like I don't know if this will work or not.

[04:30:00]If it does then I'll actually be really surprised.

[04:30:04]Okay, I have seven minutes left. Can I see if this will work or not?

[04:30:09]So, yeah. So this is 12 again.

[04:30:17]So this is x y z w x + y = 5 over 14 z + w = 12.

[04:30:52]Okay. So, I think I have a strategy. So, first pick the largest one in here to get the 1/4.

[04:31:00]Then you Then if Jang picks the small one here or the or this one here, then you just pick this. Then he just pick this one and then uh yeah, that just wins after taking the 17th. Wait, you can't necessarily take the 17th. So, oh well, I guess you can do like 12 + 5 over 28.

[04:31:45]This is so cursed.

[04:31:50]Okay. So, so therefore, if you after you choose that W, Jung must choose 17th.

[04:32:00]Then you can't choose Z otherwise you get cooked.

[04:32:09]So you have to choose the larger one in here.

[04:32:27]And then you have to end up picking one of these guys and then you're done.

[04:32:32]So in conclusion, I think that 17th 514 to 112 is a valid construction.

[04:32:41]Can you tell me if 1/2 Can you tell me if the one the 257 construction works?

[04:32:47]I'm pretty sure it does.

[04:33:15]Oh, wait. Can you tell me why?

[04:33:29]Also, how are you able to figure this out so quickly?

[04:34:00]Oh yeah, this is true. Yeah, I guess I found this intuition earlier on where splitting the sticks is a thing.

[04:34:08]Okay. All right, I have 2 minutes left, 3 minutes left on the clock, so I guess I can't really do much at all.

[04:34:15]Okay, so it's GG's.

[04:34:18]All right. So that means that my final score Okay. So that means that my final score is 77 I think zero.

[04:34:56]Do you think I got a point on problem one for getting the get a point on problem three for getting the correct answer?

[04:35:44]Wait, does getting the answer on problem three get me a point or no? Oh, yeah, probably not. Yeah.

[04:35:53]So, yeah, it's probably 770. That's fine.

[04:35:57]We still got the second day tomorrow.

[04:36:08]Okay, I'm going to go ahead and uh and I'm going to see if it's fine if I stay up a bit late because if I if I can stay up a lot later, then I'll go ahead and uh end stream.

[04:36:23]No, no. If I can stay up a lot later, then I'll then I might live stream my me writing on my solutions onto AOPS.

[04:36:40]Oh, cool.

[04:36:46]Yeah, problem one and problem two are actually such good problems. Like, I really, really like problem one. Can I talk about this for a second?

[04:36:54]Like I really No, I'm not at math cam set. Sorry. This is why I'm actually able to stream. If I were at math camp, then this it would be really really really hard to stream cuz then I have to uh sign in and everything during my live stream.

[04:37:20]I guess there's like no point of retracting the message because like I already said math camp on the live stream.

[04:37:30]Wait, should I show my Wait, should I stream my me writing up my solutions or should I end stream?

[04:37:59]Hello, bro. Ohio. Uh, wait. I'm going to be right back.

[04:38:08]Give me uh give me a minute.

[04:39:30]Okay. So, I decided that I'm going to go ahead and end stream.

[04:39:35]All right. Also, thank you so much, Joey, for helping me out with this problem three and supporting me through this process.

[04:39:43]uh even though I didn't end up solving it, like it's I still made lots of good progress because a lot a lot of it is because of your support. So, thank you so much for supporting the stream and supporting me on my problem three process.

[04:40:00]All right.

[04:40:07]All right. I'm going to go ahead and end stream.

[04:40:10]All right. Thank you so much for watching the stream, guys. It was truly one of the greatest experiences ever, and I can't wait to see you guys tomorrow for the day two. Yeah. Thank you so much, Joey, for tuning in and helping me out with problem three. All right. Bye.