I am always so impressed both with the Olympians and the mathematicians who created such intriguing problems. I would enjoy seeing a breakdown of the other problems.
I always understand each step he explains on this channel. I would love to see other difficult problems explained by him rather than anyone else. Hopefully he will solve another IMO problem soon
I have always loved geometry, even though I haven't 'used' this kind of math all that much in life. I find the whole set of concept fascinating, and making proofs was my favorite part of it all. Using rule x to show y is just so fulfilling. All that to say, I had to pause and rewatch a LOT in this particular video, not just because some of the concepts I was either never fully taught, or simply don't remember. Once I finally followed along with all of the details, it all made perfect sense, and while I certainly could never have come up with this in my own ability, I still got the joy from seeing it all come together. On thing though, in the future, when doing all of the angles and showing how they are the same, if you could color or somehow label the angles other than the strict
Do the students get these questions in English only? As a Finnish math teacher I did not know what "circumcircle" meant, so I probably would second guess myself and end up not solving this. With the help of Google I found out it's "ulkoympyrä" which directly translates to "outer circle" or "outcircle". With this knowledge I was able to solve this in roughly 1 hour. Great problem, will be giving this out in Finnish to gifted students as a challenge. Now, I have to find the rest of these questions and see if I get stumped with the 2,3,5 and 6 :)
I can tell how expert geometrists can easily think of this. Firstly, they would see that L and K are midpoints, and immediately take a homothety of 2 centered at A on triangle LIK. After that, they will immediately notice the parallelogram and complete it, since it would fit with point I after the homothety. After that, label the angles on the parallelogram, find which angles are same to the angles that are labeled, and the problem kind of just falls apart. I’ve been into Olympiad geometry for a while, so I somehow just recognize this kind of stuff lol. Also, this is an easy geometry problem compared to most out there in Olympiads. Any expert in olympiads will probably agree with this.
@13:15 I don't think you need to prove that length AI is equal to length IQ since as AC is parallel to XQ and AB is parallel to YQ (by the question), the formed quadrilateral is a rhombus with inscribed circle omega thus I is the center of it and divides the diagonal exactly in two same lengths (no proof needed per se) Looking forward in the video, I see you're having Q' that "maybe" has some connection to Q, and you proved it actually is the same using the prior proof of lengths AI and IQ / IQ'... BUT... why prove all that if you have a circle inscribed in a triangle, then two parallels of two triangle sides that tangent that circle form a rhombus AX'QY' that has the same circle inscribed?!? The definition of a rhombus is exactly a quadrilateral with four same length sides of which two opposite are parallel! And the center of an inscribed circle into a rhombus always divides the two diagonals in halfs! That is by definition!
this...points deducted for being too complicated and not using the most elegant solution...by definiton/construction we have 2 pairs of parallel lines with equal distances, so its a rhombus, so the circle centre lies on both diagonals, so all the length and angles match, thats super straight forward...using 2/3 of the video time of the proof on the most obvious part and only 1/3 on the actually difficult parts is a bit ridiculous...
DEAR GOD surely you can solve this without construcijg BCQ which I don't see Anyone any math grnius thinking of..or inscribed angle theorem..so why do this?? Please solve a way ppl.will actually think of..
@@BigAsciiHappyStar please observe with more concentration: You've a triangle and it's inscribed circle with the center I You now translate the sides AC and AB so that they tangent that inscribed circle also on the other side of it. They intersect in a point Q. If you draw a line between the two tangent points you WILL get a straight line that has the length of 2 radii of the inscribed circle, and that is also the length of the sides of the new quadrilateral, as they are parallel in pairs. And all four are the same length and equal 2 radii of the inscribed circle. By doing this you get a rhombus BY DEFINITION, as the two pairs of tangents/sides are PARALLEL and also are of the SAME LENGTH. These two characteristics are the definition of a rhombus. A rhombus also has one inscribed circle by known proofs in Euclidian geo, and it is THE SAME circle as the triangle inscribed circle. Also by prior euclidian proofs you know that the rhombus has 2 diagonals (that are perpendicular one to each other, but this not used here, so no need), and that one of them is longer and the other is shorter. Also you know that by prior proofs the two diagonals intersect exactly in the point of the center of its inscribed circle. Also the diagonals are split in half by their point of intersection which is also the center of the inscribed circle. Thus, due to proofs done in Euclidian geometry already 2500 years ago, you KNOW that Q = Q' and AI = IQ
At around 15:30 you concluded that AI = QI with some good calculations, But couldn't you also say that becouse: X'Q -> AY' and: AX'-> Y'Q, With a circle midpoint I, And because the circle is in the middle of the square, Which should mean that: AI = QI And X'I = IY'?
Although I also found the geometric solution, i used trigonometry and found an elegant property that the angle bisector splits KIL into two angles that are equal to the base angles of ∆XPY
sweet sweet IMO. I remember the guys from math classes i went with flew to other countries to win awards and i couldn't because of some paperwork problems
I solved it differently. First I labelled the intereection point of xxprime and yyprime as D. Said AXprimeDYprime is a parallelogram. Proved XprimeIYprime ate collinear and thus it is a rhombus. Then the solution is basically the same
We would definitely appreciate if you also bring vids for rest of the 6. This one is commendable, great explanation, even a 7th grader can understand this
I feel like the colour coding and diagram itself could have been made clearer. I got stumped when the opposite inscribed quadrilateral angles were seemingly interchanged with the triangle angles and wasn’t sure if the two inscribed circles were being used simultaneously, if that makes sense
what he means to say that either one of the criteria imply the other so you just need to prove any one of them, it is the standar mathematical definition of that theorem
Instead of showing that Q and Q' are the same you could have described Q as the intersection of XX' and YY' and show that IA part of a parallelogram AX'Y'Q because it bisects X'AY' and therefore Q is on AP.
P1 was just really ugly in the solution - there are nice ones but you qickly realise the ugly approach works and under contest pressure you write that up. P4 at least uses some ideas.
At 1:50, it is heartening to see India finish 4th this year. We have been doing great over the last few years, thanks in large part to the teachers in the month long training camp. It is only a matter of time, before we dominate IMO like we dominate the International Chess (expect for that odd Norwegian😀 )
@@DhoklaAboveVadapav - People used to wonder the same when Fischer was in his prime and when Kasparov was unbeatable for over two decades. I am pretty sure, someone better will come along. I wish it is from India 😀
This is very useful for gears inside of a motors engine creating torques. So the pistons could keep moving and they burst the gas, so you're able to drive your vehicle.
In ∆ABQ , L and I are midpoints of AB and AQ respectively. By the theorem, in a triangle the line joining two midpoints of any two sides is half of the 3rd side and PARALLEL TO THE 3rd SIDE. Hence, LI || BQ and taking AQ as transversal Angle AIL is congruent to angle ABQ Similarly on other side Hope you get it
Without a compass and straight edge, I would have a hard time making a legible diagram and some of the points might fall on top of each other. I assume that contestants aren't allowed to use drawing tools or software.
They are provided all the tools needed such as geometrical tools, scientific calculators etc. They are even given snacks like cookies, bananas, energy/fruit drinks, dark chocolates etc too! Since they have to sit and think hard for 4.5hours straight.
2nd easiest or easiest, and the difficulty ramping is brutal, even the 2nd problems on each day are significantly harder (except for problem 5 which was really easy this year)
@@placeholderfornow4766 it must've been a nice class. But that outcome was kinda of what happened in the Olympiad based on the results and some comments I read.
The problem with the problem, initially, is the word jargon that math uses instead of modern english, like latin or Ole English, ya just don't know what they hell they saying. I talked this out with a 16 year ole and they got it right in about 15 minutes. Because when ya talk with someone ... you can use your hands and form a mental image in 3D rather than 2D ... even if it is a 2D issue ... you have no problem rotating the image in space and turning in a way to advantage you. But if you have to create the image you have to create the next D. It's easier to see it at the higher D and work backwards. Which goes to say what they did at this comp was harder IMO. and lets not forget the pressure they were under. me and the neighbor ... we were just chit chatting.
This is the International Math Olympiad. Every country selects its 6 best students to compete each year. They can't just have some simple, straight forward solution that anyone can think of. These students are the best of the best!
It doesn't feelgood seeing that there are people who can solve this question while I cant get through half of the process, even though the process seems easy.
If you don't study for math Olympiads like we do that's completely normal lmao. The amount of work, and study we have to do for this problems is insane but interesting, fun and rewarding.
The only way USA can compete is to not show Russian team 🤣. In individual results a Russian got silver. Previous year Russia got gold in individual results. 💪 China and Russia are the strongest
@@wintersfan i googled team members. china team members in 2023: Shi, Wang, Zhang, Liang, Sun, Jiang Usa team members in 2023: Liu, Wang, Lin, Zhao, Lu, Shen. major race in usa is white (around 60%), and asians are like 9-10%. I believe there were two asian teams and none from America or you are another state of china. From 60% of the nation there is no white who is smart anough? This is pathetic. Usa - United states of asia😂 P. S. Russian team in 2023 : Алиса Волкова, Роман Кузнецов и Эльдар Хисамутдинов, Александр Гнусов, Ратибор Коптилин, Павел Прозоров. At least 4 members are with clear Russian names.
@@wintersfan because I can😋 I didn't say they don't have American citizenship. America can "buy" and import smart people, it is not a secret, it does it all the time. But the fact that it needs smart asians (10 % of population) to compete with china while the major race can't even represent the country in IMO tells a lot about education system. This is big respect for asian Americans and humiliation for the rest. Also, the fact that the only question you have is "why you mentioned race? " also tells a lot about western culture of the offended by anything. My country has not less nationalities than in US, probably even more (190+ nationalities, also no segregation zones for any of them 😉) , but i still can name races freely and this is not offencive to people (except for cases of true offenciveness).
@@TheBurzhuy i am not offended by you saying about race. But IMO is about Math and bringing up race is irrelevant since the IMO is meant to represent the 6 best students in a country.
Its very intuitive, it encompases both lines given and additionally lies on the angle bisector, so introducing the point Q should be the first thing to do
@@tollspiller2043I'm sorry but it's not intuitive at all even if you complete the quadrilateral..i don't see how..intuition is subjective maybe?..and what angle bjsectpr are you referring to? ..and furthermore can't you solve without completing that quadrilateral with the parallel lines..thats something else thats nkt at all intuitive or organically logical so please solve without it. Wouldn't you agree with that? Thanks for answering.
@@leif1075 there are other solutions that dont involve Q but other points, e.g. a point R such that BIR are collibear and AR is perpendicular to BR. Point Q is the easiest and modt obvious one to introduce
@@whitemountain4851 that is exactly what I did first. I call this the Principle Of Most Disgusting - focus on the most disgusting part of the problem and work out how to make it less disgusting 😀
Only after receiving the formalized statement of the problem in AlphaGeometry’s own language. They essentially cheated by translating the problems from english into machine manageable versions
@@redsolaris1 yes but that doesn't change the fact that it gets it right in 19 seconds. Its just how computers work, and they're faster than us humans, accept that.
@@wowyok4507 The ai got 0 points on problems 5 and 3 though, it’s because they were more logic oriented problems and the speed of computers is basically useless when it comes to that
@@wowyok4507I’m not saying it won’t get better, just that it’s not as powerful as you think. Till now in the field of combinatorics, it’s gotten nowhere(it got 0 points on both of the combinatorics problems), since theres something inherently different about this field which makes it difficult for ai to grasp. Though i am curious to see if it will improve significantly in the near future
Surely there is a way to make it cleaer and a way tonsolve WITHOUT the inscribed a gle theirem since a lot of ppl no matter how genius will NOT recall that.
Presh you have a mistake. You did not use all letters in alphabet. My conclusion: it was not an Olympiad, it was linear work in shop for cashier. Competition should be interesting. I'm sorry Presh, be happy! 🤷♀️
I'm not sure whether to believe AI is already smarter or can be smarter than humans..if Presh couldn't solve it,does thatmean AI is smarter than him..and if most ppl.have so much difficulty with math are humans really as smart as we tout and aren't AI already smarter than a lot of ppl?
Glad to see Presh back to posting challenging and ponder-worthy problems !
I am always so impressed both with the Olympians and the mathematicians who created such intriguing problems. I would enjoy seeing a breakdown of the other problems.
Please could you upload a video in which you describe the hardest one?
Great video as always.
I always understand each step he explains on this channel. I would love to see other difficult problems explained by him rather than anyone else. Hopefully he will solve another IMO problem soon
Wow! I could follow all the steps of the solution, but am glad I quickly gave up on trying to solve this one. I never would have come close!
You NEED to cover problem 5!!! The solution is so easy, and I think it would be a perfect video for your channel.
for someone whow only knows the perpendicular formula developed from the coordinate product instead of the bisector of angles, the centerpoint of the inner circle of a triangle will be the result of a repeated calculation:
10 print "mind your decisions-geometry question to the worlds best math"
20 xg1=0:yg1=0:xg2=1:yg2=1:xpu=1:ypu=0:gosub 120:gosub 570:print"test", xlu,ylu:
30 lab=0.8:lac=0.7:lbc=1:la=lab:lb=lac:lc=lbc:lh=(la^2+lc^2-lb^2)/2/lc
40 h=sqr(la^2-lh^2):xa=lh:ya=h:xb=0:yb=0:xc=lbc:yc=0:dim x(2,3),y(2,3):rem den innenkreis berechnen***
50 x(0,0)=0:y(0,0)=0:x(0,1)=lh:y(0,1)=0:x(0,2)=lbc:y(0,2)=0:x(0,3)=lh:y(0,3)=h
60 xm1=0:ym1=0:r1=0:xvi=1000:yvi=850:n1=2:n2=3:xm2=0:ym2=0:r2=0:gosub 630
70 x1=x(0,0):y1=y(0,0):x2=x(0,2):y2=y(0,2):x3=x(0,3):y3=y(0,3):gosub 550:xm1=xlu:ym1=ylu
80 r1=sqr((x1-xlu)^2+(y1-ylu)^2):gosub 630:@zoom%=@zoom%*1.1
90 sw=sqr(la^2+lb^2+lc^2)/100:swx=sw/10:ym=sw:xg1=lh:yg1=h:xg2=lbc:yg2=0:goto 230
100 xg1=0:yg1=0:xg2=lh:yg2=h:r2=ym:
110 xpu=xm:ypu=ym:gosub 120:goto 150
120 a11=xg2-xg1:a12=yg2-yg1:a131=xpu*(xg2-xg1):a132=ypu*(yg2-yg1)
130 a21=yg1-yg2:a22=xg2-xg1:a231=yg1*(xg2-xg1):a232=xg1*(yg1-yg2)
140 a23=a231+a232:a13=a131+a132:return
150 gosub 570:dgu1=sqr((xlu-xpu)^2+(ylu-ypu)^2)/lab
160 dgu2=r2/lab:dg=dgu1-dgu2:return
170 xm=swx:gosub 100
180 dg1=dg:xmu1=xm:xm=xm+sw:xmu2=xm:gosub 100:if dg1*dg>0 then 180
190 xm=(xmu1+xmu2)/2:gosub 100:if dg1*dg>0 then xmu1=xm else xmu2=xm
200 if abs(dg)>1E-10 then 190 else return
210 gosub 170:xg1=lbc:yg1=0:xg2=lh:yg2=h:yp=ym
220 gosub 120:gosub 570:ru=sqr((xlu-xpu)^2+(ylu-ypu)^2):df=(ym-ru)/ym:return
230 gosub 210:rem print xm,ym:stop
240 df1=df:ymu1=ym:ym=ym+sw:ymu2=ym:gosub 210:if df1*df>0 then 240
250 ym=(ymu1+ymu2)/2:gosub 210:if df1*df>0 then ymu1=ym else ymu2=ym
260 if abs(df)>1E-9 then 250
270 print xm,"%",ym:xm2=xm:ym2=ym:gosub 630:rem l1 und l2 durch drehung um 180 grad berechnen
280 xcu=xm2:ycu=ym2:xu=xa:yu=ya:gosub 300:x(1,0)=xv:y(1,0)=yv:rem print xv,yv:stop
290 xg11=lh:yg11=h:xg12=lbc:yg12=0:xu=0:yu=0:gosub 300:xg21=xv:yg21=yv:goto 310
300 xv=2*xcu-xu:yv=2*ycu-yu:return
310 xu=lh:yu=h:gosub 300:xg22=xv:yg22=yv:gosub 350:x(1,1)=xlu:y(1,1)=ylu:x(1,2)=xa:y(1,2)=ya
320 xu=xa:yu=ya:gosub 300:xg21=xv:yg21=yv
330 xu=lbc:yu=0:gosub 300:xg22=xv:yg22=yv:xg11=0:yg11=0:xg12=lh:yg12=h
340 gosub 350:x(1,3)=xlu:y(1,3)=ylu:cls:gosub 630:goto 380
350 a11=yg12-yg11:a12=xg11-xg12:a131=xg11*(yg12-yg11):a132=yg11*(xg11-xg12)
360 a21=yg22-yg21:a22=xg21-xg22:a231=xg21*(yg22-yg21):a232=yg21*(xg21-xg22)
370 a13=a131+a132:a23=a231+a232:gosub 570:return
380 x1=xa:y1=ya:x2=xm1:y2=ym1:dx=x2-x1:dy=y2-y1:xm=xm1:ym=ym1:dl=dx^2+dy^2
390 px=(x1-xm1)*dx:py=(y1-ym1)*dy:p=2*(px+py)/dl:q=((x1-xm1)^2+(y1-ym1)^2-r1^2)/dl
400 dis=p*p/4-q:if dis
What's the name of the routine? SeaWolf? Cause it's a gosub?
@@ejrupp9555 thanks for your aattention but what do you really want?
I have always loved geometry, even though I haven't 'used' this kind of math all that much in life. I find the whole set of concept fascinating, and making proofs was my favorite part of it all. Using rule x to show y is just so fulfilling. All that to say, I had to pause and rewatch a LOT in this particular video, not just because some of the concepts I was either never fully taught, or simply don't remember. Once I finally followed along with all of the details, it all made perfect sense, and while I certainly could never have come up with this in my own ability, I still got the joy from seeing it all come together.
On thing though, in the future, when doing all of the angles and showing how they are the same, if you could color or somehow label the angles other than the strict
Do the students get these questions in English only? As a Finnish math teacher I did not know what "circumcircle" meant, so I probably would second guess myself and end up not solving this. With the help of Google I found out it's "ulkoympyrä" which directly translates to "outer circle" or "outcircle". With this knowledge I was able to solve this in roughly 1 hour. Great problem, will be giving this out in Finnish to gifted students as a challenge. Now, I have to find the rest of these questions and see if I get stumped with the 2,3,5 and 6 :)
I'm sure there are translated versions of the test
We also get it in our own languages.
You may request (and get) the exam in up to 3 languages
I think the name Internacional math Olympiad speaks for itself no? Each country is given the test in their own native language.
if you participate in IMO, the test will actually be in your own language.
I can tell how expert geometrists can easily think of this. Firstly, they would see that L and K are midpoints, and immediately take a homothety of 2 centered at A on triangle LIK. After that, they will immediately notice the parallelogram and complete it, since it would fit with point I after the homothety. After that, label the angles on the parallelogram, find which angles are same to the angles that are labeled, and the problem kind of just falls apart. I’ve been into Olympiad geometry for a while, so I somehow just recognize this kind of stuff lol.
Also, this is an easy geometry problem compared to most out there in Olympiads. Any expert in olympiads will probably agree with this.
I think its much easier to prove that AX'QY' is parallelogram and AI as bisect is also a diagonal and I is the center
Not only a parallelogram, but a specific one, a rhombus. All sides have the same length. Thus center of inscribed circle is half of both diagonals.
@13:15 I don't think you need to prove that length AI is equal to length IQ since as AC is parallel to XQ and AB is parallel to YQ (by the question), the formed quadrilateral is a rhombus with inscribed circle omega thus I is the center of it and divides the diagonal exactly in two same lengths (no proof needed per se)
Looking forward in the video, I see you're having Q' that "maybe" has some connection to Q, and you proved it actually is the same using the prior proof of lengths AI and IQ / IQ'...
BUT... why prove all that if you have a circle inscribed in a triangle, then two parallels of two triangle sides that tangent that circle form a rhombus AX'QY' that has the same circle inscribed?!? The definition of a rhombus is exactly a quadrilateral with four same length sides of which two opposite are parallel! And the center of an inscribed circle into a rhombus always divides the two diagonals in halfs! That is by definition!
this...points deducted for being too complicated and not using the most elegant solution...by definiton/construction we have 2 pairs of parallel lines with equal distances, so its a rhombus, so the circle centre lies on both diagonals, so all the length and angles match, thats super straight forward...using 2/3 of the video time of the proof on the most obvious part and only 1/3 on the actually difficult parts is a bit ridiculous...
DEAR GOD surely you can solve this without construcijg BCQ which I don't see Anyone any math grnius thinking of..or inscribed angle theorem..so why do this?? Please solve a way ppl.will actually think of..
Proving AI = IQ is a must. Otherwise how do you prove that angle LIK = angle BQC?
@@BigAsciiHappyStar please observe with more concentration:
You've a triangle and it's inscribed circle with the center I
You now translate the sides AC and AB so that they tangent that inscribed circle also on the other side of it. They intersect in a point Q. If you draw a line between the two tangent points you WILL get a straight line that has the length of 2 radii of the inscribed circle, and that is also the length of the sides of the new quadrilateral, as they are parallel in pairs. And all four are the same length and equal 2 radii of the inscribed circle.
By doing this you get a rhombus BY DEFINITION, as the two pairs of tangents/sides are PARALLEL and also are of the SAME LENGTH. These two characteristics are the definition of a rhombus.
A rhombus also has one inscribed circle by known proofs in Euclidian geo, and it is THE SAME circle as the triangle inscribed circle.
Also by prior euclidian proofs you know that the rhombus has 2 diagonals (that are perpendicular one to each other, but this not used here, so no need), and that one of them is longer and the other is shorter.
Also you know that by prior proofs the two diagonals intersect exactly in the point of the center of its inscribed circle.
Also the diagonals are split in half by their point of intersection which is also the center of the inscribed circle.
Thus, due to proofs done in Euclidian geometry already 2500 years ago, you KNOW that Q = Q' and AI = IQ
my first thought: too many trees, so i guess someone didn't see the forrest ;-)
At around 15:30 you concluded that AI = QI with some good calculations,
But couldn't you also say that becouse:
X'Q -> AY' and: AX'-> Y'Q,
With a circle midpoint I,
And because the circle is in the middle of the square,
Which should mean that:
AI = QI
And
X'I = IY'?
“wrong solution, just use barycentric instead” - my roommate at mathcamp
Hi Epicbird!
@@wintersfanhello
BAN
Skull
I honestly feel like this is more of a reading comprehension and visualization excersize than a math excersize
Not really, for if you give this to your English prof/teacher I’m pretty sure they won’t be able to solve it
@@reveal-n7zgood point
Reading, compréhension and visualization are parts of mathematics imo
@@julienchalimon4643 fair enough, so this is still very much a math challenge 👍
That's the easy part. IMO problems are brutal
Although I also found the geometric solution, i used trigonometry and found an elegant property that the angle bisector splits KIL into two angles that are equal to the base angles of ∆XPY
?
@@Advith12987 what you don't understand? I can explain a bit
@@vitalsbat2310 the property
Random question what do you think is harder making it to IMO or beating tidal wave
@@vitalsbat2310 the property you found
THANKS SIR CONTINUED UPTO 8 DAYS AND I FEEL REALL A BETTER IMPROVEMENT IN MYSELF ❤❤ LOVE MATHEMATICS ♾️🔥
sweet sweet IMO. I remember the guys from math classes i went with flew to other countries to win awards and i couldn't because of some paperwork problems
Everyone:Cyclic Meanwhile him:seaclick.
I solved it differently. First I labelled the intereection point of xxprime and yyprime as D. Said AXprimeDYprime is a parallelogram. Proved XprimeIYprime ate collinear and thus it is a rhombus. Then the solution is basically the same
We would definitely appreciate if you also bring vids for rest of the 6. This one is commendable, great explanation, even a 7th grader can understand this
I feel like the colour coding and diagram itself could have been made clearer. I got stumped when the opposite inscribed quadrilateral angles were seemingly interchanged with the triangle angles and wasn’t sure if the two inscribed circles were being used simultaneously, if that makes sense
At 6:17, it should be A+C=180 AND B+D=180 (instead of OR).
what he means to say that either one of the criteria imply the other so you just need to prove any one of them, it is the standar mathematical definition of that theorem
Please provide us insights on harder questions from IMO
Please share how you approach a problem you haven't seen it before
Critical thinking, math isn't about memorising problems
You will build intuition by doing many problems
You gotta tell us about the harder questions
sir, can you please do for SINGA maths competition? competiton is on 31 august. deeply appreciated 🙏🏻🙏🏻
Instead of showing that Q and Q' are the same you could have described Q as the intersection of XX' and YY' and show that IA part of a parallelogram AX'Y'Q because it bisects X'AY' and therefore Q is on AP.
How should I approach these questions?
Somehow I think this was even easier than P1, which was pretty trivial if you have general knowledge of the floor function
P1 was just really ugly in the solution - there are nice ones but you qickly realise the ugly approach works and under contest pressure you write that up. P4 at least uses some ideas.
At 1:50, it is heartening to see India finish 4th this year. We have been doing great over the last few years, thanks in large part to the teachers in the month long training camp. It is only a matter of time, before we dominate IMO like we dominate the International Chess (expect for that odd Norwegian😀 )
I wonder If we will ever find a player better than magnus, or he will remain the undisputed king for infinity and beyond.
@@DhoklaAboveVadapav - People used to wonder the same when Fischer was in his prime and when Kasparov was unbeatable for over two decades. I am pretty sure, someone better will come along. I wish it is from India 😀
@@agytjax True. Surely someone will come. But I doubt in next 10 years.
🔥
This is very useful for gears inside of a motors engine creating torques. So the pistons could keep moving and they burst the gas, so you're able to drive your vehicle.
If you are serious, can you explain how or point to some resource available in public domain ?
12:52 13:04 Lol XPB instead of XBP.
Also, it'd have been faster if you'd noticed that AX'QY' is a rhombus.
At 10:16 you made a mistake.it's angle YQX NOT angle YPX right?
why is BQC=KIL 9:56 is there a rule that Im missing?
Midpont Theorem
In ∆ABQ ,
L and I are midpoints of AB and AQ respectively.
By the theorem, in a triangle the line joining two midpoints of any two sides is half of the 3rd side and PARALLEL TO THE 3rd SIDE.
Hence, LI || BQ and taking AQ as transversal
Angle AIL is congruent to angle ABQ
Similarly on other side
Hope you get it
Sir can you make video on geometry concepts
Without a compass and straight edge, I would have a hard time making a legible diagram and some of the points might fall on top of each other. I assume that contestants aren't allowed to use drawing tools or software.
They are provided all the tools needed such as geometrical tools, scientific calculators etc. They are even given snacks like cookies, bananas, energy/fruit drinks, dark chocolates etc too! Since they have to sit and think hard for 4.5hours straight.
who creates these questions? 🤔😮😥
i would love to see one of the harder problems even tho im only in year nine :DD
That’s the easy problem on the exam? Wow.
That's Olympiad math problems for ya
2nd easiest or easiest, and the difficulty ramping is brutal, even the 2nd problems on each day are significantly harder (except for problem 5 which was really easy this year)
@@placeholderfornow4766 P5 was so troll xD
@@iMíccoli ikr, i was in a group class where we solved 2024 imo for fun and half of us got it instantly while the other half were STRUGGLING
@@placeholderfornow4766 it must've been a nice class. But that outcome was kinda of what happened in the Olympiad based on the results and some comments I read.
The problem with the problem, initially, is the word jargon that math uses instead of modern english, like latin or Ole English, ya just don't know what they hell they saying. I talked this out with a 16 year ole and they got it right in about 15 minutes. Because when ya talk with someone ... you can use your hands and form a mental image in 3D rather than 2D ... even if it is a 2D issue ... you have no problem rotating the image in space and turning in a way to advantage you.
But if you have to create the image you have to create the next D. It's easier to see it at the higher D and work backwards.
Which goes to say what they did at this comp was harder IMO. and lets not forget the pressure they were under. me and the neighbor ... we were just chit chatting.
Can't you solve this without the sickly quadrilateral theorem or imbibed angle theorem since I don't think a lot of ppl will think of them?
This is the International Math Olympiad. Every country selects its 6 best students to compete each year. They can't just have some simple, straight forward solution that anyone can think of. These students are the best of the best!
Make a series with ALL these problems, and you can rename this video with the text in brackets first
It doesn't feelgood seeing that there are people who can solve this question while I cant get through half of the process, even though the process seems easy.
Don't worry, 99,9% of us can't either ;)
If you don't study for math Olympiads like we do that's completely normal lmao. The amount of work, and study we have to do for this problems is insane but interesting, fun and rewarding.
I don’t know the answer to this question
…but I do like seeing kittens playing in hay barns.
lol, massimo gobbino is one of my professors at unipi (calculus 2)
Complicado. Sobre todo el poder dibujar el problema. Muchas gracias
Proof that spiders instinctively know geometry. 😉
Pozdrawiam z Polski! 😊
Witam serdecznie i również pozdrawiam 😊
@@barteqw napisałem ten komentarz po polsku w kontekście tego zadania.
@@marcinbednara3825 wiem że pochodzi z naszego kraju
8:01 oh... thats AI in MindYourDecisions...
The only way USA can compete is to not show Russian team 🤣. In individual results a Russian got silver. Previous year Russia got gold in individual results. 💪 China and Russia are the strongest
In this IMO 2024, the total score of Russian students are 185, China 190 and USA 192
So i don't know what you're talking about 😂😂
@@wintersfan i googled team members. china team members in 2023: Shi, Wang, Zhang, Liang, Sun, Jiang
Usa team members in 2023: Liu, Wang, Lin, Zhao, Lu, Shen.
major race in usa is white (around 60%), and asians are like 9-10%. I believe there were two asian teams and none from America or you are another state of china. From 60% of the nation there is no white who is smart anough? This is pathetic. Usa - United states of asia😂
P. S. Russian team in 2023 : Алиса Волкова, Роман Кузнецов и Эльдар Хисамутдинов, Александр Гнусов, Ратибор Коптилин, Павел Прозоров. At least 4 members are with clear Russian names.
@@TheBurzhuy why are you bringing race tho? They are still American.
@@wintersfan because I can😋 I didn't say they don't have American citizenship. America can "buy" and import smart people, it is not a secret, it does it all the time. But the fact that it needs smart asians (10 % of population) to compete with china while the major race can't even represent the country in IMO tells a lot about education system. This is big respect for asian Americans and humiliation for the rest.
Also, the fact that the only question you have is "why you mentioned race? " also tells a lot about western culture of the offended by anything. My country has not less nationalities than in US, probably even more (190+ nationalities, also no segregation zones for any of them 😉) , but i still can name races freely and this is not offencive to people (except for cases of true offenciveness).
@@TheBurzhuy i am not offended by you saying about race. But IMO is about Math and bringing up race is irrelevant since the IMO is meant to represent the 6 best students in a country.
I don't see why anyone would construct triangle BQC..right would anyone really, even Chat GPT SOB? So why not solve without that please?
Its very intuitive, it encompases both lines given and additionally lies on the angle bisector, so introducing the point Q should be the first thing to do
@@tollspiller2043I'm sorry but it's not intuitive at all even if you complete the quadrilateral..i don't see how..intuition is subjective maybe?..and what angle bjsectpr are you referring to? ..and furthermore can't you solve without completing that quadrilateral with the parallel lines..thats something else thats nkt at all intuitive or organically logical so please solve without it. Wouldn't you agree with that? Thanks for answering.
@@leif1075 there are other solutions that dont involve Q but other points, e.g. a point R such that BIR are collibear and AR is perpendicular to BR.
Point Q is the easiest and modt obvious one to introduce
I feel like the idea here is foresight - you know that the midpoint and the incenter are very badly connected so you want to “move” the angle
@@whitemountain4851 that is exactly what I did first. I call this the Principle Of Most Disgusting - focus on the most disgusting part of the problem and work out how to make it less disgusting 😀
Test for top 0.0001% from top 0.001% students. Not for most of regular human.
Hi presh!!
This one gave me a headache
20 minutes to do an hour program. Slow down and call out the points.
For 4 hour's it's doable
10:16 u said angle QYC instead of QCY!!!
Good catch 😊 This was an alphabet soup so it was a challenge to say everything correctly.
What a monstrosity
You lost me reading it...
I have a simpler solution, no knowledge of cyclic quadrilaterals required
Share it then
😮😊❤🎉
🥹🥹savebangladeshistudents
Do for class 7 mtg sof
Are you allowed a ruler and compass on the IMO? Otherwise this is just cruel.
Of course lol, you can't even do this problems without it.
ai solved this problem in 19 seconds :/
Only after receiving the formalized statement of the problem in AlphaGeometry’s own language. They essentially cheated by translating the problems from english into machine manageable versions
@@redsolaris1 yes but that doesn't change the fact that it gets it right in 19 seconds. Its just how computers work, and they're faster than us humans, accept that.
@@wowyok4507 The ai got 0 points on problems 5 and 3 though, it’s because they were more logic oriented problems and the speed of computers is basically useless when it comes to that
@@MariusGjika ah yes, an ai that was created half a year ago won't get any better, yes agreed
@@wowyok4507I’m not saying it won’t get better, just that it’s not as powerful as you think. Till now in the field of combinatorics, it’s gotten nowhere(it got 0 points on both of the combinatorics problems), since theres something inherently different about this field which makes it difficult for ai to grasp. Though i am curious to see if it will improve significantly in the near future
KHAO PEEYO😂😂😂
ALMAA HII KAAFI HAIN ....
5th
Early Again.... WTH is going on 😅
u wot m8?
10:12 Extremely complicated with a jungle of letters and lines.
Unfortunately...
Surely there is a way to make it cleaer and a way tonsolve WITHOUT the inscribed a gle theirem since a lot of ppl no matter how genius will NOT recall that.
Presh you have a mistake. You did not use all letters in alphabet. My conclusion: it was not an Olympiad, it was linear work in shop for cashier. Competition should be interesting. I'm sorry Presh, be happy! 🤷♀️
I'm not sure whether to believe AI is already smarter or can be smarter than humans..if Presh couldn't solve it,does thatmean AI is smarter than him..and if most ppl.have so much difficulty with math are humans really as smart as we tout and aren't AI already smarter than a lot of ppl?
Proofread what you write and spell out your words.
Were you dropped in the head as a baby? This is why dependence on AI makes you completely useless
Based on how you wrote that, yes they most certainly are
@@JakeMarley-k6glol.come on i was lazy and writing fast and autocorrect sucks so surely you don't mean that?
@@leif1075 I don't, I'm just messing
Sorry pall, the worst part you ever recorded.
First to see