Geometry question to test the world's best math students (IMO 2024 problem 4)

The 2024 International Mathematical Olympiad has just wrapped up. Let's work out this geometry problem.

0:00 intro
2:15 problem
4:02 review
6:33 diagram
9:22 proof

Thanks to Nestor and Tony for spotting an error in a draft video about the incircle property definition!

IMO official
Dedekind cuts video
IMO 2024 solutions
concyclic

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[agytjax]

Glad to see Presh back to posting challenging and ponder-worthy problems !

[beopqtu]

Please could you upload a video in which you describe the hardest one?
Great video as always.

[deerho]

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.

[KupoPallo]

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 :)

[alexandrubivolaru]

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

[waheisel]

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!

[zmaj]

You NEED to cover problem 5!!! The solution is so easy, and I think it would be a perfect video for your channel.

[thegreatgreenarkleseizure]

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 <ABC format, it really helps with those of us who are visual learners. Seeing 𝜃 or 𝜌 or 𝛼 used as labels or the colors like you did a little of (blue, green, red, etc.) is mush easier to comprehend the compared angles and sums. I know it isn't 'math standard' for notation, and I know it isn't an easy task, but I know I cannot be alone in knowing it would accelerate my full understanding and lessen the number of pause/rewind cycles.

Thanks for your hard work in keeping all of our minds sharp!

[zey]

@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!

[Doktor_Vem]

I honestly feel like this is more of a reading comprehension and visualization excersize than a math excersize

[zdrastvutye]

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"
20 120:gosub 570:print"test", xlu, ylu:
30
40 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
60 630
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
90 230
100
110 xpu=xm:ypu=ym:gosub 120:goto 150
120
130
140
150 gosub
160
170 xm=swx:gosub 100
180 100: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
220 gosub 120:gosub
230 gosub 210:rem print xm, ym:stop
240 210: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
280 300:x(1, 0)=xv:y(1, 0)=yv:rem print xv, yv:stop
290 300:xg21=xv:yg21=yv:goto 310
300
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
340 gosub 350:x(1, 3)=xlu:y(1, 3)=ylu:cls:gosub 630:goto 380
350
360
370 570:return
380
390
400 dis=p*p/4-q:if dis<0 then stop else sd=sqr(dis)
410
420
430 print quo
440 xg11=0:yg11=0:xg12=lbc:yg12=0:xg21=x(1, 3):yg21=y(1, 3):xg22=x(1, 0):yg22=y(1, 0):gosub 350:xx=xlu:yx=ylu
450 xg21=x(1, 0):yg21=y(1, 0):xg22=x(1, 1):yg22=y(1, 1):gosub 350:xy=xlu:yy=ylu
460 x=xx:y=yx:gosub 670:xbn=xbu:ybn=ybu:gosub 830
470 x=xy:y=yy:gosub 670:xbn=xbu:ybn=ybu:gosub 830:rem parallelverschiebung und berechnung der
480 xg11=xm2:yg11=ym2:xg12=xg11+x(0, 0)-x(1, 0):yg12=yg11+y(0, 0)-y(1, 0)
490 350:xl=xlu:yl=ylu
500 xg11=xa:yg11=ya:xg12=lbc:yg12=0:xg21=xm2:yg21=ym2:xg22=xg21+x(1, 0)-x(0, 2):yg22=yg21+y(1, 0)-y(0, 2)
510 gosub 670:xbn=xbu:ybn=ybu
520 gosub 830:x=xk:y=yk:gosub 670:xbn=xbu:ybn=ybu:gosub 830:x=0:y=0:gosub 670:xba=xbu:yba=ybu
530 x=x(1, 0):y=y(1, 0):gosub 670:xbn=xbu:ybn=ybu:gosub 830:x=x(0, 2):y=y(0, 2):gosub 670:xbn=xbu:ybn=ybu
540 gosub 830:print "run in bbc basic sdl and hit ctrl tab to copy from the results window":end
550
560
570 ngl1=a12*a21:ngl2=a22*a11
580 ngl=ngl1-ngl2:if ngl=0 then print "keine loesung":end
590
600
610 xlu=zx/ngl:ylu=zy/ngl:rem print "x=";xl;"y=";yl
620 return
630 sux=0:suy=0:anz=0:rem die grafische darstellung ***
640 for a=0 to n1:for b=0 to n2:x=x(a, b):y=y(a, b):xb=x:yb=y
650 b:next a
660 680
670
680
690 for a=0 to n1:for b=0 to n2:xb=x(a, b):yb=y(a, b)
700 if xb<xmin then xmin=xb
710 if xb>xmax then xmax=xb
720 if yb<ymin then ymin=yb
730 if yb>ymax then ymax=yb
740 next b:next a:if xmin=xmax or ymin=ymax then else 780
750 if xmin=xmax then else 770
760 mass=yvi/(ymin-ymax):goto 800
770 mass=xvi/(xmin-xmax):goto 800
780
790 if masx<masy then mass=masx else mass=masy
800 for a=0 to n1:gcol8+a:x=x(a, 0):y=y(a, 0):gosub 670:xba=xbu:yba=ybu
810 for b=1 to n2+1:ib=b:if ib=n2+1 then ib=0
820 x=x(a, ib):y=y(a, ib):gosub 670:xbn=xbu:ybn=ybu:gosub 830:goto 840
830 line xba, yba, xbn, ybn:xba=xbn:yba=ybn:return
840 next b:next a:gcol8:x=xm1:y=ym1:gosub 670:circle xbu, ybu, r1*mass
850 x=xm2:y=ym2:gosub 670:circle xbu, ybu, r2*mass
860 a$=inkey$(0):if a$="" then 860 else return
thanks for everyone's attention.

[viktor.tsonev]

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

[kaspervanoverbeek]

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'?

[epicbird]

“wrong solution, just use barycentric instead” - my roommate at mathcamp

[opendstudio]

Proof that spiders instinctively know geometry. 😉

[AAAEA]

You gotta tell us about the harder questions

[vitalsbat]

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

[ggritans]

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

[sanjeevkumarsingh]

Please share how you approach a problem you haven't seen it before

[youtuger]

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