My journey to IMO
by MTA_2024, May 30, 2025, 6:18 PM
Note to moderators: I had no idea if this is the ideal forum for this or not, feel free to move it wherever you want 
Hi everyone,
I am a random 14 years old 9th grader, national olympiad winner, and silver medalist in the francophone olympiad of maths (junior section) Click here to see the test in itself.
While on paper, this might seem like a solid background (and tbh it kinda is); but I only have one problem rn: an extreme lack of preparation (You'll understand very soon just keep reading
).
You see, when the francophone olympiad, the national olympiad and the international kangaroo ended (and they where in the span of 4 days!!!) I've told myself :"aight, enough math, take a break till summer" (and btw, summer starts rh in July and ends in October) and from then I didn't seriously study maths.
That was until yesterday, (see, none of our senior's year students could go because the bachelor's degree exam and the IMO's dates coincide). So they replaced them with us, junior students. And suddenly, with no previous warning, I found myself at the very bottom of the IMO list of participants. And it's been months since I last "seriously" studied maths.
I'm really looking forward to this incredible journey, and potentially winning a medal
. But regardless of my results I know it'll be a fantastic journey with this very large and kind community.
Any advices or help is more than welcome <3 .Thank yall for helping me reach and surpass a ton of my goals.
Sincerely.
Hi everyone,
I am a random 14 years old 9th grader, national olympiad winner, and silver medalist in the francophone olympiad of maths (junior section) Click here to see the test in itself.
While on paper, this might seem like a solid background (and tbh it kinda is); but I only have one problem rn: an extreme lack of preparation (You'll understand very soon just keep reading
).You see, when the francophone olympiad, the national olympiad and the international kangaroo ended (and they where in the span of 4 days!!!) I've told myself :"aight, enough math, take a break till summer" (and btw, summer starts rh in July and ends in October) and from then I didn't seriously study maths.
That was until yesterday, (see, none of our senior's year students could go because the bachelor's degree exam and the IMO's dates coincide). So they replaced them with us, junior students. And suddenly, with no previous warning, I found myself at the very bottom of the IMO list of participants. And it's been months since I last "seriously" studied maths.
I'm really looking forward to this incredible journey, and potentially winning a medal
. But regardless of my results I know it'll be a fantastic journey with this very large and kind community.Any advices or help is more than welcome <3 .Thank yall for helping me reach and surpass a ton of my goals.
Sincerely.
This post has been edited 3 times. Last edited by MTA_2024, 6 hours ago
be an acute triangle with 
and let 
be its orthocenter. Let 
be a point on the perpendicular bisector of 
such that 
and 
are on different sides of 
,
a point on the perpendicular bisector of 
such that 
and 
a point on the perpendicular bisector of 
such that 
and 
lie on the opposite side of 
w.r.t 
.
and 
,
and 
lie on 
and 
.
and 
are the reflections of 
such that the circumcenter of triangle 
coincides with the circumcenter of triangle 
.
square is filled with numbers 
.
squares is summed,and arranged in an increasing order. Is it possible to obtain the following sequences as a result of this operation?
and 
Find the value of 
be a triangle with a circle 
with center 
tangent to sides 
at 
respectively. Suppose the circle with diameter 
intersects the circumcircle of 
is the reflection of 
lie on 
are parallel to 
lie on the circumcircle of 
lie on the perpendicular bisector of 
lie on a circle. Suppose 
are perpendicular to sides 
at points 
The tangent lines from points 
intersects at point 
Prove that: 
are parallel.
that satisfy the equation 
?
.
be a tangent line to 
and 
be the lines obtained by reflecting 
and ![[asy]
import graph; size(10cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(9); pointfontpen=fp; real xmin=2.5,xmax=14,ymin=-2,ymax=8.8;
pen wwwwww=rgb(0.4,0.4,0.4), wwzzcc=rgb(0.4,0.6,0.8), yqqqyq=rgb(0.5019607843137255,0.,0.5019607843137255), qqcccc=rgb(0,0.8,0.8), cqcqcq=rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882);
pair A=(9.,3.2), B=(7.6,1.), C=(11.06,1.), P=(10.9037267872876,2.2894742445215552), A_1=(4.225302839837132,4.713218228557546), B_1=(13.318947812589142,4.209009023926156), C_1=(11.23751359726076,0.33222222307412996), M=(9.699442104579619,-0.3030337832920753), P_a=(10.9037267872876,-0.28947424452155524), P_b=(9.783568496159925,1.2405987537383683), P_c=(7.368886499833012,4.538918063810838), H=(9.,2.310909090909089), I_1=(10.610893030990937,2.689468626599);
D(A--B--C--cycle,wwwwww+linewidth(1)); D(A_1--B_1--C_1--cycle,wwzzcc+linewidth(1)); D(A--A_1,linetype("4 4")+wwzzcc);
D(CR((9.33,1.4445454545454548),1.786202861143444),green); D((xmin,-1.8625555265220795*xmin+22.598270831870913)--(xmax,-1.8625555265220795*xmax+22.598270831870913),wwzzcc); D(CR((8.755537154895718,4.161938324953828),4.563653413817887),yqqqyq); D(CR((9.33,0.5554545454545452),1.786202861143444),red); D(CR((10.73,2.7554545454545445),1.786202861143444),red); D(CR((7.27,2.7554545454545463),1.786202861143444),red); D(P_a--P_c,linetype("4 4")+wwzzcc); D(CR((5.579166076673129,0.695625707960863),4.239574899162703),linetype("0 2")+qqcccc+linewidth(1)); D(CR((10.26502172862502,2.9510482106080516),3.3028669264989134),linetype("0 2")+qqcccc+linewidth(1)); D(CR((10.293309185404135,0.43870973889632964),0.9501902769422019),linetype("0 2")+qqcccc+linewidth(1)); D(M--A_1,wwzzcc); D(M--B_1,wwzzcc);
D(A); MP("A",(9.078267017098327,3.3218831364461052),NE*lsf); D(B); MP("B",(7.208843909967847,0.5368242217415146),NE*lsf); D(C); MP("C",(11.233826314095717,0.7657331736350426),NE*lsf); D(P); MP("P",(10.985841616211061,2.406247328871993),NE*lsf); D(C_1); MP("C_1",(11.367356536033608,0.11715780993671321),NE*lsf); D(A_1); MP("A_1",(3.698906647600416,4.981473037674183),NE*lsf); D(B_1); MP("B_1",(13.484764341048743,4.046761484108944),NE*lsf); D(M); MP("M",(9.688690888814401,-0.7030992676817621),NE*lsf); D(P_a); MP("P_a",(10.966765870219934,-0.5886447917349982),NE*lsf); D(P_b); MP("P_b",(9.898524094716802,1.1663238394487165),NE*lsf); D(P_c); MP("P_c",(7.11346518001221,4.752564085780656),NE*lsf); D(H); MP("H",(9.078267017098327,2.4253230748631207),NE*lsf); D(I_1); MP("I_1",(10.680629680353023,2.8068379946856674),NE*lsf);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* Labels, by liberator */
defaultpen(fontsize(9pt));
label("$\Gamma_a$",(10,-1.2),S*lsf);
label("$\Gamma_b$",(12,4),dir(30)*lsf);
label("$\Gamma_c$",(6,1),N*lsf);
label("$\Gamma_1$",(5,7),N*lsf);
[/asy]](http://latex.artofproblemsolving.com/f/b/8/fb89592e2b7f740039a8d16e7184e085a07a93d6.png)
Let 
,
analogously; let 
be the circumcircle of 
.
,
.
is tangent to 
,
,
is the Steiner line of 
lie on 
concur at 
be the incenter of 
are the centers of 
then 
.
;
.
:
.
,
is the internal bisector of 
.
are the internal bisectors of 
respectively. Hence 
concur at 
as the Miquel point of the line 
.
is a 
![[asy]
import graph; size(10cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(9); pointfontpen=fp; real xmin=2.5,xmax=14,ymin=-2,ymax=8.8;
pen wwwwww=rgb(0.4,0.4,0.4), wwzzcc=rgb(0.4,0.6,0.8), yqqqyq=rgb(0.5019607843137255,0.,0.5019607843137255), cqcqcq=rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882);
pair A=(9.,3.2), B=(7.6,1.), C=(11.06,1.), P=(10.9037267872876,2.2894742445215552), A_1=(4.225302839837132,4.713218228557546), B_1=(13.318947812589142,4.209009023926156), C_1=(11.23751359726076,0.33222222307412996), M=(9.699442104579619,-0.3030337832920753), P_a=(10.9037267872876,-0.28947424452155524), P_b=(9.783568496159925,1.2405987537383683), P_c=(7.368886499833012,4.538918063810838), H=(9.,2.310909090909089), I_1=(10.610893030990937,2.689468626599), A_2=(7.556877292936604,1.6603150697678517), B_2=(11.116107846985981,1.4629688117894197), K=(12.028150798496817,7.3426508899567695);
D(A--B--C--cycle,wwwwww+linewidth(1)); D(A_1--B_1--C_1--cycle,wwzzcc+linewidth(1)); D(I_1--A_2--B_2--cycle,cqcqcq); D(A_1--K--B_1--cycle,wwzzcc); D(A--A_1,linetype("4 4")+wwzzcc);
D(CR((9.33,1.4445454545454548),1.786202861143444),green); D((xmin,-1.8625555265220795*xmin+22.598270831870913)--(xmax,-1.8625555265220795*xmax+22.598270831870913),wwzzcc); D(CR((8.755537154895718,4.161938324953828),4.563653413817887),yqqqyq); D(P_a--P_c,linetype("4 4")+wwzzcc); D(CR((5.579166076673129,0.695625707960863),4.239574899162703),linetype("0 2")+red+linewidth(1)); D(CR((10.26502172862502,2.9510482106080516),3.3028669264989134),linetype("0 2")+red+linewidth(1)); D(CR((10.293309185404135,0.43870973889632964),0.9501902769422019),linetype("0 2")+red+linewidth(1)); D(M--A_1,wwzzcc); D(M--B_1,wwzzcc); D((xmin,0.33698371098991503*xmin+3.289359997533001)--(xmax,0.33698371098991503*xmax+3.289359997533001),wwzzcc); D((xmin,-2.4276798224809775*xmin+36.54314988522595)--(xmax,-2.4276798224809775*xmax+36.54314988522595),wwzzcc); D(I_1--A_2--B_2--cycle,wwzzcc);
D(A); MP("A",(9.078267017098327,3.3218831364461052),NE*lsf); D(B); MP("B",(7.208843909967847,0.5368242217415146),NE*lsf); D(C); MP("C",(11.233826314095717,0.7657331736350426),NE*lsf); D(P); MP("P",(10.985841616211061,2.406247328871993),NE*lsf); D(C_1); MP("C_1",(11.367356536033608,0.11715780993671321),NE*lsf); D(A_1); MP("A_1",(3.698906647600416,4.981473037674183),NE*lsf); D(B_1); MP("B_1",(13.484764341048743,4.046761484108944),NE*lsf); D(M); MP("M",(9.688690888814401,-0.7030992676817621),NE*lsf); D(P_a); MP("P_a",(10.966765870219934,-0.5886447917349982),NE*lsf); D(P_b); MP("P_b",(9.898524094716802,1.1663238394487165),NE*lsf); D(P_c); MP("P_c",(7.11346518001221,4.752564085780656),NE*lsf); D(H); MP("H",(9.078267017098327,2.4253230748631207),NE*lsf); D(I_1); MP("I_1",(10.680629680353023,2.8068379946856674),NE*lsf); D(A_2); MP("A_2",(7.266071147941229,1.9484294250849374),NE*lsf); D(B_2); MP("B_2",(11.386432282024735,1.5096872672890087),NE*lsf); D(K); MP("K",(12.07315913770532,7.5376230004852465),NE*lsf);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* Labels, by liberator */
defaultpen(fontsize(9pt));
label("$\Gamma_1$",(5,7),N*lsf);
[/asy]](http://latex.artofproblemsolving.com/3/0/c/30cdce277f8b9fc56322d3d0a0baa20ca56c49e3.png)
Suppose the tangents to 
at 
respectively intersect at 
.
;
.
,
.
.
,
are collinear.
,
and 
are homothetic.
By a converse of Reim's theorem, 
are tangent at ![\[f(xf(y) + x) = xy + f(x)\]](http://latex.artofproblemsolving.com/9/9/f/99f580ebc50846e6bc2c004667559922749a4dfa.png)
be a positive integer larger than 
.
of subsets of 
having the following properties:
distinct sets of 
October 2018
April 2018