fun factory or boredom buster!
by jkim0656, Apr 4, 2025, 4:13 AM
Loading poll details...
\end{table}2025 MATHCOUNTS State Hub
by SirAppel, Apr 1, 2025, 12:22 PM
This post has been edited 59 times. Last edited by SirAppel, Yesterday at 9:39 PM
real math problems
by Soupboy0, Mar 25, 2025, 11:40 PM
Ill be posting questions once in a while. Here's the first question:
What fraction of numbers from
to 
have the digit 
and are divisible by 
?
What fraction of numbers from
to 
have the digit 
and are divisible by 
?The daily problem!
by Leeoz, Mar 21, 2025, 10:01 PM
Every day, I will try to post a new problem for you all to solve! If you want to post a daily problem, you can! 
Please hide solutions and answers, hints are fine though!
The first problem is:
Please hide solutions and answers, hints are fine though!
The first problem is:
March 21st Problem wrote:
Alice flips a fair coin until she gets 2 heads in a row, or a tail and then a head. What is the probability that she stopped after 2 heads in a row? Express your answer as a common fraction.
and 
intersect at the point 
.
?
,
.
are chosen from 0 to 1. What is the probability that their positive difference is more than 
?
?
This post has been edited 4 times. Last edited by Leeoz, Mar 30, 2025, 12:08 AM
2024
by Technodoggo, Jan 1, 2024, 2:00 PM
Happy New Year!
To celebrate the start of 2024, I've decided to put up a few facts about the year. I don't have many, so y'all should also
(1 and 2 are the same :sob: just distribute)
Post more cool facts here! Keep a running chain with quote boxes for facts. (I don't know if this technically qualifies as a marathon, but I hope this is allowed lol)
To celebrate the start of 2024, I've decided to put up a few facts about the year. I don't have many, so y'all should also
![\[(2+0+2+4)\left((2\cdot2\cdot4)^2-\dfrac{2^2+0^2+2^2+4^2}{2+0+2+4}\right)=2024.\]](http://latex.artofproblemsolving.com/0/0/7/0074cd2c9c2f8327db52e113ef2beefe98ceca90.png)
![\[(2+0+2+4)(2\cdot2\cdot4)^2-\left(2^2+0^2+2^2+4^2\right)=2024.\]](http://latex.artofproblemsolving.com/d/5/4/d54152fae9e6e259dc464dbef3ed39272eb9e599.png)
(1 and 2 are the same :sob: just distribute)
Post more cool facts here! Keep a running chain with quote boxes for facts. (I don't know if this technically qualifies as a marathon, but I hope this is allowed lol)
IMO 2011 Problem 6
by liberator, Jul 21, 2015, 4:57 PM
Problem: Let 
be an acute triangle with circumcircle 
. Let 
be a tangent line to , and let 
and 
be the lines obtained by reflecting in the lines 
, 
and 
, respectively. Show that the circumcircle of the triangle determined by the lines and is tangent to the circle .
Proposed by Japan
be an acute triangle with circumcircle 
. Let 
be a tangent line to , and let 
and 
be the lines obtained by reflecting in the lines 
, 
and 
, respectively. Show that the circumcircle of the triangle determined by the lines and is tangent to the circle .Proposed by Japan
![[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 
w.r.t 
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 
.
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 This post has been edited 2 times. Last edited by liberator, Jul 22, 2015, 1:35 PM
IMO 2015 Problem 3
by liberator, Jul 14, 2015, 2:48 PM
Problem: Let be an acute triangle with 
. Let 
be its cirumcircle, 
its orthocenter, and 
the foot of the altitude from 
. Let 
be the midpoint of . Let 
be the point on such that 
and let 
be the point on such that 
. Assume that the points , 
, 
, and are all different and lie on in this order.
Prove that the circumcircles of triangles
and 
are tangent to each other.
Proposed by Ukraine
be an acute triangle with 
. Let 
be its cirumcircle, 
its orthocenter, and 
the foot of the altitude from 
. Let 
be the midpoint of . Let 
be the point on such that 
and let 
be the point on such that 
. Assume that the points , 
, 
, and are all different and lie on in this order.Prove that the circumcircles of triangles
and 
are tangent to each other.Proposed by Ukraine
![[asy]
/* IMO 2015 Problem 3, free script by liberator, 14 July 2015 */
import olympiad;
import cse5;
unitsize(3.2cm);
pen ddd = rgb(0.4,0.6,0.8);
pointpen=black;
defaultpen(fontsize(10pt));
pair A = dir(75);
pair B = dir(205);
pair C = dir(-25);
pair O = origin;
pair H = orthocenter(A,B,C);
pair F = foot(A,B,C);
pair D = IP(L(H,F,0,2015),unitcircle);
pair M = midpoint(B--C);
pair Ap = B+C-H;
pair Q = IP(L(M,H,0,2015), unitcircle);
pair Qp = A+Ap-Q;
pair K = IP(L(Qp,H,-0.1,2015), unitcircle);
pair P = circumcenter(D,H,K);
D(unitcircle, heavygreen);
D(circumcircle(K,Q,H), red+linewidth(1));
D(circumcircle(F,K,M), heavymagenta);
D(circumcircle(D,H,K), grey+linetype("0 2")+linewidth(1));
D(A--B--C--cycle, ddd+linewidth(1));
D(A--D, ddd);
D(A--Qp, ddd);
D(Q--Ap, ddd+dashed);
D(K--Qp, ddd+dashed);
D(H--P--K, ddd);
D(rightanglemark(C,F,A,2), ddd);
D("A",A,dir(A)*2);
D("B",B,dir(B));
D("C",C,dir(C));
D("D",D,dir(-110));
D("F",F,dir(F));
D("H",H,dir(-65)*2);
D("K",K,dir(-15)*1.6);
D("M",M,dir(-70));
D("P",P,dir(P));
D("Q",Q,dir(Q));
D("A'",Ap,dir(Ap));
D("Q'",Qp,dir(Qp));
D("\Gamma",O,dir(160)*28);
[/asy]](http://latex.artofproblemsolving.com/a/a/6/aa6ae4ba390a833b56195faffc62aaf837297443.png)
be the antipodes of 
respectively w.r.t 
are collinear. Also observe that 
are collinear, since 
,
:
.
to 
at 
;
from rectangle 
,
,
,
is tangent to 
.
,
is tangent to 
:
,This post has been edited 2 times. Last edited by liberator, Jul 14, 2015, 8:58 PM
Coaxal circles in incenter/excenter configuration
by liberator, Apr 15, 2015, 7:18 PM
Problem: Let be a triangle, whose excircle (opposite ) touches 
at 
respectively. Denote 
as the intersection of the lines 
and , and 
as the intersection of the lines 
and . If 
is the excenter of 
, opposite , prove that the circumcircles of triangles 
and 
are coaxal.
Commentary: We may replace "excircle" with "incircle", "excenter" with "incenter", and the result still holds.
See interactive diagram here.
be a triangle, whose excircle (opposite ) touches 
at 
respectively. Denote 
as the intersection of the lines 
and , and 
as the intersection of the lines 
and . If 
is the excenter of 
, opposite , prove that the circumcircles of triangles 
and 
are coaxal.Commentary: We may replace "excircle" with "incircle", "excenter" with "incenter", and the result still holds.
See interactive diagram here.
denote the second intersection of circles 
and 
.
.
is cyclic, from perpendicularity of tangents.
is tangent to 
.
and 
,
.
and 
,
.
at 
,
.
,Attachments:
October 2018
April 2018
April 2016
December 2014
October 2014
September 2014
August 2014