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Cup of Combinatorics
M11100111001Y1R   7
N an hour ago by MathematicalArceus
Source: Iran TST 2025 Test 4 Problem 2
There are \( n \) cups labeled \( 1, 2, \dots, n \), where the \( i \)-th cup has capacity \( i \) liters. In total, there are \( n \) liters of water distributed among these cups such that each cup contains an integer amount of water. In each step, we may transfer water from one cup to another. The process continues until either the source cup becomes empty or the destination cup becomes full.

$a)$ Prove that from any configuration where each cup contains an integer amount of water, it is possible to reach a configuration in which each cup contains exactly 1 liter of water in at most \( \frac{4n}{3} \) steps.

$b)$ Prove that in at most \( \frac{5n}{3} \) steps, one can go from any configuration with integer water amounts to any other configuration with the same property.
7 replies
M11100111001Y1R
May 27, 2025
MathematicalArceus
an hour ago
Inequality
knm2608   17
N an hour ago by Adywastaken
Source: JBMO 2016 shortlist
If the non-negative reals $x,y,z$ satisfy $x^2+y^2+z^2=x+y+z$. Prove that
$$\displaystyle\frac{x+1}{\sqrt{x^5+x+1}}+\frac{y+1}{\sqrt{y^5+y+1}}+\frac{z+1}{\sqrt{z^5+z+1}}\geq 3.$$When does the equality occur?

Proposed by Dorlir Ahmeti, Albania
17 replies
knm2608
Jun 25, 2017
Adywastaken
an hour ago
Circumcircle of XYZ is tangent to circumcircle of ABC
mathuz   39
N 2 hours ago by zuat.e
Source: ARMO 2013 Grade 11 Day 2 P4
Let $ \omega $ be the incircle of the triangle $ABC$ and with centre $I$. Let $\Gamma $ be the circumcircle of the triangle $AIB$. Circles $ \omega $ and $ \Gamma $ intersect at the point $X$ and $Y$. Let $Z$ be the intersection of the common tangents of the circles $\omega$ and $\Gamma$. Show that the circumcircle of the triangle $XYZ$ is tangent to the circumcircle of the triangle $ABC$.
39 replies
mathuz
May 22, 2013
zuat.e
2 hours ago
Arc Midpoints Form Cyclic Quadrilateral
ike.chen   57
N 2 hours ago by cj13609517288
Source: ISL 2022/G2
In the acute-angled triangle $ABC$, the point $F$ is the foot of the altitude from $A$, and $P$ is a point on the segment $AF$. The lines through $P$ parallel to $AC$ and $AB$ meet $BC$ at $D$ and $E$, respectively. Points $X \ne A$ and $Y \ne A$ lie on the circles $ABD$ and $ACE$, respectively, such that $DA = DX$ and $EA = EY$.
Prove that $B, C, X,$ and $Y$ are concyclic.
57 replies
ike.chen
Jul 9, 2023
cj13609517288
2 hours ago
Complex number
ronitdeb   0
2 hours ago
Let $z_1, ... ,z_5$ be vertices of regular pentagon inscribed in a circle whose radius is $2$ and center is at $6+i8$. Find all possible values of $z_1^2+z_2^2+...+z_5^2$
0 replies
ronitdeb
2 hours ago
0 replies
Elementary Problems Compilation
Saucepan_man02   29
N 2 hours ago by Electrodynamix777
Could anyone send some elementary problems, which have tricky and short elegant methods to solve?

For example like this one:
Solve over reals: $$a^2 + b^2 + c^2 + d^2  -ab-bc-cd-d +2/5=0$$
29 replies
Saucepan_man02
May 26, 2025
Electrodynamix777
2 hours ago
Generic Real-valued FE
lucas3617   4
N 2 hours ago by GreekIdiot
$f: \mathbb{R} -> \mathbb{R}$, find all functions where $f(2x+f(2y-x))+f(-x)+f(y)=2f(x)+f(y-2x)+f(2y)$ for all $x$,$y \in \mathbb{R}$
4 replies
lucas3617
Apr 25, 2025
GreekIdiot
2 hours ago
Find all possible values of BT/BM
va2010   54
N 3 hours ago by lpieleanu
Source: 2015 ISL G4
Let $ABC$ be an acute triangle and let $M$ be the midpoint of $AC$. A circle $\omega$ passing through $B$ and $M$ meets the sides $AB$ and $BC$ at points $P$ and $Q$ respectively. Let $T$ be the point such that $BPTQ$ is a parallelogram. Suppose that $T$ lies on the circumcircle of $ABC$. Determine all possible values of $\frac{BT}{BM}$.
54 replies
va2010
Jul 7, 2016
lpieleanu
3 hours ago
A Familiar Point
v4913   52
N 3 hours ago by SimplisticFormulas
Source: EGMO 2023/6
Let $ABC$ be a triangle with circumcircle $\Omega$. Let $S_b$ and $S_c$ respectively denote the midpoints of the arcs $AC$ and $AB$ that do not contain the third vertex. Let $N_a$ denote the midpoint of arc $BAC$ (the arc $BC$ including $A$). Let $I$ be the incenter of $ABC$. Let $\omega_b$ be the circle that is tangent to $AB$ and internally tangent to $\Omega$ at $S_b$, and let $\omega_c$ be the circle that is tangent to $AC$ and internally tangent to $\Omega$ at $S_c$. Show that the line $IN_a$, and the lines through the intersections of $\omega_b$ and $\omega_c$, meet on $\Omega$.
52 replies
v4913
Apr 16, 2023
SimplisticFormulas
3 hours ago
Tangential quadrilateral and 8 lengths
popcorn1   72
N 3 hours ago by cj13609517288
Source: IMO 2021 P4
Let $\Gamma$ be a circle with centre $I$, and $A B C D$ a convex quadrilateral such that each of the segments $A B, B C, C D$ and $D A$ is tangent to $\Gamma$. Let $\Omega$ be the circumcircle of the triangle $A I C$. The extension of $B A$ beyond $A$ meets $\Omega$ at $X$, and the extension of $B C$ beyond $C$ meets $\Omega$ at $Z$. The extensions of $A D$ and $C D$ beyond $D$ meet $\Omega$ at $Y$ and $T$, respectively. Prove that \[A D+D T+T X+X A=C D+D Y+Y Z+Z C.\]
Proposed by Dominik Burek, Poland and Tomasz Ciesla, Poland
72 replies
popcorn1
Jul 20, 2021
cj13609517288
3 hours ago
Equal Distances in an Isosceles Setting
mojyla222   3
N Apr 20, 2025 by sami1618
Source: IDMC 2025 P4
Let $ABC$ be an isosceles triangle with $AB=AC$. The circle $\omega_1$, passing through $B$ and $C$, intersects segment $AB$ at $K\neq B$. The circle $\omega_2$ is tangent to $BC$ at $B$ and passes through $K$. Let $M$ and $N$ be the midpoints of segments $AB$ and $AC$, respectively. The line $MN$ intersects $\omega_1$ and $\omega_2$ at points $P$ and $Q$, respectively, where $P$ and $Q$ are the intersections closer to $M$. Prove that $MP=MQ$.

Proposed by Hooman Fattahi
3 replies
mojyla222
Apr 20, 2025
sami1618
Apr 20, 2025
Equal Distances in an Isosceles Setting
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G H BBookmark kLocked kLocked NReply
Source: IDMC 2025 P4
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mojyla222
103 posts
#1 • 1 Y
Y by sami1618
Let $ABC$ be an isosceles triangle with $AB=AC$. The circle $\omega_1$, passing through $B$ and $C$, intersects segment $AB$ at $K\neq B$. The circle $\omega_2$ is tangent to $BC$ at $B$ and passes through $K$. Let $M$ and $N$ be the midpoints of segments $AB$ and $AC$, respectively. The line $MN$ intersects $\omega_1$ and $\omega_2$ at points $P$ and $Q$, respectively, where $P$ and $Q$ are the intersections closer to $M$. Prove that $MP=MQ$.

Proposed by Hooman Fattahi
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Retemoeg
59 posts
#2 • 1 Y
Y by sami1618
Not really a fan of this solution but.. whatever I guess :-D

Let $MN$ intersect $\omega_1$ and $\omega_2$ again at $X, Y$ such that $M$ lies between $P$ and $X$ at the same time, between $Q$ and $Y$. Denote $E, F$ the midpoints of $PX$ and $QY$.
As per the definition of $\omega_2$, We have that $BE$ passes through the center of $\omega_2$, which means $\triangle BPX$ is isoceles. By symmetry $\triangle AQY$ is also isoceles. We have that $AEBF$ is a parallelogram, thus $\overline{ME} + \overline{MF} = 0$. Or in other words:
\[\overline{MP} + \overline{MQ} + \overline{MX} + \overline{MY} = 0\]But then, by power of a point: $\overline{MP}\cdot\overline{MX} = \overline{MK}\cdot\overline{MB} = \overline{MQ}\cdot\overline{MY}$
Let $\overline{MP} = a, \overline{MQ} = b, \overline{MX} = c, \overline{MY} = d$.
At this point: $a + b + c + d = 0$ and $ac = bd$. That being said, $bd + (b + c + d)\cdot c = 0 \leftrightarrow (c + b)(c + d) = 0$.
The case $c = -b$ is obviously ruled out as $P, M, X$ are collinear in that very same order.
Hence, $c = -d$, we'll have $a = -b$ or $MP = MQ$, thus proving the problem.
This post has been edited 2 times. Last edited by Retemoeg, Apr 20, 2025, 6:20 AM
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Mahdi_Mashayekhi
697 posts
#3 • 1 Y
Y by sami1618
Let $PQ$ meet $\omega_1$ and $\omega_2$ again at $X$ and $Y$. Let $MX = \alpha, MY = \beta, MP = \theta, MQ = \lambda$ Now let $f(X) = XP^2-XQ^2$. By linearity of PoP we have $f(M)=\frac{\theta (\theta+\lambda)^2-\lambda (\theta+\lambda)^2}{\theta+\lambda}=(\theta+\lambda)(\theta-\lambda)$ and $f(M)=\frac{\alpha f(X) + \beta f(Y)}{\alpha+\beta}=\frac{\alpha (\theta+\lambda)(\theta-\lambda+2\alpha)-\beta (\theta+\lambda)(\lambda-\theta+2\beta))}{\alpha+\beta}$ so we have $\frac{\alpha(\theta-\lambda+2\alpha)-\beta(\lambda-\theta+2\beta)}{\alpha+\beta}=(\theta-\lambda)$ which implies $\alpha^2-\beta^2=0 \implies \alpha=\beta$. Now note that $MP.MX=MK.MB=MQ.MY$ so $\alpha\theta=\beta\lambda$ so now we have $\theta=\lambda$ as wanted.
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sami1618
920 posts
#4 • 1 Y
Y by Retemoeg
Let line $MN$ cut $\omega_1$ and $\omega_2$ for the second time at $P'$ and $Q'$, respectively. Let $O_1$ and $O_2$ be the centers of $\omega_1$ and $\omega_2$ respectively. Finally, let $M_p$ be the midpoint of $PP'$ and $M_q$ be the midpoint $QQ'$.
[asy]
import geometry;
size(10cm);

pair A=dir(90);
pair B=dir(220);
pair C=dir(320);
pair K=B+.6(A-B);
pair O_1=circumcenter(B,K,C);
circle w_1=circle(K,B,C);
pair M=.5(A+B);
pair N=.5(A+C);
point[] T=intersectionpoints(w_1,line(N,M));
pair P=T[0];
pair Pp=T[1];
line b=line(.5(B+K),.5(B+K)+rotate(90)*(B-K));
line l=line(B,B+rotate(90)*(C-B));
pair O_2=intersectionpoint(b,l);
circle w_2=circle(B,K,O_2+rotate(90)*(K-O_2));
point[] X=intersectionpoints(w_2,line(M,N));
pair Q=X[1];
pair Qp=X[0];
pair M_p=.5(P+Pp);
pair M_q=.5(Q+Qp);
fill(A--B--C--cycle, palered);
draw(A--B--C--cycle);
draw(w_1,deepblue);
draw(w_2,deepgreen);
draw(Pp--Qp);
draw(B--M_q, dashed); draw(O_1--A, dashed);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$K$",K,2*dir(90));
dot("$O_1$",O_1,dir(270));
dot("$M$",M,dir(120));
dot("$N$",N,dir(60));
dot("$P$",P,dir(120));
dot("$P'$",Pp,dir(60));
dot("$O_2$",O_2,dir(180));
dot("$Q$",Q,dir(60));
dot("$Q'$",Qp,dir(120));
dot("$M_p$",M_p,dir(60));
dot("$M_q$",M_q,dir(120));




[/asy]
Since $O_1M_p\perp MN\parallel BC$, it follows that $M_p$ is also the midpoint of $MN$. Similarly, $M_q$ is the foot of $B$ onto $MN$, so $|M_pM|=|M_qM|$. Then $$|M_pP|^2-|M_pM|^2=|MP|\cdot|MP'|=-\mathbb{P}(M,\omega_1)=|MK|\cdot |MB|=$$$$-\mathbb{P}(M,\omega_2)=|MQ|\cdot|MQ'|=|M_qQ|^2-|M_qM|^2,$$so $|M_pP|=|M_qQ|$. The result now easily follows (if $K$ does not lie on segment $AM$ then the signs would need to be reversed).
This post has been edited 6 times. Last edited by sami1618, Apr 20, 2025, 10:06 PM
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