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functional inequality with equality
miiirz30   0
40 minutes ago
Source: 2025 Euler Olympiad, Round 2
Find all functions \( f : \mathbb{R} \to \mathbb{R} \) such that the following two conditions hold:

1. For all real numbers $a$ and $b$ satisfying $a^2 + b^2 = 1$, We have $f(x) + f(y) \geq f(ax + by)$ for all real numbers $x, y$.

2. For all real numbers $x$ and $y$, there exist real numbers $a$ and $b$, such that $a^2 + b^2 = 1$ and $f(x) + f(y) = f(ax + by)$.

Proposed by Zaza Melikidze, Georgia
0 replies
miiirz30
40 minutes ago
0 replies
two lines passsing through the midpoint
miiirz30   0
an hour ago
Source: 2025 Euler Olympiad, Round 2
Points $A$, $B$, $C$, and $D$ lie on a line in that order, and points $E$ and $F$ are located outside the line such that $EA=EB$, $FC=FD$ and $EF \parallel AD$. Let the circumcircles of triangles $ABF$ and $CDE$ intersect at points $P$ and $Q$, and the circumcircles of triangles $ACF$ and $BDE$ intersect at points $M$ and $N$. Prove that the lines $PQ$ and $MN$ pass through the midpoint of segment $EF$.

Proposed by Giorgi Arabidze, Georgia
0 replies
miiirz30
an hour ago
0 replies
Intertwined numbers
miiirz30   1
N an hour ago by JARP091
Source: 2025 Euler Olympiad, Round 2
Let a pair of positive integers $(n, m)$ that are relatively prime be called intertwined if among any two divisors of $n$ greater than $1$, there exists a divisor of $m$ and among any two divisors of $m$ greater than $1$, there exists a divisor of $n$. For example, pair $(63, 64)$ is intertwined.

a) Find the largest integer $k$ for which there exists an intertwined pair $(n, m)$ such that the product $nm$ is equal to the product of the first $k$ prime numbers.
b) Prove that there does not exist an intertwined pair $(n, m)$ such that the product $nm$ is the product of $2025$ distinct prime numbers.
c) Prove that there exists an intertwined pair $(n, m)$ such that the number of divisors of $n$ is greater than $2025$.

Proposed by Stijn Cambie, Belgium
1 reply
miiirz30
an hour ago
JARP091
an hour ago
Find the value
sqing   11
N an hour ago by mathematical-forest
Source: 2024 China Fujian High School Mathematics Competition
Let $f(x)=a_6x^6+a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0,$ $a_i\in\{-1,1\} ,i=0,1,2,\cdots,6 $ and $f(2)=-53 .$ Find the value of $f(1).$
11 replies
sqing
Jun 22, 2024
mathematical-forest
an hour ago
Self-evident inequality trick
Lukaluce   20
N an hour ago by ytChen
Source: 2025 Junior Macedonian Mathematical Olympiad P4
Let $x, y$, and $z$ be positive real numbers, such that $x^2 + y^2 + z^2 = 3$. Prove the inequality
\[\frac{x^3}{2 + x} + \frac{y^3}{2 + y} + \frac{z^3}{2 + z} \ge 1.\]When does the equality hold?
20 replies
Lukaluce
May 18, 2025
ytChen
an hour ago
three circles
barasawala   8
N an hour ago by FrancoGiosefAG
Source: Mexico 2003
$A, B, C$ are collinear with $B$ betweeen $A$ and $C$. $K_{1}$ is the circle with diameter $AB$, and $K_{2}$ is the circle with diameter $BC$. Another circle touches $AC$ at $B$ and meets $K_{1}$ again at $P$ and $K_{2}$ again at $Q$. The line $PQ$ meets $K_{1}$ again at $R$ and $K_{2}$ again at $S$. Show that the lines $AR$ and $CS$ meet on the perpendicular to $AC$ at $B$.
8 replies
barasawala
Mar 2, 2007
FrancoGiosefAG
an hour ago
d(2025^{a_i}-1) divides a_{n+1}
navi_09220114   3
N an hour ago by quacksaysduck
Source: TASIMO 2025 Day 2 Problem 5
Let $a_n$ be a strictly increasing sequence of positive integers such that for all positive integers $n\ge 1$
\[d(2025^{a_n}-1)|a_{n+1}.\]Show that for any positive real number $c$ there is a positive integers $N_c$ such that $a_n>n^c$ for all $n\geq N_c$.

Note. Here $d(m)$ denotes the number of positive divisors of the positive integer $m$.
3 replies
navi_09220114
May 19, 2025
quacksaysduck
an hour ago
Interesting inequalities
sqing   6
N 2 hours ago by sqing
Source: Own
Let $ a,b,c,d\geq  0 , a+b+c+d \leq 4.$ Prove that
$$a(bc+bd+cd)  \leq \frac{256}{81}$$$$ ab(a+2c+2d ) \leq \frac{256}{27}$$$$  ab(a+3c+3d )  \leq \frac{32}{3}$$$$ ab(c+d ) \leq \frac{64}{27}$$
6 replies
sqing
Yesterday at 1:25 PM
sqing
2 hours ago
Trapezium with two right-angles: prove < AKB = 90° and more
Leonardo   6
N 2 hours ago by FrancoGiosefAG
Source: Mexico 2002
Let $ABCD$ be a quadrilateral with $\measuredangle DAB=\measuredangle ABC=90^{\circ}$. Denote by $M$ the midpoint of the side $AB$, and assume that $\measuredangle CMD=90^{\circ}$. Let $K$ be the foot of the perpendicular from the point $M$ to the line $CD$. The line $AK$ meets $BD$ at $P$, and the line $BK$ meets $AC$ at $Q$. Show that $\angle{AKB}=90^{\circ}$ and $\frac{KP}{PA}+\frac{KQ}{QB}=1$.

[Moderator edit: The proposed solution can be found at http://erdos.fciencias.unam.mx/mexproblem3.pdf .]
6 replies
Leonardo
May 8, 2004
FrancoGiosefAG
2 hours ago
Interesting inequalities
sqing   5
N 2 hours ago by sqing
Source: Own
Let $ a,b,c\geq  0 ,a+b+c\leq 3. $ Prove that
$$a^2+b^2+c^2+2ab+2bc  +  abc \leq \frac{244}{27}$$$$a^2+b^2+c^2+\frac{1}{2}ab +2ca+2bc +  abc \leq \frac{73}{8}$$$$ a^2+b^2+c^2+ab+2ca+2bc  + \frac{1}{2}abc  \leq \frac{487}{54}$$$$a^2+b^2+c^2+a+b+ab+2ca+2bc+2abc\leq 12$$
5 replies
sqing
Yesterday at 12:52 PM
sqing
2 hours ago
Two circles, a tangent line and a parallel
Valentin Vornicu   105
N May 15, 2025 by Fly_into_the_sky
Source: IMO 2000, Problem 1, IMO Shortlist 2000, G2
Two circles $ G_1$ and $ G_2$ intersect at two points $ M$ and $ N$. Let $ AB$ be the line tangent to these circles at $ A$ and $ B$, respectively, so that $ M$ lies closer to $ AB$ than $ N$. Let $ CD$ be the line parallel to $ AB$ and passing through the point $ M$, with $ C$ on $ G_1$ and $ D$ on $ G_2$. Lines $ AC$ and $ BD$ meet at $ E$; lines $ AN$ and $ CD$ meet at $ P$; lines $ BN$ and $ CD$ meet at $ Q$. Show that $ EP = EQ$.
105 replies
Valentin Vornicu
Oct 24, 2005
Fly_into_the_sky
May 15, 2025
Two circles, a tangent line and a parallel
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G H BBookmark kLocked kLocked NReply
Source: IMO 2000, Problem 1, IMO Shortlist 2000, G2
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MagicalToaster53
159 posts
#97 • 1 Y
Y by cubres
Notice that $MN$ bisects $\overline{AB},$ as the point where they intersect has equivalent powers to the respective circles. From this we discover that $M$ is the midpoint of $\overline{PQ}$ due to the homothety at $N$ mapping $\overline{PQ} \mapsto \overline{AB}.$ It then suffices to show $EM \perp PQ$.

Claim: $\triangle EAB \cong \triangle MAB.$
Proof: Observe that $AB$ bisects $\angle EAM:$ \[\angle EAB = \angle ECM = \angle ACM = \angle BAM. \]Similarly by symmetry, we find that $AB$ bisects $\angle EBM.$ Then by SAS, we conclude that $\triangle EAB \cong \triangle MAB,$ as was to be shown. $\square$

Now $EAMB$ is a kite, so that $EM \perp AB \implies EM \perp PQ,$ as $AB \parallel PQ$. Hence $\triangle EPQ$ is isosceles, as desired. $\blacksquare$
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lnzhonglp
120 posts
#98 • 1 Y
Y by cubres
Observe that $$\measuredangle EBA = \measuredangle EDC = \measuredangle ABM$$and $$\measuredangle EAB = \measuredangle ECM = \measuredangle MAB,$$so $\triangle EAB \cong \triangle MAB$ and $EM \perp AB$. From radical axis we see that $MN$ bisects $AB$, so by homothety at $N$ we find that $M$ is the midpoint of $PQ$. Then since $EM \perp PQ$ it follows that $EP = EQ$.
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Likeminded2017
391 posts
#99 • 1 Y
Y by cubres
Let $MN$ and $AB$ intersect at $K.$ As $K$ lies on the radical axis we have $KA=KB.$ Taking the homothety centered at $N$ that maps $M$ to $K$ we find that $MP=MQ.$ Then $\angle BAM=\angle MCA=\angle BAE$ and similarly $\angle EBA=\angle ABM$ so by ASA congruence $\triangle EAB \cong \triangle MAB.$ Thus $EM$ is perpendicular to $AB$ and thus $PQ$ so $EP=EQ.$
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ezpotd
1285 posts
#100 • 1 Y
Y by cubres
Letting $AB$ meet $MN$ at $K$, we see $KA^2 = KM \cdot KN = KB^2$, so $K$ is the midpoint of $AB$, taking the homothety from $AB$ to $PQ$ this gives $MP = MQ$, so it suffices to prove $EM \perp PQ$, or $EM \perp AB$. However, angle chasing gives $\angle EAB = \angle ECM = \angle AMC = \angle BAM, \angle EBA = \angle EDM = \angle BMD = \angle ABM$, so $\triangle EAB \cong \angle MAB$, so $E,M$ are reflections about $AB$, done.
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optimusprime154
23 posts
#101 • 2 Y
Y by SorPEEK, cubres
let \(Z \)= \(MN  \cap AB\) from PoP we know \(ZB^2 = ZM * ZN\) and \(ZA^2 = ZM * ZN\) which means \(ZA = ZB\) and since \(PQBA\) is a trapezoid, \(PM = MQ\)
now we prove \(EM \perp CD\) and we're done. label \(\angle ACM = x, \angle BDM = y\) we know from \(AB \parallel CD\) that \(\angle EAB = x\) and \(\angle EDC = y\)
we also know from \(AB\) being tangent that \(\angle MBA\ = y\) and \(\angle MAB = x\) the above facts indicate the similarity of \(\triangle MAB , \triangle EAB \) so \(EM \perp AB\) => \(EM \perp CD\) and we're done
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Maximilian113
575 posts
#102 • 1 Y
Y by cubres
Extend $MN$ to intersect $AB$ at $K,$ then clearly $AK=BK.$ Therefore $PM=MQ$ by the parallel lines.

Now, notice that $\angle EAB = \angle ACM = \angle BAM = \angle AMC.$ Therefore $AB$ bisects $\angle EAM$ and $AC=AM.$ Similarly $AB$ bisects $EBM$ and $BD=BM.$ Therefore, $\triangle EAB \cong \triangle MAB$ so $A, B$ are the circumcenters of $\triangle EMC, \triangle EMD$ respectively so by Thales' Theorem it follows that $\angle EMP = \angle EMQ = 90^\circ.$ Hence $EM$ is the perpendicular bisector of $PQ,$ and we are done. QED
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ehuseyinyigit
839 posts
#104 • 1 Y
Y by cubres
WELL-KNOWN. Since $AB\parallel CD$, we have $$\angle ACM=\angle AMC=\angle MAB=\angle BAE$$Thus, $AC=AM$ and $AB$ bisects $\angle EAM$. Similarly $BD=BM$ and $AB$ bisects $\angle EBM$ hence $ABME$ is a kite $\Rightarrow AE=AM=AC$ implying $EM\perp CD$. On the other hand, for point $F$ being intersection of lines $AB=MN$, $AF=BF$ implies $PM=MQ$. We obtain $EP=EQ$.

WELL-KNOWN. We will show $EN$ bisects $\angle CND$. This is true because $$\angle ENB=\angle EAB=\angle ENC$$and $\angle ANE=\angle ABE=\angle BND$. Thus, $EN$ bisects $\angle CND$.
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mathwiz_1207
103 posts
#105 • 1 Y
Y by cubres
We have
\[\measuredangle EAB = \measuredangle ACM = \measuredangle ANM = \measuredangle BAM = \measuredangle CMA\]\[\measuredangle EBA = \measuredangle BDM = \measuredangle BNM = \measuredangle ABM = \measuredangle DMB\], implying that $\triangle AMB \cong \triangle AEB$, and that $\triangle AMC, \triangle BDM$ are isosceles. Therefore,
\[AE = AM = AC\]so $A$ is the center of $(EMC)$, and similarly $B$ is the cetner of $(EMD)$. Thus, $EM \perp CD$. Now, let $L = MN \cap AB$, by Radical Axis,
\[LA = LB\]and since $\triangle NPQ \sim \triangle NAB$, we have $MP = MQ \implies EQ = EP$, so we are done.
This post has been edited 1 time. Last edited by mathwiz_1207, Feb 23, 2025, 11:53 PM
Reason: forgot to add last line
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study1126
569 posts
#106 • 1 Y
Y by cubres
All angles will be directed in this solution.

First, $\angle EAB=\angle ECB=\angle BAM$, since $AB||CD$ and $AB$ is tangent to the two circles. Similarly, $\angle EBA=\angle ABM$. Combined with $AB=AB$ gives $\triangle EAB\cong \triangle MAB$.

This implies $EAMB$ is a kite, so $EM\perp AB$, which means $EM\perp CD$. Since $M$ is on the radical axis, extending $NM$ to $X$ on $AB$ gives $X$ is the midpoint of $AB$. Homothety shows $M$ is the midpoint of $PQ$. Thus, $MP=MQ$, so by congruent triangle $\triangle EPM\cong \triangle EQM$, $EP=EQ$, as desired.
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joshualiu315
2534 posts
#107 • 1 Y
Y by cubres
It is well-known that $\overline{MN}$ bisects $\overline{AB}$, and since $\overline{AB} \parallel \overline{CD}$, we also know that $\overline{MN}$ bisects $\overline{PQ}$. This implies $M$ is the midpoint of $\overline{PQ}$.

We have $\angle EAB = \angle ACM = \angle MAB$ and similarly, $\angle EBA = \angle BDM = \angle MBA$. This means $\triangle EAB \cong \triangle MAB$, which implies that $\overline{EM} \perp \overline{AB}$. Because $\overline{AB} \parallel \overline{CD}$, we have $\overline{EM} \perp \overline{PQ}$, so $\overline{EM}$ is the perpendicular bisector of $\overline{PQ}$. Thus, $EP=EQ$, as desired. $\blacksquare$
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LeYohan
54 posts
#108 • 1 Y
Y by cubres
First IMO #1 solve and it's geo!! :D

Claim: $PM = MQ$.
Let $NM$ intersect $AB$ at $X$. It's well known that the radical axis intersects the common tangent, hence $AX = XB$. By the Thales Theorem, we inmediately get the result. $\square$

Now doing some angle chase using the fact that $CD \parallel AB$ and $AB$ is tangent to both circles, we get:
$\angle EBA = \angle BDM = \angle BNM = \angle ABM = \angle BMD$, and $\angle EAB = \angle ACM = \angle ANM = \angle BAM = \angle AMC$. As a result, $\triangle ACM$ and $\triangle BMC$ are isosceles. Furthermore, $\triangle MAB \cong \triangle EAB \implies EB = BD, EA = AC$. This means that $\angle EMC = \angle EMD = 90$, and notice that we originally proved that $PM = MQ$, so $\triangle EPQ$ is isosceles and we're done. $\square$
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Avron
37 posts
#109 • 1 Y
Y by cubres
$MN$ is the radical axis of $G_1,G_2$ so it bisects $AB$ so we also know that $M$ is the midpoint of $PQ$, thus it is enough to prove $ME\perp CD$.
Now notice that $\angle MAB=\angle ACM = \angle EAB$ and similarly $\angle MBA = \angle ABE$ so $EAB \cong MAB$ and we're done
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zuat.e
66 posts
#110 • 1 Y
Y by cubres
Easy angle chasing yields the cyclicity of $(BEAN)$. We claim the following:

Claim: $EA=AM=AC$ and similarly $EB=BM=BD$
Proof: Let $O_1$ be the center of $G_1$. It is clear that $O_1A$ is the side bisector of $MC$, hence $AM=AC$.
Furthermore, note that $\measuredangle BAE=\measuredangle DCE= \measuredangle BMC=\measuredangle MAB$ and similarly $\measuredangle EBA =\measuredangle ABM$, hence $\triangle EAB\cong MAB$ (because they also share $AB$).

It now follows that $AC=AM=AE$ and analogously $BD=BM=BE$
Finally, as $\measuredangle MCA= \measuredangle MNA=\measuredangle ANC$, we have $AP\cdot AN=AC^2=AE^2$, hence $\measuredangle CEP =\measuredangle ENA=\measuredangle EBA=\measuredangle EDC$ and similarly $\measuredangle QED=\measuredangle PCE$, yielding $\measuredangle CPE =\measuredangle EQD$, as desired.
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cubres
119 posts
#111
Y by
ACGN
MOHS Hardness Scale
IMO Problem #1 (May 2) - IMO 2001 p1
This post has been edited 1 time. Last edited by cubres, May 7, 2025, 8:15 PM
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Fly_into_the_sky
3 posts
#112
Y by
Smooth one :pilot:
Claim 1: $EM \perp PQ$
Proof: Notice that $\angle{EAB}=\angle{ACM}=\angle{MAB}$ and $\angle{EBA}=\angle{BDM}=\angle{MBA}$ and $AB$ is common side
So $\triangle AEB \cong \triangle AMB \implies EM \perp AB \implies EM \perp PQ$
Claim 2: $MP=MQ$
Proof: Let $R=(MN) \cap (AB)$ then by PoP, $R$ is the midpoint of $AB$ and so $M$ is the midpoint of $(PQ)$ by playing with Thales theorem on $\triangle RNB$ and $\triangle RAN$
Conclude
This post has been edited 2 times. Last edited by Fly_into_the_sky, May 15, 2025, 5:26 PM
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