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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
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0 replies
jlacosta
Apr 2, 2025
0 replies
NT random problem with tau function
CrazyInMath   16
N 3 minutes ago by luutrongphuc
Source: 2022 IMOC N6
Find all integer coefficient polynomial $P(x)$ such that for all positive integer $x$, we have $$\tau(P(x))\geq\tau(x)$$Where $\tau(n)$ denotes the number of divisors of $n$. Define $\tau(0)=\infty$.
Note: you can use this conclusion. For all $\epsilon\geq0$, there exists a positive constant $C_\epsilon$ such that for all positive integer $n$, the $n$th smallest prime is at most $C_\epsilon n^{1+\epsilon}$.

Proposed by USJL
16 replies
CrazyInMath
Sep 5, 2022
luutrongphuc
3 minutes ago
Sum of complex numbers over plus/minus
Miquel-point   1
N 16 minutes ago by removablesingularity
Source: RNMO 1980 10.2
Show that if $z_1,z_2,z_3\in\mathbb C$ then
\[\sum |\pm z_1\pm z_2\pm z_3|^2=2^3\sum_{i=1}^3|z_k|^2.\]Generalize the problem.

1 reply
Miquel-point
Yesterday at 6:07 PM
removablesingularity
16 minutes ago
isogonal geometry
Tuguldur   0
17 minutes ago
Let $P$ and $Q$ be isogonal conjugates with respect to $\triangle ABC$. Let $\triangle P_1P_2P_3$ and $\triangle Q_1Q_2Q_3$ be their respective pedal triangles. Let\[ X_1=P_2Q_3\cap P_3Q_2,\quad X_2=P_1Q_3\cap P_3Q_1,\quad X_3=P_1Q_2\cap P_2Q_1 \]Prove that the points $X_1$, $X_2$ and $X_3$ lie on the line $PQ$.
0 replies
Tuguldur
17 minutes ago
0 replies
Cute inequality in equilateral triangle
Miquel-point   1
N 20 minutes ago by Quantum-Phantom
Source: Romanian IMO TST 1981, Day 3 P5
Let $ABC$ be an equilateral triangle, $M$ be a point inside it, and $A',B',C'$ be the intersections of $AM,\; BM,\; CM$ with the sides of $ABC$. If $A'',\; B'',\; C''$ are the midpoints of $BC$, $CA$, $AB$, show that there is a triangle with sides $A'A''$, $B'B''$ and $C'C''$.

Laurențiu Panaitopol
1 reply
Miquel-point
Yesterday at 6:44 PM
Quantum-Phantom
20 minutes ago
No more topics!
Hard geometry
TheOverlord   2
N Jun 5, 2020 by amar_04
Source: Iran TST 2015 ,exam 1, day 2 problem 3
$ABCD$ is a circumscribed and inscribed quadrilateral. $O$ is the circumcenter of the quadrilateral. $E,F$ and $S$ are the intersections of $AB,CD$ , $AD,BC$ and $AC,BD$ respectively. $E'$ and $F'$ are points on $AD$ and $AB$ such that $A\hat{E}E'=E'\hat{E}D$ and $A\hat{F}F'=F'\hat{F}B$. $X$ and $Y$ are points on $OE'$ and $OF'$ such that $\frac{XA}{XD}=\frac{EA}{ED}$ and $\frac{YA}{YB}=\frac{FA}{FB}$. $M$ is the midpoint of arc $BD$ of $(O)$ which contains $A$.
Prove that the circumcircles of triangles $OXY$ and $OAM$ are coaxal with the circle with diameter $OS$.
2 replies
TheOverlord
May 11, 2015
amar_04
Jun 5, 2020
Hard geometry
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G H BBookmark kLocked kLocked NReply
Source: Iran TST 2015 ,exam 1, day 2 problem 3
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TheOverlord
97 posts
#1 • 4 Y
Y by Dadgarnia, little-fermat, Adventure10, Mango247
$ABCD$ is a circumscribed and inscribed quadrilateral. $O$ is the circumcenter of the quadrilateral. $E,F$ and $S$ are the intersections of $AB,CD$ , $AD,BC$ and $AC,BD$ respectively. $E'$ and $F'$ are points on $AD$ and $AB$ such that $A\hat{E}E'=E'\hat{E}D$ and $A\hat{F}F'=F'\hat{F}B$. $X$ and $Y$ are points on $OE'$ and $OF'$ such that $\frac{XA}{XD}=\frac{EA}{ED}$ and $\frac{YA}{YB}=\frac{FA}{FB}$. $M$ is the midpoint of arc $BD$ of $(O)$ which contains $A$.
Prove that the circumcircles of triangles $OXY$ and $OAM$ are coaxal with the circle with diameter $OS$.
This post has been edited 1 time. Last edited by TheOverlord, May 11, 2015, 2:10 PM
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Luis González
4146 posts
#2 • 4 Y
Y by Generic_Username, little-fermat, Adventure10, XX-math-XX
If $I$ is the incenter of $ABCD,$ then $E' \equiv IE \cap AD$ and $F' \equiv IF \cap AB.$ Condition $\tfrac{XA}{XD}=\tfrac{EA}{ED}$ means that $X$ is on the E-Apollonius circle of $\triangle EAD,$ which is orthogonal to any circle through $A,D,$ thus particularly orthogonal to the circumcircle $(O,R)$ of $ABCD$ $\Longrightarrow$ $R^2=OE' \cdot OX$ and similarly $R^2=OF' \cdot OY.$

Inversion WRT $(O,R)$ takes $X,Y$ into $E',F',$ takes the circle with diameter $\overline{OS}$ into $EF$ (as $EF$ is the polar of $S$ WRT $(O)$) and fixes $A,M.$ If $E'F',IA$ cut $EF$ at $T,K,$ then from the complete quadrilateral $EE'F'F,$ it follows that $(E,F,K,T)=-1.$ But since $AM,AI$ bisect $\angle BAD,$ we have $A(E,F,K,M)=-1$ $\Longrightarrow$ $M \in AT$ $\Longrightarrow$ $E'F',AM$ and $EF$ concur at $T$ $\Longrightarrow$ their inverses $\odot(OXY),$ $\odot(OAM)$ and the circle with diameter $\overline{OS}$ intersect at a second point, i.e. they are coaxal.
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amar_04
1915 posts
#3 • 4 Y
Y by a_simple_guy, GeoMetrix, mueller.25, Bumblebee60
Nothing too different from above.
TheOverlord wrote:
$ABCD$ is a circumscribed and inscribed quadrilateral. $O$ is the circumcenter of the quadrilateral. $E,F$ and $S$ are the intersections of $AB,CD$ , $AD,BC$ and $AC,BD$ respectively. $E'$ and $F'$ are points on $AD$ and $AB$ such that $A\hat{E}E'=E'\hat{E}D$ and $A\hat{F}F'=F'\hat{F}B$. $X$ and $Y$ are points on $OE'$ and $OF'$ such that $\frac{XA}{XD}=\frac{EA}{ED}$ and $\frac{YA}{YB}=\frac{FA}{FB}$. $M$ is the midpoint of arc $BD$ of $(O)$ which contains $A$.
Prove that the circumcircles of triangles $OXY$ and $OAM$ are coaxal with the circle with diameter $OS$.



Let $\{I,O\}$ be the Incenter and Circumcenter of $ABCD$ respectively as $\frac{FA}{FB}=\frac{AF'}{F'B}$. So, by Angle -Bisector Theorem we get that $\overline{F-F'-I}$ and similarly $\overline{E-E'-I}$. Now as $\frac{YA}{YB}=\frac{FA}{FB}$. Hence, $Y\in F-\text{ Appolonius Circle } (\omega_1)$ of $\triangle AFB$. Let $\omega_1\cap AB=X_1$. So, $X_1\in\text{ Polar of }F'$ WRT $\odot(ABCD)$.So, By Self-Orthogonality Lemma $\omega_1\perp\odot(ABCD)$. So, $OY\cdot OF'=R^2$ where $R$ is the circumradius of $\odot(ABCD)$. Similarly $OX\cdot OE'=R^2$. Now Consider an Inversion $(\Psi)$ around $\odot(ABCD)$. $\Psi$ swaps $\{S,(\overline{O-I-S})\cap EF\}$ (by Brocard), $\{F',Y\},\{E',X\}$ and $\{A,M\}$ remains fixed. So, $\Psi$ swaps $\{\odot(OS),EF\},\{\odot(OAM),AM\},\{\odot(OXY),E'F'\}$. So, the Problem becomes equivalent as follows.
Inverted Problem wrote:
$ABCD$ be a bicentric quadrilateral with $I$ as the center of Inscribed Circle and $AB\cap CD=E$ and $AD\cap BC=F$. Let $\{IF\cap AB=F'\}$ and $\{IE\cap AD=E'\}$ and let $M$ be the midpoint of Minor Arc $BD$. Then $\{EF,AM,E'F'\}$ concurs.

Let $EF\cap E'F'=T$ and $TA\cap\{\odot(ABCD),IE\}=\{M',K\}$ respectively. So, $-1=(E,E';K,I)\overset{A}{=}(B,D;M',AI\cap\odot(ABCD))$. Now notice that $AI\cap\odot(ABCD)$ is the Midpoint of $\widehat{BCD}$. Hence, $M'$ must be the midpoint of Minor Arc $BD$. Hence, $M'\equiv M$. Hence $\{EF,AM,E'F'\}$ are concurrent. So, Inverting back we get that $\{\odot(OXY),\odot(OAM),\odot(OS)\}$ are coaxial. $\blacksquare$
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