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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
sequence positive
malinger   36
N 23 minutes ago by Maximilian113
Source: ISL 2006, A2, VAIMO 2007, P4, Poland 2007
The sequence of real numbers $a_0,a_1,a_2,\ldots$ is defined recursively by \[a_0=-1,\qquad\sum_{k=0}^n\dfrac{a_{n-k}}{k+1}=0\quad\text{for}\quad n\geq 1.\]Show that $ a_{n} > 0$ for all $ n\geq 1$.

Proposed by Mariusz Skalba, Poland
36 replies
malinger
Apr 22, 2007
Maximilian113
23 minutes ago
lots of perpendicular
m4thbl3nd3r   1
N 24 minutes ago by Captainscrubz
Let $\omega$ be the circumcircle of a non-isosceles triangle $ABC$ and $SA$ be a tangent line to $\omega$ ($S\in BC$). Let $AD\perp BC,I$ be midpoint of $BC$ and $IQ\perp AB,AH\perp SO,AH\cap QD=K$. Prove that $SO\parallel CK$
1 reply
m4thbl3nd3r
Yesterday at 4:44 PM
Captainscrubz
24 minutes ago
Old hard problem
ItzsleepyXD   0
39 minutes ago
Source: IDK
Let $ABC$ be a triangle and let $O$ be its circumcenter and $I$ its incenter.
Let $P$ be the radical center of its three mixtilinears and let $Q$ be the isogonal conjugate of $P$.
Let $G$ be the Gergonne point of the triangle $ABC$.
Prove that line $QG$ is parallel with line $OI$ .
0 replies
ItzsleepyXD
39 minutes ago
0 replies
Funny function that there isn't exist
ItzsleepyXD   1
N an hour ago by ItzsleepyXD
Source: Own, Modified from old problem
Determine all functions $f\colon\mathbb{Z}_{>0}\to\mathbb{Z}_{>0}$ such that, for all positive integers $m$ and $n$,
$$ m^{\phi(n)}+n^{\phi(m)} \mid f(m)^n + f(n)^m$$
1 reply
ItzsleepyXD
Apr 10, 2025
ItzsleepyXD
an hour ago
No more topics!
Bisector, orthogonal projection
mruczek   13
N Apr 14, 2025 by home245
Source: 69 Polish MO 2018 Second Round - Problem 3
Bisector of side $BC$ intersects circumcircle of triangle $ABC$ in points $P$ and $Q$. Points $A$ and $P$ lie on the same side of line $BC$. Point $R$ is an orthogonal projection of point $P$ on line $AC$. Point $S$ is middle of line segment $AQ$. Show that points $A, B, R, S$ lie on one circle.
13 replies
mruczek
Apr 28, 2018
home245
Apr 14, 2025
Bisector, orthogonal projection
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Source: 69 Polish MO 2018 Second Round - Problem 3
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mruczek
67 posts
#1 • 5 Y
Y by Mathuzb, buratinogigle, qwertyboyfromalotoftime, Adventure10, Mango247
Bisector of side $BC$ intersects circumcircle of triangle $ABC$ in points $P$ and $Q$. Points $A$ and $P$ lie on the same side of line $BC$. Point $R$ is an orthogonal projection of point $P$ on line $AC$. Point $S$ is middle of line segment $AQ$. Show that points $A, B, R, S$ lie on one circle.
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Arkmmq
214 posts
#2 • 4 Y
Y by Michcio, Mmqark, Adventure10, Mango247
Set $(ABC)$ as the unit circle and WLOG $q=-1,p=1$ .
$\longrightarrow$$c=\frac{1}{b}$ ,$r=1/2(a+c+1-ac) and s=\frac{a-1}{2}$ .
Now it is easy to find that :
$S=\frac{b-a}{r-a}.\frac{r-s}{b-s} \in R$.
This post has been edited 1 time. Last edited by Arkmmq, Apr 28, 2018, 6:18 PM
Reason: ..
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Wizard_32
1566 posts
#3 • 1 Y
Y by Adventure10
Key claim: $SR=SB$
Proof
Now, since $AS$ bisects $\angle BAR$ and $SB=SR$, hence $S$ is the midpoint fo arc $ARB$ of $\odot(ARB)$, and hence lies on this circle, as desired. $\blacksquare$
This post has been edited 3 times. Last edited by Wizard_32, Apr 28, 2018, 6:47 PM
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timon92
224 posts
#4 • 2 Y
Y by Adventure10, Mango247
This problem was proposed by Burii.
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matinyousefi
499 posts
#5 • 2 Y
Y by Adventure10, Mango247
we just need to prove $$ARsin(\frac{A}{2})+c.sin(\frac{A}{2})=ASsin(A)$$which is obvious using $$bsin(\frac{A}{2})+csin(\frac{A}{2})=AQsin(A) , AR=\frac{|b-c|}{2}$$
This post has been edited 1 time. Last edited by matinyousefi, Apr 28, 2018, 8:27 PM
Reason: b-c-----> |b-c|
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mathenthusiastic
168 posts
#6 • 2 Y
Y by Adventure10, Mango247
have anyone thought about the problem in this way?
please help me to prove synthetically that $RS$ is the $R$-symmedian in $\triangle BRC$
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ABCCBA
237 posts
#7 • 2 Y
Y by Adventure10, Mango247
My solution:
Let $BR$ cuts $(ABC)$ again at $X,$ $A'$ is the reflection of $A$ wrt $BX$
$\Rightarrow \angle PXR = 180^{\circ} - \angle PAB = \angle PAC = \angle PYR \Rightarrow P, X, Y, R$ are concyclic (lazy to use direct angle here)
$\Rightarrow BXY = \angle RPY = \angle APR =\frac{1}{2} \angle A = \angle BXQ,$ so $X, Y, Q$ are collinear
Since $RS \parallel YQ,$ by Reim theorem$, A, B, R, S$ are concyclic
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PROF65
2016 posts
#8 • 4 Y
Y by mathenthusiastic, Adventure10, Mango247, Mango247
Let $A',B'$ be the midpoints of $BC,AC$ ,$D$ foot of the $A$-bisector .
Remark that $RA'$ is the simson line of $P$ thus $A'R\parallel AQ$ ; we have $ABQ\sim  ADC$ then $ABS \sim ADB'$ hence to show that $ARSB$ cyclic it suffices to show $\angle BRA=\angle DB'A$ or $BR \parallel DB'$ but $\frac{CD}{CB'}=2\cdot\frac{CD}{CA}=2\cdot\frac{CA'}{CR}=\frac{CB}{CR}$ then the result follows .
RH HAS
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khanhnx
1618 posts
#9 • 3 Y
Y by AlastorMoody, Adventure10, Mango247
Here is my solution for this problem
Solution
Let $M$ be midpoint of $BC$
Since: $\widehat{AQP}$ = $\widehat{RCP}$ and $\widehat{PAQ}$ = $\widehat{PRC}$ = $90^o$, we have: $\triangle$ $PAQ$ $\sim$ $\triangle$ $PRC$
Hence: $\dfrac{AQ}{RC}$ = $\dfrac{PQ}{PC}$ = $\dfrac{CQ}{CM}$ = $\dfrac{BQ}{BM}$ or $\dfrac{SQ}{RC}$ = $\dfrac{BQ}{BC}$
But: $\widehat{BQS}$ = $\widehat{BRC}$, so $\triangle$ $BSQ$ $\sim$ $\triangle$ $BRC$
Then: $\widehat{SBQ}$ = $\widehat{RBC}$ or $\widehat{RBS}$ = $\widehat{CBQ}$ = $\widehat{CAQ}$ = $\widehat{RAS}$
So: $A$, $B$, $S$, $R$ lie on a circle
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Euler365
143 posts
#10 • 1 Y
Y by Adventure10
Consider $Hom\left(A , \frac{1}{2}\right)$
Let $B$ go to $X$, $R$ go to $Y$. Also $S$ goes to $Q$. Then the problem is equivalent to proving that $A$, $X$, $Q$, $Y$ lie on a circle.
By Archimedes theorem, $CY = AB = BX$.
Also $\angle QBX =B+\frac{A}{2} = \angle QCY$ and $QB = QC$.
$\therefore \triangle QBX \cong \triangle QCY$
$\therefore \angle AXQ = \angle BXQ = \angle CYQ \implies AXYQ$ is cyclic as desired.
This post has been edited 2 times. Last edited by Euler365, Feb 25, 2022, 8:02 PM
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Mahdi_Mashayekhi
694 posts
#12
Y by
Claim1 : $CPR$ and $QPA$ are similar.
Proof : $\angle PCR = \angle PQA$ and $\angle PRC = \angle 90 = \angle PAQ$.

Claim2 : $POC$ and $BQC$ are similar.
Proof : $PO = OC$, $BQ = QC$ and $\angle CPO = \angle CBQ$.

Claim3 : $BSQ$ and $BRC$ are similar.
Proof : $\angle RCB = \angle SQB$ and $\frac{RC}{SQ} = \frac{2RC}{AQ} = \frac{2PC}{PQ} = \frac{PC}{PO} = \frac{BC}{BQ}$.

Now we have $\angle ARB = \angle ASB$.
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sunken rock
4384 posts
#13
Y by
Another nice synthetic solution at https://stanfulger.blogspot.com/2022/02/polish-mo-2018-2nd-round.html

Best regards,
sunken rock
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Pinionrzek
54 posts
#14
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Consider an inversion in $A$ with radius $\sqrt{AB \cdot AC}$ composed with the symmetry along the $AQ$. Let $D = BC \cap AQ,  P'= BC \cap AP$. Then $P'$ is an image of $P$ in this inversion.
Moreover, the image of $S$ is some point $S'$ so that $D$ is a midpoint of $AS'$. Denote $R'$ the image of $R$ and $X = AP \cap S'C,   R^{''} = AB \cap S'C$. The statement is now equivalent to $R'=R^{''}$. Notice:
$-1 = (B, C; D, P') = (R^{''}, C, S', X) = P'(R^{''},  D; S', A)$. However, since $D$ is a midpoint of $AS'$ and $R'P' \parallel AS'$, we also have: $P'(R', D; S', A)= -1$, so $R'=R^{''}$ follows.
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home245
95 posts
#15
Y by
It is also All Russian Olympiad Regional round, 2012-2013, Problem 4 for 11 grade.
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