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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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[list][*]April 3rd (Webinar), 4pm PT/7:00pm ET, Learning with AoPS: Perspectives from a Parent, Math Camp Instructor, and University Professor
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April 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 2, 2025
0 replies
powers of 2
ErTeeEs06   1
N 2 minutes ago by Davdav1232
Source: BxMO 2025 P4
Let $a_0, a_1, \ldots, a_{10}$ be integers such that, for each $i \in \{0,1,\ldots,2047\}$, there exists a subset $S \subseteq \{0,1,\ldots,10\}$ with
\[
\sum_{j \in S} a_j \equiv i \pmod{2048}.
\]Show that for each $i \in \{0,1,\ldots,10\}$, there is exactly one $j \in \{0,1,\ldots,10\}$ such that $a_j$ is divisible by $2^i$ but not by $2^{i+1}$.

Note: $\sum_{j \in S} a_j$ is the summation notation, for instance, $\sum_{j \in \{2,5\}} a_j = a_2 + a_5$, while for the empty set $\varnothing$, one defines $\sum_{j \in \varnothing} a_j = 0$.
1 reply
ErTeeEs06
Yesterday at 11:18 AM
Davdav1232
2 minutes ago
Parallel lines with incircle
buratinogigle   0
12 minutes ago
Source: Own, test for the preliminary team of HSGS 2025
Let $ABC$ be a triangle with incircle $(I)$, which touches sides $CA$ and $AB$ at points $E$ and $F$, respectively. Choose points $M$ and $N$ on the line $EF$ such that $BM = BF$ and $CN = CE$. Let $P$ be the intersection of lines $CM$ and $BN$. Define $Q$ and $R$ as the intersections of $PN$ and $PM$ with lines $IC$ and $IB$, respectively. Assume that $J$ is the intersection of $QR$ and $BC$. Prove that $PJ \parallel MN$.
0 replies
buratinogigle
12 minutes ago
0 replies
poly roots equation
NicoN9   0
19 minutes ago
Source: My acquaintance
Let $a, b, c, d$ be positive integers. Let $\alpha$, $\beta$, $\gamma$ be three roots of the polynomial $P(x)=ax^3+bx^2+cx+d$. Given that $\alpha^2+\beta^2+\gamma^2=-1$, $\alpha^3+\beta^3+\gamma^3=-3$, and $2a^2+c^2=27$, find the values of $a$, $b$, $c$, $d$.
0 replies
NicoN9
19 minutes ago
0 replies
BMO 2021 problem 3
VicKmath7   20
N 20 minutes ago by Grasshopper-
Source: Balkan MO 2021 P3
Let $a, b$ and $c$ be positive integers satisfying the equation $(a, b) + [a, b]=2021^c$. If $|a-b|$ is a prime number, prove that the number $(a+b)^2+4$ is composite.

Proposed by Serbia
20 replies
VicKmath7
Sep 8, 2021
Grasshopper-
20 minutes ago
No more topics!
configurational geometry as usual
GorgonMathDota   11
N Apr 1, 2025 by ratavir
Source: Indonesia National Math Olympiad 2021 Problem 7 (INAMO 2021/7)
Given $\triangle ABC$ with circumcircle $\ell$. Point $M$ in $\triangle ABC$ such that $AM$ is the angle bisector of $\angle BAC$. Circle with center $M$ and radius $MB$ intersects $\ell$ and $BC$ at $D$ and $E$ respectively, $(B \not= D, B \not= E)$. Let $P$ be the midpoint of arc $BC$ in $\ell$ that didn't have $A$. Prove that $AP$ angle bisector of $\angle DPE$ if and only if $\angle B = 90^{\circ}$.
11 replies
GorgonMathDota
Nov 9, 2021
ratavir
Apr 1, 2025
configurational geometry as usual
G H J
Source: Indonesia National Math Olympiad 2021 Problem 7 (INAMO 2021/7)
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GorgonMathDota
1063 posts
#1 • 1 Y
Y by Rounak_iitr
Given $\triangle ABC$ with circumcircle $\ell$. Point $M$ in $\triangle ABC$ such that $AM$ is the angle bisector of $\angle BAC$. Circle with center $M$ and radius $MB$ intersects $\ell$ and $BC$ at $D$ and $E$ respectively, $(B \not= D, B \not= E)$. Let $P$ be the midpoint of arc $BC$ in $\ell$ that didn't have $A$. Prove that $AP$ angle bisector of $\angle DPE$ if and only if $\angle B = 90^{\circ}$.
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BolaOne
20 posts
#2
Y by
$DMEP$ is cyclic. Then it's just simple angle chasing.
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somebodyyouusedtoknow
259 posts
#3
Y by
I heard that this is the most difficult problem on the test, by far.
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Timmy456
92 posts
#5 • 3 Y
Y by PRMOisTheHardestExam, Siddharth03, Rounak_iitr
Since $AM$ is the angle bisector of $\angle BAC$ and $P$ is the midpoint of arc $BC$, it is obvious that $P$ is located at the second intersection of $AM$ and $\ell$.

Case $1$, If we know that $\angle B = 90^{\circ}$
By angle chasing,
$\angle DEM = \frac{180^{\circ}-\angle DME}{2} = 90^{\circ}-\frac{\angle DME}{2} = 90^{\circ}-\angle DBE$ $= 90^{\circ}-(180^{\circ}-\angle DPC) = 90^{\circ}-(180^{\circ}-(\angle DPM + 90^{\circ}))$
$= 90^{\circ}-(90^{\circ}-\angle DPM) = \angle DPM$
Hence, $DPEM$ is a cyclic quadrilateral. Since $DM = ME$ and $DPEM$ is a cyclic quadrilateral, it is obvious that $\angle DPM = \angle MPE \implies AP$ is the angle bisector of $\angle DPE$.

Case $2$, If we know that $AP$ is the angle bisector of $\angle DPE$
That means, $DPEM$ is a cyclic quadrilateral. By angle chasing,
$\angle ABC = \angle APC = 180^{\circ}-(\angle DBE + \angle DPM) = 180^{\circ}-(\frac{\angle DME}{2} + \angle DEM)$
$= 180^{\circ}-(\frac{180^{\circ}-\angle MDE-\angle DEM}{2} + \angle DEM) = 180^{\circ}-(\frac{180^{\circ}-2\angle DEM}{2} + \angle DEM)$
$= 180^{\circ}-(90^{\circ}-\angle DEM + \angle DEM) = 180^{\circ}-90^{\circ} = 90^{\circ}$, as desired.

Hence, we have proven that $AP$ is the angle bisector of $\angle DPE$ if and only if $\angle B = 90^{\circ}$.
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Wildabandon
507 posts
#6
Y by
For case 2, can you explain why $DPEM$ cyclic? I think thats doesn't work if $DP=PE$.
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Timmy456
92 posts
#7
Y by
@above I think I fake solved case 2. Do you have a solution?
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Timmy456
92 posts
#8
Y by
The following picture matches the criteria for case 2, but why isn't $\angle ABC = 90^{\circ}$
https://i.imgur.com/YGY21Tk.png
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Timmy456
92 posts
#9
Y by
Did I miss any given information?
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Wildabandon
507 posts
#10
Y by
I also didn't find a solution if $DP=PE$. I also tried using geogebra with $15$ digit accuracy, but couldn't find the correct construction .-.
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Timmy456
92 posts
#11
Y by
This problem only works for case 1
I wonder what the official solution says for case 2
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somebodyyouusedtoknow
259 posts
#12
Y by
Here's a fixed version of this problem.

Let the circumcircle of $\triangle{ABC}$ be $\ell$. There exists a point $M$ in $\triangle{ABC}$ such that $AM$ is the angle bisector of $\angle{BAC}$. The circle centred at $M$ with radius $MB$ intersects $\ell$ and $BC$ at $D$ and $E$ respectively ($B \neq D, B\neq E$). Let $P$ be the midpoint of arc $BC$ in $\ell$ not containing $A$. Prove that if $\triangle{DPE}$ is not isosceles, line $AP$ acts as the angle bisector of $\angle{DPE}$ if and only if $\angle{B} = 90^{\circ}$.
This post has been edited 1 time. Last edited by somebodyyouusedtoknow, Jan 20, 2022, 4:23 PM
Reason: I forgot that AoPS doesn't use \textbf
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ratavir
1 post
#13
Y by
@above
was this the one written in the official problem?
This post has been edited 1 time. Last edited by ratavir, Apr 1, 2025, 10:22 PM
Reason: tagging people
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