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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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[*]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
inquequality
ngocthi0101   10
N 14 minutes ago by MS_asdfgzxcvb
let $a,b,c > 0$ prove that
$\frac{a}{b} + \sqrt {\frac{b}{c}}  + \sqrt[3]{{\frac{c}{a}}} > \frac{5}{2}$
10 replies
ngocthi0101
Sep 26, 2014
MS_asdfgzxcvb
14 minutes ago
Functional Equation
AnhQuang_67   5
N 18 minutes ago by jasperE3
Find all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying $$2\cdot f\Big(\dfrac{-xy}{2}+f(x+y)\Big)=xf(y)+yf(x), \forall x, y \in \mathbb{R} $$
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AnhQuang_67
Yesterday at 4:50 PM
jasperE3
18 minutes ago
Special line through antipodal
Phorphyrion   8
N 21 minutes ago by optimusprime154
Source: 2025 Israel TST Test 1 P2
Triangle $\triangle ABC$ is inscribed in circle $\Omega$. Let $I$ denote its incenter and $I_A$ its $A$-excenter. Let $N$ denote the midpoint of arc $BAC$. Line $NI_A$ meets $\Omega$ a second time at $T$. The perpendicular to $AI$ at $I$ meets sides $AC$ and $AB$ at $E$ and $F$ respectively. The circumcircle of $\triangle BFT$ meets $BI_A$ a second time at $P$, and the circumcircle of $\triangle CET$ meets $CI_A$ a second time at $Q$. Prove that $PQ$ passes through the antipodal to $A$ on $\Omega$.
8 replies
Phorphyrion
Oct 28, 2024
optimusprime154
21 minutes ago
PoP+Parallel
Solilin   1
N an hour ago by sansgankrsngupta
Source: Titu Andreescu, Lemmas in Olympiad Geometry
Let ABC be a triangle and let D, E, F be the feet of the altitudes, with D on BC, E on CA, and F on AB. Let the parallel through D to EF meet AB at X and AC at Y. Let T be the intersection of EF with BC and let M be the midpoint of side BC. Prove that the points T, M, X, Y are concyclic.
1 reply
Solilin
an hour ago
sansgankrsngupta
an hour ago
School Math Problem
math_cool123   6
N 5 hours ago by anduran
Find all ordered pairs of nonzero integers $(a, b)$ that satisfy $$(a^2+b)(a+b^2)=(a-b)^3.$$
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math_cool123
Apr 2, 2025
anduran
5 hours ago
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titaniumfalcon   3
N Yesterday at 11:16 PM by mathprodigy2011
Post any solutions you have, with explanation or proof if possible, good luck!
3 replies
titaniumfalcon
Thursday at 10:40 PM
mathprodigy2011
Yesterday at 11:16 PM
Geo Mock #10
Bluesoul   2
N Yesterday at 8:26 PM by Bluesoul
Consider acute $\triangle{ABC}$ with $AB=10$, $AC<BC$ and area $135$. The circle $\omega$ with diameter $AB$ meets $BC$ at $E$. Let the orthocenter of the triangle be $H$, connect $CH$ and extend to meet $\omega$ at $N$ such that $NC>HC$ and $NE$ is the diameter of $\omega$. Draw the circumcircle $\Gamma$ of $\triangle{AHB}$, chord $XY$ of $\Gamma$ is tangent to $\omega$ and it passes through $N$, compute $XY$.
2 replies
Bluesoul
Apr 1, 2025
Bluesoul
Yesterday at 8:26 PM
Inequalities
sqing   4
N Yesterday at 3:28 PM by sqing
Let $ a, b,c\geq 0 $ and $ 2a+3b+ 4c=11.$ Prove that
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sqing
Apr 1, 2025
sqing
Yesterday at 3:28 PM
Geo Mock #9
Bluesoul   1
N Yesterday at 3:19 PM by vanstraelen
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1 reply
Bluesoul
Apr 1, 2025
vanstraelen
Yesterday at 3:19 PM
Geo Mock #6
Bluesoul   1
N Yesterday at 1:59 PM by vanstraelen
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Bluesoul
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vanstraelen
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Congruence
Ecrin_eren   1
N Yesterday at 1:39 PM by Ecrin_eren
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for 0 ≤ x, y < 41.

1 reply
Ecrin_eren
Thursday at 10:34 AM
Ecrin_eren
Yesterday at 1:39 PM
Probability
Ecrin_eren   1
N Yesterday at 1:38 PM by Ecrin_eren
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1 reply
Ecrin_eren
Thursday at 11:21 AM
Ecrin_eren
Yesterday at 1:38 PM
Excalibur Identity
jjsunpu   9
N Yesterday at 12:21 PM by fruitmonster97
proof is below
9 replies
jjsunpu
Thursday at 3:27 PM
fruitmonster97
Yesterday at 12:21 PM
.problem.
Cobedangiu   2
N Yesterday at 12:06 PM by Lankou
Find the integer coefficients after expanding Newton's binomial:
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2 replies
Cobedangiu
Yesterday at 6:20 AM
Lankou
Yesterday at 12:06 PM
DM and BL are parallel
FelixD   5
N Nov 26, 2011 by littletush
Source: MEMO 2009, problem 6, team competition
Suppose that $ ABCD$ is a cyclic quadrilateral and $ CD=DA$. Points $ E$ and $ F$ belong to the segments $ AB$ and $ BC$ respectively, and $ \angle ADC=2\angle EDF$. Segments $ DK$ and $ DM$ are height and median of triangle $ DEF$, respectively. $ L$ is the point symmetric to $ K$ with respect to $ M$. Prove that the lines $ DM$ and $ BL$ are parallel.
5 replies
FelixD
Oct 1, 2009
littletush
Nov 26, 2011
Source: MEMO 2009, problem 6, team competition
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FelixD
588 posts
#1 • 1 Y
Y by Adventure10
Suppose that $ ABCD$ is a cyclic quadrilateral and $ CD=DA$. Points $ E$ and $ F$ belong to the segments $ AB$ and $ BC$ respectively, and $ \angle ADC=2\angle EDF$. Segments $ DK$ and $ DM$ are height and median of triangle $ DEF$, respectively. $ L$ is the point symmetric to $ K$ with respect to $ M$. Prove that the lines $ DM$ and $ BL$ are parallel.
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livetolove212
859 posts
#2 • 2 Y
Y by Adventure10, Mango247
$ BD$ is the bisector of angle $ ABC$ and $ \angle EDF=\frac{1}{2}\angle ADC=90^o-\frac{1}{2}\angle EBF$ therefore $ D$ is the B-excenter of triangle $ EBF$.
Thus $ L$ is the tangency of incenter $ (I)$ of triangle $ EBF$ and $ EF$. Let $ J=BD\cap EF$
$ \frac{r_b}{r}=\frac{DK}{IL}=\frac{KJ}{LJ}=\frac{ML}{LJ}+\frac{MJ}{LJ}=1+2\frac{MJ}{LJ}$
$ \frac{r_b}{r}=\frac{DJ}{IJ}=\frac{DB}{IB}$ then it easy to show that $ \frac{MJ}{LJ}=\frac{DJ}{BJ}$
$ \Rightarrow BL//DM$
Attachments:
picture32.pdf (15kb)
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Palina.41
2 posts
#3 • 2 Y
Y by Adventure10, Mango247
Let $ G \in AC$. $ CG=EA \Rightarrow \triangle EAD = \triangle GCD$, $ BEDG$ is cyclic, $ DF$ is a bisector of $ \angle EDF$.
$ DF \cap (BEDG)=M$,
$ BM \cap DE=N$,
$ ME \cap BD=W$,
$ FE \cap WD=S$,
$ I$-midpoint of $ WD$.
$ DM$ is a diameter of $ (BEDG) \Rightarrow ME$ and $ DB$ are altitudes of $ \triangle MND$ and $ BWEN$ is cyclic $ \Rightarrow \angle NDM=\angle NBE=\angle NWE$.
$ \angle EWD=\angle DFG=\angle DFE \Rightarrow FWFD$ is cyclic $ \Rightarrow \angle EWF=180^0 - \angle NDM$.
Then $ \angle NWE+\angle EWF=180^0$ and $ N, W, F$ lie on the same line and $ NF$ is an altitude.

We shall prove:
For an arbitrary quadriateral $ XYZT$ such that $ \angle Y=\angle T=90^0$, if $ X'$ and $ Z'$ are the projections of $ X$ and $ Z$ on $ YT$, then $ YX'=TZ'$.
$ \triangle X'XY \sim \triangle Z'YZ$
$ \triangle XX'T \sim \triangle TZ'Z$
Then,
$ \frac {XX'}{YX'}= \frac {YZ'}{ZZ'}$ and $ \frac {XX'}{TX'}= \frac {TZ'}{ZZ'} \Rightarrow \frac{TX'}{YX'}= \frac{YZ'}{TZ'} \Rightarrow YX'=TZ'$

Then $ WL \perp FE$ and $ \triangle SWL \sim \triangle SIM \Rightarrow \frac{SW}{SI}=\frac{SL}{SM}$.
$ \angle FIE=2 \angle FDE$
$ \angle FBE=\angle FBW+\angle EBW=\angle EMD+\angle FND=180^0-2\angle FDE$.
Then, $ BEIF$ is cyclic. We know that $ FWED$ is cyclic, so
$ BS \cdot SI=FS \cdot SE=WS\cdot SD \Rightarrow \frac{SW}{SI}=\frac{BS}{SD} \Rightarrow\frac{SL}{SM}=\frac{BS}{SD}\Rightarrow \triangle SBL\sim\triangle SDM \Rightarrow BL \parallel MD$.
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stergiu
1648 posts
#4 • 2 Y
Y by Adventure10, Mango247
livetolove212 wrote:
................................

then it easy to show that $ \frac {MJ}{LJ} = \frac {DJ}{BJ}$
$ \Rightarrow BL//DM$

Can someone explain how we arrive to the first relation ?

Babis
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Palina.41
2 posts
#5 • 1 Y
Y by Adventure10
stergiu wrote:
livetolove212 wrote:
................................

then it easy to show that $ \frac {MJ}{LJ} = \frac {DJ}{BJ}$
$ \Rightarrow BL//DM$

Can someone explain how we arrive to the first relation ?

Babis
$ 2\frac{MJ}{LJ}+1=\frac{DB}{IB}=\frac{EB+BF+EF}{EB+BF-EF}=\frac{2EF}{EB+BF-EF}+1$
$ =2\frac{1}{\frac{EB+BF-EF}{EF}}+1=2\frac{1}{\frac{EB+BF}{EF}-1}+1$
$ =2\frac{1}{\frac{BF}{\frac{BF.EF}{EB+BF}}-1}+1=2\frac{1}{\frac{BF}{FJ}-1}+1$
$ =2\frac{1}{\frac{BD}{DJ}-1}+1=2\frac{1}{\frac{BJ}{DJ}}+1=2\frac{DJ}{BJ}+1$
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littletush
761 posts
#6 • 2 Y
Y by Adventure10, Mango247
it's easy to prove that $D$ is the excenter of triangle $BEF$
hence the result is trivialized by ex-Boone theorem.
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