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Reflections and midpoints in triangle
TUAN2k8   2
N 2 minutes ago by Double07
Source: Own
Given an triangle $ABC$ and a line $\ell$ in the plane.Let $A_1,B_1,C_1$ be reflections of $A,B,C$ across the line $\ell$, respectively.Let $D,E,F$ be the midpoints of $B_1C_1,C_1A_1,A_1B_1$, respectively.Let $A_2,B_2,C_2$ be the reflections of $A,B,C$ across $D,E,F$, respectively.Prove that the points $A_2,B_2,C_2$ lie on a line parallel to $\ell$.
2 replies
TUAN2k8
5 hours ago
Double07
2 minutes ago
Integer-Valued FE comes again
lminsl   208
N 27 minutes ago by megahertz13
Source: IMO 2019 Problem 1
Let $\mathbb{Z}$ be the set of integers. Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that, for all integers $a$ and $b$, $$f(2a)+2f(b)=f(f(a+b)).$$Proposed by Liam Baker, South Africa
208 replies
lminsl
Jul 16, 2019
megahertz13
27 minutes ago
2025 Beijing High School Mathematics Competition Q9
sqing   2
N 28 minutes ago by sqing
Source: China
Let $ a,b,c,d >0  $ and $ (a^2+b^2+c^2)(b^2+c^2+d^2)=36. $ Find the maximum value of $a^2d.$
2 replies
sqing
2 hours ago
sqing
28 minutes ago
2025 Guangdong High School Mathematics Competition Q14
sqing   2
N 28 minutes ago by sqing
Source: China
Let $ x_1, x_2, x_3, x_4, x_5\geq 0 $ and $ x^2_1+x^2_2+x^2_3+ x^2_4+ x^2_5=4. $ Find the maximum value of
$\sum_{i=1}^5 \frac{1}{x_i+1} \sum_{i=1}^5 x_i .$
2 replies
sqing
2 hours ago
sqing
28 minutes ago
JBMO Shortlist 2023 G7
Orestis_Lignos   6
N 36 minutes ago by MR.1
Source: JBMO Shortlist 2023, G7
Let $D$ and $E$ be arbitrary points on the sides $BC$ and $AC$ of triangle $ABC$, respectively. The circumcircle of $\triangle ADC$ meets for the second time the circumcircle of $\triangle BCE$ at point $F$. Line $FE$ meets line $AD$ at point $G$, while line $FD$ meets line $BE$ at point $H$. Prove that lines $CF, AH$ and $BG$ pass through the same point.
6 replies
Orestis_Lignos
Jun 28, 2024
MR.1
36 minutes ago
Inspired by 2025 Xinjiang
sqing   0
an hour ago
Source: Own
Let $ a,b,c >0  . $ Prove that
$$  \left(1+\frac {a} { b}\right)\left(2+\frac {a}{ b}+\frac {b}{ c}\right) \left(1+\frac {a}{b}+\frac {b}{ c}+\frac {c}{ a}\right)  \geq 12+8\sqrt 2 $$
0 replies
sqing
an hour ago
0 replies
2025 Xinjiang High School Mathematics Competition Q11
sqing   1
N an hour ago by sqing
Source: China
Let $ a,b,c >0  . $ Prove that
$$  \left(1+\frac {a} { b}\right)\left(1+\frac {a}{ b}+\frac {b}{ c}\right) \left(1+\frac {a}{b}+\frac {b}{ c}+\frac {c}{ a}\right)  \geq 16 $$
1 reply
sqing
2 hours ago
sqing
an hour ago
Nice "if and only if" function problem
ICE_CNME_4   6
N an hour ago by BBNoDollar
Let $f : [0, \infty) \to [0, \infty)$, $f(x) = \dfrac{ax + b}{cx + d}$, with $a, d \in (0, \infty)$, $b, c \in [0, \infty)$. Prove that there exists $n \in \mathbb{N}^*$ such that for every $x \geq 0$
\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\](For $n \in \mathbb{N}^*$ and $x \geq 0$, the notation $f_n(x)$ represents $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )

Please do it at 9th grade level. Thank you!
6 replies
ICE_CNME_4
Yesterday at 7:23 PM
BBNoDollar
an hour ago
D1037 : The super irreductibilty
Dattier   0
an hour ago
Source: les dattes à Dattier
Let $P \in \mathbb Q[x]$, $P$ is super irreductible if $\forall Q \in \mathbb [x], \deg(Q)>0, P(Q(x))$ is irreductible.

Are there some $P \in \mathbb Q[x]$ super irreductible?
0 replies
Dattier
an hour ago
0 replies
Inspired by 2025 Beijing
sqing   1
N an hour ago by sqing
Source: Own
Let $ a,b,c,d >0  $ and $ (a^2+b^2+c^2)(b^2+c^2+d^2)=36. $ Prove that
$$ab^2c^2d \leq 8$$$$a^2bcd^2 \leq 16$$$$ ab^3c^3d \leq \frac{2187}{128}$$$$ a^3bcd^3 \leq \frac{2187}{32}$$
1 reply
sqing
2 hours ago
sqing
an hour ago
Arrange marbles
FunGuy1   5
N 2 hours ago by FunGuy1
Source: Own?
Anna has $200$ marbles in $25$ colors such that there are exactly $8$ marbles of each color. She wants to arrange them on $50$ shelves, $4$ marbles on each shelf such that for any $2$ colors there is a shelf that has marbles of those colors.
Can Anna achieve her goal?
5 replies
FunGuy1
Today at 12:23 PM
FunGuy1
2 hours ago
Question 2
Valentin Vornicu   88
N Apr 20, 2025 by Nari_Tom
Consider five points $ A$, $ B$, $ C$, $ D$ and $ E$ such that $ ABCD$ is a parallelogram and $ BCED$ is a cyclic quadrilateral. Let $ \ell$ be a line passing through $ A$. Suppose that $ \ell$ intersects the interior of the segment $ DC$ at $ F$ and intersects line $ BC$ at $ G$. Suppose also that $ EF = EG = EC$. Prove that $ \ell$ is the bisector of angle $ DAB$.

Author: Charles Leytem, Luxembourg
88 replies
Valentin Vornicu
Jul 25, 2007
Nari_Tom
Apr 20, 2025
Question 2
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TechnoLenzer
55 posts
#75
Y by
Let $\Gamma = (BCED)$, $P = DA \cap \Gamma$, $Q = BA \cap \Gamma$. We have $F = \lambda \cap CD$, and define $X = QF \cap \Gamma$.

By Pascal on $PXQBCD$, we have $A = BQ \cap DP$, $F = XQ \cap CD$, and $BC \cap XP$ collinear. Hence they are collinear on $\lambda$, and since $G = \lambda \cap BC$ then $G$ lies on $XP$.

Claim 1: $X$ is the $C$-dumpty point of $\triangle CGF$.
Proof: $\angle XGC = \angle PGC = \angle GPD = \angle XPD = \angle XCD = \angle XCF$. Similarly, $\angle GCX = \angle BPX = \angle BQX = \angle CFX$. Hence, by AA we have $\triangle XCG \sim \triangle XFC$, and $X$ is the $C$-dumpty point. $\square$

It's well-known that $CX \perp EX$. Hence, since $C, X, E$ lie on $\Gamma$, either $X = E$ or $E$ is the $C$-antipode in $\Gamma$.

Case 1: $X = E \Rightarrow \angle ECG = \angle EFC = \angle FCE \Rightarrow CE$ is the angle bisector of $\angle FCG$. Hence $\triangle CGF$ is isosceles. Since $\triangle FDA \sim \triangle FCG$ by AA we have $\angle FAB = \angle AFD = \angle DAF$, and $\lambda$ is the bisector of $\angle DAB$.

Case 2: $E$ is the $C$-antipode. Then $(CGF)$ is the image of $\Gamma$ under homothety scale factor 2 at $C$, and so no other point on $(CGF)$ lies inside $\Gamma$. But $F \in (CGF)$ lies on segment $CD$ within $\Gamma$, contradiction.
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JAnatolGT_00
559 posts
#76
Y by
Note that the Simson line of $E$ wrt $\triangle BCD$ is homothetic to $\ell$ wrt $C$ with factor $1/2,$ and therefore passes through the center of $ABCD$. This yields $|EB|=|ED|\implies \angle FCE=\angle DBE=\angle BDE=\angle GCE,$ so it's trivial that $|CF|=|CG|$ and hence $\angle BAF=\angle CFG=\angle CGF=\angle DAF,$ as desired.
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HamstPan38825
8868 posts
#77
Y by
The key is to construct the Simson Line of $E$ with respect to triangle $BCD$. By taking a homothety at $C$ with ratio $2$, it follows that this line is parallel to $\ell$.

On the other hand, it is well-known that the the Simson Line is parallel to $\overline{CJ}$, where $J$ is the point at which the altitude at $E$ to $\overline{BD}$ meets $(BCD)$. Thus, it suffices to show $J$ is the arc midpoint.

But the aforementioned homothety implies that the Simson Line bisects $\overline{AC}$, so it bisects $\overline{BD}$ too. This common midpoint $O$ is thus both the midpoint and foot from $E$ in triangle $EAB$, implying that it is isosceles. Thus, $\overline{EJ}$ bisects $\widehat{BD}$, which is enough.
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awesomeming327.
1732 posts
#78
Y by
Let $M,N,O$ be midpoints of $CG$, $CF$, and $BD$, respectively. Note that $EN\perp CD$ and $EM\perp BC$. Note that the perpendicular from $E$ onto $BD$ must lie on $MN$. Since $O,M,N$ are midpoints of $AC$, $FC$, and $GC$ respectively, $O$ lies on $MN$. Therefore, $EO\perp BD$.

Now, $EB=ED$, $\angle EMB=\angle END$ and $\angle NDE=\angle EBM$ so $\triangle BME\cong \triangle DNE$. Therefore, $EM=EN$ which implies $CG=CF$. Thus, \[\angle FAD=\angle CGF=\angle CFG=\angle BAF\]as desired.
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Marinchoo
407 posts
#79
Y by
Let $H_{b}$ and $H_{d}$ be the midpoints of $CG$ and $CF$, respectively. Notice that $\triangle ADF \sim \triangle GCF$ and $\triangle EH_{d}D \sim \triangle EH_{b}B$. The first follows from $ABCD$ being a parallelogram and the second follows from $\angle EDC = \angle EBC$ combined with $\overline{EH_{d}} \perp \overline{FC}$ and $\overline{EH_{b}} \perp \overline{CG}$. These give:
\[\frac{\frac{1}{2}CF\cdot\tan\angle EFC}{\frac{1}{2}CG\cdot\tan\angle ECG} = \frac{EH_{d}}{EH_{b}} = \frac{DH_{d}}{BH_{b}} = \frac{DF + FH_{d}}{BC + CH_{b}} = \frac{DF + \frac{1}{2}CF}{BC + \frac{1}{2}CG} = \frac{CF}{CG}\cdot \frac{\frac{DF}{CF}+\frac{1}{2}}{\frac{AD}{CG}+\frac{1}{2}} = \frac{CF}{CG}\]where the last equality follows from $\triangle ADF \sim \triangle GCF$. Now $\tan\angle EFC = \tan\angle ECG$ and the angles are acute, hence $\angle EFC = \angle ECG$. This implies that $\triangle EFC \cong \triangle EGC$, so $CF = CG$ and therefore $AD = DF$, implying that $\ell$ is the angle bisector of $\angle BAC$ as desired.
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huashiliao2020
1292 posts
#80
Y by
Got a hint to construct midpoints, but I think the finish is pretty unique!
Let J,K,L be the midpoints of GC,FC,BD. Notice that by a homothety of 1/2 ratio at C we get AFG is a line implies LKJ is a line; Since EJ perp. BC, EK perp. CD, by converse of Simson line we get that EL perp. BD. Now note by pop we get $$\frac{DF}{FC}=\frac{AD}{CG}=\frac{BC}{CG}\implies\frac{DF}{BC}=\frac{FC}{CG}=\frac{AB}{BG}=\frac{CD}{BG}\implies BC\cdot CD=DF\cdot BG\implies^{BC\cdot BG=DF\cdot CD}CD=BG,BC=DF\implies BAG=BGA=GAD,$$as desired. $\blacksquare$
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IAmTheHazard
5003 posts
#81 • 2 Y
Y by centslordm, ike.chen
Let $M$ be the midpoint of $\overline{BD}$, so by a homothety at $C$, $M$ is collinear with the feet of the altitudes from $E$ to $\overline{CB}$ and $\overline{CD}$. On the other hand, by Simson lines, these two feet are also collinear with the foot from $E$ to $\overline{BD}$, hence this foot is just $M$, so $E$ is the midpoint of arc $BCD$.

Then $\angle FCE=\angle GCE$, so $CF=CG$ and thus $\angle GFC=\angle FGC \implies \angle GAB=\angle FAD$, which finishes the problem. $\blacksquare$

EDIT: bro ive done this problem before too whatttttttttttt
This post has been edited 1 time. Last edited by IAmTheHazard, Aug 30, 2023, 8:57 PM
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asdf334
7585 posts
#82
Y by
had to get hint ;-; my brain

Just draw Simson line of $E$ to $BCD$, which passes through midpoint of $BD$. So $E$ is arc midpoint, finish since $\triangle ECF\sim\triangle EBD\sim \triangle EGC$
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EpicBird08
1755 posts
#83 • 1 Y
Y by GeoKing
I have absolutely no clue how I even conjured this solution.

We start with a key claim:

Claim: Let $XYZ$ be a triangle, and let $P$ and $Q$ be on lines $XY$ and $XZ$ such that $PQ \parallel YZ.$ Let $R$ be on $YZ$ such that $YPQR$ is a parallelogram. Then regardless of the position of points $P$ and $Q$, the circumcircle of $\triangle YPR$ will pass through a fixed point other than $Y.$

Proof: It suffices to show that all circles of this type share a radical axis. To do so, we will use barycentric coordinates with $\triangle XYZ$ as the reference triangle. Let $X = (1,0,0), Y = (0,1,0),$ and $Z = (0,0,1).$ Also let $P = (1 - \lambda, \lambda, 0)$ and $Q = (1 - \lambda, 0, \lambda),$ where $\lambda$ is an arbitrary real number, so that $P - Q = \lambda(0,1,-1) = \lambda \cdot (Y-Z)$ and therefore $PQ \parallel YZ.$ Then $R = Q + Y - P = (0,1-\lambda, \lambda).$

Now we find the circumcircle of $\triangle YPR.$ A point $(x,y,z)$ on this circle must satisfy an equation of the form
\[
a^2 yz + b^2 zx + c^2 xy = ux + vy + wz,
\]where $a = YZ, b = ZX,$ and $c = XY.$ Plugging in $Y = (0,1,0)$ gives $v = 0,$ so
\[
a^2 yz + b^2 zx + c^2 xy = ux + wz.
\]Plugging in $P = (1 - \lambda, \lambda, 0)$ gives $u(1-\lambda) = c^2 \lambda(1-\lambda),$ so $u = c^2 \lambda.$ Similarly, plugging in $R = (0, 1-\lambda, \lambda)$ gives $w\lambda = a^2 \lambda(1-\lambda),$ so $w = a^2 (1-\lambda).$

Therefore, our circumcircle is
\[
\boxed{a^2 yz + b^2 zx + c^2 xy = c^2 \lambda x + a^2 (1-\lambda) z}.
\]If we let $\lambda_1$ and $\lambda_2$ be two distinct real values of $\lambda,$ then the radical axis of the two corresponding circles is given by
\[
(c^2 \lambda_1 - c^2 \lambda_2) x + (a^2 (1 - \lambda_1) - a^2 (1-\lambda_2))z = 0.
\]Dividing both sides by $\lambda_1 - \lambda_2$ gives $c^2 x - a^2 z = 0,$ which is clearly fixed. This concludes our (lengthy) proof of the claim.

Now apply the claim with $(X,Y,Z,P,Q,R) = (G,C,F,B,A,D).$ Since $E$ is the circumcenter of $GCF,$ it has no relevance to $B,A,$ or $D,$ so $E$ is the fixed point as shown to exist by the claim. Taking $\lambda$ to be arbitrarily close to $0,$ we discover that $(FEC)$ is tangent to $CG.$ Hence $\measuredangle GCF = \measuredangle CEF = 2 \measuredangle CGF.$ However, by sum of angles in a triangle, $\measuredangle GCF = \measuredangle CGF + \measuredangle GFC,$ so $\measuredangle CGF = \measuredangle GFC.$ Since $CF \parallel AB$ and $DA \parallel CG,$ we get $\measuredangle CGF = \measuredangle DAF$ and $\measuredangle GFC = \measuredangle FAB.$ Thus $\measuredangle DAF = \measuredangle FAB,$ and we conclude.

Ok reading the solutions above there is a Simson line??????
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Z4ADies
64 posts
#84
Y by
Different and short solution :D
Assume that,the angle bisector of $\angle DAB \cap CD=G$ and $\angle DAB \cap BC=F$ call triangle $CGF$'s center $E$ we will prove $BCED$ cyclic.Let,$\angle DAG=\angle GAB=\angle AGD=\angle CGF=\angle CFG=\alpha$ from Ptolemy's sine lemma that is equivalent to show $CE\sin(2\alpha)+CB\sin(90-\alpha)=CD\sin(90+\alpha)$.We know $AD=DG=BC=x$ and $GC=CF=y$ ,we also know $AB=BF$.Sine Thrm to $\triangle CEG$ $\implies$ $CE\sin(2\alpha)=y\sin(90-\alpha)$.So, $\sin(90-\alpha)(x+y)=CD\sin(90+\alpha)=(x+y)\sin(90+\alpha)$ which is indeed true.
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cursed_tangent1434
640 posts
#86
Y by
Magical solution, needed a pointer to look at Simson Lines. Let $M$ and $N$ denote the midpoints of $GC$ and $FC$. Now, let $P = \overline{AC} \cap \overline{MN}$. By the Midpoint Theorem, it follows that $P$ is the midpoint of $AC$. Since diagonals of a parallelogram bisect each other, it also follows that $P$ is the midpoint of $BD$. But now, consider the Simson Line of $E$ with respect to $\triangle PCD$. Clearly the feet of the perpendicular from $E$ to $GC$ and $FC$ and $M$ and $N$ and thus, $EP\perp BD$. But note that this means $\overline{EP}$ is in fact the perpendicular bisector of segment $BD$, which implies that $EB=ED$. Thus,
\[\measuredangle BAD = \measuredangle DCB = \measuredangle DEB = 2\measuredangle DBE =  2 \measuredangle FCE = \measuredangle FEC = 2 \measuredangle FGC = 2\measuredangle FAD \]which indeed implies that $\ell$ is the internal $\angle BAD-$bisector.
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AngeloChu
471 posts
#87
Y by
denote the perpendiculars from $E$ to $BC$, $CD$, and $BD$ as $P_1$, $P_2$, and $P_3$ respectively, and they are collinear by simson line
it is easy to tell that $P_1$ is the midpoint of $CG$ and $P_2$ is the midpoint of $CF$, so $P_1P_2P_3$ is parallel to $AFG$ so $P_3$ is the midpoint of $BD$ and $AC$
$DBE=DCE$ since $BCDE$ is cyclic, and $GCE=BDE$ also since $BCDE$ is cyclic
thus $FCE=GCE$ and we can prove $CE$ is perpendicular to $FG$ and $P_1P_2$
thus, $CP_1=CP_2$, and angle chasing gives $BAG=DAG=CP_1P_2$
thus, $ \ell$ is the bisector of angle $DAB$
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ihatemath123
3449 posts
#88
Y by
Let $\ell'$ be the image of $\ell$ under a homothety of factor $\tfrac{1}{2}$ centered at $C$. Simson's theorem on $E$ WRT $(BCD)$ tells us that the foot from $E$ to $\overline{BD}$ lies on $\ell'$. But the intersection of $\ell'$ with $\overline{BD}$ is always the midpoint of $\overline{BD}$, so $E$ is one of the arc midpoints. This determines $\ell$, as desired.
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ItsBesi
146 posts
#89
Y by
My idea is the same but I am posting it for storage, solved it on 10.10.2024
Here is my long solution:

Let $K$,$M$ and $N$ be the midpoints of the sides $CG$,$CF$ and $AC$ respectively.

Claim: Points $\overline{N-M-K}$ are collinear
Proof:


Claim: $ED=EB$
Proof:

Claim: $\ell$ is the angle bisector of $\angle DAB$
Proof:
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This post has been edited 1 time. Last edited by ItsBesi, Mar 30, 2025, 7:21 PM
Reason: typo
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Nari_Tom
117 posts
#90
Y by
My approach was also nice.

Of course we have $\frac{BC}{CG}=\frac{DF}{FC}$, and $\angle CDE= \angle GBE$. Since we have $E$ lies on perpendicular bisectors of $CG$ and $CE$, we get $EBCG \sim EDFG$ which concludes the result instantly.
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