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Number of real roots
girishpimoli   5
N 4 hours ago by mrgenius000
If $f(x)=x^2-2x$. Then number of real roots of $f(f(f(f(x))))=3$
5 replies
girishpimoli
Today at 3:44 AM
mrgenius000
4 hours ago
Proving that the line passes through the midpoint.
MTA_2024   2
N 6 hours ago by Royal_mhyasd
Let $ABC$ be a triangle of orthocenter $H$. The circle of diameter $AC$ and the circumcircle of triangle $AHB$ intersect a second time in $K$.
Prove that the line $(CK)$ passes through the midpoint of segment $HB$.
2 replies
MTA_2024
May 7, 2025
Royal_mhyasd
6 hours ago
Square number
linkxink0603   4
N Today at 3:29 AM by pooh123
Find m is positive interger such that m^4+3^m is square number
4 replies
linkxink0603
Yesterday at 11:20 AM
pooh123
Today at 3:29 AM
Inequalities
sqing   7
N Today at 2:01 AM by sqing
Let $ a,b>0, a^2+ab+b^2 \geq 6  $. Prove that
$$a^4+ab+b^4\geq 10$$Let $ a,b>0, a^2+ab+b^2 \leq \sqrt{10}  $. Prove that
$$a^4+ab+b^4  \leq 10$$Let $ a,b>0,  a^2+ab+b^2 \geq \frac{15}{2}  $. Prove that
$$ a^4-ab+b^4\geq 10$$Let $ a,b>0,  a^2+ab+b^2 \leq \sqrt{10}  $. Prove that
$$-\frac{1}{8}\leq  a^4-ab+b^4\leq 10$$
7 replies
sqing
Thursday at 2:42 PM
sqing
Today at 2:01 AM
Compilation of functions problems
Saucepan_man02   2
N Today at 12:45 AM by Saucepan_man02
Could anyone post some handout/compilation of problems related to functions (difficulty similar to AIME/ARML/HMMT etc)?

Thanks..
2 replies
Saucepan_man02
May 7, 2025
Saucepan_man02
Today at 12:45 AM
How many triangles
Ecrin_eren   5
N Today at 12:10 AM by jasperE3


"Inside a triangle, 2025 points are placed, and each point is connected to the vertices of the smallest triangle that contains it. In the final state, how many small triangles are formed?"


5 replies
Ecrin_eren
May 2, 2025
jasperE3
Today at 12:10 AM
Triangle on a tetrahedron
vanstraelen   2
N Yesterday at 7:51 PM by ReticulatedPython

Given a regular tetrahedron $(A,BCD)$ with edges $l$.
Construct at the apex $A$ three perpendiculars to the three lateral faces.
Take a point on each perpendicular at a distance $l$ from the apex such that these three points lie above the apex.
Calculate the lenghts of the sides of the triangle.
2 replies
vanstraelen
Yesterday at 2:43 PM
ReticulatedPython
Yesterday at 7:51 PM
shadow of a cylinder, shadow of a cone
vanstraelen   2
N Yesterday at 6:33 PM by vanstraelen

a) Given is a right cylinder of height $2R$ and radius $R$.
The sun shines on this solid at an angle of $45^{\circ}$.
What is the area of the shadow that this solid casts on the plane of the botom base?

b) Given is a right cone of height $2R$ and radius $R$.
The sun shines on this solid at an angle of $45^{\circ}$.
What is the area of the shadow that this solid casts on the plane of the base?
2 replies
vanstraelen
Yesterday at 3:08 PM
vanstraelen
Yesterday at 6:33 PM
2023 Official Mock NAIME #15 f(f(f(x))) = f(f(x))
parmenides51   3
N Yesterday at 5:13 PM by jasperE3
How many non-bijective functions $f$ exist that satisfy $f(f(f(x))) = f(f(x))$ for all real $x$ and the domain of f is strictly within the set of $\{1,2,3,5,6,7,9\}$, the range being $\{1,2,4,6,7,8,9\}$?

Even though this is an AIME problem, a proof is mandatory for full credit. Constants must be ignored as we dont want an infinite number of solutions.
3 replies
parmenides51
Dec 4, 2023
jasperE3
Yesterday at 5:13 PM
Geometry
AlexCenteno2007   3
N Yesterday at 4:18 PM by AlexCenteno2007
Let ABC be an acute triangle and let D, E and F be the feet of the altitudes from A, B and C respectively. The straight line EF and the circumcircle of ABC intersect at P such that F is between E and P, the straight lines BP and DF intersect at Q. Show that if ED = EP then CQ and DP are parallel.
3 replies
AlexCenteno2007
Apr 28, 2025
AlexCenteno2007
Yesterday at 4:18 PM
Cyclic points and concurrency [1st Lemoine circle]
shobber   10
N Apr 24, 2025 by Ilikeminecraft
Source: China TST 2005
Let $\omega$ be the circumcircle of acute triangle $ABC$. Two tangents of $\omega$ from $B$ and $C$ intersect at $P$, $AP$ and $BC$ intersect at $D$. Point $E$, $F$ are on $AC$ and $AB$ such that $DE \parallel BA$ and $DF \parallel CA$.
(1) Prove that $F,B,C,E$ are concyclic.

(2) Denote $A_{1}$ the centre of the circle passing through $F,B,C,E$. $B_{1}$, $C_{1}$ are difined similarly. Prove that $AA_{1}$, $BB_{1}$, $CC_{1}$ are concurrent.
10 replies
shobber
Jun 27, 2006
Ilikeminecraft
Apr 24, 2025
Cyclic points and concurrency [1st Lemoine circle]
G H J
Source: China TST 2005
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shobber
3498 posts
#1 • 1 Y
Y by Adventure10
Let $\omega$ be the circumcircle of acute triangle $ABC$. Two tangents of $\omega$ from $B$ and $C$ intersect at $P$, $AP$ and $BC$ intersect at $D$. Point $E$, $F$ are on $AC$ and $AB$ such that $DE \parallel BA$ and $DF \parallel CA$.
(1) Prove that $F,B,C,E$ are concyclic.

(2) Denote $A_{1}$ the centre of the circle passing through $F,B,C,E$. $B_{1}$, $C_{1}$ are difined similarly. Prove that $AA_{1}$, $BB_{1}$, $CC_{1}$ are concurrent.
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yetti
2643 posts
#2 • 2 Y
Y by Adventure10, Mango247
AP is the A-symmedian of the triangle $\triangle ABC.$ Let O be the triangle circumcenter and K the symmedian point.

(1) AEDF is a parallelogram, hence its diagonals AD, EF cut each other in half. Since the midpoint of EF lies on the A-symmedian AD, EF is antiparallel to BC with respect to the angle $\angle A,$ wich means that the points B, C, E, F are concyclic.

(2) Let parallels to the B-, C-symmedians BK, CK through the foot $D \in BC$ of the A-symmedian $AK \equiv AD \equiv AP$ meet the rays (AB, (AC at B', C'. The triangles $\triangle AB'C' \sim \triangle ABC$ are centrally similar with the similarity center A and D is the symmedian point of the triangle $\triangle AB'C'.$ It immediately follows that the circumcircle $(A_{1})$ of the quadrilateral BCEF is the 1st Lemoine circle of the triangle $\triangle AB'C'$ centered at the midpoint X' of the segment DO', where O' is the circumcenter of this triangle. Therefore, $AA_{1}$ intersects the segment KO of the original triangle $\triangle ABC$ also at its midpoint X, the center of the 1st Lemoine circle of the original triangle. Simiarly, $BB_{1}, CC_{1}$ cut KO at its midpoint X, hence all three are concurrent at X.
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alpha-beta
20 posts
#3 • 2 Y
Y by Adventure10, Mango247
can someone define 1st Lemoine circle or give some links?
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mihai miculita
666 posts
#4 • 1 Y
Y by Adventure10
$ \mbox{The three parallels to the sides of a triangle ABC through the Lemoine point of the triangle ABC, }$
$ \mbox{ determine on the sides of triangle ABC, 6 concyclic points.}$
$ \mbox{The circle of the 6 points is the 1-st Lemoine circle of triangle ABC.}$
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Sardor
804 posts
#5 • 2 Y
Y by Adventure10, Mango247
What's Lamoine point?
Please help me .
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Dilshodbek
115 posts
#6 • 1 Y
Y by Adventure10
alpha-beta wrote:
can someone define 1st Lemoine circle or give some links?

can you explain me about Lemoin circle please
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ak12sr99
156 posts
#7 • 1 Y
Y by Adventure10
Here is my solution with some angle and length chasing

Disclaimer: This is definitely not as elegant as yetti's beautiful solution (:thumbup:), but it is much neater than I had originally expected it to be, which is the reason I decided to mention it anyway.

$(1):$

By Thales' Theorem, $\frac{BD}{BC} = \frac{BF}{BA}$ and $\frac{CD}{CB} = \frac{CE}{CA}$. As $ADP$ is the symmedian, $\frac{BD}{DP} = \frac{AB^2}{AC^2}$ (as the symmedian is the reflection of the median over the angle bisector).

This yields the following, where $a=BC$ etc. (we will use these in part $(2)$ as well):
$BF=\frac{c^3}{b^2+c^2} ...(1)\\ \\AF=\frac{cb^2}{b^2+c^2} ...(2)\\ \\AE=\frac{bc^2}{b^2+c^2} ...(3)\\ \\CE=\frac{b^3}{b^2+c^2}...(4)$

From here we get $AF.AB = AE.AC = \frac{b^2c^2}{b^2+c^2}$ and concyclicity follows.


$(2):$

Let the radius of circle $BFEC$ be $r$.

Let $\angle BCF=\alpha \implies \angle BA_1F=2\alpha \implies \angle A_1BF=\angle A_1FB=90-\alpha \implies \angle A_1BC = B+\alpha-90 = \angle A_1CB \implies \angle A_1CE= 90-\alpha-B+C = \angle A_1EC \implies \angle CA_1E = 2(\alpha+B-C)$.

Now, in $\Delta sA_1BF$ and $A_1CE$ we get, using equations $(1)$ and $(4)$ above,
$2r sin \alpha = \frac{c^3}{b^2+c^2}$ and $2r sin (\alpha+B-C) = \frac{b^3}{b^2+c^2}$
$\implies \frac{sin \alpha}{sin (\alpha+B-C)} = \frac{c^3}{b^3}    ...(5)$

Now we observe that,
$\frac {[ABA_1]}{ACA_1]} = \frac{\frac{1}{2}AB. AA_1 sin \angle BAA_1}{\frac{1}{2}AC.AA_1 sin \angle CAA_1} = \frac{c}{b}.\frac{sin \angle BAA_1}{sin \angle CAA_1} ...(6)$
and
$\frac {[ABA_1]}{ACA_1]} = \frac{\frac{1}{2} AB. BA_1 sin \angle ABA_1}{\frac{1}{2} AC.CA_1 sin \angle ACA_1} =  \frac{c}{b}.\frac{sin(90-\alpha)}{sin(90-\alpha-B+C)}  = \frac{c}{b}.\frac{cos \alpha}{cos (\alpha-B+C)}  ...(7)$

$(6)$ and $(7)$ together imply
$\frac{sin \angle BAA_1}{sin \angle CAA_1} = \frac{cos \alpha}{cos (\alpha-B+C)}  ...(8)$

Now after some elementary manipulations on relation $(5)$ we get,
$\frac{cos \alpha}{cos (\alpha-B+C)} = \frac{\frac{b^3}{c^3} - cos (B-C)}{\frac{b^3}{c^3}cos (B-C) - 1}    ...(9)$

Finally we use $cos \theta = cos^2 \frac{\theta}{2} - sin^2 \frac{\theta}{2} = \frac{1-tan^2 \frac{\theta}{2}}{1+tan^2 \frac{\theta}{2}}$ (on $\theta = B-C$ duh :P ) and $tan \frac{B-C}{2} = \frac{b-c}{b+c}.cot\frac{A}{2}$ in relations $(8)$ and $(9)$ to finish the proof by the trigonometric form of Ceva's theorem.
This post has been edited 5 times. Last edited by ak12sr99, Sep 16, 2017, 2:40 PM
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Sanjana42
21 posts
#8 • 1 Y
Y by kamatadu
(1) Define $E,F$ as follows. Let the line passing through the midpoint of $AD$ which is antiparallel to $BC$ w.r.t $AB,AC$ intersect $AB,AC$ at $F,E\implies FBCE$ cyclic. Since $AD$ is isogonal to the $A$-median in $\triangle ABC$, it must be the $A$-median in $\triangle AEF\implies$ the midpoint of $AD$ (which is on $FE$) is also the midpoint of $FE$, so $AFDE$ is a parallelogram, so $E,F$ are the same $E,F$ in the problem statement.

(2) Let $EF=a_A,AF=b_A,AE=c_A$. By similarity we get $a=BC=\frac{a_A(b_A^2+c_A^2)}{b_Ac_A}$ and $FB=\frac{c_A^2}{b_A}$.

Let $\angle FBE = \angle FCE = \theta_A$. Similarly define $\theta_B,\theta_C$. Sine rule in $\triangle FEB$ gives us $$\frac{\sin (C-\theta_A)}{\sin \theta_A}=\frac{c_A^2}{a_Ab_A}=\frac{c^2}{ab}=\frac{\sin (C-\theta_B)}{\theta_B}$$by symmetry. Therefore the corresponding $\theta$ is the same for all 3 vertices.

Let the feet from $A_1$ to $AB,AC$ be $M_a,N_a$. Note that $\angle FA_1M_a=\angle FEB=C-\theta$. $$\implies \frac{\sin \angle BAA_1}{\sin \angle CAA_1}=\frac{A_1M}{A_1N}=\frac{A_1M}{A_1F}\cdot\frac{A_1E}{A_1N}=\frac{\cos (C-\theta)}{\cos (B-\theta)}$$
Clearly the cyclic product of these is 1, so we're done by trig Ceva.
This post has been edited 1 time. Last edited by Sanjana42, Jan 5, 2025, 8:09 PM
Reason: typo
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cursed_tangent1434
625 posts
#9 • 1 Y
Y by stillwater_25
Solved with stilwater_25. Amazing problem! We realized what the concurrence point is but missed the slick Lemoine circle argument that can be done by shifting the reference triangle.

For part (1) note that since $AEDF$ is a parallelogram by definition, $\overline{AD}$ bisects $EF$. It is well known that the $A-$symmedian only bisects the antiparallels to $BC$, which implies that $BFEC$ is cyclic.

Now, we can move to the interesting part of the problem. We claim that these lines concur at $X_{182}$, the midpoint of $OK$ where $O$ and $K$ are the circumcenter and the symmedian point of $\triangle ABC$ respectively. We show that $\overline{AA_1}$ bisects segment $OK$ from which the result follows due to symmetry.

Let $M_a$ and $M$ denote the midpoints of segments $BC$ and $EF$ respectively. Let $X$ be the intersection of lines $\overline{EF}$ and $\overline{BC}$. Let $K_a$ denote the intersection of the $A-$symmedian with $(ABC)$. The key claim is the following.

Claim : Points $M$ , $A_1$ , $M_a$ and $K_a$ are concyclic.

Proof : It is clear that $XM_aA_1M$ is cyclic due to the right angles. Let $Y$ be the intersection of the $A-$tangent with $\overline{BC}$. Since any antiparallel to side $BC$ is parallel to the $A-$tangent, note that
\[-1=(EF;M\infty)\overset{A}{=}(BC;DY)\]Thus,
\[DY \cdot DM_a = DB \cdot DC \]Further, from the midpoint theorem it follows that $X$ is the midpoint of segment $YD$. Thus,
\[DM \cdot DK_a = \frac{DA\cdot DK_a}{2} = \frac{DB\cdot DC}{2} = \frac{DY \cdot DM_a}{2} = DX \cdot DM_a\]which implies that $MM_aK_aX$ is also cyclic. Putting these observations together proves the claim.

We now show the following.

Claim : Lines $\overline{OK}$ and $\overline{DA_1}$ are parallel.

Proof : This is a simple length chase. First remember that $(AK_a;DP)=-1$. Note that,
\[PA_1 \cdot PM_a = PK_a \cdot PM\]Also,
\[PM_a \cdot PO = PB^2\]This then implies,
\[\frac{PA_1}{PO} = \frac{PK_a \cdot PM}{PB^2} = \frac{PM}{PA}\]Now, let $K_c$ denote the intersection of the $C-$symmedian with $(ABC)$. Then,
\[-1=(AB;CK_a)\overset{C}{=}(AD;PK)\]Thus,
\[PD \cdot PA = PK \cdot PM\]Thus,
\[\frac{PA_1}{PO} = \frac{PM}{PA}=\frac{PD}{PK}\]which implies that $OK \parallel DA_1$ as claimed.

Now we are done since letting $X = \overline{AA_1} \cap \overline{OK}$ we have,
\[(OK;X\infty)\overset{A_1}{=}(PK;AD)=-1\]which implies that $X$ is indeed the midpoint of $OK$ and we are done.
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Batsuh
152 posts
#10
Y by
(1) Let $E' = DE \cap PB$ and $F' = DF \cap PC$. By an easy angle chase we see that $BFCF'$ and $BECE'$ are cyclic. So by PoP we have
\[FD \cdot DF' = BD \cdot DC = ED \cdot DE'\]so the points $B, F, E, C, F', E'$ are cyclic.

(2) Let $Q$ be the Symmedian point of $ABC$ and let $O$ be the center of $\omega$. We'll show that $AA_1$ passes through the midpoint of $OQ$, after which we'll be done by symmetry.

[asy]
import geometry;
import olympiad;
size(9cm);
filldraw(unitcircle, purple+white+white, blue);
pair A = dir(110); pair B = dir(225); pair C = dir(315);
pair O = (0,0);
pair M = B / 2+ C / 2;
pair P = extension(B, B+rotate(90)*(B-O),O,M);
pair D = extension(A,P,B,C);
pair E = intersectionpoint(parallel(D,line(A,B)),line(A,C));
pair Ep = extension(E,D,B,P);
pair F = intersectionpoint(parallel(D,line(A,C)),line(A,B));
pair Fp = extension(F,D,C,P);
circle BFEC = circle(B,F,E);
pair A_1 = circumcenter(B,F,E);
pair N = B / 2 + Ep / 2;
pair Q = intersectionpoint(parallel(B,line(N,D)), line(A,P));



draw(A -- B -- C -- cycle);
draw(line(P, false, B));
draw(line(P, false, C));
draw(E -- Ep);
draw(F -- Fp);
draw(O -- P);
draw(Q -- O, darkblue+1);
draw(D -- A_1, darkblue+1);
draw(B -- Q, darkblue+1);
draw(N -- D, darkblue+1);
draw(A -- P);
draw(circumcircle(B,F,E), red);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$P$", P, dir(P));
dot("$D$", D, dir(D));
dot("$Q$", Q, NW);
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$E'$", Ep, dir(Ep));
dot("$F'$", Fp, dir(Fp));
dot("$O$", O, NW);
dot("$A_1$", A_1, SE);
dot("$N$", N, dir(N));


[/asy]

Let $N$ be the midpoint of $BE'$. Observe that triangles $\triangle BDE'$ and $\triangle ABC$ are inversely similar with parallel sides. This means that the $B$-symmedian in $\triangle ABC$ and the $D$-median in $\triangle BDE'$ are parallel. In other words, $BQ \parallel ND$. Therefore,
\[\frac{PA_1}{PO} = \frac{PM}{PB} = \frac{PD}{PQ}\]which implies that $QO \parallel DA_1$. Now,
\[-1 = (A,D;Q,P) \overset{A_1}{=} (AA_1 \cap QO, QO_{\infty}; Q, O)\]implies that $AA_1 \cap QO$ is the midpoint of $QO$ as needed.
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Ilikeminecraft
626 posts
#11
Y by
For part one, we simply note that $EF$ and $AB$ are antiparallel since $AD$ is the $A$-median in $AEF.$

Let $O$ denote the center of $ABC.$ Let $L$ denote the Lemoine point(intersection of symmedians).
I claim that $AA_1$ passes through the midpoint of $LO.$

Let $E’, F’$ be the intersections of $BP, CP$ with $(BFEC).$
Observe that $\angle BE’E = \angle BCE = \angle AFE = \angle FED$ so $FB\parallel EE’,$ so $EDE’$ are collinear.
Similarly, $FDF’$ are collinear.
Let $N$ be the midpoint of $BE’.$
Next, note that $BDE’$ and $ABC$ are inversely similar, with $B$ corresponding to $D.$ Thus, the $B$ symmedian in $ABC$ must be parallel to the $D$-median in $BDE’.$ Hence, $BL\parallel ND.$
Furthermore, $BO\parallel NA_1.$
Thus, there is homothety centered at $P$ sending $BLO$ to $NDA_1.$
Thus, $LO\parallel DA_1.$
Finally, by Ceva-Menelaus, we have $-1 = (AD;LP).$ Projection through $A_1$ onto $LO$ finishes.
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