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Geometry
Lukariman   0
30 minutes ago
Given acute triangle ABC ,AB=b,AC=c . M is a variable point on side AB. The circle circumscribing triangle BCM intersects AC at N.

a)Let I be the center of the circle circumscribing triangle AMN. Prove that I always lies on a fixed line.

b)Let J be the center of the circle circumscribing triangle MBC. Prove that line segment IJ has a constant length.
0 replies
Lukariman
30 minutes ago
0 replies
Finding positive integers with good divisors
nAalniaOMliO   3
N an hour ago by nAalniaOMliO
Source: Belarusian National Olympiad 2025
For every positive integer $n$ write all its divisors in increasing order: $1=d_1<d_2<\ldots<d_k=n$.
Find all $n$ such that $2025 \cdot n=d_{20} \cdot d_{25}$.
3 replies
nAalniaOMliO
Mar 28, 2025
nAalniaOMliO
an hour ago
Concurrent lines
MathChallenger101   4
N an hour ago by oVlad
Let $A B C D$ be an inscribed quadrilateral. Circles of diameters $A B$ and $C D$ intersect at points $X_1$ and $Y_1$, and circles of diameters $B C$ and $A D$ intersect at points $X_2$ and $Y_2$. The circles of diameters $A C$ and $B D$ intersect in two points $X_3$ and $Y_3$. Prove that the lines $X_1 Y_1, X_2 Y_2$ and $X_3 Y_3$ are concurrent.
4 replies
MathChallenger101
Feb 8, 2025
oVlad
an hour ago
Find the value
sqing   7
N an hour ago by giangtruong13
Source: Own
Let $a,b,c$ be distinct real numbers such that $ \frac{a^2}{(a-b)^2}+ \frac{b^2}{(b-c)^2}+ \frac{c^2}{(c-a)^2} =1. $ Find the value of $\frac{a}{a-b}+ \frac{b}{b-c}+ \frac{c}{c-a}.$
Let $a,b,c$ be distinct real numbers such that $\frac{a^2}{(b-c)^2}+ \frac{b^2}{(c-a)^2}+ \frac{c^2}{(a-b)^2}=2. $ Find the value of $\frac{a}{b-c}+ \frac{b}{c-a}+ \frac{c}{a-b}.$
Let $a,b,c$ be distinct real numbers such that $\frac{(a+b)^2}{(a-b)^2}+ \frac{(b+c)^2}{(b-c)^2}+ \frac{(c+a)^2}{(c-a)^2}=2. $ Find the value of $\frac{a+b}{a-b}+\frac{b+c}{b-c}+ \frac{c+a}{c-a}.$
7 replies
sqing
Mar 17, 2025
giangtruong13
an hour ago
2025 HMIC-5
EthanWYX2009   0
an hour ago
Source: 2025 HMIC-5
Compute the smallest positive integer $k > 45$ for which there exists a sequence $a_1, a_2, a_3, \ldots ,a_{k-1}$ of positive integers satisfying the following conditions:[list]
[*]$a_i = i$ for all integers $1 \le i \le 45;$
[*] $a_{k-i} = i$ for all integers $1 \le i \le 45;$
[*] for any odd integer $1 \le n \le k -45,$ the sequence $a_n, a_{n+1}, \ldots  , a_{n+44}$ is a permutation of
$\{1, 2, \ldots  , 45\}.$[/list]
Proposed by: Derek Liu
0 replies
EthanWYX2009
an hour ago
0 replies
I need the technique
DievilOnlyM   14
N an hour ago by Entei
Let a,b,c be real numbers such that: $ab+7bc+ca=188$.
FInd the minimum value of: $5a^2+11b^2+5c^2$
14 replies
DievilOnlyM
May 23, 2019
Entei
an hour ago
Cyclic Quads and Parallel Lines
gracemoon124   15
N an hour ago by Adywastaken
Source: 2015 British Mathematical Olympiad?
Let $ABCD$ be a cyclic quadrilateral. Let $F$ be the midpoint of the arc $AB$ of its circumcircle which does not contain $C$ or $D$. Let the lines $DF$ and $AC$ meet at $P$ and the lines $CF$ and $BD$ meet at $Q$. Prove that the lines $PQ$ and $AB$ are parallel.
15 replies
gracemoon124
Aug 16, 2023
Adywastaken
an hour ago
Something nice
KhuongTrang   33
N 2 hours ago by NguyenVanHoa29
Source: own
Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that

$$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le 1+2\sqrt{a+b+c+abc}.$$
33 replies
KhuongTrang
Nov 1, 2023
NguyenVanHoa29
2 hours ago
Geometry hard
Lukariman   2
N 2 hours ago by Primeniyazidayi
Given triangle ABC inscribed in circle (O). The bisector of angle A intersects (O) at D. Let M, N be the midpoints of AB, AC respectively. OD intersects BC at P and AD intersects MN at S. The circle circumscribed around triangle MPS intersects BC at Q different from P. Prove that QA is tangent to (O).
2 replies
Lukariman
2 hours ago
Primeniyazidayi
2 hours ago
help!!!!!!!!!!!!
Cobedangiu   3
N 2 hours ago by sqing
help
3 replies
Cobedangiu
Mar 23, 2025
sqing
2 hours ago
Inequality
nguyentlauv   1
N 2 hours ago by NguyenVanHoa29
Source: Own
Let $a,b,c$ be positive real numbers such that $ab+bc+ca=3$ and $k\ge 0$, prove that
$$\frac{\sqrt{a+1}}{b+c+k}+\frac{\sqrt{b+1}}{c+a+k}+\frac{\sqrt{c+1}}{a+b+k} \geq \frac{3\sqrt{2}}{k+2}.$$
1 reply
nguyentlauv
Yesterday at 12:19 PM
NguyenVanHoa29
2 hours ago
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]
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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
623 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
616 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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