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jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
Inequality with a,b,c
GeoMorocco   7
N 7 minutes ago by lele0305
Source: Morocco Training
Let $   a,b,c   $ be positive real numbers such that : $   ab+bc+ca=3   $ . Prove that : $$\frac{\sqrt{1+a^2}}{1+ab}+\frac{\sqrt{1+b^2}}{1+bc}+\frac{\sqrt{1+c^2}}{1+ca}\ge \sqrt{\frac{3(a+b+c)}{2}}$$
7 replies
GeoMorocco
Apr 11, 2025
lele0305
7 minutes ago
Property of the divisors of k^3 - 2
Scilyse   2
N an hour ago by Assassino9931
Source: KoMaL A. 892
Given two integers, $k$ and $d$ such that $d$ divides $k^3 - 2$. Show that there exists integers $a$, $b$, $c$ satisfying $d = a^3 + 2b^3 + 4c^3 - 6abc$.

Proposed by Csongor Beke and László Bence Simon, Cambridge
2 replies
Scilyse
Jan 13, 2025
Assassino9931
an hour ago
trigonometric functions
VivaanKam   12
N an hour ago by aok
Hi could someone explain the basic trigonometric functions to me like sin, cos, tan etc.
Thank you!
12 replies
VivaanKam
Apr 29, 2025
aok
an hour ago
Centroid, altitudes and medians, and concyclic points
BR1F1SZ   1
N an hour ago by sami1618
Source: Austria National MO Part 1 Problem 2
Let $\triangle{ABC}$ be an acute triangle with $BC > AC$. Let $S$ be the centroid of triangle $ABC$ and let $F$ be the foot of the perpendicular from $C$ to side $AB$. The median $CS$ intersects the circumcircle $\gamma$ of triangle $\triangle{ABC}$ at a second point $P$. Let $M$ be the point where $CS$ intersects $AB$. The line $SF$ intersects the circle $\gamma$ at a point $Q$, such that $F$ lies between $S$ and $Q$. Prove that the points $M,P,Q$ and $F$ lie on a circle.

(Karl Czakler)
1 reply
BR1F1SZ
3 hours ago
sami1618
an hour ago
Something nice
KhuongTrang   31
N an hour ago by NguyenVanHoa29
Source: own
Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that

$$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le 1+2\sqrt{a+b+c+abc}.$$
31 replies
KhuongTrang
Nov 1, 2023
NguyenVanHoa29
an hour ago
A Collection of Good Problems from my end
SomeonecoolLovesMaths   8
N 4 hours ago by Math-lover1
This is a collection of good problems and my respective attempts to solve them. I would like to encourage everyone to post their solutions to these problems, if any. This will not only help others verify theirs but also perhaps bring forward a different approach to the problem. I will constantly try to update the pool of questions.

The difficulty level of these questions vary from AMC 10 to AIME. (Although the main pool of questions were prepared as a mock test for IOQM over the years)

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5
8 replies
SomeonecoolLovesMaths
Sunday at 8:16 AM
Math-lover1
4 hours ago
find number of elements in H
Darealzolt   1
N 5 hours ago by alexheinis
If \( H \) is the set of positive real solutions to the system
\[
x^3 + y^3 + z^3 = x + y + z
\]\[
x^2 + y^2 + z^2 = xyz
\]then find the number of elements in \( H \).
1 reply
Darealzolt
Yesterday at 1:50 AM
alexheinis
5 hours ago
primes and perfect squares
Bummer12345   0
Yesterday at 5:08 PM
If $p$ and $q$ are primes, then can $2^p + 5^q + pq$ be a perfect square?
0 replies
Bummer12345
Yesterday at 5:08 PM
0 replies
simple trapezoid
gggzul   0
Yesterday at 4:44 PM
Let $ABCD$ be a trapezoid. By $x$ we denote the angle bisector of angle $X$ . Let $P=a\cap c$ and $Q=b\cap d$. Prove that $ABPQ$ is cyclic.
0 replies
gggzul
Yesterday at 4:44 PM
0 replies
geometry
JetFire008   0
Yesterday at 4:14 PM
Given four concyclic points. For each subset of three points take the incenter. Show that the four incentres form a rectangle.
0 replies
JetFire008
Yesterday at 4:14 PM
0 replies
Inequalities
sqing   11
N Yesterday at 3:02 PM by sqing
Let $a,b,c> 0$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1.$ Prove that
$$  (1-abc) (1-a)(1-b)(1-c)  \ge 208 $$$$ (1+abc) (1-a)(1-b)(1-c)  \le -224 $$$$(1+a^2b^2c^2) (1-a)(1-b)(1-c)  \le -5840 $$
11 replies
sqing
Jul 12, 2024
sqing
Yesterday at 3:02 PM
A rather difficult question
BeautifulMath0926   3
N Yesterday at 2:23 PM by evankuang
I got a difficult equation for users to solve:
Find all functions f: R to R, so that to all real numbers x and y,
1+f(x)f(y)=f(x+y)+f(xy)+xy(x+y-2) holds.
3 replies
BeautifulMath0926
Apr 13, 2025
evankuang
Yesterday at 2:23 PM
The return of an inequality
giangtruong13   4
N Yesterday at 1:26 PM by sqing
Let $a,b,c$ be real positive number satisfy that: $a+b+c=1$. Prove that: $$\sum_{cyc} \frac{a}{b^2+c^2} \geq \frac{3}{2}$$
4 replies
giangtruong13
Mar 18, 2025
sqing
Yesterday at 1:26 PM
Polynomial
kellyelliee   1
N Yesterday at 1:19 PM by Jackson0423
Let the polynomial $f(x)=x^2+ax+b$, where $a,b$ integers and $k$ is a positive integer. Suppose that the integers
$m,n,p$ satisfy: $f(m), f(n), f(p)$ are divisible by k. Prove that:
$(m-n)(n-p)(p-m)$ is divisible by k
1 reply
kellyelliee
Yesterday at 3:57 AM
Jackson0423
Yesterday at 1:19 PM
Circles with same radical axis
Jalil_Huseynov   9
N Apr 17, 2025 by Nari_Tom
Source: DGO 2021, Individual stage, Day2 P3
Let $O$ be the circumcenter of triangle $ABC$. The altitudes from $A, B, C$ of triangle $ABC$ intersects the circumcircle of the triangle $ABC$ at $A_1, B_1, C_1$ respectively. $AO, BO, CO$ meets $BC, CA, AB$ at $A_2, B_2, C_2$ respectively. Prove that the circumcircles of triangles $AA_1A_2, BB_1B_2, CC_1C_2$ share two common points.

Proporsed by wassupevery1
9 replies
Jalil_Huseynov
Dec 26, 2021
Nari_Tom
Apr 17, 2025
Circles with same radical axis
G H J
G H BBookmark kLocked kLocked NReply
Source: DGO 2021, Individual stage, Day2 P3
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Jalil_Huseynov
439 posts
#1
Y by
Let $O$ be the circumcenter of triangle $ABC$. The altitudes from $A, B, C$ of triangle $ABC$ intersects the circumcircle of the triangle $ABC$ at $A_1, B_1, C_1$ respectively. $AO, BO, CO$ meets $BC, CA, AB$ at $A_2, B_2, C_2$ respectively. Prove that the circumcircles of triangles $AA_1A_2, BB_1B_2, CC_1C_2$ share two common points.

Proporsed by wassupevery1
This post has been edited 1 time. Last edited by Jalil_Huseynov, Dec 28, 2021, 12:33 PM
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Davsch
381 posts
#2 • 1 Y
Y by PRMOisTheHardestExam
We use complex numbers with $(ABC)$ as the unit circle. Then $a_1=-\frac{bc}a,a_2=\frac{a^2(b+c)}{a^2+bc}$. The condition $Z\in (AA_1A_2)$ is equivalent to $\frac{(z-a)(a^3b+a^3c+a^2bc+b^2c^2)}{a(az+bc)}=\frac{(a\bar z-1)(a^3+bc(a+b+c))}{bc\bar z+a}$, or, after diving by $a^2+bc$, \[z\bar z(b^2c^2-a^4)+za(a+b)(a+c)-\bar zabc(a+b)(a+c)-a(b+c)(a^2-bc)=0.\]Analogous equations hold for $Z\in (BB_1B_2),(CC_1C_2)$. We will now show that these equations are linearly dependent. We compute that (to evaluate the $3\times 3$-determinants, we always subtract row $1$ from rows $2,3$)
\[\det\begin{pmatrix}a(a+b)(a+c)&-abc(a+b)(a+c)\\b(b+a)(b+c)&-abc(b+a)(b+c)\end{pmatrix}=-abc(a+b)^2(b+c)(c+a)(a-b)\neq 0,\]\[\det\begin{pmatrix}a(a+b)(a+c)&-abc(a+b)(a+c)&b^2c^2-a^4\\b(b+a)(b+c)&-abc(b+a)(b+c)&c^2a^2-b^4\\c(c+a)(c+b)&-abc(c+a)(c+b)&a^2b^2-c^4\end{pmatrix}\]\[=-abc\det\begin{pmatrix}a(a+b)(a+c)&(a+b)(a+c)&b^2c^2-a^4\\(b-a)(b+a)(a+b+c)&(b-a)(b+a)&(a-b)(a+b)(a^2+b^2+c^2)\\(c-a)(c+a)(a+b+c)&(c-a)(c+a)&(a-c)(a+b)(a^2+b^2+c^2)\end{pmatrix}=0,\]\[\det\begin{pmatrix}a(a+b)(a+c)&-abc(a+b)(a+c)&a(b+c)(a^2-bc)\\b(b+a)(b+c)&-abc(b+a)(b+c)&b(c+a)(b^2-ca)\\c(c+a)(c+b)&-abc(c+a)(c+b)&c(a+b)(c^2-ab)\end{pmatrix}\]\[=-abc\det\begin{pmatrix}a(a+b)(a+c)&(a+b)(a+c)&b^2c^2-a^4\\(b-a)(b+a)(a+b+c)&(b-a)(b+a)&(b-a)(b+a)(ab+bc+ca)\\(c-a)(c+a)(a+b+c)&(c-a)(c+a)&(c-a)(c+a)(ab+bc+ca)\end{pmatrix}=0,\]as desired.
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BelieverofMaths
263 posts
#3 • 2 Y
Y by Noob_at_math_69_level, vuanhnshn
is there any geomerical proof of this one ?
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bin_sherlo
719 posts
#4 • 1 Y
Y by ehuseyinyigit
Nice problem!
Replace $A_2,B_2,C_2$ with $D,E,F$. Let $H$ be the orthocenter of $\triangle ABC$. Set $(BB_1E)\cap (CC_1F)=P,Q$.
Claim: $H$ lies on $PQ$.
Proof:
\[Pow(H,(BB_1EPQ))=HB.HB_1=HC.HC_1=Pow(H,(CC_1FPQ))\]Thus, $H$ lies on the radical axis of $(BB_1E)$ and $(CC_1F)$ which is $PQ$.$\square$
Claim: $A,A_1,P,Q$ are concyclic.
Proof:
\[HA.HA_1=HB.HB_1=HP.HQ\]Which gives the desired result.$\square$
Take $\sqrt{bc}$ inversion and reflect over the angle bisector of $\angle BAC$.
New Problem Statement: $ABC$ is a triangle whose circumcenter is $O$. $BC$ intersects the reflection of $AO$ with respect to $AB,AC$ at $B_1,C_1$ respectively. Parallel lines to $AB_1,AC_1$ through $C,B$ intersect $AB,AC$ at $E,F$ respectively. $D$ is on $(ABC)$ which holds $AD\perp BC$. Prove that $D$ lies on the radical axis of $(B_1EC)$ and $(C_1FB)$.
Lemma: $ABC$ is a triangle whose circumcenter is $O$. $AO\cap (BOC)=G,BO\cap AC=E$. $C_1$ is the intersection of the altitude from $C$ to $AB$ and $(ABC)$. Then, $C_1,E,C,G$ are concyclic.
Proof: Let $(GCE)\cap AG=S$.
\[\measuredangle ESG=180-\measuredangle GCA=\measuredangle C=90-\measuredangle GAB\]Hence $AS\perp AB\perp CC_1$. Also $\measuredangle OES=90-\measuredangle ABS=\measuredangle C=\measuredangle ESO$ which implies $OE=OS$. Combining this with $OC_1=OC,$ we conclude that $CESC_1$ is an isosceles trapezoid. Thus, $G,C,E,C_1,S$ are concyclic.$\square$
Let $B_1F\cap C_1E=T$. Let $(B_1EC)\cap EC_1=L,(C_1BF)\cap B_1F=K$. By inverting the configuration of the lemma with $\sqrt{bc}$, we get that $H,B,C_1,E$ are concyclic.
\[\measuredangle ECB_1=90-\measuredangle A=\measuredangle EC_1H=\measuredangle EC_1B+\measuredangle BC_1H=\measuredangle EC_1B+\measuredangle DC_1B_1=\measuredangle EC_1B+\measuredangle C_1B_1D=\measuredangle(EC_1,B_1D)\]Hence $B_1,D,L$ are collinear. Similarily, $C_1,D,K$ are collinear.
Claim: $TB_1=TC_1$.
Proof:
\[\frac{BB_1}{BF}=\frac{BC}{BF}.\frac{AB_1}{CE}=\frac{BC}{CE}.\frac{AB_1}{BF}=\frac{CC_1}{CE}\]And $\measuredangle FBB_1=90+\measuredangle A=\measuredangle C_1CE$ subsequently $FBB_1\sim ECC_1$. So $\measuredangle BB_1F=\measuredangle EC_1C$ which yields $TB_1=TC_1$.$\square$
Claim: $TK=TL$.
Proof: Let $M=(B_1EC)\cap B_1F,N=(C_1FB)\cap C_1E$.
\[\measuredangle EMB_1=\measuredangle ECB_1=90-\measuredangle A=\measuredangle C_1BF=\measuredangle C_1NF\]Thus, $M,N,E,F$ are concyclic. By using this,
\[TM.TF.TB_1.TK=(TK.TF)(TM.TB_1)=(TN.TC_1)(TE.TL)=(TN.TE).TC_1.TL=TM.TF.TB_1.TL\]Hence $TK=TL$.$\square$
Since $TK=TL$ and $TB_1=TC_1,$ we see that $B_1C_1LK$ is an isosceles trapezoid whose diagonals intersect at $D$.
\[Pow(D,(B_1EC))=DB_1.DL=DC_1.DK=Pow(D,(C_1FB))\]Which completes our proof as desired.$\blacksquare$
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soryn
5342 posts
#5
Y by
Very nice problem and very nices solutions!
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breloje17fr
37 posts
#6
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To continue, prove that the same is true when A1, B1 and C1 are the feet of the altitudes and A2, B2 and C2 are the antipodes of A, B and C on the circumcircle, and that the radical axis then is the Euler's line.
On the figure below, the red circles and the green circles corespond to the initial problem and the continuation, respectively.
Attachments:
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Nari_Tom
117 posts
#7
Y by
Two beautiful proof for this beautiful problem. Here i provides two lemmas that technically solves the problem. How to prove them is your choice, it can be shorter may be. But i have nicer proofs here.

Lemma 1: (just for the sake of proving lemma 2). $O$ is the circumcircle of $\triangle ABC$. $AO \cap (BC)=A_2$, and $S$ be the intersection of $A$ altitude with $(ABC)$. Let's assume $B_2$ and $C_2$ defined similarly as $A_2$. Let $B'$ and $C'$ be the antipodes of $B$ and $C$ respectively. Let $D=SA_2 \cap (ABC)$, $D'=DO \cap 
(ABC)$. And let $B'C'$ and tangent of $(ABC)$ at $A$ intersect at $F$. Then $F$ lies on $B_2C_2$.


Lemma 2: Let's assume all the points defined same as Lemma 1. Let $X$ be the projection of $A_2$ to line $B_2C_2$. Then $X$ lies on $(AA_1A_2)$, which is actually follows from $F$ lies on D'S.

Proof for the lemma 1:
For the convenience let's assume that $B_2C_2 \ AA=F$, then let's prove that $F$ lies on $B'C'$. By the converse of Menelaus theorem it suffices to prove that $\frac{FC_2}{FB_2} \cdot \frac{C'O}{C'C_2} \cdot \frac{B'B_2}{B'O}=\frac{FC_2}{FB_2} \cdot \frac{B'B_2}{C'C_2}=1$, which can turn to mess if you don't use trigonometry instantly. In $\triangle AC_2B_2$:
by extended sines theorem, $\frac{FC_2}{FB_2}=\frac{AC_2}{AB_2} \cdot \frac{sin \gamma}{sin \beta}$. Now we can express everything in terms of $\triangle ABC$, so it's not hard for you.

Proof for the lemma 2:
I wont use any analytic technique here. Once you realize $FAA_2S$ is cyclic with diameter $A_2F$, we just need to prove that $F-D'-S$ are collinear. Let's forget about $B_2, C_2$ and use lemma 1. Then new definition of $F$ will be $F=C'B' \cap AA$. Since there is too many antipodes let's use inversion which swaps them. (which is the combination of inversion with $(ABC)$, and reflection with point $O$).

Let $O$ be the circumcenter of $\triangle ABC$. Let $S$ be the intersection of $(ABC)$ and $A$ altitude. Let $A'$ and $S'$ be the antipodes of $A$ and $S$. Let $A_2=AO \cap BC$ and $D=SA_2 \cap (ABC)$. Let $F'$ be the intersection of $(OBC)$ and circle with diameter $A'O$. Prove that $OS'DF'$ are concyclic.

Let $F=OF' \cap BC$. By regular inversion at $(ABC)$ (we're not using that, just motivational thing), we know that $FA'$ is tangent to $(ABC)$. By some angle chase $DA_2A'F$ is cyclic, which proves the collinearity $F-D-S'$. And we're done.
This post has been edited 1 time. Last edited by Nari_Tom, Apr 17, 2025, 9:12 AM
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Nari_Tom
117 posts
#8
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Actually they should mention that $\triangle ABC$ is not isosceles, at least not equilateral for this one.

Let $H$ and $H_2$ be the orthocenters of $\triangle ABC, \triangle A_2B_2C_2$, respectively.
Since $Pow(H, (AA_1A_2))=Pow(H, (BB_1B_2))=Pow(H, (CC_1C_2))=HA \cdot HA_1=HB \cdot HB_1= HC \cdot HC_1$, thus $H$ is radical center of these circles. By using the lemma 2 and, easy power of a point we can deduce that $H_2$ is also a radical center of these circles. Which means circles$ (AA_1A_2)$, $(BB_1B_2)$, $(CC_1C_2)$ have two different radical centers $\implies$ these circles have common radical axis.
This post has been edited 1 time. Last edited by Nari_Tom, Apr 17, 2025, 9:58 AM
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Nari_Tom
117 posts
#9
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Another solution using very nice lemma.
Lemma: Let $ABCD$ be a convex quadrilateral. Let $E=AB \cap CD$, $F=AD \cap BC$. let $H_1, H_2, H_3, H_4$ be the four orthocenters of triangles formed by lines $AB,AD, CB,CD$. Circles with diameter $AC, BD, EF$ have common radical axis, which passes through these orthocenters.
(Proof consists some power of a power argument, which is not hard. So you will do it yourselves.)

Let's solve the main problem. Let $X, Y, Z$ be the antipodes of $A_2, B_2, C_2$ in circles $(AA_1A_2), (BB_1B_2), (CC_1C_2)$, respectively. We proved $X, Y, Z$ lies on the lines $B_2C_2, A_2C_2, A_2B_2$, respectively. So let's just prove that $X-Y-Z$ are collinear. While proving the lemma 1, we've proved that $\frac{XC_2}{XB_2}=\frac{AC_2}{AB_2} \cdot \frac{AB}{AC}$. So just by the Menelaus theorem on the $\triangle A_2B_2C_2$ and points $X, Y, Z$, we're done.
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Nari_Tom
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For the second solution some should prove these circles should intersect (not tangent and have intersections with each other), For the first solution some should prove $H$ and $H_2$ are different. May be i will comeback for these later
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