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An inequality by my friend.
jesse_is_a_crook   2
N an hour ago by jesse_is_a_crook
x,y,z are real numbers, xy+yz+zx=1.
Prove(or give an example of x,y,z that makes this ineq incorrect):
8(xyz)^2≥(1-x^2)(1-y^2)(1-z^2)

Note: This ineq is quite easy when x,y,z are positive. The main part is 2 positive and 1 negative.
2 replies
jesse_is_a_crook
Yesterday at 10:51 AM
jesse_is_a_crook
an hour ago
Interesting equation
AlexCenteno2007   0
an hour ago
The equation
\[
\frac{a}{x+a} + \frac{b}{x+b} = \frac{c}{x+c} + \frac{d}{x+d}
\]where $a, b, c, d$ are nonzero real parameters, has two equal roots. Prove that either one of the numbers $a$ or $b$ is equal to one of the numbers $c$ or $d$, or else
\[
\frac{1}{a} + \frac{1}{b} = \frac{1}{c} + \frac{1}{d}.
\]Find all the roots of the equation in terms of the parameters.
0 replies
AlexCenteno2007
an hour ago
0 replies
[PMO23 Qualifying I.6] Dice Probability
kae_3   2
N 2 hours ago by JumpingJelly
In rolling three fair twelve-sided dice simultaneously, what is the probability that the resulting numbers can be arranged to form a geometric sequence?

$\text{(a) }\dfrac{1}{72}\qquad\text{(b) }\dfrac{5}{288}\qquad\text{(c) }\dfrac{1}{48}\qquad\text{(d) }\dfrac{7}{288}$

Answer Confirmation
2 replies
kae_3
Feb 9, 2025
JumpingJelly
2 hours ago
Inequalities
sqing   8
N 2 hours ago by sqing
Let $ a, b, c $ be real numbers such that $ a + b + c = 0 $ and $ abc = -16 $. Prove that$$ \frac{a^2 + b^2}{c} + \frac{b^2 + c^2}{a} + \frac{c^2 + a^2}{b} -a^2\geq 2$$$$ \frac{a^2 + b^2}{c} + \frac{b^2 + c^2}{a} + \frac{c^2 + a^2}{b}-a^2-bc\geq -2$$Equality holds when $a=-4,b=c=2.$
8 replies
sqing
Jul 9, 2025
sqing
2 hours ago
Inequalities
sqing   22
N 2 hours ago by sqing
Let $ a,b,c\geq 0  $ and $ ab+bc+ca=2. $ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} \geq 9-3\sqrt{6}$$Let $ a,b,c\geq 0  $ and $ ab+bc+ca=4. $ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} \geq 6\sqrt{3}-9$$Let $ a,b,c\geq 0  $ and $ ab+bc+ca=6. $ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} \geq 3(\sqrt{2}-1)$$Let $ a,b,c\geq 0  $ and $ ab+bc+ 3c^2=7. $ Prove that
$$ \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{ 3c+1}\geq \frac{3}{ 2}$$
22 replies
sqing
Jul 17, 2025
sqing
2 hours ago
Inequalities
sqing   16
N 2 hours ago by sqing
Let $ a,b,c> 0,a+b+c+ab+bc+ca=6 .$ Prove that
$$(a^2+b^2+c^2)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) \geq 9$$Let $ a,b,c> 0,ab+bc+ca=3 .$ Prove that
$$(a^2+b^2+c^2)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) \geq 9$$Let $ a,b,c> 0,a+b+c=3 .$ Prove that
$$( ab+bc+ca)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) \geq 9$$
16 replies
sqing
Jul 18, 2025
sqing
2 hours ago
Two Polynomials
4everwise   23
N 3 hours ago by neeyakkid23
Consider the polynomials $P(x)=x^{6}-x^{5}-x^{3}-x^{2}-x$ and $Q(x)=x^{4}-x^{3}-x^{2}-1.$ Given that $z_{1},z_{2},z_{3},$ and $z_{4}$ are the roots of $Q(x)=0,$ find $P(z_{1})+P(z_{2})+P(z_{3})+P(z_{4}).$
23 replies
4everwise
Dec 26, 2006
neeyakkid23
3 hours ago
Complex numbers
preatsreard   2
N 3 hours ago by vincentwant
How many complex z exist, such that z,z²,z³,....,z²⁰²¹ form a perfect 2021-gon (in arbitary order) in a complex plane.
2 replies
preatsreard
Yesterday at 4:52 PM
vincentwant
3 hours ago
13th PMO National Stage (Oral Phase) #10
monsterbob   5
N 4 hours ago by mathprodigy2011

$x^2 + 4x + 8 = 0$ has roots $a$ and $b$. Find the minimum sum of coefficients of a quadratic polynomial with integer coefficients whose roots are $\dfrac{1}{a}$ and $\dfrac{1}{b}$.
5 replies
monsterbob
Yesterday at 7:10 PM
mathprodigy2011
4 hours ago
1\p=1/a^2+1/b^2 diophantine with prime p (Greece Juniors 2017 p3)
parmenides51   5
N 4 hours ago by Just1
Find all triplets $(a,b,p)$ where $a,b$ are positive integers and $p$ is a prime number satisfying: $\frac{1}{p}=\frac{1}{a^2}+\frac{1}{b^2}$
5 replies
parmenides51
Mar 16, 2020
Just1
4 hours ago
Geometry Parallel Proof Problem
CatalanThinker   5
N May 10, 2025 by Tkn
Source: No source found, just yet, please share if you find it though :)
Let M be the midpoint of the side BC of triangle ABC. The bisector of the exterior angle of point A intersects the side BC in D. Let the circumcircle of triangle ADM intersect the lines AB and AC in E and F respectively. If the midpoint of EF is N, prove that MN || AD.
I have done some constructions, but still did not quite get to the answer, see diagram attached below
5 replies
CatalanThinker
May 9, 2025
Tkn
May 10, 2025
Geometry Parallel Proof Problem
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G H BBookmark kLocked kLocked NReply
Source: No source found, just yet, please share if you find it though :)
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CatalanThinker
13 posts
#1 • 1 Y
Y by Rounak_iitr
Let M be the midpoint of the side BC of triangle ABC. The bisector of the exterior angle of point A intersects the side BC in D. Let the circumcircle of triangle ADM intersect the lines AB and AC in E and F respectively. If the midpoint of EF is N, prove that MN || AD.
I have done some constructions, but still did not quite get to the answer, see diagram attached below
Attachments:
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ItzsleepyXD
162 posts
#2
Y by
note that $D'$ is midpoint of arc $BC$
It is easy to see the line $DD'$ is perpendicular bisector of segment $EF$
tangent at $A$ to $(ABC)$ intersect $(ADM)$ and $BC$ at $X,Y$
known that $\angle YAD = \angle YDA$ implies that $DA \parallel MX$ .
will prove that $M,X,N$ collinear
Point $Z$ on $(ADM)$ such that $\angle BAZ = \angle MAC$ . so line $DD'$ is perpendicular bisector of segment $MZ$ .
$N' = MX \cap EF$
$-1=(AY,AZ;AB,AC)=(AX,AZ;AE,AF) = (X,Z;E,F) = (MX,MZ;ME,MF) = (N', \infty_{EF} ; E ,F)$ implies that $N'$ is midpoint of $EF$ .
So $MN \parallel AD$
This post has been edited 3 times. Last edited by ItzsleepyXD, May 9, 2025, 4:06 AM
Reason: typoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo
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CatalanThinker
13 posts
#3
Y by
Any other solutions..
This post has been edited 1 time. Last edited by CatalanThinker, May 9, 2025, 4:39 AM
Reason: typo
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CatalanThinker
13 posts
#4
Y by
ItzsleepyXD wrote:
note that $D'$ is midpoint of arc $BC$
It is easy to see the line $DD'$ is perpendicular bisector of segment $EF$
tangent at $A$ to $(ABC)$ intersect $(ADM)$ and $BC$ at $X,Y$
known that $\angle YAD = \angle YDA$ implies that $DA \parallel MX$ .
will prove that $M,X,N$ collinear
Point $Z$ on $(ADM)$ such that $\angle BAZ = \angle MAC$ . so line $DD'$ is perpendicular bisector of segment $MZ$ .
$N' = MX \cap EF$
$-1=(AY,AZ;AB,AC)=(AX,AZ;AE,AF) = (X,Z;E,F) = (MX,MZ;ME,MF) = (N', \infty_{EF} ; E ,F)$ implies that $N'$ is midpoint of $EF$ .
So $MN \parallel AD$

Thanks, understood
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CatalanThinker
13 posts
#5
Y by
Any other solutions?
Z K Y
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Tkn
49 posts
#6
Y by
[asy]
import graph;
import geometry;
        
size(9cm);
defaultpen(fontsize(10pt));

pair A = (0,2);
pair B = (-0.8,0);
pair C = (3,0);
pair M = (B+C)/2;
pair in = unit(B-A)+unit(C-A)+A;
pair ex = rotate(90,A)*in;
pair D = extension(A,ex,B,C);

path circ1 = circumcircle(A,B,C);
path circ2 = circumcircle(A,D,M);

pair F = intersectionpoints(A+3*(A-C)--A, circ2)[0];
pair E1 = intersectionpoints(B+3*(B-A)--A, circ2)[0];
pair N1 = (E1+F)/2;
pair P = intersectionpoints(circ1,circ2)[1];
pair Q = extension(F,E1,D,C);

path circ3 = circumcircle(Q,N1,M);
path circ4 = circumcircle(Q,B,E1);

draw(A--B--C--cycle, black);
draw(A--F, black);
draw(B--E1, black);
draw(D--B, black);
draw(E1--F, blue);
draw(D--P, orange);
draw(M--N1, royalblue);
draw(A--D, royalblue);

draw(circ1);
draw(circ2, red);
draw(circ3, deepgreen+dashed);
draw(circ4, deepgreen+dashed);

dot(A);
dot(B);
dot(C);
dot(D);
dot(M);
dot(F);
dot(E1);
dot(N1);
dot(P);
dot(Q);

label("$A$", A, N, black);
label("$B$", B, S+1.25W, black);
label("$C$", C, SE, black);
label("$D$", D, W, black);
label("$M$", M, NE, black);
label("$E$", E1, S, black);
label("$F$", F, N, black);
label("$N$", N1, SW, black);
label("$P$", P, SE, black);
label("$Q$", Q, NW, black);
[/asy]
First, note that $D$ is the midarc $FE$ because $\angle{DAE}=\angle{DAF}$.
Let $P\neq D$ be an intersection of the line $DN$ and $(ADM)$, and $Q$ be an intersection of $\overline{FE}$ and $\overline{DM}$.
Since $FPED$ is a harmonic quadrilateral with diameter $\overline{DP}$, Picking ratio from $A$ to $\overline{DC}$:
$$(D,AP\cap DC;B,C)=-1.$$So, $\overline{AP}$ bisects $\angle{BAC}$. Note that $\angle{DMP}=90^{\circ}$, so $P\in (ABC)$.
It is easy to see that $Q,N,M$ and $P$ are concyclic (since $\angle{QNP}=90^{\circ}=\angle{QMP}$).
Note that $A=BE\cap CF$ and $(AFE)$ meets $(ABC)$ again at $P$.
Therefore $P$ is a spiral center sending $\overline{BC}\mapsto \overline{EF}$. Now, we have $\triangle{PBC}\sim \triangle{PFE}$.

Next, observe that $\angle{QEP}=\angle{PBC}$. So, $Q,B,P$ and $E$ are concyclic.
Note that $Q=BM\cap NE$, and $(QNM)$ meets $(QBE)$ again at $P$.
So, $P$ is also a sprial center sending $\overline{BE}\mapsto \overline{MN}$. Therefore, we have $\triangle{PNM}\sim \triangle{PEB}$.
Since, $P$ sends $\overline{BC}\rightarrow\overline{EF}$, $P$ also sends $\overline{BE}\rightarrow\overline{CF}$. Which implies
$$\triangle{FPC}\sim \triangle{EPB}\sim\triangle{NPM}.$$We must have $$\angle{PNM}=\angle{PFC}=\angle{PFA}=\angle{PDA}.$$Therefore, $\overline{MN}\parallel\overline{AD}$ as desired.
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