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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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0 replies
jlacosta
May 1, 2025
0 replies
IMO ShortList 1999, geometry problem 2
orl   13
N 18 minutes ago by ezpotd
Source: IMO ShortList 1999, geometry problem 2
A circle is called a separator for a set of five points in a plane if it passes through three of these points, it contains a fourth point inside and the fifth point is outside the circle. Prove that every set of five points such that no three are collinear and no four are concyclic has exactly four separators.
13 replies
+1 w
orl
Nov 13, 2004
ezpotd
18 minutes ago
Brilliant Problem
M11100111001Y1R   5
N 38 minutes ago by Davdav1232
Source: Iran TST 2025 Test 3 Problem 3
Find all sequences \( (a_n) \) of natural numbers such that for every pair of natural numbers \( r \) and \( s \), the following inequality holds:
\[
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2
\]
5 replies
+1 w
M11100111001Y1R
May 27, 2025
Davdav1232
38 minutes ago
4 var inequality
SunnyEvan   0
41 minutes ago
Source: Own
Let $ x,y,z,t \in R^+ ,$ such that : $ (x+y+z+t)^2 = x+y+z+t + (x+z)(y+t) $ and $ x \geq y \geq z \geq t .$
Try to prove or disprove : $$ \frac{2 \sqrt{x+y+z+t +(x+t)(y+z)}}{x^2+y^2+z^2+t^2 +3xz+3yt+xt+yz} \geq \frac{11(x+z)(z+t)-(x+y+z+t)}{x+y+z+t +(x+z)(y+t)} $$
0 replies
SunnyEvan
41 minutes ago
0 replies
Simple inequality
sqing   60
N 42 minutes ago by Adywastaken
Source: Shortlist BMO 2018, A1
Let $a, b, c $ be positive real numbers such that $abc = \frac {2} {3}. $ Prove that:

$$\frac {ab}{a + b} + \frac {bc} {b + c} + \frac {ca} {c + a} \geqslant  \frac {a+b+c} {a^3+b ^ 3 + c ^ 3}.$$
60 replies
sqing
May 3, 2019
Adywastaken
42 minutes ago
Inequalities
sqing   21
N Today at 12:37 AM by sqing
Let $ a,b,c\geq 0 ,a+b+c\leq 3. $ Prove that
$$a^2+b^2+c^2+ab +2ca+2bc +  abc \leq \frac{251}{27}$$$$ a^2+b^2+c^2+ab+2ca+2bc  + \frac{2}{5}abc  \leq \frac{4861}{540}$$$$ a^2+b^2+c^2+ab+2ca+2bc  + \frac{7}{20}abc  \leq \frac{2381411}{26460}$$
21 replies
sqing
May 21, 2025
sqing
Today at 12:37 AM
Polar Coordinates
pingpongmerrily   4
N Today at 12:11 AM by K124659
Convert the equation $r=\tan(\theta)$ into rectangular form.
4 replies
pingpongmerrily
Today at 12:02 AM
K124659
Today at 12:11 AM
Calculus, sets
wl8418   2
N Yesterday at 11:18 PM by wl8418
Is empty set a proper set of an non empty set? Why or why not? Any clarification or insight is appreciated. Thanks in advance!
2 replies
wl8418
Yesterday at 6:11 AM
wl8418
Yesterday at 11:18 PM
Geometry
AlexCenteno2007   1
N Yesterday at 11:13 PM by ohiorizzler1434
Given triangle ABC, it is true that BD = CF where D and F are points in the same half-plane with respect to line BC and it is also known that BD is parallel to AC and CF is parallel to AB. Show that BF, CD and the interior bisector of A are concurrent.
1 reply
AlexCenteno2007
Yesterday at 10:02 PM
ohiorizzler1434
Yesterday at 11:13 PM
How do I prove this? - Sets and symmetric difference
smadadi1000   1
N Yesterday at 8:21 PM by KSH31415
For sets A, B and C, prove that (A Δ B) Δ C = (A Δ C) Δ (A \ B).

The textbook - Proofs by Jay Cummings - had this definition for symmetric difference:
A Δ B = ( A U B)/(A ∩ B)

this is exercise 3.37 (e).
1 reply
smadadi1000
Yesterday at 3:23 PM
KSH31415
Yesterday at 8:21 PM
help me ..
exoticc   2
N Yesterday at 5:00 PM by sodiumaka
Find all pairs of functions (f,g) : R->R that satisfy:
f(1)=2025;
g(f(x+y))+2x+y-1=f(x)+(2x+y)g(y) , ∀x,y∈R
2 replies
exoticc
Yesterday at 9:26 AM
sodiumaka
Yesterday at 5:00 PM
21st PMO National Orals #9
yes45   0
Yesterday at 3:22 PM
In square $ABCD$, $P$ and $Q$ are points on sides $CD$ and $BC$, respectively, such that $\angle{APQ} = 90^\circ$. If $AP = 4$ and $PQ = 3$, find the area of $ABCD$.

Answer Confirmation
Solution
0 replies
yes45
Yesterday at 3:22 PM
0 replies
Original Problem (qrxz17)
qrxz17   0
Yesterday at 1:32 PM
Problem. Suppose that
\begin{align*}
    (a^2+b^2)^2 + (b^2+c^2)^2 +(c^2+a^2)^2 &= 138 \\
    (a^2+b^2+c^2)^2 &=100.
\end{align*}Find \(a^2(a^2-1) + b^2(b^2-1)+c^2(c^2-1)\).
Answer: Click to reveal hidden text
Solution. Subtracting the two given equations, we get
\begin{align*}
        a^4+b^4+c^4=38.
    \end{align*}Taking the square root of the second equation, we get

\begin{align*}
        a^2+b^2+c^2 = 10.
    \end{align*}
Then,
\begin{align*}
        a^4+b^4+c^4-(a^2+b^2+c^2) &= a^4-a^2+b^4-b^2+c^4-c^2 \\
        &= a^2(a^2-1)+b^2(b^2-1)+c^2(c^2-1) = \boxed{28}.
    \end{align*}
0 replies
qrxz17
Yesterday at 1:32 PM
0 replies
17th PMO Area Stage #5
yes45   0
Yesterday at 1:31 PM
Triangle \(ABC\) has a right angle at \(B\), with \(AB = 3\) and \(BC = 4\). If \(D\) and \(E\) are points
on \(AC\) and \(BC\), respectively, such that \(CD = DE = \frac{5}{3}\), find the perimeter of quadrilateral
\(ABED\).
Answer Confirmation
Solution
0 replies
yes45
Yesterday at 1:31 PM
0 replies
Sipnayan 2019 SHS Orals Semifinal Round B Difficult \#3
qrxz17   0
Yesterday at 1:30 PM
Problem. Suppose that \(a\), \(b\), \(c\) are positive integers such that \((a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) = 63\). Find all possible values of \(a+b+c\).
Answer. Click to reveal hidden text
Solution. Expanding and cancelling "like" terms, we get

\begin{align*}
        (a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) &= (a^4 +b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2) - 2(a^4+b^4+c^4) \\
        &= 2a^2b^2+2a^2c^2+2b^2c^2 - a^4 -b^4-c^4
    \end{align*}
We multiply the entire equation by \(-1\) to rearrange the terms and place the higher-degree terms at the front of the expression in order to have a more recognizable form.

\begin{align*}
        a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2 = -63.
    \end{align*}
Simplifying the left-hand expression, we get

\begin{align*}
        a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2 &=a^4-2a^2(b^2+c^2)+b^4-2b^2c^2+c^4 \\
        &= a^4-2a^2(b^2+c^2)+(b^2+c^2)^2 - 4b^2c^2 \\
        &= [a^2 - (b^2+c^2)]^2-(2bc)^2 \\
        &= (a^2-b^2-c^2+2bc)(a^2-b^2-c^2-2bc)\\
        &=[a^2-(b-c)^2][a^2-(b+c)^2] \\
        &= (a+b-c)(a-b+c)(a+b+c)(a-b-c) = -63.
    \end{align*}
If \(a=b=c\), we get
\begin{align*}
     -3a^4 &=-63 \\
     a^4 &= 21
    \end{align*}

in which there are no real solutions for \(a\).

If we either have \(a=b\), \(a=c\), or \(b=c\), we get something of the form

\[
    a^2(a+2b)(a-2b)=-63.
    \]
Since \(1\) and \(9\) are the only square factors of \(63\), we have \(a = \{1, 3\}\).

In this case, the solutions for \((a, b, c)\) are \((1, 4, 4)\) and \((3, 2, 2)\), including all their permutations.

If \(a, b, c\) are all distinct positive integers, there are no solutions.

Therefore, all possible values of \(a+b+c\) are \(\boxed{7 \text{ and }9}\).
0 replies
qrxz17
Yesterday at 1:30 PM
0 replies
Problem 4, Iberoamerican Olympiad 2011
Gerard.123   19
N Feb 25, 2024 by AshAuktober
Let $ABC$ be an acute-angled triangle, with $AC \neq BC$ and let $O$ be its circumcenter. Let $P$ and $Q$ be points such that $BOAP$ and $COPQ$ are parallelograms. Show that $Q$ is the orthocenter of $ABC$.
19 replies
Gerard.123
Oct 2, 2011
AshAuktober
Feb 25, 2024
Problem 4, Iberoamerican Olympiad 2011
G H J
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Gerard.123
23 posts
#1 • 4 Y
Y by ImSh95, Adventure10, Mango247, Spiritpalm
Let $ABC$ be an acute-angled triangle, with $AC \neq BC$ and let $O$ be its circumcenter. Let $P$ and $Q$ be points such that $BOAP$ and $COPQ$ are parallelograms. Show that $Q$ is the orthocenter of $ABC$.
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WakeUp
1347 posts
#2 • 5 Y
Y by Wizard_32, ImSh95, Adventure10, Mango247, Spiritpalm
The problem is just begging for vectors!

We have $\bold{q}=\bold{p}+\overrightarrow{PQ}=\bold{b}+\overrightarrow{BP}+\overrightarrow{PQ}=\bold{b}+\bold{a}+\bold{c}$ so $Q$ is the orthocentre.
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BigSams
6591 posts
#3 • 2 Y
Y by ImSh95, Adventure10
It has been some time since I last allowed myself to indulge in geometry. Please verify the following solution.

Since $BOAP$ is a parallelogram and $BO=AO=R$, where $R$ is the circumradius of $\triangle ABC$, $BOAP$ is a rhombus of side length $R$. So $BP=AP=R$. Since $COPQ$ is a parallelogram with $CO=P$, we know that the opposite side $QP=R$. Then $QP=BP=AP=R$. Without loss of generality, suppose that $\angle B>\angle A$ (generality is not lost as the following arguments can be tailored for the case $A>B$ by replacing each $B$ with $A$, and each $A$ with $B$; the problem states that $AC\neq BC\iff \angle A\neq\angle B$, so that case needs not be considered). Note that $\angle APC,\angle BPC=\frac{\angle APB}{2}=\frac{\angle AOB}{2}=\angle C$. Also, $\angle BOC = 2\angle A\implies \angle POC = \angle C + 2\angle A\implies \angle OPQ = \angle B - \angle A$.
We can now say that $\begin{cases}\angle BPQ = \angle C + \angle A - \angle B=\pi - 2\angle B\implies BQ= \sqrt{BP^2 + PQ^2-(BP)(QP)\cos \angle BQP}=2R\cos B\\ \angle APQ = \angle C + \angle B - \angle A=\pi-2\angle A\implies AQ= \sqrt{AP^2 + PQ^2-(AP)(QP)\cos \angle AQP}=2R\cos A\end{cases}$.
Let $H$ be the orthocenter of $\triangle ABC$. It is well-known that $\begin{cases}BH = 2R\cos \angle B\\ AH =2R\cos \angle A\end{cases}$. There are up to two points that satisfy these distances from $A,B$; i.e. the interesections of the circle with center $B$ and radius $2R\cos \angle B$, and the circle with center $A$ and radius $2R\cos \angle A$. These two points are clearly the same point or on different sides of $AB$. Since $PB=PQ=PA=R$, $Q$ is inside $\triangle ABC$, and since $\triangle ABC$ is acute, $Q$ is the orthocenter.
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hatchguy
555 posts
#4 • 4 Y
Y by DominicanAOPSer, ImSh95, Adventure10, and 1 other user
Let $M$ be midpoint of $AB$. We have $CQ \perp AB$ since $CQ \parallel OP$ and clearly $OP$ is the perpendicular bisector of $AB$.

Also $CQ = OP$ and $OP = 2OM = CH$ and therefore $CQ = CH$. Hence $Q=H$.
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daniel73
253 posts
#5 • 2 Y
Y by ImSh95, Adventure10
Clearly $OA=OB$, hence $P$ is on the perpendicular to $AB$ through its midpoint $M$, and $P$ is the symmetric of $O$ with respect to $AB$. Let $N$ be the midpoint of $CP$, clearly inside $ABC$ (otherwise $O$ would be outside $ABC$), because $ABC$ is acute. Note now that
\[MN^2=\frac{ON^2+PN^2}{2}-\frac{OP^2}{4}=\frac{OC^2+OP^2}{4}-\frac{PC^2}{8}+\frac{PN^2}{2}-\frac{OP^2}{4}=\frac{OC^2}{4},\]
hence $N$ is at a distance equal to $\frac{R}{2}$ from $M$, on the line halfway between the perpendicular bisector of $AB$ and the altitude from $C$ (by Thales' theorem and because $N$ is the midpoint of $PC$), and inside $ABC$. Hence $N$ is the nine-point center, and $Q$ the symmetric of $O$ with respect to it, thus the orthocenter.
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semy
20 posts
#6 • 4 Y
Y by Vladimir_Djurica, ImSh95, Adventure10, Mango247
This problem is just Servois' theorem itself.
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Gerard.123
23 posts
#7 • 3 Y
Y by ImSh95, Adventure10, Mango247
semy wrote:
This problem is just Servois' theorem itself.
I've looked for this, but couldn't find it, where can I find a source which talks about this theorem?
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Gregorio
14 posts
#8 • 2 Y
Y by ImSh95, Adventure10
This problem is actually trivial if you know that $2MO=CH$. Because of this $COPH$ is a parallelogram and so is $COPQ$, thus $H=Q$
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RMACR7LP
40 posts
#9 • 4 Y
Y by ImSh95, Adventure10, Mango247, AlexCenteno2007
Solution
Attachments:
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Arkmmq
214 posts
#10 • 4 Y
Y by Vladimir_Djurica, ImSh95, Adventure10, Spiritpalm
Easy by complex numbers.....just set ($ABC$) as the unit circle and notice that $p=a+b$ and $q=a+b+c$ ..done.
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jayme
9801 posts
#11 • 5 Y
Y by ImSh95, Mango247, Mango247, Mango247, Spiritpalm
Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%207.pdf p. 54...

Sincerely
Jean-Louis
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blackbluecar
303 posts
#12 • 1 Y
Y by ImSh95
Let $H$ be the orthocenter of $ABC$. Let us prove $COPH$ is a parallelogram.

Claim 1: $HP = OC$

Notice that $BP=OA=OC=OB=PA$. Thus $O$ and $P$ are reflections of each other over line $AB$. Realizing the circle centered at $P$ passing through $A$ (say, $\omega)$ has the same radius as the circle centered at $P$ passing through $A$ (say, $\omega')$. The conclusion follows $\omega$ and $\omega'$ are reflections of each other about $AB$. Recall, the reflection of $H$ about $AB$ lies on $\omega$ so, $H \in \omega'$. Hence, $OH = OA = OB$. Proving the claim.

Claim 2: $CH \parallel OP$

Clearly, $O$ and $P$ lie on the perpendicular bisector of $AB$ so, $OP \perp AB$. Clearly, $CH \perp AB$ so $CH \parallel OP$. Proving the claim.

Claim 1 and Claim 2 are sufficient to ensure $COPH$ is a parallelogram. So, we are done.
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primesarespecial
364 posts
#13 • 3 Y
Y by Vladimir_Djurica, ImSh95, Spiritpalm
Well Just use complex numbers
Let $a,b,c$ lie on the unit circle .
I don't like vectors (Alone) , so we use the fact that the diagonals of a parallelogram bisect each other to get

$p=a+b$ and $q=a+b+c$,
Thus $q$ is the orthocenter .
Moral of the story , Don't be sticky with Synthetic solutions.
Like complex numbers finish this in a second or two.
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407420
2113 posts
#14 • 3 Y
Y by ImSh95, Mango247, Spiritpalm
We have $OA=OB=OC$ because $O$ is the circumcenter of $ABC$, so the paralellograms imply that $PA=PB=PQ$, hence $P$ is the circumcenter of $ABQ$, this gives$$\angle AQB=180^\circ -\frac{\angle BPA}{2}=180^\circ -\frac{\angle AOB}{2}=180^\circ -\angle ACB.$$Observe that $OA=OB$ and $PA=PB$ imply that $OP\perp AB$, so $CQ\perp AB$. Hence $Q$ is the orthocenter of $ABC$.
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Tsikaloudakis
982 posts
#15 • 2 Y
Y by ImSh95, Spiritpalm
βλέπε σχήμα:
Attachments:
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Ianis
418 posts
#16 • 2 Y
Y by ImSh95, Spiritpalm
$$Q=P+C=A+B+C$$
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Bigtaitus
75 posts
#17 • 2 Y
Y by Spiritpalm, Vahe_Arsenyan
Clarly $P$ is the reflection of $O$ wrt $AB.$ So $Q$ is a point in the $C-altitude$ such that $OP=CQ.$ Clarly the orthocenter satisfies this lenght condition, and thus we are donne.
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CrazyInMath
460 posts
#18
Y by
So we use complex numbers...
Set $(ABC)$ as the unit circle.
We got $p+0=a+b$ so $p=a+b$.
We got $c+p=0+q$ so $q=c+p=a+b+c$ which means $Q$ is the orthocenter.
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cursed_tangent1434
649 posts
#19
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Complex is actually quite immediate but we give a synthetic solution anyways. First, we show $Q$ lies on $\overline{CH}$.

Claim : The line $\overline{CQ}$ is infact the $C$-altitude of $\triangle ABC$.
We note that,
\begin{align*}
\measuredangle QCO &= \measuredangle POC \\
&=  \measuredangle POB + \measuredangle BOC\\
&= \measuredangle OPA + \measuredangle BOC \\
&= \measuredangle AOP + \measuredangle BOC\\
&= 90 + \measuredangle OAB + \measuredangle BOC \\
&= \measuredangle ACB + 2\measuredangle BAC
\end{align*}Also, we can verify that
\begin{align*}
\measuredangle HCO & = \measuredangle BCA + 2 \measuredangle ACO\\
&= \measuredangle BCA + \measuredangle AOC\\
&= \measuredangle BCA + 2 \measuredangle ABC \\
&= \measuredangle ACB + 2 \measuredangle BAC
\end{align*}from which it is clear that $\measuredangle QCO = \measuredangle HCO$ which implies that $Q$ lies on $\overline{CH}$ as claimed.

Now, it is well known that (and easy to prove) if $M_C$ is the midpoint of $AB$ and $N_C$ is the midpoint of $CH$, then $CN_CM_CO$ is a parallelogram. First note that since in a parallelogram, diagonals bisect each other, $PO$ passes through the midpoint of $AB$ (which is $M_C$).

Thus,
\[QC = OP = 2OM_C = 2CN_C=CH\]But combining this with the previous fact, concludes that indeed $Q$ is the orthocenter of $\triangle ABC$ and we are done.
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AshAuktober
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#20 • 1 Y
Y by Spiritpalm
Clearly the point $Q$ is unique. Let $H$ be the orthocentre of $\Delta ABC$. Our goal is to prove that $COPH$ is a parallelogram.
Observe that if $F$ is the midpoint of $AB$, then $OF \perp AB \implies P \text{ is the reflection of } O \text{about line} AB$. Now clearly $OP \parallel CH$, and also $OP = 2R\cos C = CH$ where $R$ is the circumradius of $\Delta ABC$, from standard trigonometry results, so we are done.
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