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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
Mount Inequality erupts in all directions!
BR1F1SZ   2
N 4 minutes ago by sqing
Source: Austria National MO Part 1 Problem 1
Let $a$, $b$ and $c$ be pairwise distinct nonnegative real numbers. Prove that
\[
(a + b + c) \left( \frac{a}{(b - c)^2} + \frac{b}{(c - a)^2} + \frac{c}{(a - b)^2} \right) > 4.
\](Karl Czakler)
2 replies
1 viewing
BR1F1SZ
May 5, 2025
sqing
4 minutes ago
Inequality, inequality, inequality...
Assassino9931   6
N 4 minutes ago by NO_SQUARES
Source: Al-Khwarizmi Junior International Olympiad 2025 P6
Let $a,b,c$ be real numbers such that \[ab^2+bc^2+ca^2=6\sqrt{3}+ac^2+cb^2+ba^2.\]Find the smallest possible value of $a^2 + b^2 + c^2$.

Binh Luan and Nhan Xet, Vietnam
6 replies
Assassino9931
5 hours ago
NO_SQUARES
4 minutes ago
inequality involving GCD and square roots
gaussious   1
N 6 minutes ago by Lil_flip38
how to even approach this?
1 reply
gaussious
4 hours ago
Lil_flip38
6 minutes ago
f(f(x+y)) = f(x)+f(y)
NamelyOrange   3
N 12 minutes ago by jasperE3
Source: Evan's FE handout
Find all continuous $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(f(x+y)) = f(x)+f(y)$ for all real $x,y$.
3 replies
1 viewing
NamelyOrange
Mar 14, 2025
jasperE3
12 minutes ago
No more topics!
AP is perpendicular to PC
WakeUp   12
N Mar 27, 2024 by anantmudgal09
Source: China TST 2011 - Quiz 1 - D2 - P1
Let one of the intersection points of two circles with centres $O_1,O_2$ be $P$. A common tangent touches the circles at $A,B$ respectively. Let the perpendicular from $A$ to the line $BP$ meet $O_1O_2$ at $C$. Prove that $AP\perp PC$.
12 replies
WakeUp
May 19, 2011
anantmudgal09
Mar 27, 2024
AP is perpendicular to PC
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G H BBookmark kLocked kLocked NReply
Source: China TST 2011 - Quiz 1 - D2 - P1
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WakeUp
1347 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let one of the intersection points of two circles with centres $O_1,O_2$ be $P$. A common tangent touches the circles at $A,B$ respectively. Let the perpendicular from $A$ to the line $BP$ meet $O_1O_2$ at $C$. Prove that $AP\perp PC$.
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goodar2006
1347 posts
#2 • 1 Y
Y by Adventure10
It's somehow intresting to me that our teacher gave us this problem about $2$ years ago! for solving it, extend $PC$ to intersect the circle with center $O_1$...
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Luis González
4148 posts
#3 • 1 Y
Y by Adventure10
Hmm, this is basically question 2 of 15 IBMO-Venezuela. See the following threads

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=83793
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=336494
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=288&t=80971
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math154
4302 posts
#4 • 5 Y
Y by ZacPower123, trying_to_solve_br, Adventure10, Mango247, and 1 other user
This solution assumes $AB$ is the closer tangent to $P$, which I think the original problem states (it doesn't matter though).

Let $(O_1)\cap(O_2)=\{P,Q\}$, $M=AB\cap PQ$, and $A'$ be the reflection of $A$ over $O_1$. Clearly $\angle{CPA}=90^\circ\Longleftrightarrow\angle{AA'C}=\angle{AA'P}$. But $\angle{AA'P}=\angle{BAP}$, so it suffices to show $\triangle{AA'C}\sim\triangle{BAP}\Longleftrightarrow\triangle{AO_1C}\sim\triangle{BMP}$ (equivalent since $O_1,M$ are the midpoints of $AA',BA$ respectively), which is just a simple angle chase.
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RaleD
118 posts
#5 • 4 Y
Y by yttrerium, ZacPower123, Adventure10, Mango247
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yttrerium
53 posts
#6 • 2 Y
Y by Adventure10, Mango247
RaleD wrote:
Click to reveal hidden text

This solution by RaleD is a very nice solution. At first glance, I thought it is similar to mine. In fact, it's way simpler and nicer. In my opinion, it's very clever to use the radical axis.

My solution also uses something like the equation $CN*CA=CP^2$ to get a geometric relation.

Here is how it goes:
Let the line through $C$ perpendicular to $AB$ intersect $AB$ at point $S$.
A direct calculation of side lengths (starting with $BC^2-PC^2=AB^2-AP^2$) yields $AS\cdot AB=AP^2$
Now, it is easy to find that $A,S,C,P$ are concyclic, which leads to $\angle{APC}=90^\circ$
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junioragd
314 posts
#7 • 2 Y
Y by Adventure10, Mango247
Denote $Q$ the other intersection point of two circles,let $M$ be the intersection point of $PQ$ and $AB$ and let $O$ be the circumcenter of $ABQ$.Now,let $C'$ be the intersection of the perpendicular from $A$ to $BP$ and the perpendicular from $P$ to $AP$.Now,we have that $AOM$ is similar to $APC'$ so we get $APM$ is similar to $AC'P$.Now,from an easy angle chase and using the previos similarity we get $QC'O$ is similar to $QPB$ and from this we get $QC'=PC'$ so $C'=C$ and that is it.
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Dukejukem
695 posts
#8 • 2 Y
Y by Adventure10, Mango247
Here's another solution:

Define $Z \equiv AB \cap O_1O_2$ and let $ZP, AC$ cut $(O_1)$ for a second time at $Q, C'$, respectively. Let $H$ be the projection of $A$ onto $O_1O_2.$ Note that $Z$ is the external center of homothety that maps $(O_2) \mapsto (O_1).$ It is clear that this homothety takes $B \mapsto A$ and $P \mapsto Q$, so $AC \perp BP \implies AC \perp AQ.$ Hence, $\overline{QC'}$ is a diameter of $(O_1) \implies Q, C', O_1$ are collinear. Furthermore, note that $\angle ZHA = \angle ZAO_1 = 90^{\circ} \implies \triangle ZAH \sim \triangle ZO_1A \implies ZH \cdot ZO_1 = ZA^2 = ZP \cdot ZQ$, where the last step follows from Power of a Point. Therefore, $H, O_1, P, Q$ are concyclic, so $\measuredangle PHC = \measuredangle PHO_1 = \measuredangle PQO_1 = \measuredangle PQC' = \measuredangle PAC' = \measuredangle PAC$, where the angles are directed. Hence, $P, H, C, A$ are concyclic, so $\measuredangle APC = \measuredangle AHC = 90^{\circ}$, as desired. $\square$

http://farm6.staticflickr.com/5469/18090868695_ce94a8844d_z.jpg
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AlastorMoody
2125 posts
#9 • 4 Y
Y by amar_04, DPS, GeoMetrix, Adventure10
Solution
This post has been edited 2 times. Last edited by AlastorMoody, Feb 16, 2020, 6:34 PM
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Arefe
65 posts
#10
Y by
with using the special point on the median ( X ) , the problem sounds easy :)
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MagicalToaster53
159 posts
#11
Y by
My solution is pretty much the identical solution outlined through by @math154.

Let $P, Q$ be the points of intersection between circles with circumcenters $O_1, O_2$ with $P$ closer to line $AB$ than $Q$. Let $X = PQ \cap O_1O_2$, and $A_1, B_1$ denote the reflections of $A, B$ over $O_1, O_2$, respectively. Also let $M$ denote the midpoint of line $AB$. We then make the following claim:

Claim: $\triangle A_1O_1C \sim \triangle AMP$.
Proof: We have that as $AA_1 \parallel BB_1$, that \[\measuredangle A_1OC = \measuredangle BO_2C = \measuredangle AMP, \]as $BMXO_2$ is also cyclic. $\square$

We now show that $A_1, C, P$ are collinear:

Claim: $A_1, C, P$ are collinear.
Proof: We wish to show $\measuredangle AA_1C = \measuredangle AA_1P$. It then suffices to equivalently show $\triangle A_1O_1C \sim \triangle AMP$. Indeed, we have that \[\measuredangle AA_1C = \measuredangle MAP, \]and by the above claim/lemma, we obtain the desired similarlity. $\square$

Hence we have that $A_1, C, P$ are collinear, in particular $AP \perp PC$, as desired. $\blacksquare$
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PROF65
2016 posts
#12
Y by
Let $T$ the symmetric of $A$ wrt $O_1O_2$ ;
$S$ be the ex-similicenter ; $Q$ the intersection of $(O_1)$ and $ SP$;
$P',A'$ the antipodes of $P,A$;
$D$ the intersection of $A'P'$ and $O_1O_2$ ;
$C'$ the intersection of $ PP'$ and $O_1O_2$ we will show that $C'=C$ :
by butterfly we deduce that $O_1C'=O_1D$ i.e. $PAP'A'$ is parallelogram.
More $A'(PQ,AT)=-1$ but $A'(C'D,O_1 T)=-1$ hence $A',D,Q$ are collinear.
Since $A'Q\parallel AC'$ then $AC'\perp AQ\parallel BP$
therefore $C'=C$ .
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anantmudgal09
1980 posts
#13
Y by
WakeUp wrote:
Let one of the intersection points of two circles with centres $O_1,O_2$ be $P$. A common tangent touches the circles at $A,B$ respectively. Let the perpendicular from $A$ to the line $BP$ meet $O_1O_2$ at $C$. Prove that $AP\perp PC$.

By drawing the missing orthocentre in $\triangle APB$, the problem becomes equivalent to a statement regarding the queue-config which is quite neat: In triangle $ABC$ with orthocentre $H$ and point $Q$ on arc $\widehat{BAC}$ such that $\angle AQH =90^{\circ}$ (hence $QH$ bisects $BC$), the line through $H$ parallel to $AB$, the line $AC$, and the perpendicular bisector of line $QH$ concur. Let $X, Y$ be points on $AB, AC$ such that $AXHY$ is a parallelogram. Then line $XY$ passes through the midpoint of $AH$, i.e, the centre of the circumcircle of $\triangle AQH$. Also, $(AX, AY, AH, AQ)=-1$ by projecting all on line $BC$, hence $XY \parallel AQ$ or $XY \perp QH$, hence $XY$ is the perpendicular bisector of $QH$, and proving $Y$ lies on it was all we needed.
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