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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.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Non-linear Recursive Sequence
amogususususus   4
N 2 minutes ago by GreekIdiot
Given $a_1=1$ and the recursive relation
$$a_{i+1}=a_i+\frac{1}{a_i}$$for all natural number $i$. Find the general form of $a_n$.

Is there any way to solve this problem and similar ones?
4 replies
+1 w
amogususususus
Jan 24, 2025
GreekIdiot
2 minutes ago
Russian Diophantine Equation
LeYohan   2
N 4 minutes ago by RagvaloD
Source: Moscow, 1963
Find all integer solutions to

$\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} = 3$.
2 replies
LeYohan
Yesterday at 2:59 PM
RagvaloD
4 minutes ago
PQ = r and 6 more conditions
avisioner   41
N 8 minutes ago by ezpotd
Source: 2023 ISL G2
Let $ABC$ be a triangle with $AC > BC,$ let $\omega$ be the circumcircle of $\triangle ABC,$ and let $r$ be its radius. Point $P$ is chosen on $\overline{AC}$ such taht $BC=CP,$ and point $S$ is the foot of the perpendicular from $P$ to $\overline{AB}$. Ray $BP$ mets $\omega$ again at $D$. Point $Q$ is chosen on line $SP$ such that $PQ = r$ and $S,P,Q$ lie on a line in that order. Finally, let $E$ be a point satisfying $\overline{AE} \perp \overline{CQ}$ and $\overline{BE} \perp \overline{DQ}$. Prove that $E$ lies on $\omega$.
41 replies
avisioner
Jul 17, 2024
ezpotd
8 minutes ago
Functional equation from R^2 to R
k.vasilev   19
N 9 minutes ago by megahertz13
Source: All-Russian Olympiad 2019 grade 10 problem 1
Each point $A$ in the plane is assigned a real number $f(A).$ It is known that $f(M)=f(A)+f(B)+f(C),$ whenever $M$ is the centroid of $\triangle ABC.$ Prove that $f(A)=0$ for all points $A.$
19 replies
k.vasilev
Apr 23, 2019
megahertz13
9 minutes ago
Functional equations in IMO TST
sheripqr   50
N 17 minutes ago by megahertz13
Source: Iran TST 1996
Find all functions $f: \mathbb R \to \mathbb R$ such that $$ f(f(x)+y)=f(x^2-y)+4f(x)y $$ for all $x,y \in \mathbb R$
50 replies
sheripqr
Sep 14, 2015
megahertz13
17 minutes ago
JBMO TST Bosnia and Herzegovina 2023 P4
FishkoBiH   1
N 33 minutes ago by Maths_VC
Source: JBMO TST Bosnia and Herzegovina 2023 P4
Let $n$ be a positive integer. A board with a format $n*n$ is divided in $n*n$ equal squares.Determine all integers $n$3 such that the board can be covered in $2*1$ (or $1*2$) pieces so that there is exactly one empty square in each row and each column.
1 reply
FishkoBiH
Today at 1:38 PM
Maths_VC
33 minutes ago
IMO Shortlist 2009 - Problem C3
nsato   25
N an hour ago by popop614
Let $n$ be a positive integer. Given a sequence $\varepsilon_1$, $\dots$, $\varepsilon_{n - 1}$ with $\varepsilon_i = 0$ or $\varepsilon_i = 1$ for each $i = 1$, $\dots$, $n - 1$, the sequences $a_0$, $\dots$, $a_n$ and $b_0$, $\dots$, $b_n$ are constructed by the following rules: \[a_0 = b_0 = 1, \quad a_1 = b_1 = 7,\] \[\begin{array}{lll}
	a_{i+1} = 
	\begin{cases}
		2a_{i-1} + 3a_i, \\
		3a_{i-1} + a_i, 
	\end{cases} & 
        \begin{array}{l} 
                \text{if } \varepsilon_i = 0, \\  
                \text{if } \varepsilon_i = 1, \end{array} 
         & \text{for each } i = 1, \dots, n - 1, \\[15pt]
        b_{i+1}= 
        \begin{cases}
		2b_{i-1} + 3b_i, \\
		3b_{i-1} + b_i, 
	\end{cases} & 
        \begin{array}{l} 
                \text{if } \varepsilon_{n-i} = 0, \\  
                \text{if } \varepsilon_{n-i} = 1, \end{array} 
         & \text{for each } i = 1, \dots, n - 1.
	\end{array}\] Prove that $a_n = b_n$.

Proposed by Ilya Bogdanov, Russia
25 replies
nsato
Jul 6, 2010
popop614
an hour ago
JBMO TST Bosnia and Herzegovina 2024 P4
FishkoBiH   1
N an hour ago by TopGbulliedU
Source: JBMO TST Bosnia and Herzegovina 2024 P4
Let $m$ and $n$ be natural numbers. Every one of the $m*n$ squares of the $m*n$ board is colored either black or white, so that no 2 neighbouring squares are the same color(the board is colored like in chess").In one step we can pick 2 neighbouring squares and change their colors like this:
- a white square becomes black;
-a black square becomes blue;
-a blue square becomes white.
For which $m$ and $n$ can we ,in a finite sequence of these steps, switch the starting colors from white to black and vice versa.
1 reply
FishkoBiH
5 hours ago
TopGbulliedU
an hour ago
Van der Corput Inequality
EthanWYX2009   1
N an hour ago by grupyorum
Source: en.wikipedia.org/wiki/Van_der_Corput_inequality
Let $V$ be a real or complex inner product space. Suppose that ${\displaystyle v,u_{1},\dots ,u_{n}\in V} $ and that ${\displaystyle \|v\|=1}.$ Then$${\displaystyle \displaystyle \left(\sum _{i=1}^{n}|\langle v,u_{i}\rangle |\right)^{2}\leq \sum _{i,j=1}^{n}|\langle u_{i},u_{j}\rangle |.}$$
1 reply
EthanWYX2009
Today at 3:36 AM
grupyorum
an hour ago
two sequences of positive integers and inequalities
rmtf1111   52
N an hour ago by Kappa_Beta_725
Source: EGMO 2019 P5
Let $n\ge 2$ be an integer, and let $a_1, a_2, \cdots , a_n$ be positive integers. Show that there exist positive integers $b_1, b_2, \cdots, b_n$ satisfying the following three conditions:

$\text{(A)} \ a_i\le b_i$ for $i=1, 2, \cdots , n;$

$\text{(B)} \ $ the remainders of $b_1, b_2, \cdots, b_n$ on division by $n$ are pairwise different; and

$\text{(C)} \ $ $b_1+b_2+\cdots b_n \le n\left(\frac{n-1}{2}+\left\lfloor \frac{a_1+a_2+\cdots a_n}{n}\right \rfloor \right)$

(Here, $\lfloor x \rfloor$ denotes the integer part of real number $x$, that is, the largest integer that does not exceed $x$.)
52 replies
rmtf1111
Apr 10, 2019
Kappa_Beta_725
an hour ago
side ratio in triangle, angle bisectors form cyclic quadrilateral
jasperE3   3
N 2 hours ago by lpieleanu
Source: Mongolia MO 2000 Grade 10 P6
In a triangle $ABC$, the angle bisector at $A,B,C$ meet the opposite sides at $A_1,B_1,C_1$, respectively. Prove that if the quadrilateral $BA_1B_1C_1$ is cyclic, then
$$\frac{AC}{AB+BC}=\frac{AB}{AC+CB}+\frac{BC}{BA+AC}.$$
3 replies
jasperE3
Apr 22, 2021
lpieleanu
2 hours ago
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   5
N 2 hours ago by JARP091
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
5 replies
OgnjenTesic
May 22, 2025
JARP091
2 hours ago
JBMO TST Bosnia and Herzegovina 2024 P3
FishkoBiH   2
N 2 hours ago by Ianis
Source: JBMO TST Bosnia and Herzegovina 2024 P3
Let $ABC$ be a right-angled triangle where $ACB$=90°.Let $CD$ be an altitude of that triangle and points $M$ and $N$ be the midpoints of $CD$ and $BC$, respectively.If $S$ is the circumcenter of the triangle $AMN$, prove that $AS$ and $BC$ are paralel.
2 replies
FishkoBiH
6 hours ago
Ianis
2 hours ago
A line parallel to the asymptote of a cubic
kosmonauten3114   0
2 hours ago
Source: My own, but uninspiring...
Given a scalene triangle $\triangle{ABC}$, let $P$ be a point ($\neq \text{X(4)}$). Let $P'$ be the anticomplement of $P$, and let $Q$ be the $\text{X(1)}$-anticomplementary conjugate of $P$. Prove that the line $P'Q$ is parallel to the real asymptote of the circular pivotal isocubic with pivot $P$.
0 replies
kosmonauten3114
2 hours ago
0 replies
Concyclic with B,C
dibyo_99   16
N Apr 20, 2024 by Number1048576
Source: China Team Selection Test 3 Day 1 P1
$\triangle{ABC}$ is isosceles with $AB = AC >BC$. Let $D$ be a point in its interior such that $DA = DB+DC$. Suppose that the perpendicular bisector of $AB$ meets the external angle bisector of $\angle{ADB}$ at $P$, and let $Q$ be the intersection of the perpendicular bisector of $AC$ and the external angle bisector of $\angle{ADC}$. Prove that $B,C,P,Q$ are concyclic.
16 replies
dibyo_99
Mar 25, 2015
Number1048576
Apr 20, 2024
Concyclic with B,C
G H J
Source: China Team Selection Test 3 Day 1 P1
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dibyo_99
487 posts
#1 • 6 Y
Y by tenplusten, Davi-8191, nguyendangkhoa17112003, Adventure10, Mango247, Rounak_iitr
$\triangle{ABC}$ is isosceles with $AB = AC >BC$. Let $D$ be a point in its interior such that $DA = DB+DC$. Suppose that the perpendicular bisector of $AB$ meets the external angle bisector of $\angle{ADB}$ at $P$, and let $Q$ be the intersection of the perpendicular bisector of $AC$ and the external angle bisector of $\angle{ADC}$. Prove that $B,C,P,Q$ are concyclic.
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TelvCohl
2312 posts
#2 • 6 Y
Y by langkhach11112, dagezjm, mathisreal, SMSGodslayer, Adventure10, Mango247
My solution:

Let $ P^*=PD \cap AB, Q^*=QD \cap AC $ and $ AB=AC=\delta $ .

Easy to see $ P, Q $ is the midpoint of arc $ AB $ in $ \odot (ABD), $ arc $ AC $ in $ \odot (ACD) $, respectively .

From the condition $ \Longrightarrow  \frac{BD}{AD}+\frac{CD}{AD}=1 \Longrightarrow \frac{BP^*}{\delta + BP^*}+\frac{CQ^*}{\delta + CQ^*}=1 \Longrightarrow \delta^2=BP^* \cdot CQ^* $ ,

so from $ \frac{P^*B}{P^*A}=\frac{DB}{DA} \Longrightarrow P^*B \cdot DA=P^*A \cdot DB = \delta \cdot DB+P^*B \cdot DB $

$  \Longrightarrow P^*B \cdot DC= \delta \cdot DB \Longrightarrow ( P^*B \cdot DC )^2=\delta ^2 \cdot DB^2=P^*B \cdot Q^*C \cdot DB^2 \Longrightarrow \frac{DC^2}{Q^*C}=\frac{DB^2}{P^*B} $ . ... $ (\star) $

If we denote $ R_1=BP \cap AD, R_2=CQ \cap AD $ , then from $ (\star) $ we get

$ \frac{AR_1}{DR_1}=\frac{[ABP]}{[DBP]}=\frac{ \delta \cdot AP}{DB \cdot DP}=\frac{ \delta \cdot BP^*}{DB^2}=\frac{ \delta \cdot CQ^*}{DC^2}=\frac{\delta \cdot AQ}{DC \cdot DQ}=\frac{[ACQ]}{[DCQ]}=\frac{AR_2}{DR_2} \Longrightarrow R_1 \equiv R_2 \equiv R $ ,

so we get $ BR\cdot RP=AR \cdot RD=CR \cdot RQ \Longrightarrow B, C, P, Q $ are concyclic .

Q.E.D
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andria
824 posts
#3 • 4 Y
Y by Virgil Nicula, Davi-8191, Adventure10, Mango247
Trivial problem for China
Solution: apply an inversion with center $D$ and arbitrary power $K$. Let $X'$ inverse of $X$ note that points $Q$ and $P$ are midpoints of arcs $ADC$ and $ADB$ of triangles $ADC$ and $ADB$ so $Q$' and $P'$ are feet of external angle bisectors of $A'DC'$ and $A'DB'$. also from the condition we have $1/DA'=1/DB'+1/DC'$ using this fact and external bisector theorem in triangles $A'DB'$ and $A'DC'$ with easy calculation we get $A'B'/A'P'=A'Q'/A'C'$ so $B'Q'||C'P'$ and quadrilateral $B'Q'P'C'$ is trapezoid so to prove the problem we have to show that B'A'=Q'A' from the condition that $AB=AC$ and external bisector theorem we get $DC'/DB'$=$Q'C'/Q'A'$×$P'A'/P'B'$=$A'C'/A'B'$ using this fact and $Q'A'B'\sim P'A'C'$ after an easy calculation we get the result that $A'Q'=A'B'$.
This post has been edited 1 time. Last edited by andria, Mar 27, 2015, 6:14 PM
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Wolstenholme
543 posts
#4 • 3 Y
Y by tenplusten, Adventure10, Mango247
It is clear that $ P $ is the midpoint of arc $ \overarc{BDA} $ of the circumcircle of $ \triangle{BDA} $ and similarly $ Q $ is the midpoint of $ \overarc{CQA} $ of the circumcircle of $ \triangle{CQA}. $ Consider the inversion with center $ D $ and arbitrary radius. I will denote inverted points by their original letter (hopefully the use will be clear). Because of the first observation, we have that $ C, A, Q $ and $ B, A, P $ are collinear (in that order). Now we proceed with four metric observations:

$ (1): \frac{1}{AD} = \frac{1}{BD} + \frac{1}{CD} $ - this follows from the given condition

$ (2): \frac{AB}{AC} = \frac{BD}{CD} $ - this follows from the fact that in the original diagram, $ \triangle{ABC} $ was isosceles.

$ (3): \frac{AP}{BP} = \frac{AD}{BD} $ - this follows from the fact that in the original diagram, $ \triangle{APB} $ was isosceles.

$ (4): \frac{AQ}{CQ} = \frac{AD}{CD} $ - this follows from the fact that in the original diagram, $ \triangle{AQC} $ was isosceles.

Now we want to show that quadrilateral $ BCPQ $ is cyclic, which is equivalent to showing that $ AC \cdot AQ = AB \cdot AP. $ But by (3) and (4) we have that $ \frac{AQ}{BP} = \frac{AB \cdot BP}{AC \cdot CQ} $ so by (2) it suffices to show that $ CQ = BP. $ But by (3) and (4) again and the facts that $ AC + AQ = CQ $ and $ AB + AP = BP $ we find that $ CQ = \frac{AB}{1 - \frac{AD}{CD}} $ and $ BP = \frac{AC}{1 - \frac{AD}{BD}} $ and now by using (1) and (2) we can simplify to obtain the desired result.
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Dukejukem
695 posts
#5 • 4 Y
Y by Devesh14, Adventure10, Mango247, Rounak_iitr
We begin with a well-known lemma that a few above have cited:
Lemma
Now, applying the lemma to the problem at hand, we find that $P, Q$ are the midpoints of arcs $\widehat{ADB}, \widehat{ADC}$ in $\odot(ADB), \odot(ADC)$, respectively. We will now prove that the lines $AD, BP, CQ$ are concurrent, whence Power of a Point will finish the proof. Let $X$ be the point on $\overline{AD}$ for which $AX = DB$, and let $PB, QC$ cut $AD$ at $P', Q'$, respectively. We will show that $AP / DP' = AQ' / DQ'$, whence it will follow that $P' \equiv Q'$; but first, a few synthetic observations: Since $DX = DA - AX = DC$, it follows that $\triangle DCX$ is isoceles. Furthermore, we have $\angle CDX = \angle CDA = \angle CQA$, so by side-angle-side similarity, $\triangle CDX \sim \triangle CQA.$ By spiral similarity, we also know that $\triangle CDQ \sim \triangle CXA.$ In addition, since $PA = PB, AX = DB, \angle PAX = \angle PAD = \angle PBD$, it follows by side-angle-side similarity that $\triangle PAX \sim \triangle PBD.$ By spiral similarity, we also have $\triangle PAB \sim \triangle PXD.$ Now, we are ready:

By the Ratio Lemma (Law of Sines) applied to $\triangle ADB, ADC$, it follows that \[\frac{AP'}{DP'} = \frac{BA}{BD} \cdot \frac{\sin\angle ABP}{\sin\angle DBP}, \quad \quad \quad \frac{AQ'}{DQ'} = \frac{CA}{CD} \cdot \frac{\sin\angle ACQ}{\sin\angle DCQ}.\] Keeping in mind that $BA = CA$, to prove that these two ratios are equal, we need only show that \[\frac{BD}{CD} = \frac{\sin\angle DCQ}{\sin\angle ACQ} \cdot \frac{\sin\angle ABP}{\sin\angle DBP}.\] Recalling our similar triangles in mind and applying the Ratio Lemma to $\triangle ADC$, we have \[\frac{BD}{CD} = \frac{AX}{DX} = \frac{CA}{CD} \cdot \frac{\sin\angle ACX}{\sin\angle DCX} = \frac{BA}{DX} \cdot \frac{\sin\angle DCQ}{\sin\angle ACQ}.\] Because $\triangle PAB \sim \triangle PXD$, one final calculation yields \[\frac{BA}{DX} = \frac{PB}{PD} = \frac{PA}{PD} = \frac{\sin\angle PDA}{\sin\angle PAD} = \frac{\sin\angle ABP}{\sin\angle DBP}.\] Hence, $AP' / DP' = AQ' / DQ'$, so $P' \equiv Q'$, as desired. Now, let us denote the intersection of $AD, BP, CQ$ by $Y.$ Then since $Y$ lies on the radical axis of $\odot(APDB)$ and $\odot(AQDC)$, it follows by Power of a Point that $YP \cdot YB = YQ \cdot YC.$ By Power of a Point once again, this implies that $B, C, P, Q$ are concyclic, as desired. $\square$
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yugrey
2326 posts
#6 • 5 Y
Y by TripteshBiswas, tenplusten, Adventure10, Mango247, Infinityfun
We easily see that $AQDC$ and $BDPA$ are cyclic.

Now, pick $Y$ on $AD$ with $DY=DC$, and note the angle $DCY$ is the complement of half the angle $ADC$ by the isosceles triangle. As angle $ACQ$ is the complement of half of $AQC$ and also of $ADC$, we have that in $DCA$ that $CQ$ and $CY$ are isogonal. So if $CQ$ meets $AD$ at $X$, we have $\frac {AY} {YD}\frac {AX} {XD}=\frac {AC} {DC}^2$.

So $\frac {AX} {XD}=(\frac {AC} {DC})(\frac {AC} {DC})(\frac {YD} {AY})=(\frac {AC} {DC})(\frac {AC} {DC})(\frac {DC} {DB})$, using $DA=DB+DC$. This works out to $\frac {(AB)(AC)} {(DB)(DC)}$. So then this expression is symmetric, and looking at where $BP$ meets $AD$ gives the same result. Thus $BP$, $CQ$, $AD$ meet at $X$. So then we have $(XB)(XP)=(XA)(XD)=(XQ)(XC)$ using POP and this finishes the problem.
This post has been edited 2 times. Last edited by yugrey, Apr 4, 2015, 6:31 PM
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JuanOrtiz
366 posts
#8 • 2 Y
Y by Adventure10, Mango247
With an inversion at D radius 1, the problem becomes: "Let 1/DC+1/DB=1/DA and AC/DC = AB/DB and P,Q be the intersections of AC, AB with the external bisector of ADC and ADB. Prove BPQC is cyclic".

Notice CP/PA=CD/DA=CD*(1/DC+1/DB)=1+(CD/DB) Thus CA/PA=CD/DB and this implies PA*CA=(CA^2)/(CD/DB) and similarly QA*BA=(BA^2)(CD/DB) and thus all that remains is to prove CA^2/(CD/DB)=BA^2(CD/DB), which is equivalent to AC/DC=AB/DB, which is given in the problem statement.
This post has been edited 1 time. Last edited by JuanOrtiz, Apr 18, 2015, 6:36 PM
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drmzjoseph
445 posts
#9 • 7 Y
Y by tranquanghuy7198, langkhach11112, Uagu, GGPiku, k12byda5h, Adventure10, Mango247
First off $Q$ is the midpoint of the arc $CDA$ because is unique point that satisfies this.

Let $K$ be a point on the extension of $CD$, such that $KD=DB \Rightarrow  \triangle KCQ \cong \triangle DAQ$.
So $Q$ is the spiral similarity of $KD \mapsto CA$ and send $B \mapsto X$, so $A$ is circumcenter of $\triangle BXC$

Now only angle-chasing; since $\triangle QBX \sim \triangle QDA$ we get $\angle QBX=\angle QCA$ if $N \equiv AC \cap BX ( BQNC$ is cyclical) so $\angle BNC=\angle BXC + \angle ACX =\frac{\angle BAC}{2}+\frac{\angle BDC}{2}= \angle BQC$ analogously $\angle BPC$ is equal.

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This post has been edited 1 time. Last edited by drmzjoseph, May 25, 2015, 12:26 AM
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tranquanghuy7198
253 posts
#10 • 5 Y
Y by langkhach11112, thedragon01, Adventure10, Mango247, Infinityfun
My solution:
The inversion $I_A$ gives us the equivalent problem: $\triangle{ABC}$ has $AB = AC$. $D$ is the point such that $DB+DC = AB = AC$. $P, Q$ are on $DB, DC$ such that $BP = BA = CA = CQ$. Prove that: $B, C, P, Q$ are concyclic.

Proof.
We have: $DP = BP-BD = BA-BD = DC$ and analogously, we have: $DQ = DB$
$\Rightarrow BCPQ$ is the isoceles trapezium, and the conclusion follows.
Q.E.D
Attachments:
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utkarshgupta
2280 posts
#11 • 4 Y
Y by langkhach11112, anantmudgal09, Adventure10, Mango247
It is well known that $P$ is the midpoint of the arc $ADB$ of $\odot ABD$
Simialrly, $Q$ is the midpoint of arc $ADC$ of $\odot ACD$

Now invert with centre $D$and radius $DA$.

Let the image of $X$ be denoted by $X'$.

Now since $A=A'$,
$P'$ lies on line $AB'$
Also, $Q'$ lies on line $AC'$

Now we have to prove that $B'C'P'Q'$ are concyclic.

That is we have to show that, $AB' \cdot AP' = AC' \cdot AQ'$


Now we know that
$AP' = \frac{AD \cdot AP}{PD}$,
$AQ' = \frac{AD \cdot AQ}{QD}$
$AC' = \frac{AD \cdot AC}{CD}$
$AB' = \frac{AD \cdot AB}{BD}$

Thus we are left to prove that
$$\frac{AP}{PD}\cdot \frac{AB}{BD} = \frac{AQ}{QD} \cdot \frac{AC}{CD}$$That is $$AP \cdot CD \cdot DQ = AQ \cdot PD \cdot BD$$
Applying Ptolemy's theorem to cyclic quadrilateral $APDB$,
$$AP \cdot BD + PD \cdot AB = AD \cdot BP$$Using, $DA = DB+DC$ and $PA=PB$,
$$PD \cdot AB = CD \cdot AP$$
Similarly,
$$QD \cdot AC = BD \cdot AQ$$
Since $AB=AC$,
$$\frac{PD}{QD}=\frac{CD \cdot AP}{BD \cdot AQ}$$That is $$AP \cdot CD \cdot DQ = BD \cdot AQ \cdot PD$$
QED
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XmL
552 posts
#12 • 2 Y
Y by parola, Adventure10
Lemma(Extention of Archemedes Midpoint Theorem): If $M$ is the midpoint of arc $BAC$, then \[\frac {MP}{MA}=\frac {|AB-AC|}{BC}\]
Main Proof: We quickly see that $ABDP, ACDQ$ are cyclic quadrilaterals. By radical axis theorem, $B,C,Q,P$ are concyclic $\iff QC,BP,AD$ are concurrent. By letting $BP\cap AD=E$, $CQ\cap AD=F$, it suffices to show \[\frac {DE}{AE}=\frac {DF}{AF}\].

Applying the lemma on $\triangle ADB$ and midpoint $P$, we have \[\frac {PD}{AP}=\frac {PD}{BP}=\frac {AD-DB}{AB}=\frac {DC}{AB}\]Hence we have \[\frac {DE}{AE}=\frac {BD}{AB}\cdot \frac {\sin \angle DBP}{\sin \angle ABP}=\frac {BD}{AB}\cdot \frac {PD}{AP}=\frac {BD\cdot DC}{AB^2}\]
By symmetry, we can similarly derive $\frac {DF}{AF}=\frac {BD\cdot DC}{AC^2}=\frac {BD\cdot DC}{AB^2}=\frac {DE}{AE}$, and the ratio equality is proven.
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WizardMath
2487 posts
#13 • 2 Y
Y by Adventure10, Mango247
Invert around $A$ with radius $AB$. Denote the inverse by a $^*$.
$D^*A$ = $AB.DB/AD$ etc imply $D^*B + D^*C = AB$.
We recognise $P$ as the midpoint of arc $ADB$, so it is on the intersection of it and the perpendicular bisector of $BC$. Thus its inverse is the intersection of $BD^*$ and circle with $B$ as center and $AB$ as radius.
Now we compute $D^*Q^* . D^*C = (AB-D^*C)D^*C = AB.D^*C - D^*C^2$. By a similar computation for the other product get $B, C, P^*, Q^*$ are concyclic. Inverting back, we have the conclusion.
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dagezjm
88 posts
#14 • 2 Y
Y by Adventure10, Mango247
先证明A、P、D、B四点共圆
过P做AD、BD垂线,垂足为X、Y
显然PX=PY、AP=BP、∠AXP=∠AYP
∴AXP、BYP全等
∠PAX=∠PBY
∴A、P、D、B共圆
同理 A、Q、D、C共圆
下证BP、CQ、AD共点
设BP交AD于T1,CQ交AD于T2
只需证AT1/T1D=AT2/T2D
延长DB至K使得KB=DC,连接AK
显然APB、ADK相似
立得APD、ABK相似
∴AT1/T1D=(AB/BD)*(sin∠ABT1/sin∠DBT1)=(AB/BD)*(sin∠ADP/sin∠PAD)=(AB/BD)*(AP/PD)=(AB/BD)*(AB/BK)=(AB*AC)/(BD*CD)
同理AT2/T2D=(AB*AC)/(BD*CD)
∴T1=T2(称这点为T)
∴BP、CQ、AD交于同一点T
∴BT*TP=AT*TD=CT*TQ
即B、P、C、Q四点共圆
Q.E.D.
This post has been edited 1 time. Last edited by dagezjm, Feb 22, 2017, 3:33 PM
Reason: 打错了
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anantmudgal09
1980 posts
#15 • 2 Y
Y by Adventure10, Mango247
I wonder if a solution along the lines of "...cevians $\overline{BP}, \overline{CQ}, \overline{DA}$ concur hence by radical axis..." exists.
dibyo_99 wrote:
$\triangle{ABC}$ is isosceles with $AB = AC >BC$. Let $D$ be a point in its interior such that $DA = DB+DC$. Suppose that the perpendicular bisector of $AB$ meets the external angle bisector of $\angle{ADB}$ at $P$, and let $Q$ be the intersection of the perpendicular bisector of $AC$ and the external angle bisector of $\angle{ADC}$. Prove that $B,C,P,Q$ are concyclic.

Invert at $D$. Then $D$ lies inside $\triangle A'B'C'$ with $\frac{A'B'}{A'C'}=\frac{DB'}{DC'}$ and $\frac{1}{DA'}=\frac{1}{DB'}+\frac{1}{DC'}$; external bisectors of angles $A'DB', A'DC'$ meet lines $A'B'$ and $A'C'$ again at $P',Q'$. It suffices to show $B',C',P',Q'$ are concyclic. Equivalently, we want $A'B' \cdot A'P'=A'C' \cdot A'Q'$.

By external bisector theorem, $\frac{A'P'}{A'B'}=\frac{A'D}{B'D-A'D}=\frac{DC'}{DB'}$ and similarly $\frac{A'Q'}{A'C'}=\frac{DB'}{DC'}$. The claim now follows.
This post has been edited 1 time. Last edited by anantmudgal09, Mar 25, 2018, 10:29 PM
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MarkBcc168
1595 posts
#16 • 2 Y
Y by Adventure10, Mango247
This problem is trivial with inversion but very nice to solve without it :). Here is my synthetic solution.

Clearly quadrilaterals $APDB$ and $AQDC$ are cyclic. Extend $DB$ beyond $B$ and $DC$ beyond $C$ to $X,Y$ respectively so that $DA=DX=DY$. Then $\triangle ADX\stackrel{+}{\sim}\triangle APB$ $\implies \triangle APD\stackrel{+}{\sim}\triangle ABX$. Similarly, we get $\triangle AQD\stackrel{+}{\sim}\triangle ACY$.

Now construct point $U$ such that $\triangle DAU\stackrel{+}{\sim} \triangle DPC$. Hence $\triangle DCU\stackrel{+}{\sim} \triangle DPA \sim\triangle XBA$. But $BX=DC$ so $\triangle DCU\cong\triangle XBA$ or $CU=CA$. Similarly if we construct point $V$ such that $\triangle DAV\stackrel{+}{\sim} \triangle DQB$, we also get $BV=BA$. Moreover,
$$\angle BPC = \angle BPD + \angle DPC = \angle BAD + \angle DAU = \angle BAU$$Similarly $\angle BQC = \angle CAV$. Hence it suffices to show that $\triangle ABV\cong\triangle ACU$. But
$$\angle ACU = \angle DCU - \angle ACD = \angle DPA - \angle ACD = 180^{\circ} - \angle ABD - \angle ACD$$which is symmetric w.r.t. $B,C$ so we are done.
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Arefe
65 posts
#17
Y by
First intersect perpendicular from A , B to AD , BD in K , perpendicular from A , C to AD , CD in J , perpendicular from B , C to BD , CD in I .
Bisector of K , J meet IJ , IK in M ,L .
It's easy to find that D is on the line LM (because AD=DB+DC)
We know that P , K , M and Q , J , L are collinear ( because ADBK , ADCJ are cyclic )
It's easy to see that DPMC , DDQLB are cyclic .
Now we have BPC=BAD+DMC , BQC=DAC+DLB .
we should just prove that BAD+DMC=DAC+DLB and it's equal to prove that JMD=KDL .
If LM intersect KJ at X , we know that XJ.KI=XK.IJ and we want to say that DJ.XK=DK.XJ so it's enough to prove IJ.DK=DJ.IK .
Because AC=AB we can easily prove IJ.DK=DJ.IK so the problem is solved :)
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Number1048576
91 posts
#18
Y by
Long solution but with some nice other properties of the config
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