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jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
Factorials divide
va2010   37
N 3 minutes ago by ND_
Source: 2015 ISL N2
Let $a$ and $b$ be positive integers such that $a! + b!$ divides $a!b!$. Prove that $3a \ge 2b + 2$.
37 replies
va2010
Jul 7, 2016
ND_
3 minutes ago
IMO Shortlist 2011, Number Theory 2
orl   24
N 6 minutes ago by ezpotd
Source: IMO Shortlist 2011, Number Theory 2
Consider a polynomial $P(x) =  \prod^9_{j=1}(x+d_j),$ where $d_1, d_2, \ldots d_9$ are nine distinct integers. Prove that there exists an integer $N,$ such that for all integers $x \geq N$ the number $P(x)$ is divisible by a prime number greater than 20.

Proposed by Luxembourg
24 replies
orl
Jul 11, 2012
ezpotd
6 minutes ago
Inequality in triangle
Nguyenhuyen_AG   3
N 14 minutes ago by Nguyenhuyen_AG
Let $a,b,c$ be the lengths of the sides of a triangle. Prove that
\[\frac{1}{(a-4b)^2}+\frac{1}{(b-4c)^2}+\frac{1}{(c-4a)^2} \geqslant \frac{1}{ab+bc+ca}.\]
3 replies
Nguyenhuyen_AG
Today at 6:17 AM
Nguyenhuyen_AG
14 minutes ago
Problem 1
randomusername   73
N 26 minutes ago by ND_
Source: IMO 2015, Problem 1
We say that a finite set $\mathcal{S}$ of points in the plane is balanced if, for any two different points $A$ and $B$ in $\mathcal{S}$, there is a point $C$ in $\mathcal{S}$ such that $AC=BC$. We say that $\mathcal{S}$ is centre-free if for any three different points $A$, $B$ and $C$ in $\mathcal{S}$, there is no points $P$ in $\mathcal{S}$ such that $PA=PB=PC$.

(a) Show that for all integers $n\ge 3$, there exists a balanced set consisting of $n$ points.

(b) Determine all integers $n\ge 3$ for which there exists a balanced centre-free set consisting of $n$ points.

Proposed by Netherlands
73 replies
randomusername
Jul 10, 2015
ND_
26 minutes ago
No more topics!
Midpoint of base in isosceles triangle [<APM + <BPC = 180°]
warut_suk   22
N Jul 26, 2024 by Grasshopper-
Source: Poland National Olympiad 2000, Day 1, Problem 2
Let a triangle $ABC$ satisfy $AC = BC$; in other words, let $ABC$ be an isosceles triangle with base $AB$. Let $P$ be a point inside the triangle $ABC$ such that $\angle PAB = \angle PBC$. Denote by $M$ the midpoint of the segment $AB$. Show that $\angle APM + \angle BPC = 180^{\circ}$.
22 replies
warut_suk
Jan 25, 2005
Grasshopper-
Jul 26, 2024
Midpoint of base in isosceles triangle [<APM + <BPC = 180°]
G H J
Source: Poland National Olympiad 2000, Day 1, Problem 2
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warut_suk
803 posts
#1 • 3 Y
Y by Adventure10, jhu08, Mango247
Let a triangle $ABC$ satisfy $AC = BC$; in other words, let $ABC$ be an isosceles triangle with base $AB$. Let $P$ be a point inside the triangle $ABC$ such that $\angle PAB = \angle PBC$. Denote by $M$ the midpoint of the segment $AB$. Show that $\angle APM + \angle BPC = 180^{\circ}$.
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grobber
7849 posts
#2 • 5 Y
Y by Adventure10, jhu08, Mango247, and 2 other users
By an angle chase, we can see that $P$ moves on a circle tangent in $A,B$ to $CA,CB$ respectively. Let $Q$ be the second intersection of $CP$ with this circle. $PAQB$ is a harmonic cyclic quadrilateral, so, if $C'$ is the intersection between the tangent at $P$ to the circle and $AB$, we find $(PA,PB;PQ,PC')=-1$, meaning that $PQ$ is the symmedian of $PAB$, which is precisely what we want.

I was in a hurry while writing this, but I'll clarify if needed.
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darij grinberg
6555 posts
#3 • 3 Y
Y by Adventure10, jhu08, Mango247
Rewriting your problem with different notations:

Let ABC be an isosceles triangle with AB = AC. Let P be a point inside the triangle ABC such that < PBC = < PCA. Let M be the midpoint of the segment BC. Prove that < BPM + < APC = 180°.

Now, this problem was solved in posts #8 and #9 of http://www.mathlinks.ro/Forum/viewtopic.php?t=18258 (see the Lemma in post #8), with the only difference being that an additional assumption was used, namely the assumption that < PCB = < PBA; but this assumption is actually unnecessary, since it follows from the other assumptions of the problem (namely, since the triangle ABC is isosceles with AB = AC, we have < ACB = < ABC, and since < PBC = < PCA, we have < PCB = < ACB - < PCA = < ABC - < PBC = < PBA, so that < PCB = < PBA comes out as a consequence of the other assumptions of the problem).

Darij
This post has been edited 2 times. Last edited by darij grinberg, Mar 15, 2007, 9:35 AM
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Arne
3660 posts
#4 • 7 Y
Y by LoveMath4ever, r31415, Adventure10, jhu08, Mango247, and 2 other users
warut_suk wrote:
Let a triangle $ABC$ satisfy $AC = BC$; in other words, let $ABC$ be an isosceles triangle with base $AB$. Let $P$ be a point inside the triangle $ABC$ such that $\angle PAB = \angle PBC$. Denote by $M$ the midpoint of the segment $AB$. Show that $\angle APM + \angle BPC = 180^{\circ}$.

Take point Q such that APBQ is a parallellogram.

Take point R on AQ such that ACB, APR are similar. (I assume R lies between A and Q, the other case is similar) Then BPRQ is cyclic, and ACP, ABR are similar.

Hence BPQ = BRQ = 180 - ARB = 180 - APC. Hence result.
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Virgil Nicula
7054 posts
#5 • 7 Y
Y by LoveMath4ever, Abdollahpour, Adventure10, jhu08, Mango247, and 2 other users
$S\in AB\cap CP$, $CA=CB=a$, $AB=b$, $m(\widehat {PCA})=u$, $m(\widehat {PCB})=v$,
$m(\widehat {PAB})=m(\widehat {PBC})=x$, $m(\widehat {PAC})=m(\widehat {PBA})=y$.

$\blacktriangle$ The theorem of Sinus in the triangle $APB\Longrightarrow \frac{PA}{PB}=\frac{\sin y}{\sin x}\ \ (1)$.
$\blacktriangle$ The trigonometrical form of the Menelaus' theorem in
the triangle $ABC$ for the inner point $P\Longrightarrow$
$\sin u\sin^2x=\sin v\sin^2y$ $\Longrightarrow$ $\frac{\sin u}{\sin v}=\left(\frac{\sin y}{\sin x}\right)^2\ \ (2)$.
$\blacktriangle\ \frac{SA}{SB}=\frac{CA}{CB}\cdot \frac{\sin \widehat {ACS}}{\sin \widehat {BCS}}$ $\Longrightarrow$ $\frac{SB}{SC}=\frac{\sin u}{\sin v}\ \ (3).$

From the relations $(1)$, $(2)$, $(3)$ results: $\frac{SA}{SB}=\left(\frac{PA}{PB}\right)^2$, i.e. the ray $[PS$ is
the symmedian from the vertex $P$ in the triangle $APB$, meaning

$\widehat {APM}\equiv \widehat {BPS}\Longrightarrow$ $\boxed {\ m(\widehat {APM})+m(\widehat {BPC})=180^{\circ}\ }.$
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treegoner
637 posts
#6 • 2 Y
Y by Adventure10, jhu08
The problem can be generalized as follow (which is well-known)
Let $PAB$ be a triangle and $M$ be the midpoint of $AB$. Suppose $C$ is the intersection of the tangents of the circumcircle at $A$ and $B$. Then $\measuredangle{BPM}= \measuredangle{CPB}$.
[Moderator edit: This is basically the assertion of http://www.mathlinks.ro/Forum/viewtopic.php?t=99571 .]
Back to the problem , it is easy to see that $CA$ and $CB$ are the tangents of $(PAB)$ at $A$ and $B$.
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Virgil Nicula
7054 posts
#7 • 3 Y
Y by Adventure10, jhu08, Mango247
See http://www.mathlinks.ro/Forum/viewtopic.php?t=55581
http://www.mathlinks.ro/Forum/viewtopic.php?t=43202
where are mentioned the properties of the harmonical quadrilateral.
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zbghj
1 post
#8 • 3 Y
Y by Adventure10, jhu08, Mango247
Let us see the circumcircle of APB. Let v=<ABC.
Use the center of the circumcircle as origin of coordinate and give the coordinates as:

C (0,1)
M (0, (cos v)^2)
P (cos v*cos u, cos v*sin u)

It is easy to obtain: PM/PC=cos v

So, we can get: BC/PC=AM/PM.
So, <BPC=<APM or <BPC+<APM=180.

When <BPC=<APM, then <AMP=<BCP. So, <BPC=<BMC=90. As a result, <BPC+<APM=180 too.
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yetti
2643 posts
#9 • 3 Y
Y by Adventure10, jhu08, Mango247
P is on circle (U) tangent to BC at B and passing through A(then $ \angle PAB = \angle PBC$). $ \triangle ABC$ is isosceles $ \Longrightarrow$ (U) is also tangent to CA at A, AB is polar of C WRT the circle (U). CP meets (U) at P and again at Q. If P' is a reflection of P in $ CM \perp AB,$ $ P' \in (U)$ and CP' meets (U) again at Q'. PP'Q'Q is isosceles trapezod symmetrical WRT CM, the diagonals PQ', P'Q meet on CM and on the polar AB of C, $ M \equiv PQ' \cap P'Q \Longrightarrow$ $ \angle APM \equiv \angle APQ' = \angle BP'Q = \angle BPQ = 180^\circ - \angle BPC.$
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Leonhard Euler
247 posts
#10 • 3 Y
Y by Adventure10, jhu08, Mango247
We have $ \angle PBA=\angle PAC$. Let $ X$ be on $ PM$ such $ PAXB$ is parallogram. Then $ \angle XAB=\angle PAC,\angle XBA=\angle PBC$. Hence $ C,X$ are isogonal points wrt $ \triangle PAB$. Hence $ PC,PX$ is isogonal line $ \triangle PAB$. This imply $ \angle APM+\angle BPC=180$.
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Learner94
634 posts
#11 • 4 Y
Y by samirka259, jhu08, Adventure10, Mango247
Lemma: Let $ABC$ be a triangle and $\Gamma $ its circumcircle. Let the tangent to $\Gamma $ at $B$ and $C$ intersect at $D$. Then $AD$ coincides with the $A$- symmedian of $\triangle ABC$. The proof can be found here.

By angle chasing we can show that $AC$ and $BC$ are tangents to the circumcircle of $\triangle PAB$ at points $A$ and $B$ respectively. The tangents $AC$ and $BC$ intersect at $C$. So $PC$ coincides with the $C$- symmedian of $\triangle ABC$. Since $PM$ is the median of $\triangle ABC$, $\angle APM = \angle BPE$. or $\angle APM = 180^{\circ} - \angle BPC$. So $\angle APM +  \angle BPC = 180^{\circ} $.
Attachments:
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DaChickenInc
418 posts
#12 • 3 Y
Y by jhu08, Adventure10, Mango247
Translation of Official Solution
Official Solutions
Problem Significance
Attachments:
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Shauryajain123
152 posts
#13 • 2 Y
Y by AllanTian, jhu08
A nice problem for symmedians :)
We see that AC and AB are tangents to the circumcircle of $\triangle{APB} \implies CP $ is the symmedian.
Then the rest easily follows from how symmedians are defined
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PCChess
548 posts
#14 • 1 Y
Y by jhu08
Since $\triangle ABC$ is isosceles, we know that $\angle PAC=\angle PBA$. Consider $(APB)$. Since $\angle PAB=\angle PBC$ and $\angle PAC=\angle PBA$, lines $AC$ and $BC$ are tangent to $(APB)$. This implies that $PC$ lies on the $P$-symmedian. Let $CP$ intersect $AB$ at point $E$. By the definition of the symmedian, $\angle APM=\angle BPE$. Clearly $\angle BPE+\angle BPC=180^{\circ}$, so $\angle APM + \angle BPC = 180^{\circ}$.
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JustKeepRunning
2958 posts
#15 • 1 Y
Y by jhu08
Let $CP\cap AB=G$. Then the problem is equivalent to $\angle APM=\angle BPG$, aka. $CG$ is a symmedian in $\triangle APG$. Notice that $CB$ and $CA$ are the tangents to $(APB),$ so we are done.
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Mogmog8
1080 posts
#16 • 2 Y
Y by centslordm, jhu08
Since $\angle CBP=\angle BAP$ and $\angle PAC=\angle PBA,$ $C$ is the intersection of the tangents to $(PAB)$ at $A$ and $B.$ Hence, $\overline{PC}$ is the $P$-symmedian point of triangle $PAB.$ Notice that $$\measuredangle BPC=-\measuredangle(\overline{CP},\overline{PB})=-\angle APM.$$$\square$
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JustKeepRunning
2958 posts
#17 • 1 Y
Y by jhu08
Nice problem!

It is well known that $P$ is the $A$ humpty point, and so $M_1:=AP\cap BC$ is the midpoint of $BC$. Furthermore, we have that $\angle PCB=\angle PAC=\angle PBA$ and $\angle PAB=\angle PBC$. Hence, we get that $\angle APM+\angle BPC=\angle APM+\angle CPM_1+\angle M_1PB=\angle APM+\angle MPB+\angle BPM_1=180^{\circ},$ where this follows from similar triangles $\triangle PMB$ and $\triangle PM_1C$.

EDIT: oops just realized I already solved this before. At least this is a slightly different/more motivated solution.
This post has been edited 1 time. Last edited by JustKeepRunning, Dec 11, 2021, 1:48 AM
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Albert123
204 posts
#18 • 1 Y
Y by jhu08
Let $(APB) \cap CP=L$
As well: $CA$ and $CB$ are tangent of $(APB)$
Then: $PQ$ is symedian of $(APB)$
$\implies \angle APM=\angle LPB \implies \angle APM + \angle BPC=180$
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Mahdi_Mashayekhi
698 posts
#19
Y by
Let $CP$ meet circumcircle of $APB$ at S. Since $\angle PAB = \angle PBC$ then $CB$ is tangent to $APB$ and since $CB = CA$ then $SP$ is symmedian of $ASB$ so $\angle APM = \angle BPS$ and so $\angle APM + \angle BPC = \angle 180$.
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EulersTurban
386 posts
#20
Y by
Let $CP \cap AB = D$. The angle condition implies that $AC$ and $AB$ are tangent to the circumcircle of $APB$, then we know that $PD$ is the $P$-symmedian of $APB$, which implies that $\angle TPC = \angle APD = \angle MPB$ thus we have that $\angle BPC = \angle MPT$, which clearly gives us that $\angle APM + \angle BPC = 180$
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HamstPan38825
8869 posts
#21
Y by
Let $\overline{PM'}$ be a symmedian in triangle $BPC$, so it suffices to show that $A$ lies along the symmedian. But this is evident by the tangent intersection definition.
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Ianis
418 posts
#22
Y by
Complex


Synthetic
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Grasshopper-
10 posts
#23
Y by
Let CP intersect AB at N, the circumradius of tri. ACP(center O) and tri. BCP(center O') be r, r' respectively. We have to prove that CP is P -symmedian in tri. APB and to show that we need to prove AN/BN=AP^2/BP^2 ,
We have [tri. APN]/[tri. BPN] = AN/BN = [tri. APC]/[tri. BPC]=(AP × sin<CAP)/(BP × sin<CBP)
by sine rule,
sin<CAP/sin<CBP=r'/r, by angle chasing we will get that tri. OO'P and tri. APB are similar so, r'/r = AP/BP then, we get that AN/BN=AP^2/BP^2
so, PN is P-symmedian in tri. APB. This implies <APM = <BPN; <BPN + <CPB=180°=> <APM + <BPC=180° .
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