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Three concurrent circles
jayme   4
N 41 minutes ago by jayme
Source: own?
Dear Mathlinkers,

1. ABC a triangle
2. 0 the circumcircle
3. Tb, Tc the tangents to 0 wrt. B, C
4. D the point of intersection of Tb and Tc
5. B', C' the symmetrics of B, C wrt AC, AB
6. 1b, 1c the circumcircles of the triangles BB'D, CC'D.

Prove : 1b, 1c and 0 are concurrents.

Sincerely
Jean-Louis
4 replies
jayme
Yesterday at 3:08 PM
jayme
41 minutes ago
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angle relations in a convex ABCD given, double segment wanted
parmenides51   12
N 43 minutes ago by Nuran2010
Source: Iranian Geometry Olympiad 2018 IGO Intermediate p2
In convex quadrilateral $ABCD$, the diagonals $AC$ and $BD$ meet at the point $P$. We know that $\angle DAC = 90^o$ and $2 \angle ADB = \angle ACB$. If we have $ \angle DBC + 2 \angle ADC = 180^o$ prove that $2AP = BP$.

Proposed by Iman Maghsoudi
12 replies
parmenides51
Sep 19, 2018
Nuran2010
43 minutes ago
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USAMO 2014 Day 1
liberator   5
N Dec 27, 2014 by liberator
Even though I am not an American citizen, I decided to try several USAMO questions for practice. Here is my work for day 1.

Problem 1: Let $a$, $b$, $c$, $d$ be real numbers such that $b-d \ge 5$ and all roots $x_1, x_2, x_3,x_4$ of the polynomial $P(x)=x^4+ax^3+bx^2+cx+d$ are real. Find the smallest value the product $(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)$ can take.

[b]My solution[/b]

Problem 2: Let $\mathbb{Z}$ be the set of integers. Find all functions $f : \mathbb{Z} \rightarrow \mathbb{Z}$ such that \[xf(2f(y)-x)+y^2f(2x-f(y))=\frac{f(x)^2}{x}+f(yf(y))\] for all $x, y \in \mathbb{Z}$ with $x \neq 0$.

[b]My solution[/b]

Problem 3: Prove that there exists an infinite set of points \[ \dots, \; P_{-3}, \; P_{-2},\; P_{-1},\; P_0,\; P_1,\; P_2,\; P_3,\; \dots \] in the plane with the following property: For any three distinct integers $a,b,$ and $c$, points $P_a$, $P_b$, and $P_c$ are collinear if and only if $a+b+c=2014$.

[b]My solution[/b]
5 replies
liberator
Aug 18, 2014
liberator
Dec 27, 2014
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PAMO Problem 4: Perpendicular lines
DylanN   11
N Apr 23, 2025 by ATM_
Source: 2019 Pan-African Mathematics Olympiad, Problem 4
The tangents to the circumcircle of $\triangle ABC$ at $B$ and $C$ meet at $D$. The circumcircle of $\triangle BCD$ meets sides $AC$ and $AB$ again at $E$ and $F$ respectively. Let $O$ be the circumcentre of $\triangle ABC$. Show that $AO$ is perpendicular to $EF$.
11 replies
DylanN
Apr 9, 2019
ATM_
Apr 23, 2025
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PAMO Problem 4: Perpendicular lines
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Source: 2019 Pan-African Mathematics Olympiad, Problem 4
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DylanN
194 posts
DylanN
#1 h  Apr 9, 2019, 6:54 AM • 2 Y
Y by Adventure10, Mango247
The tangents to the circumcircle of $\triangle ABC$ at $B$ and $C$ meet at $D$. The circumcircle of $\triangle BCD$ meets sides $AC$ and $AB$ again at $E$ and $F$ respectively. Let $O$ be the circumcentre of $\triangle ABC$. Show that $AO$ is perpendicular to $EF$.
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pad
1671 posts
pad
#3 h  Apr 9, 2019, 7:06 AM • 2 Y
Y by Adventure10, Mango247
We know $AD$ is the $A$-symmedian of $\triangle ABC$, so it is well known that $B,C,D,O$ are concyclic. So $(BCD)=(BCO)$. Then $\angle EAO = \angle CAO=90-B$, and $\angle AEF=180-\angle CEF=180-\angle CBF=B$ since $E,C,F,B$ are concyclic. Therefore, $AO\perp EF$.
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DylanN
194 posts
DylanN
#4 h  Apr 9, 2019, 7:29 AM • 2 Y
Y by Adventure10, Mango247
It's not important that $D$ is defined as the intersection of the tangents. We can replace the circumcircle of $\triangle BCD$ with any circle that passes through $B$ and $C$. Indeed, we know that $\angle CAO = 90^\circ - \angle B$, and since $BCEF$ is cyclic, we have that $\angle B = \angle CEF$, from which the result follows.
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khanhnx
1618 posts
khanhnx
#5 h  Apr 9, 2019, 9:35 AM • 1 Y
Y by Adventure10
This problem has been posted in here
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AshAuktober
1007 posts
AshAuktober
#6 h  Mar 5, 2024, 12:29 PM
Y by
Observe that we can obtain $\Delta AEF$ by reflecting $\Delta ABC$ about its $A-$angle bisector and taking a homothety at $A$. Now since the circumcentre and orthocentre are isogonal conjugates, $O$ has to lie on the $A-$altitude of $\Delta AEF$, and therefore we are done. $\square$
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Aiden-1089
293 posts
Aiden-1089
#7 h  Mar 6, 2024, 2:06 AM
Y by
Let $H$ be the orthocenter of $ABC$.
$\measuredangle BEC = \measuredangle BDC = \measuredangle BFC \implies BCEF$ is concyclic $\implies \Delta ABC \sim \Delta AEF$.
Consider taking a suitable homothety about A then reflecting across the angle bisector of $\angle BAC$, such that $BC$ goes to $EF$. Under this transformation, line $AH$ goes to line $AO$ because they are isogonal in $\angle BAC$. Since $AH \perp BC$, we have $AO \perp EF$.
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Jishnu4414l
155 posts
Jishnu4414l
#8 h  Apr 23, 2024, 4:01 PM
Y by
We know that $\measuredangle OAC=90^{\circ}-\measuredangle CBA=\measuredangle KAE$
Also, $\measuredangle CBA=\measuredangle CEF=\measuredangle AEK$
Now, $\measuredangle KAE+\measuredangle AEK+\measuredangle EKA=0$
$\implies 90^{\circ}-\measuredangle CBA+\measuredangle CBA +\measuredangle EKA=0$
$\implies \measuredangle EKA=90^{\circ}$
And we are done!
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fasttrust_12-mn
118 posts
fasttrust_12-mn
#9 h  Jul 19, 2024, 1:26 AM
Y by
Claim: $O$ lies on the circumcircle of $\triangle BDC$
proof:
$$\measuredangle BAC =\measuredangle BCD \implies \measuredangle BDC= 180^{\circ}- 2\times \measuredangle BAC$$, we know that
$$\measuredangle BOC = 2\times \measuredangle BAC$$so
$$OBDC \text{ is cyclic quad }$$
$$\measuredangle FBC=\measuredangle AEF $$$$\measuredangle OAC=\measuredangle OCA$$$$\measuredangle OCB=90^{\circ}$$we know that
$$\measuredangle AEG=90^{\circ}-\measuredangle BAO-\measuredangle OAE + \measuredangle BAO\implies 90^{\circ}-\measuredangle OAE$$$$\measuredangle AGE =180^{\circ}-(90^{\circ}-\measuredangle OAE+\measuredangle OAE)=90^{\circ}$$hence we are done $\blacksquare$
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Tony_stark0094
69 posts
Tony_stark0094
#10 h  Apr 11, 2025, 12:48 PM
Y by
$$\angle EBF=\angle BAF+\angle AFB=\angle A + \angle BDC=\angle A + 180 - 2\angle A = 180 - \angle A$$$$\implies FA=FB$$$$by\ symmetry\ we \ can\ get \ that\ EA=EC$$hence $F$ lies on perpendicular bisector of $AB$ and $E$ lies on perpendicular bisector of $AC$ moreover $G$ also lies on these lines
$\implies G$ is the orthocentre of $\Delta AEF$ hence $AG$ is perpendicular to $EF$
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Rayanelba
15 posts
Rayanelba
#11 h  Apr 23, 2025, 11:04 PM • 1 Y
Y by ATM_
To prove that $AO$ perpendicular to $EF$ , it suffies to prove that: $OAC=\frac{\pi}{2}-\measuredangle CFE$
We know that : $\measuredangle OAC=\frac{\pi}{2}-\measuredangle CBA$
And : $\measuredangle CBA=\pi-\measuredangle CBE=\measuredangle CFE$ (because CFEB is cyclic)
$\implies \measuredangle OAC=\frac{\pi}{2}-\measuredangle CFE$
Q.E.D
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ATM_
23 posts
ATM_
#12 h  Apr 23, 2025, 11:07 PM
Y by
nice solution*
This post has been edited 1 time. Last edited by ATM_, Apr 23, 2025, 11:08 PM
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ATM_
23 posts
ATM_
#13 h  Apr 23, 2025, 11:10 PM
Y by
here is a cute sketch
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