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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
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0 replies
jlacosta
Apr 2, 2025
0 replies
ARMO 2025
Oksutok   0
5 minutes ago
All-Russian MO 2025 Day 1
0 replies
Oksutok
5 minutes ago
0 replies
general form
pennypc123456789   1
N 7 minutes ago by sqing
If $a,b,c$ are positive real numbers, $k \ge 3$ then
$$
\frac{a + b}{a + kb + c} + \dfrac{b + c}{b + kc + a}+\dfrac{c + a}{c + ka + b} \geq \dfrac{6}{k+2}$$
1 reply
pennypc123456789
3 hours ago
sqing
7 minutes ago
JBMO Shortlist 2023 A4
Orestis_Lignos   6
N 9 minutes ago by Novmath
Source: JBMO Shortlist 2023, A4
Let $a,b,c,d$ be positive real numbers with $abcd=1$. Prove that

$$\sqrt{\frac{a}{b+c+d^2+a^3}}+\sqrt{\frac{b}{c+d+a^2+b^3}}+\sqrt{\frac{c}{d+a+b^2+c^3}}+\sqrt{\frac{d}{a+b+c^2+d^3}} \leq 2$$
6 replies
Orestis_Lignos
Jun 28, 2024
Novmath
9 minutes ago
2 var inquality
sqing   5
N 21 minutes ago by sqing
Source: Own
Let $ a,b\geq 0 $ and $ a+ b+2ab=4 . $ Prove that
$$ 3(a+b-2)(2 -  ab) \ge (a-1)(b-1)(a-b)$$$$ 9 (a+b-2)(3 - 2ab) \ge 2\sqrt 5(a-1)(b-1)(a-b)$$$$9(a+b-2)(6 - 5ab) \ge2\sqrt {14} (a-1)(b-1)(a-b)$$
5 replies
sqing
Today at 1:50 AM
sqing
21 minutes ago
No more topics!
Yet another cyclic quadrilateral
Matematika   16
N Dec 12, 2024 by lelouchvigeo
Source: European Mathematical Cup 2012, Senior Division, Problem 2
Let $ABC$ be an acute triangle with orthocenter $H$. Segments $AH$ and $CH$ intersect segments $BC$ and $AB$ in points $A_1$ and $C_1$ respectively. The segments $BH$ and $A_1C_1$ meet at point $D$. Let $P$ be the midpoint of the segment $BH$. Let $D'$ be the reflection of the point $D$ in $AC$. Prove that quadrilateral $APCD'$ is cyclic.

Proposed by Matko Ljulj.
16 replies
Matematika
Jul 27, 2013
lelouchvigeo
Dec 12, 2024
Yet another cyclic quadrilateral
G H J
Source: European Mathematical Cup 2012, Senior Division, Problem 2
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Matematika
139 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $ABC$ be an acute triangle with orthocenter $H$. Segments $AH$ and $CH$ intersect segments $BC$ and $AB$ in points $A_1$ and $C_1$ respectively. The segments $BH$ and $A_1C_1$ meet at point $D$. Let $P$ be the midpoint of the segment $BH$. Let $D'$ be the reflection of the point $D$ in $AC$. Prove that quadrilateral $APCD'$ is cyclic.

Proposed by Matko Ljulj.
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pepat
20 posts
#2 • 3 Y
Y by Adventure10, Mango247, and 1 other user
I suppose segments should be lines. The following solution is probably not the easiest one but anyway:

Solution
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sunken rock
4384 posts
#3 • 2 Y
Y by Adventure10, Mango247
Let $B_1=BH\cap AC$. It is known that $BB_1$ is the bisector of $\angle A_1B_1C_1$ and $A_1B_1C_1P$ is cyclic (and all those points lie onto the 9-point circle of $\Delta ABC$), hence $B_1D\cdot B_1P=A_1B_1\cdot C_1B_1\ (\ 1\ )$. This comes easily from 2 similar triangles.
But $AC$ is diameter of $CAC_1A_1$ and $A_1B_1$ is symmetrical of $B_1C_1$ w.r.t. $BB_1$, hence $C_1B_1$ intersect circle $\odot (CAC_1A_1)$ at $C_2$ and $B_1C_2=B_1A_1$, consequently $(1)$ becomes $B_1D\cdot B_1P=B_1C_1\cdot B_1C_2$, but from power of point $B_1$ w.r.t. $(CAC_1A_2)$ we have $B_1C_1\cdot B_1C_2=B_1C\cdot B_1A$; under the circumstances $B_1C\cdot B_1A=B_1D\cdot B_1P=B_1D'\cdot B_1P$, done.

Hope I did not mess-up the indices.

Best regards,
sunken rock
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Panoz93
61 posts
#4 • 3 Y
Y by electrovector, Adventure10, Mango247
It suffices to prove that $D$ is the orthocenter of triangle $APC$. For this just notice that $AD,DC$ are the polars of $C,A$ with respect to the circumcircle of $C_{1}HA_{1}B$.
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bojler
16 posts
#5 • 2 Y
Y by Adventure10, Mango247
Let $BB_1$ be the third altitude in triangle $ABC$. Note that triangles $AB_1H$ and $BB_1C$ are similar thus
$AB_1\cdot B_1C=B_1H\cdot B_1B$ .
$H(B,H,D,B_1)$ so $PH^2=PD\cdot PB_1$ .
$B_1P\cdot B_1D'=B_1P\cdot B_1D=B_1P\cdot (B_1P-DP)=PB_1^2-PH^2=B_1H\cdot B_1B=AB_1\cdot B_1C$.
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jayme
9775 posts
#6 • 2 Y
Y by Adventure10, Jacob._07
Dear Mathlinkers,
for #, see http://www.artofproblemsolving.com/community/c6h1069884_orthocenter

Sincerely
Jean-Louis
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orestis100
1 post
#7 • 2 Y
Y by Adventure10, Mango247
Can someone please prove why H(B,H,D,B1)?
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AlastorMoody
2125 posts
#8 • 2 Y
Y by Pluto1708, Adventure10
Apply Brokard's Theorem on $\odot (BHC_1A_1) $, $D $ is the orthocenter of $\Delta PCA $, hence the conclusion
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itslumi
284 posts
#9 • 1 Y
Y by Maxito12345
AlastorMoody wrote:
Apply Brokard's Theorem on $\odot (BHC_1A_1) $, $D $ is the orthocenter of $\Delta PCA $, hence the conclusion

Same solution,Can this be done whith inverzion???
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jayme
9775 posts
#10
Y by
Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%207.pdf p. 73...

Sincerely
Jean-Louis
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Olympikus
87 posts
#11
Y by
orestis100 wrote:
Can someone please prove why H(B,H,D,B1)?

Take perspective at $C_1$ and map to $AC$. Use the Ceva-Menelaus lemma : $H(A,C;B_1,A_1C_1\cap AC)$.
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Mikeglicker
258 posts
#12 • 3 Y
Y by Mango247, Mango247, Mango247
Why showing that D is the orthocenter of APC is suffices?
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Mikeglicker
258 posts
#13
Y by
Mikeglicker wrote:
Why showing that D is the orthocenter of APC is suffices?

Anybody?
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arzhang2001
248 posts
#14
Y by
do you ignore it:
if we let $M$ Be the midpoint of $A_1C_1$ Then should $MDAC$ be Cyclic.
This post has been edited 3 times. Last edited by arzhang2001, Jun 29, 2020, 7:11 AM
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miltosk
9 posts
#15
Y by
Mikeglicker wrote:
Mikeglicker wrote:
Why showing that D is the orthocenter of APC is suffices?

Anybody?

Because then $$DAC=DPC$$as acute angles with perpendicular sides and since $$DAC=D'AC$$due to the isosceles triangle $$DAD'$$, $$DPC=D'AC$$and then its obvious.
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iman007
270 posts
#17
Y by
$\text{Brokard kills this problem}$ :D
$\text{basically we should prove that} \thickspace \angle{ADC}=180^{\circ} -\angle{APC}$
$\text{which means to prove that} \thickspace P\thickspace \text{is the orthocenter of the triangle} \thickspace ADC$
$\textbf{Brokard theorem} \thickspace \text{on}\thickspace (A_1BC_1H) \thickspace\text{finishes the problem}$
\(\boxed{\mathcal{Q.E.D}}\)
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lelouchvigeo
179 posts
#18
Y by
Nice
Since $ACD$ are autopolar by Brocard's and $P$ is the center of circle $BA_1HC_1.$ We can observe that $D$ is the orthocenter of $\triangle ACP.$
Since the reflection of the orthocenter lies on the circle. We are done
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