Stay ahead of learning milestones! Enroll in a class over the summer!

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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 2, 2025
0 replies
Beautiful geometry
m4thbl3nd3r   1
N 5 minutes ago by m4thbl3nd3r
Let $\omega$ be the circumcircle of triangle $ABC$, $M$ is the midpoint of $BC$ and $E$ be the second intersection of $AM$ and $\omega$. Tangent line of $\omega$ at $E$ intersects $BC$ at $P$, let $PKL$ be a transversal of $\omega$ and $X,Y$ be intersections of $AK,AL$ with $BC$. Let $PF$ be a tangent line of $\omega$. Prove that $LYFP$ is cyclic
1 reply
m4thbl3nd3r
Today at 4:41 PM
m4thbl3nd3r
5 minutes ago
Another factorisation problem
kjhgyuio   1
N 10 minutes ago by martianrunner
........
1 reply
kjhgyuio
20 minutes ago
martianrunner
10 minutes ago
integers inequality
azzam2912   0
40 minutes ago

3. Let $a, b, c, d, x, y$ be positive integers that satisfy the inequality
\[
\frac{a}{b} < \frac{x}{y} < \frac{c}{d}
\]with $bc - ad = 5$. If $b + d = 2025$, determine the minimum value of $y$.


0 replies
azzam2912
40 minutes ago
0 replies
MONT pg 31 example 1.10.40
Jaxman8   0
2 hours ago
Can somebody explain why it works.
0 replies
Jaxman8
2 hours ago
0 replies
No more topics!
Nine points, three pentagons, and one triangle in space
Shu   2
N Feb 6, 2012 by anonymouslonely
Source: XVIII Olimpíada Matemática Rioplatense (2009)
Let $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, $I$ be nine points in space such that $ABCDE$, $ABFGH$, and $GFCDI$ are each regular pentagons with side length $1$. Determine the lengths of the sides of triangle $EHI$.
2 replies
Shu
Jul 23, 2011
anonymouslonely
Feb 6, 2012
Nine points, three pentagons, and one triangle in space
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Source: XVIII Olimpíada Matemática Rioplatense (2009)
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Shu
316 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, $I$ be nine points in space such that $ABCDE$, $ABFGH$, and $GFCDI$ are each regular pentagons with side length $1$. Determine the lengths of the sides of triangle $EHI$.
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vanstraelen
8967 posts
#2 • 2 Y
Y by Adventure10, Mango247
Picture of this three pentagons

$\triangle FCB$ is equilateral; $F(0,0,0),C(0,1,0),B(-\frac{\sqrt{3}}{2},\frac{1}{2},0)$.
$G(x_{G},y_{G},z_{G}); \angle GFC = 108^{o}; y_{G}= \cos 108^{o}$.
$G'(x_{G},y_{G},0)$ lies on the bisector of $\angle BFC$ with equation $ y=-\sqrt{3}x \rightarrow x_{G}=-\frac{\cos 108^{o}}{\sqrt{3}}$.
$x^{2}_{G}+y^{2}_{G}+z^{2}_{G}=1 \rightarrow z_{G}=\sqrt{1-\frac{4\cos^{2} 108^{o}}{3}}$
Equation of the plane through $F,C$ and $G : x \cdot \sqrt{1-\frac{4\cos^{2} 108^{o}}{3}} + z \cdot \frac{\cos 108^{o}}{\sqrt{3}}=0$.
Normal of the $xy-$plane: $(0,0,1)$; normal of the $FCG-$plane:$(\sqrt{1-\frac{4\cos^{2} 108^{o}}{3}},0,\frac{\cos 108^{o}}{\sqrt{3}})$.
Angle $\theta$ between both planes:
\[\cos \theta = \left|\frac{\frac{\cos 108^{o}}{\sqrt{3}}}{\sqrt{1-\frac{4\cos^{2} 108^{o}}{3}+\frac{\cos^{2} 108^{o}}{3}}}\right|=\left|\frac{1}{\sqrt{3}\tan 108^{o}}\right|=\frac{\tan 18^{o}}{\sqrt{3}}\]
Midpoint $M$ of $\left[FC\right]: MI = \frac{1}{2} \cot 18^{o}$.
$II' \bot xy-$plane, $I'\in xy-$plane: $MI' = MI \cdot \cos \theta = \frac{\sqrt{3}}{6}$.
$O$ centre of the circumcircle of $\triangle FCB : MO = \frac{\sqrt{3}}{6}$.

Circumcircle of $\triangle HIE$ with radius $R=OI' = \frac{\sqrt{3}}{3} \rightarrow \triangle HIE$ equilateral with side $HI=IE=EH=1$.
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anonymouslonely
1142 posts
#3 • 2 Y
Y by Adventure10, Mango247
the quadrilaterals $ HFCE,IFBE,IHBC $ are parallelograms because have a pair of opposite sides parallel and congruent.
from this we obtain immediately that $ EH=IH=HE=1 $.
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