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Imtersecting two regular pentagons
Miquel-point   1
N an hour ago by Edward_Tur
Source: KoMaL B. 5093
The intersection of two congruent regular pentagons is a decagon with sides of $a_1,a_2,\ldots ,a_{10}$ in this order. Prove that
\[a_1a_3+a_3a_5+a_5a_7+a_7a_9+a_9a_1=a_2a_4+a_4a_6+a_6a_8+a_8a_{10}+a_{10}a_2.\]
1 reply
Miquel-point
4 hours ago
Edward_Tur
an hour ago
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P,Q,B are collinear
MNJ2357   28
N an hour ago by Ilikeminecraft
Source: 2020 Korea National Olympiad P2
$H$ is the orthocenter of an acute triangle $ABC$, and let $M$ be the midpoint of $BC$. Suppose $(AH)$ meets $AB$ and $AC$ at $D,E$ respectively. $AH$ meets $DE$ at $P$, and the line through $H$ perpendicular to $AH$ meets $DM$ at $Q$. Prove that $P,Q,B$ are collinear.
28 replies
MNJ2357
Nov 21, 2020
Ilikeminecraft
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Easy geo
oVlad   3
N Apr 21, 2025 by Primeniyazidayi
Source: Romania EGMO TST 2019 Day 1 P1
A line through the vertex $A{}$ of the triangle $ABC{}$ which doesn't coincide with $AB{}$ or $AC{}$ intersectes the altitudes from $B{}$ and $C{}$ at $D{}$ and $E{}$ respectively. Let $F{}$ be the reflection of $D{}$ in $AB{}$ and $G{}$ be the reflection of $E{}$ in $AC{}.$ Prove that the circles $ABF{}$ and $ACG{}$ are tangent.
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oVlad
Apr 21, 2025
Primeniyazidayi
Apr 21, 2025
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Source: Romania EGMO TST 2019 Day 1 P1
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oVlad
1746 posts
oVlad
#1 h  Apr 21, 2025, 1:45 PM
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A line through the vertex $A{}$ of the triangle $ABC{}$ which doesn't coincide with $AB{}$ or $AC{}$ intersectes the altitudes from $B{}$ and $C{}$ at $D{}$ and $E{}$ respectively. Let $F{}$ be the reflection of $D{}$ in $AB{}$ and $G{}$ be the reflection of $E{}$ in $AC{}.$ Prove that the circles $ABF{}$ and $ACG{}$ are tangent.
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kokcio
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kokcio
#2 h  Apr 21, 2025, 3:14 PM
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Let $X$ be intersection of $AF$ and $CE$. Then we have that $\angle AXC = \angle XEA =\pi - \angle AEC = \pi - \angle AGC$. Hence, $AXCG$ is cyclic. We can also see that $\angle ACX = \angle ABD = \angle ABF$, so by theorem about angle between chord and tangent, we know that circles on $AXC$, $ABF$ have common tangent.
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Gggvds1
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Gggvds1
#3 h  Apr 21, 2025, 3:37 PM
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Let the line passing through the vertex A intersect BC at M and angleBAM is greater than angleCAM.
Let angle BAM be x ,then angleFAB become x and angle CAM becomes A-x. It is easy to see that angle FBA is equals to angle GCA are 90-A. Then angle AFB is 90+A-x and angleAGC is 90+x.
Let l be a line tangent to circumcircle of triangle AGC. Let P be a point on l ahead of A in direction of G. Then angleGAF becomes 90 -A. Also angle BAF is angle BAC +angle GAC+angle GAF. Therefore angle BAF becomes 90+A-x that is equal to angle BFA.

Hence the line l is also tangent to circumcircle of triangle BFA. This implies that these two circle are tangent to each other.
This post has been edited 1 time. Last edited by Gggvds1, Apr 21, 2025, 3:38 PM
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Primeniyazidayi
106 posts
Primeniyazidayi
#4 h  Apr 21, 2025, 3:37 PM
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Let H be the intersection of BD and AG. Note that H is the reflection of D on AC. Then by angle chasing it can be proven that (AFBH) exists(A, F, B, H are concyclic). Let X be the intersection point of BD and the line which is tangent to (AFB)and $H_C$ be foot of perpendicular line of C. Then $$\angle XAH=\angle ABH=\angle ACH_C=\angle ACG$$done.
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