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Inequality involving square root cube root and 8th root
bamboozled   1
N 33 minutes ago by arqady
If $a,b,c,d,e,f,g,h,k\in R^+$ and $a+b+c=d+e+f=g+h+k=8$, then find the minimum value of $\sqrt{ad^3 g^4} +\sqrt[3]{be^3 h^4} + \sqrt[8]{cf^3 k^4}$
1 reply
bamboozled
3 hours ago
arqady
33 minutes ago
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Two equal angles
jayme   3
N an hour ago by jayme
Dear Mathlinkers,

1. ABCD a square
2. I the midpoint of AB
3. 1 the circle center at A passing through B
4. Q the point of intersection of 1 with the segment IC
5. X the foot of the perpendicular to BC from Q
6. Y the point of intersection of 1 with the segment AX
7. M the point of intersection of CY and AB.

Prove : <ACI = <IYM.

Sincerely
Jean-Louis
3 replies
jayme
May 2, 2025
jayme
an hour ago
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Parallelograms and concyclicity
Lukaluce   31
N 3 hours ago by Ihatecombin
Source: EGMO 2025 P4
Let $ABC$ be an acute triangle with incentre $I$ and $AB \neq AC$. Let lines $BI$ and $CI$ intersect the circumcircle of $ABC$ at $P \neq B$ and $Q \neq C$, respectively. Consider points $R$ and $S$ such that $AQRB$ and $ACSP$ are parallelograms (with $AQ \parallel RB, AB \parallel QR, AC \parallel SP$, and $AP \parallel CS$). Let $T$ be the point of intersection of lines $RB$ and $SC$. Prove that points $R, S, T$, and $I$ are concyclic.
31 replies
Lukaluce
Apr 14, 2025
Ihatecombin
3 hours ago
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Concurrency from isogonal Mittenpunkt configuration
MarkBcc168   17
N Today at 12:25 AM by Ilikeminecraft
Source: Fake USAMO 2020 P3
Let $\triangle ABC$ be a scalene triangle with circumcenter $O$, incenter $I$, and incircle $\omega$. Let $\omega$ touch the sides $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$ at points $D$, $E$, and $F$ respectively. Let $T$ be the projection of $D$ to $\overline{EF}$. The line $AT$ intersects the circumcircle of $\triangle ABC$ again at point $X\ne A$. The circumcircles of $\triangle AEX$ and $\triangle AFX$ intersect $\omega$ again at points $P\ne E$ and $Q\ne F$ respectively. Prove that the lines $EQ$, $FP$, and $OI$ are concurrent.

Proposed by MarkBcc168.
17 replies
MarkBcc168
Apr 28, 2020
Ilikeminecraft
Today at 12:25 AM
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Centroid, altitudes and medians, and concyclic points
BR1F1SZ   1
N Yesterday at 11:02 PM by sami1618
Source: Austria National MO Part 1 Problem 2
Let $\triangle{ABC}$ be an acute triangle with $BC > AC$. Let $S$ be the centroid of triangle $ABC$ and let $F$ be the foot of the perpendicular from $C$ to side $AB$. The median $CS$ intersects the circumcircle $\gamma$ of triangle $\triangle{ABC}$ at a second point $P$. Let $M$ be the point where $CS$ intersects $AB$. The line $SF$ intersects the circle $\gamma$ at a point $Q$, such that $F$ lies between $S$ and $Q$. Prove that the points $M,P,Q$ and $F$ lie on a circle.

(Karl Czakler)
1 reply
BR1F1SZ
Yesterday at 9:45 PM
sami1618
Yesterday at 11:02 PM
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Nordic 2025 P3
anirbanbz   8
N Yesterday at 10:33 PM by lksb
Source: Nordic 2025
Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$. Let $E$ and $F$ be points on the line segments $AC$ and $AB$ respectively such that $AEHF$ is a parallelogram. Prove that $\vert OE \vert = \vert OF \vert$.
8 replies
anirbanbz
Mar 25, 2025
lksb
Yesterday at 10:33 PM
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Aime type Geo
ehuseyinyigit   0
Yesterday at 9:04 PM
Source: Turkish First Round 2024
In a scalene triangle $ABC$, let $M$ be the midpoint of side $BC$. Let the line perpendicular to $AC$ at point $C$ intersect $AM$ at $N$. If $(BMN)$ is tangent to $AB$ at $B$, find $AB/MA$.
0 replies
ehuseyinyigit
Yesterday at 9:04 PM
0 replies
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IMO Shortlist 2011, G4
WakeUp   126
N Yesterday at 6:12 PM by NuMBeRaToRiC
Source: IMO Shortlist 2011, G4
Let $ABC$ be an acute triangle with circumcircle $\Omega$. Let $B_0$ be the midpoint of $AC$ and let $C_0$ be the midpoint of $AB$. Let $D$ be the foot of the altitude from $A$ and let $G$ be the centroid of the triangle $ABC$. Let $\omega$ be a circle through $B_0$ and $C_0$ that is tangent to the circle $\Omega$ at a point $X\not= A$. Prove that the points $D,G$ and $X$ are collinear.

Proposed by Ismail Isaev and Mikhail Isaev, Russia
126 replies
WakeUp
Jul 13, 2012
NuMBeRaToRiC
Yesterday at 6:12 PM
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<DPA+ <AQD =< QIP wanted, incircle circumcircle related
parmenides51   42
N Yesterday at 6:09 PM by AR17296174
Source: IMo 2019 SL G6
Let $I$ be the incentre of acute-angled triangle $ABC$. Let the incircle meet $BC, CA$, and $AB$ at $D, E$, and $F,$ respectively. Let line $EF$ intersect the circumcircle of the triangle at $P$ and $Q$, such that $F$ lies between $E$ and $P$. Prove that $\angle DPA + \angle AQD =\angle QIP$.

(Slovakia)
42 replies
parmenides51
Sep 22, 2020
AR17296174
Yesterday at 6:09 PM
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Help my diagram has too many points
MarkBcc168   28
N Yesterday at 6:06 PM by AR17296174
Source: IMO Shortlist 2023 G6
Let $ABC$ be an acute-angled triangle with circumcircle $\omega$. A circle $\Gamma$ is internally tangent to $\omega$ at $A$ and also tangent to $BC$ at $D$. Let $AB$ and $AC$ intersect $\Gamma$ at $P$ and $Q$ respectively. Let $M$ and $N$ be points on line $BC$ such that $B$ is the midpoint of $DM$ and $C$ is the midpoint of $DN$. Lines $MP$ and $NQ$ meet at $K$ and intersect $\Gamma$ again at $I$ and $J$ respectively. The ray $KA$ meets the circumcircle of triangle $IJK$ again at $X\neq K$.

Prove that $\angle BXP = \angle CXQ$.

Kian Moshiri, United Kingdom
28 replies
MarkBcc168
Jul 17, 2024
AR17296174
Yesterday at 6:06 PM
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A lot of circles
ryan17   8
N Yesterday at 6:05 PM by AR17296174
Source: 2019 Polish MO Finals
Denote by $\Omega$ the circumcircle of the acute triangle $ABC$. Point $D$ is the midpoint of the arc $BC$ of $\Omega$ not containing $A$. Circle $\omega$ centered at $D$ is tangent to the segment $BC$ at point $E$. Tangents to the circle $\omega$ passing through point $A$ intersect line $BC$ at points $K$ and $L$ such that points $B, K, L, C$ lie on the line $BC$ in that order. Circle $\gamma_1$ is tangent to the segments $AL$ and $BL$ and to the circle $\Omega$ at point $M$. Circle $\gamma_2$ is tangent to the segments $AK$ and $CK$ and to the circle $\Omega$ at point $N$. Lines $KN$ and $LM$ intersect at point $P$. Prove that $\sphericalangle KAP = \sphericalangle EAL$.
8 replies
ryan17
Jul 9, 2019
AR17296174
Yesterday at 6:05 PM
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geometry problem
Medjl   4
N Apr 10, 2025 by tigerBoss101
Source: Netherlands TST for IMO 2017 day 3 problem 1
A circle $\omega$ with diameter $AK$ is given. The point $M$ lies in the interior of the circle, but not on $AK$. The line $AM$ intersects $\omega$ in $A$ and $Q$. The tangent to $\omega$ at $Q$ intersects the line through $M$ perpendicular to $AK$, at $P$. The point $L$ lies on $\omega$, and is such that $PL$ is tangent to $\omega$ and $L\neq Q$.
Show that $K, L$, and $M$ are collinear.
4 replies
Medjl
Feb 1, 2018
tigerBoss101
Apr 10, 2025
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geometry problem
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Reveal topic tags
geometry
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G H BBookmark kLocked kLocked NReply
Source: Netherlands TST for IMO 2017 day 3 problem 1
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Medjl
757 posts
Medjl
#1 h  Feb 1, 2018, 2:43 PM • 3 Y
Y by Muradjl, Adventure10, Mango247
A circle $\omega$ with diameter $AK$ is given. The point $M$ lies in the interior of the circle, but not on $AK$. The line $AM$ intersects $\omega$ in $A$ and $Q$. The tangent to $\omega$ at $Q$ intersects the line through $M$ perpendicular to $AK$, at $P$. The point $L$ lies on $\omega$, and is such that $PL$ is tangent to $\omega$ and $L\neq Q$.
Show that $K, L$, and $M$ are collinear.
This post has been edited 1 time. Last edited by Medjl, Feb 1, 2018, 3:04 PM
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sbealing
308 posts
sbealing
#2 h  Feb 1, 2018, 4:48 PM • 3 Y
Y by lolm2k, AlastorMoody, Adventure10
Let $C=PM \cap AK$ and $O$ be the midpoint of $AK$ then as $\angle OCP=\angle OLP=\angle OQP=90^{\circ}$ we have $OCLPQ$ is cyclic.
$$\angle LCM=\angle LCP=\angle LQP=\angle LAQ=\angle LAM$$So $LACM$ is cyclic and hence $\angle ALM=\angle ACM=90^{\circ}$ so $K,L,M$ are colinear as $AK$ is a diameter.
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Cycle
79 posts
Cycle
#3 h  May 12, 2020, 3:58 AM
Y by
Redefine $L$ to be the intersection of $KM$ with $\omega$ so that it suffices to show $PL$ is tangent to the circle. Letting $D=PM\cap AK$ and $O$ the center, we have
$$\angle PQL=\angle QKM=\angle LAQ=\angle PDL$$by alternate segment and cyclic quadrilaterals $LMDA$, and $ALQK$. Hence $PQDL$ is cyclic. But $\angle QDM=\angle QKM$ implying $\angle LDQ=2\angle QKL=\angle LOQ$ so $LODQP$ is cyclic. Then $\angle OLP=180^\circ-\angle OQP=90^\circ $, proving the claim.
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llplp
191 posts
llplp
#4 h  Aug 13, 2020, 10:43 PM
Y by
We use complex numbers with $a = 1, k = -1$. We redefine $L,Q = \ell, q$ to be arbitrary points on $\omega$ and $p = \frac{2 \ell q}{\ell + q}$, $M = KL \cap AS$ and we wish to show $PM \perp AK$. From the intersection formula we have $m = \frac{2\ell q + \ell - q}{\ell + q}$. Finally $m - p = \frac{\ell - q}{\ell + q} \in i \mathbb{R}$, so $p-m$ is perpendicular to the real axis, i.e $AK$.
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tigerBoss101
5 posts
tigerBoss101
#5 h  Apr 10, 2025, 10:53 AM
Y by
Let $T = PM \cap AK$ and $R = PM \cap KQ$. As $\angle MTK = 90^\circ = \angle MQK$, we have that $MQKT$ is cyclic. Power of a Point gives $RM \cdot RT = RQ \cdot RK = RL \cdot RA$ which means $ALMT$ is cyclic, and $R$ is the radical center of the three circles. Since $AM \perp KR$ and $RM \perp AK$, we get that $M$ is the orthocenter of $\triangle AKR$, which means $KM \perp RA$, so $K$, $L$, $M$ are collinear as $KL \perp RA$.
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