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2024 BxMO P3
beansenthusiast505   4
N 3 minutes ago by GeorgeMetrical123
Source: 2024 BxMO P3
Let $ABC$ be a triangle with incentre $I$ and circumcircle $\Omega$ such that $\left|AC\right|\neq\left|BC\right|$. The internal angle bisector of $\angle CAB$ intersects side $BC$ at $D$ and the external angle bisectors of $\angle ABC$ and $\angle BCA$ intersect $\Omega$ at $E$ and $F$ respectively. Let $G$ be the intersection of lines $AE$ and $FI$ and let $\Gamma$ be the circumcircle of triangle $BDI$. Show that $E$ lies on $\Gamma$ if and only if $G$ lies on $\Gamma$.
4 replies
beansenthusiast505
Apr 28, 2024
GeorgeMetrical123
3 minutes ago
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Length Condition on Circumcenter Implies Tangency
ike.chen   41
N 7 minutes ago by ravengsd
Source: ISL 2022/G4
Let $ABC$ be an acute-angled triangle with $AC > AB$, let $O$ be its circumcentre, and let $D$ be a point on the segment $BC$. The line through $D$ perpendicular to $BC$ intersects the lines $AO, AC,$ and $AB$ at $W, X,$ and $Y,$ respectively. The circumcircles of triangles $AXY$ and $ABC$ intersect again at $Z \ne A$.
Prove that if $W \ne D$ and $OW = OD,$ then $DZ$ is tangent to the circle $AXY.$
41 replies
ike.chen
Jul 9, 2023
ravengsd
7 minutes ago
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Probably appeared before
steven_zhang123   2
N 20 minutes ago by whwlqkd
In the plane, there are two line segments $AB$ and $CD$, with $AB \neq CD$. Prove that there exists and only exists one point $P$ such that $\triangle PAB \sim \triangle PCD$.($P$ corresponds to $P$, $A$ corresponds to $C$)
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steven_zhang123
Today at 2:29 AM
whwlqkd
20 minutes ago
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A metrical relation in a cyclical quadrilateral.
Virgil Nicula   5
N Mar 10, 2016 by PROF65
PP. Let a convex $ABCD$ what is inscribed in the circle $w$ and $P\in AC\cap BD$ . For a point $M\in (AB)$

denote $N\in MP\cap CD$ . Prove that $\boxed{\left(PM^2+MA\cdot MB\right)\left(PN^2+NC\cdot ND\right)=PA\cdot PB\cdot PC\cdot PD}$ .
5 replies
Virgil Nicula
Mar 9, 2016
PROF65
Mar 10, 2016
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A metrical relation in a cyclical quadrilateral.
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Virgil Nicula
7054 posts
Virgil Nicula
#1 h  Mar 9, 2016, 1:53 PM • 2 Y
Y by Adventure10, Mango247
PP. Let a convex $ABCD$ what is inscribed in the circle $w$ and $P\in AC\cap BD$ . For a point $M\in (AB)$

denote $N\in MP\cap CD$ . Prove that $\boxed{\left(PM^2+MA\cdot MB\right)\left(PN^2+NC\cdot ND\right)=PA\cdot PB\cdot PC\cdot PD}$ .
This post has been edited 2 times. Last edited by Virgil Nicula, Mar 9, 2016, 1:58 PM
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uvwmethod
43 posts
uvwmethod
#2 h  Mar 9, 2016, 3:27 PM • 1 Y
Y by Adventure10
Hint:Easy by applying steward to $\triangle PAB$ and $\triangle PCD$ and using the fact $PA*PB=PC*PD$
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Virgil Nicula
7054 posts
Virgil Nicula
#3 h  Mar 10, 2016, 1:38 AM • 2 Y
Y by Adventure10, Mango247
Why is easily and $PA\cdot PB=PC\cdot PD$ ?!
This post has been edited 2 times. Last edited by Virgil Nicula, Mar 10, 2016, 1:39 AM
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suli
1498 posts
suli
#4 h  Mar 10, 2016, 3:52 AM • 2 Y
Y by Adventure10, Mango247
$PA * PB = PC * PD$ is false. $PA * PC = PB * PD$ is true by Power of a Point.

But instead of listening to the Steward, let's try a synthetic solution.

Let the circumcircles of $DPC$ and $APB$ intersect line $MN$ at $R$ and $S$ respectively. Then
$$\angle DRP = \angle DCP = \angle DBM,$$so $DRBM$ is cyclic. Thus $\angle DP \cdot PB = RP \cdot PM$. Similarly, $ASCN$ is cyclic, so $AP \cdot PC = SP \cdot PN$.

Thus
$$AP \cdot BP \cdot CP \cdot DP = NP \cdot RP \cdot MP \cdot SP = (PN^2 + PN \cdot NR)(PM^2 + PM \cdot MS) = (PN^2 + ND \cdot NC)(PM^2 + MA \cdot MB)$$by use of Power of a Point.
This post has been edited 1 time. Last edited by suli, Mar 10, 2016, 4:29 AM
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Virgil Nicula
7054 posts
Virgil Nicula
#5 h  Mar 10, 2016, 2:02 PM • 2 Y
Y by Adventure10, Mango247
@ Suli. Thank you very much for your very nice proof! See PP11 and its easy extension from here
This post has been edited 1 time. Last edited by Virgil Nicula, Mar 10, 2016, 2:03 PM
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PROF65
2016 posts
PROF65
#6 h  Mar 10, 2016, 6:16 PM • 2 Y
Y by Adventure10, Mango247
Virgil Nicula wrote:
PP. Let a convex $ABCD$ what is inscribed in the circle $w$ and $P\in AC\cap BD$ . For a point $M\in (AB)$

denote $N\in MP\cap CD$ . Prove that $\boxed{\left(PM^2+\underline{AM}\cdot MB\right)\left(PN^2+\underline{CN}\cdot ND\right)=PA\cdot PB\cdot PC\cdot PD}$ .
the distances are signed .
@ Suli. your approch is good apart from ,i think just typo , the last equality which should be $(PN^2+DN\cdot NC)(PM^2+AM\cdot MB)$
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