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3 lines concurrent with 1 circle
parmenides51   4
N an hour ago by LeYohan
Source: 2021 Irish Mathematical Olympiad P8
A point $C$ lies on a line segment $AB$ between $A$ and $B$ and circles are drawn having $AC$ and $CB$ as diameters. A common tangent to both circles touches the circle with $AC$ as diameter at $P \ne C$ and the circle with $CB$ as diameter at $Q \ne C$.
Prove that $AP, BQ$ and the common tangent to both circles at $C$ all meet at a single point which lies on the circumference of the circle with $AB$ as diameter.
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parmenides51
May 30, 2021
LeYohan
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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
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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
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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
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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
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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
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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
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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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