Lines pass through a common point

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Lines pass through a common point

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Source: Baltic Way 2008, Problem 18

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April

#1 h Nov 23, 2008, 4:25 PM • 1 Y

Y by Adventure10

Let $AB$ be a diameter of a circle $S$ , and let $L$ be the tangent at $A$ . Furthermore, let $c$ be a fixed, positive real, and consider all pairs of points $X$ and $Y$ lying on $L$ , on opposite sides of $A$ , such that $|AX|\cdot |AY| = c$ . The lines $BX$ and $BY$ intersect $S$ at points $P$ and $Q$ , respectively. Show that all the lines $PQ$ pass through a common point.

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#2 h Nov 27, 2008, 6:37 PM • 4 Y

Y by Mathasocean, Adventure10, Mango247, and 1 other user

$[asy]defaultpen(fontsize(8)); pair A=(0,0), B=(10,0), X=(0,10), Y=(0,-5), H, P, Q; path S = Circle(midpoint(A--B),abs(A-B)/2); P = intersectionpoints(S, X--B)[1]; Q = intersectionpoints(S, Y--B)[1]; H = extension(P,Q,A,B); draw(S); draw(A--Y--B--X--A--P--Q--A--B); dot(P^^Q^^A^^B^^X^^Y^^H); label("A",A,(-1,0));label("B",B,(1,0));label("P",P,(1,1));label("Q",Q,(1,-1));label("X",X,(-1,0));label("Y",Y,(-1,0));label("H",H,(1,-1));[/asy]$ Let $H=PQ\cap AB$ .
Claim: As $X$ and $Y$ vary on $L$ such that $AX\cdot AY = c$ , $H$ remains fixed on $AB$ .
It is sufficient to show that $\frac{[PAQ]}{[PBQ]}$ remains constant.
$\frac{AX}{AB}=\frac{AP}{BP}$ , $\frac{AY}{AB}=\frac{AQ}{BQ}$
$\implies\frac{c}{AB^2}=\frac{AX\cdot AY}{AB^2}=\frac{AP\cdot AQ}{BP\cdot BQ}=\frac{\frac{1}{2}AP\cdot AQ\sin{PAQ}}{\frac{1}{2}BP\cdot BQ\sin{PBQ}}=\frac{[PAQ]}{[PBQ]}$

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#3 h Nov 27, 2008, 9:58 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

Let $X',\ Y'$ be, the points on the tangent line $L$ of the given circle $(O)$ at point $A,$ as the problem states and also such that $AX' = AY'$ $,(1)$

So, we have that $(AX)\cdot (AY) = u^{2} = (AX')^{2} = (AY')^{2}$ $,(2)$

We denote the points $P'\equiv (O)\cap BX',\ Q'\equiv (O)\cap BY'$ and let be the point $H\equiv AB\cap P'Q'.$

We will prove that the line-segment $PQ,$ always passes through the point $H.$

From the right angled triangle $\bigtriangleup ABX',$ we have that $(BP')\cdot (BX') = (AB)^{2}$ $,(3)$

Similarly from the right angled triangle $\bigtriangleup ABY',$ we have that $(BQ')\cdot (BY') = (AB)^{2}$ $,(4)$

From $(3),$ $(4),$ we conclude that the points $X',\ P',\ Q',\ Y'$ are concyclic, with their circumcircle so be it $(O'),$ which intersects the line-segment $AB,$ at points $M,\ N$ such that $(BP')\cdot (BX') = (BM)\cdot (BN) = (AB)^{2}$ $,(5)$ and $(AM)\cdot (AN) = u^{2}$ $,(6)$

By the same way as before, we can say that $(BP)\cdot (BX) = (AB)^{2} = (BQ)\cdot (BY)$ $,(7)$ from the right angled triangles $\bigtriangleup ABX,\ \bigtriangleup ABY.$

From $(5),$ $(7),$ we conclude that the points $X,\ P,\ M,\ N$ are concyclic, with circumcircle so be it $(O_{1}).$

Because of $(AM)\cdot (AN) = u^{2} = (AX)\cdot (AY)$ $,(8)$ we conclude that the point $Y,$ lies on $(O_{1}).$

The point $Q$ now, lies also on the circle $(O_{1}),$ because of $(7).$

That is, the circle $(O_{1}),$ always passes through the constant points $M,\ N$ and then, we conclude that the line segment $PQ,$ always passes through the fixed point $H\equiv AB\cap P'Q',$ as the radical center of the circles $(O),\ (O'),\ (O_{1})$ and the proof is completed.

Kostas Vittas.

Attachments:

t=241323.pdf (12kb)

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#4 h Oct 3, 2024, 11:45 AM

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Consider inversion around $A$ with radius $\sqrt{AX\times AY}$
Now, we need to solve the following problem:

Quote:

Let $A$ be a point on line $l$ , and let $B$ be a point such that $AB\perp l$ . let $X,Y$ be points such that $AX\times AY=c$ for some positive real constant. If circles $(ABY),(ABX)$ intersect the line through $B$ parallel to $l$ at $P,Q$ , show that $(APQ)$ passes through a fixed point.

let $Z=(APQ)\cap AB$ . We claim $Z$ is the fixed point. Note that $XYPQ$ is a rectangle, so we have that $PB\times BQ=c$ . Thus, by power of a point $Z$ is the point such that $AB\times BZ=c$ , which is independent of $X,Y$ , as desired.

This post has been edited 2 times. Last edited by Lil_flip38, Oct 3, 2024, 11:46 AM

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#5 h Apr 7, 2025, 4:15 PM

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Nice solutions above, mine is most contest like one. We assume here $A,B$ are fixed and $R$ be the radi of circle.

Let's say $\angle PBA=a$ and $\angle QBA=b$ . Then we have $tg(a)*tg(b)$ is constant. Let $Z=AB \cap PQ$ . It's suffices to prove that length of $CZ$
is constant. In $\triangle CZP$ , by the law of sines we have $CZ=\frac{R*sin(90-b-a)}{sin(90-a+b)}=\frac{R*cos(a+b)}{cos(a-b)}=\frac{cos(a)cos(b)-sin(a)sin(b)}{cos(a)cos(b)+sin(a)sin(b)}=\frac{1-tg(a)tg(b)}{1+tg(a)tg(b)}$ , which is clearly constant.

This post has been edited 1 time. Last edited by Nari_Tom, Apr 7, 2025, 4:25 PM

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