there are two circles intersecting each at two points

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there are two circles intersecting each at two points

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Source: VietNam TST 2004, problem 3

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orl

#1 h May 9, 2004, 12:35 PM • 4 Y

Y by AhirGauss, DashTheSup, Adventure10, Mango247

In the plane, there are two circles $\Gamma_1, \Gamma_2$ intersecting each other at two points $A$ and $B$ . Tangents of $\Gamma_1$ at $A$ and $B$ meet each other at $K$ . Let us consider an arbitrary point $M$ (which is different of $A$ and $B$ ) on $\Gamma_1$ . The line $MA$ meets $\Gamma_2$ again at $P$ . The line $MK$ meets $\Gamma_1$ again at $C$ . The line $CA$ meets $\Gamma_2$ again at $Q$ . Show that the midpoint of $PQ$ lies on the line $MC$ and the line $PQ$ passes through a fixed point when $M$ moves on $\Gamma_1$ .

[Moderator edit: This problem was also discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=21414 .]

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#2 h May 9, 2004, 4:49 PM • 3 Y

Y by AhirGauss, Adventure10, Mango247

The pencil of lines through $K$ determines an involution on the pencil of lines through $A$ . I'm referring to the mapping $AM\leftrightarrow AC$ . This involution is hyperbolic (it has two fixed elements, the lines $AB,\ AK$ ).

Take $D=AK\cap \Gamma_2$ and let $L$ be the intersection point of the tangents to $\Gamma_2$ in $B$ and $D$ . The pencil of lines through $L$ determines an involution on the pencil of lines through $A$ , having fixed elements $AD=AK$ and $AB$ , so it's the same involution as the one in the first paragraph, so it maps $AP=AM$ to $AQ=AC$ , so $PQ$ passes through $L$ , which is one of the conclusions we were trying to reach.

I didn't show that the midpoint of $PQ$ is on $MC$ . I'll keep trying

.

I hope I didn't make any stupid mistakes up there

.

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psykyk

#3 h May 9, 2004, 8:54 PM • 3 Y

Y by AhirGauss, Adventure10, Mango247

hi

your idea seems great... though i can't understand what do you mean by pencils and involutions, could yo explain a little or tell me where can i find the theoretical foundations to understand it?

also i found solution using rotations and homotheties, denote rotomothety as a composition of a rotation and a homothety, then M and C are carried through a rotomothety centered at B to P and Q respectively, from there it follows that PQ passes through the antirotomothetical point of K.

Also for the second part we can use Menelaus in triangle APQ and line KCM, so that it rests to prove something about A C Q and M which also should follow from the rotomothety

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#4 h May 10, 2004, 1:56 AM • 3 Y

Y by AhirGauss, Adventure10, Mango247

"A pencil of lines through $A$ " is just the fancy name for the set of lines passing through $A$

.

An involution on this set is a function defined on the pencil with values in the same set, which maps the line $d$ to the line $d'$ and vice-versa (meaning that $f\circ f=1_P$ , where $P$ is the pencil) s.t. $(d_1,d_2;d_3,d_4)=(d'_1,d'_2;d'_3,d'_4)$ for any four lines $d_i\in P$ .

However, your solution seems to be much shorter

.

Also, I think there's a standard name for your "rotomothety"

. It's usually called "spiral similarity" or "similitude".

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psykyk

#5 h May 10, 2004, 2:11 AM • 3 Y

Y by AhirGauss, Adventure10, Mango247

it's a very interesting transformation!
but what's (d1,d2;d3,d4) ? double ratio or something like that?

psykyk

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#6 h May 10, 2004, 2:23 AM • 3 Y

Y by AhirGauss, Adventure10, Mango247

It's the cross-ratio. Assume the four lines $d_i$ cut another line in the points $A,B,C,D$ . Then $(A,B;C,D)=(\frac{\overline{CA}}{\overline{CB}}):(\frac{\overline{DA}}{\overline{DB}})$ , where $\overline{XY}$ is the oriented segment $XY$ .

When we change the line which cuts $d_i$ this value doesn't change, so we can speak about the cross-ratio of the four lines (this happens for four concurrent lines).

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2125 posts

#7 h Jun 9, 2019, 4:28 PM • 2 Y

Y by AhirGauss, Adventure10

Excellent Problem!

Vietnam TST 2004 P3 wrote:

In the plane, there are two circles $\Gamma_1, \Gamma_2$ intersecting each other at two points $A$ and $B$ . Tangents of $\Gamma_1$ at $A$ and $B$ meet each other at $K$ . Let us consider an arbitrary point $M$ (which is different of $A$ and $B$ ) on $\Gamma_1$ . The line $MA$ meets $\Gamma_2$ again at $P$ . The line $MK$ meets $\Gamma_1$ again at $C$ . The line $CA$ meets $\Gamma_2$ again at $Q$ . Show that the midpoint of $PQ$ lies on the line $MC$ and the line $PQ$ passes through a fixed point when $M$ moves on $\Gamma_1$ .

Solution: Let $AK \cap \Gamma_2=E$ . Let the tangents to $\Gamma_2$ at $B,E$ intersect at $D$
$-1=(M, ~C ; ~ A, ~B) \overset{A}{=} (P, ~Q ; ~E, ~B)$ Hence, $PQ$ passes through the fixed point $D$ . Let $\odot (BDE) \cap PQ=F$ . Let $BF \cap \Gamma_2=L$
$\implies \angle BLE=\angle BAE=\angle BED=\angle BFD \implies FD||EL$ Hence,
$-1=(B, ~E ; ~Q, ~P) \overset{L}{=} (F, ~ \infty_{PQ} ; ~Q, ~P) \implies F \text{ is the midpoint of } PQ$ $\angle DEB=\angle BAK \implies \Delta BAK \stackrel{+}{\sim} \Delta BED$ $\implies$ $K \in \odot (BDE)$ $\implies$ $\angle BKD=\angle BED=\angle BAK$ $\implies$ $FD$ tangent to $\odot (BAK)$ ,
$\angle PFL=\angle BFD=\angle BKD=\angle BAK=\angle BMP \implies MBFP \text{ is cyclic}$ Hence, $\angle BAC=\angle BPF=\angle BMF \implies$ $\overline{MCK}$ bisects $PQ$

This post has been edited 5 times. Last edited by AlastorMoody, Jun 9, 2019, 4:32 PM

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1107 posts

#8 h Jun 9, 2019, 8:43 PM • 2 Y

Y by AhirGauss, Adventure10

Solution
EDIT-This is completely wrong just a round about :wallbash:

:wallbash:

This post has been edited 3 times. Last edited by Pluto1708, Jun 9, 2019, 8:55 PM

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#9 h Jul 7, 2019, 4:18 PM • 2 Y

Y by Adventure10, Mango247

Here is my solution for this problem
Solution
Let $O_1$ , $O_2$ be center of $\Omega_1$ , $\Omega_2$ , respectively; $N$ , $U$ be midpoint of $PQ$ , $CM$ , respectively
We have: $(QP; QB) \equiv (AP; AB) \equiv (AM; AB)$ (mod $\pi$ ) and $(PB; PQ) \equiv (AB; AQ) \equiv (AB; AC) \equiv (MB; MC)$ (mod $\pi$ )
Then: $\triangle{BCM}$ $\stackrel{+}{\sim}$ $\triangle{BQP}$
But: $U$ , $N$ are midpoint of $CM$ , $PQ$ , respectively so: $\triangle{BCU}$ $\stackrel{+}{\sim}$ $\triangle{BQN}$
Hence: $(NB; NQ) \equiv (UB; UC) \equiv (MB; MA) \equiv (CB; CA) \equiv (CB; CQ)$ (mod $\pi$ ) or $B$ , $C$ , $N$ , $Q$ lie on a circle
Then: $(CM; CN) \equiv (CM; CB) + (CB; CN) \equiv (QP; QB) + (CB; CN) \equiv 0$ (mod $\pi$ ) or $CM$ passes through midpoint $N$ of $PQ$
Let $S$ be a point which satisfies $KS$ $\perp$ $O_1O_2$ and $BS$ tangents $(O_2)$ at $B$
We have: $(SK; SB) \equiv (BA; BS) \equiv (PA; PB) \equiv (O_1O_2; O_2B)$ (mod $\pi$ ) and $(KB; KO_2) \equiv (KB; KO_1) \equiv \dfrac{\pi}{2} - (O_1K; O_1B) \equiv \dfrac{\pi}{2} - (MA; MB) \equiv \dfrac{\pi}{2} - (CA; CB) \equiv \dfrac{\pi}{2} - (CQ; CB)$ $\equiv \dfrac{\pi}{2} - (NQ; NB) \equiv (NB; NO_2)$ (mod $\pi$ )
So: $B$ , $K$ , $S$ , $O_2$ , $N$ lie on a circle
But: $BS$ $\perp$ $BO_2$ , $BK$ $\perp$ $O_1O_2$ then: $O_2N$ $\perp$ $SN$
Combine with: $O_2N$ $\perp$ $NQ$ , we have: $P$ , $N$ , $Q$ , $S$ are collinear
Hence: $PQ$ passes through fixed point $S$

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