Quadratic concurrence in circumscribed quad

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62861

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62861

#1 h Sep 6, 2018, 2:54 AM • 3 Y

Y by aopsuser305, Adventure10, Mango247

A convex quadrilateral $ABCD$ is circumscribed about a circle $\omega$ . Let $PQ$ be the diameter of $\omega$ perpendicular to $AC$ . Suppose lines $BP$ and $DQ$ intersect at point $X$ , and lines $BQ$ and $DP$ intersect at point $Y$ . Show that the points $X$ and $Y$ lie on the line $AC$ .

Géza Kós

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rmtf1111

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rmtf1111

#2 h Sep 6, 2018, 4:04 AM • 2 Y

Y by Adventure10, Mango247

Suppose that P is closer to D. Let I1, I2, J1, J2 be the incircle of ACD, incircle of ACB, D-excircle of ACD and B-excircle of ACB, respectively, by 2017G7, J1, J2 and AC are concurrent at U, and I1, I2 and AC are also concurrent. By homothety, D, P, U are colinear, so done.

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MarkBcc168

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MarkBcc168

#3 h Sep 6, 2018, 9:56 AM • 2 Y

Y by Adventure10, Mango247

Let $\omega$ touches $AB, BC, CD, DA$ at $E, F, G, H$ respectively. Clearly $EH, FG, BD, PQ$ are concurrent at pole $T$ of $AC$ . Let $BX, DX$ intersect $\omega$ at $U, V$ respectively. Notice that
$(EF;PU) = -1 = (GH;QV)$ so projecting through $T$ , points $U, V, T$ are colinear. Hence by Brokard's theorem we can conclude that $X=PU\cap QV$ lies on polar of $T$ w.r.t. $\omega$ .

This post has been edited 1 time. Last edited by MarkBcc168, Sep 6, 2018, 9:56 AM

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MilosMilicev

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MilosMilicev

#4 h Sep 7, 2018, 1:59 PM • 2 Y

Y by Adventure10, Mango247

Let $M,N,R,S$ be the tangency points on $AB,BC,CD,DA,$ respectively. Well-known that $MS, RN, PQ$ concur (at $E, AC$ is polar of $E$ , $B,D,E$ are also collinear because $BD$ is a radical axis of the circles $MIS,RIN$ ...). Also $AC,MP,QS$ concur and also $AC,MQ,PS$ because of well-known properties of the
cyclic quadrilateral $MQPS$ . The end follows by Desargues on $\Delta BQM,\Delta DPS$ and on $\Delta BPM,\Delta DQS$ .

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anantmudgal09

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anantmudgal09

#5 h Sep 9, 2018, 9:03 PM • 4 Y

Y by Wizard_32, mmathss, RodSalgDomPort, Adventure10

Notice that $P$ is the top-point of $\odot(I)$ and $\odot(I)$ is similar to incircle of $\triangle ABC$ under dilation at $B$ . Consequently, line $\overline{BP}$ passes through the touch-point of $B$ -excircle on $\overline{AC}$ . Likewise line $\overline{BQ}$ passes through $B$ -intouch point on $\overline{AC}$ . Finally, remark that $AB-BC=AD-DC$ ; concluding the proof.

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FISHMJ25

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FISHMJ25

#6 h Feb 18, 2019, 3:37 PM • 1 Y

Y by Adventure10

MarkBcc168 wrote:

Let $\omega$ touches $AB, BC, CD, DA$ at $E, F, G, H$ respectively. Clearly $EH, FG, BD, PQ$ are concurrent at pole $T$ of $AC$ . Let $BX, DX$ intersect $\omega$ at $U, V$ respectively. Notice that
$(EF;PU) = -1 = (GH;QV)$ so projecting through $T$ , points $U, V, T$ are colinear. Hence by Brokard's theorem we can conclude that $X=PU\cap QV$ lies on polar of $T$ w.r.t. $\omega$ .

Can you use perspectivity like that? I think that you have to have point $T$ on the circle ?

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rmtf1111

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rmtf1111

#7 h Feb 18, 2019, 4:21 PM • 1 Y

Y by Adventure10

FISHMJ25 wrote:

MarkBcc168 wrote:

Let $\omega$ touches $AB, BC, CD, DA$ at $E, F, G, H$ respectively. Clearly $EH, FG, BD, PQ$ are concurrent at pole $T$ of $AC$ . Let $BX, DX$ intersect $\omega$ at $U, V$ respectively. Notice that
$(EF;PU) = -1 = (GH;QV)$ so projecting through $T$ , points $U, V, T$ are colinear. Hence by Brokard's theorem we can conclude that $X=PU\cap QV$ lies on polar of $T$ w.r.t. $\omega$ .

Can you use perspectivity like that? I think that you have to have point $T$ on the circle ?

Yes you can, and no, $T$ does not have to be on circle if you project from circle to the same circle

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FISHMJ25

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FISHMJ25

#9 h Feb 18, 2019, 6:18 PM • 2 Y

Y by Adventure10, Mango247

rmtf1111 wrote:

FISHMJ25 wrote:

MarkBcc168 wrote:

Let $\omega$ touches $AB, BC, CD, DA$ at $E, F, G, H$ respectively. Clearly $EH, FG, BD, PQ$ are concurrent at pole $T$ of $AC$ . Let $BX, DX$ intersect $\omega$ at $U, V$ respectively. Notice that
$(EF;PU) = -1 = (GH;QV)$ so projecting through $T$ , points $U, V, T$ are colinear. Hence by Brokard's theorem we can conclude that $X=PU\cap QV$ lies on polar of $T$ w.r.t. $\omega$ .

Can you use perspectivity like that? I think that you have to have point $T$ on the circle ?

Yes you can, and no, $T$ does not have to be on circle if you project from circle to the same circle

My mistake , you just fix circle with inversion and thats it.

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TheUltimate123

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TheUltimate123

#10 h Jul 23, 2021, 12:35 AM

Y by

Solved with Jeffrey Chen, Max Lu, and Raymond Feng.

Without loss of generality $P$ is on the same side of $\overline{AC}$ as $B$ , and $Q$ is one the same side of $\overline{AC}$ as $D$ .

Let the incircles of $\triangle ABC$ and $\triangle ADC$ touch $\overline{AC}$ at a common point $T$ by Pitot. Then since $\overline{PP}\parallel\overline{QQ}\parallel\overline{AC}$ , by homothety we have $B$ , $T$ , $Q$ are collinear Analogously $D$ , $T$ , $P$ are collinear, so $T=\overline{BQ}\cap\overline{DP}\in\overline{AC}$ .

Analogously $\overline{BP}\cap\overline{DQ}\in\overline{AC}$ , as desired.

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Jalil_Huseynov

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Jalil_Huseynov

#11 h Aug 6, 2022, 1:58 PM • 1 Y

Y by farhad.fritl

Let $AB,BC,CD,DA$ touch to $\omega$ at $X,Y,Z,T$ . Let $K=TX\cap YZ, R=AC\cap PQ, F=QC\cap AD, G=PC\cap AB$ .
At first not that all poles and polars are taken wrt $\omega$ .It's well-known that $K\in BD$ . $RITX,RIYZ, TXYZ$ are cyclic, so from Radical Axis theroem $K\in PQ$ . $K$ lies on polar of $A$ and $C$ , so from La-Hire we get $AC$ is polar of $K$ . Applying Brokard's theorem on $PQXT$ , gives that $TQ, XP,AC$ are concurrent. Since $TQ\cap XP, TF\cap XG, FQ\cap GP$ are collinear, we get triangles $TFQ$ and $XGP$ are perspective. So $TX, FG, QP$ are concurrent, i.e $K,F,G$ are collinear. So $DBA$ and $QPC$ are perspective $\implies DQ\cap BP\in AC \implies X\in AC$ .
Since $AC$ is polar of $K$ , $(K,R;P,Q)=-1$ . Projecting it to $AC$ through $D$ and $B$ gives that $(BD\cap AC,R;DP\cap AC,X)=-1$ and $(BD\cap AC,R; BQ\cap AC,X)=-1$ , respectively. So $BQ\cap AC\equiv DP\cap BC \implies Y\in AC$ . So we are done.

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VicKmath7

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VicKmath7

#12 h Jan 13, 2023, 4:26 PM

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It is easy to see, using the length condition, that the incircles in $\triangle ACD$ and $\triangle ABC$ touch at a point $X$ on $AC$ . Notice that the tangent at $Q$ to $\omega$ is parallel to $AC$ , so the homothety at $D$ taking the incircle of $\triangle ACD$ to $\omega$ takes $X$ to $Q$ , so $X$ lies $DQ$ . Similarly $X$ lies on $BP$ . The same approach can be applied for $Y$ - the common point of the excircles of the two triangles.

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SomeonesPenguin

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SomeonesPenguin

#13 h Sep 16, 2024, 6:54 PM

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Here is a simple solution using Pascal and pole-polar duality.

Setup

The condition is equivalent to proving that the pole of $AC$ lies on the polar of $X$ . It is well know that the pole of $AC$ is $M$ . Now notice that $QQ$ is the polar of $Q$ and $HG$ is the polar of $D$ , hence $QQ\cap GH=\{T\}$ lies on the polar of $X$ and similarly, $PP\cap EF=\{S\}$ lies on the polar of $X$ . Therefore it suffices to prove that $M$ , $S$ and $T$ are collinear.

Note that since $AC$ is the polar of $M$ , we have that $M$ , $P$ and $Q$ are collinear by Brokard. Pascal on $QFGPHE$ gives $I$ , $J$ and $M$ collinear and Pascal on $QQFGHP$ gives $T$ , $I$ and $M$ collinear, hence $T$ lies on $\overline{M-I-J}$ . We can similarly deduce that $S$ lies on $\overline{M-I-J}$ , therefore $\overline{M-S-T}$ . $\blacksquare$

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