ZI^2 = ZQ x ZB if A, C, Y, X are concyclic, tangent ot incircle related

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ZI^2 = ZQ x ZB if A, C, Y, X are concyclic, tangent ot incircle related

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Source: 2011 Thailand TST 13.2

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#1 h Oct 31, 2020, 6:09 AM • 1 Y

Y by Mango247

Let $I$ be the incenter of $\vartriangle ABC$ , and let $\ell$ be a tangent line to the incircle of $\vartriangle ABC$ , which is not one of its sides. The line $\ell$ meets the sides $AB, BC$ and the extension of $CA$ at points $X, Y$ and $Z$ respectively. Let $AY$ intersect $CX$ at $P$ , and let lines $IP$ and $BZ$ intersect at $Q$ . Prove that if $A, C, Y, X$ are concyclic, then $ZI^2 = ZQ \cdot ZB$ .

This post has been edited 2 times. Last edited by parmenides51, Jan 22, 2021, 12:38 PM
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#2 h Jan 5, 2021, 12:25 AM

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Supercalafragalistic nice one

$\color{black}\rule{25cm}{1pt}$
First let's add some points.
Let the tangency points with the incircle at $BC,CA$ and $AB$ be $D,E$ and $F$ , respectively.
Let $H$ be the intersection of $TF$ and $DE$ , let $K$ be the intersection of $FE$ and $TD$ and let $O$ be the circumcenter of $(ACYX)$ .

Now we shall use the following lemma.
$\color{red}\rule{25cm}{0.5pt}$
Lemma: The points $K,B,H$ and $Z$ are all on the same line.
Proof:
We use Pascal's theorem on $FFTDDE$ to get that $B,H$ and $K$ are colinear, then we use it on $TTDEEF$ , to get that $Z,H$ and $K$ are colinear.
As a nice corollary we have that $O,I$ and $P$ are colinear since of the Miquel point of $(TDEF)$ and $(ACYX)$ , which is by definition the point $Q$ and we must have that $IP \perp BC$ .
$\color{red}\rule{25cm}{0.5pt}$

The next lemma we are going to use in proving a later lemma in the solution.
$\color{red}\rule{25cm}{0.5pt}$
Lemma: We have that $FD \perp TE$ .
Proof:
Let $\alpha =\angle XAC$ .

Since we have that $ACYX$ is cyclic we have that $\angle XYC = 180 - \alpha$ and since we have that $ITYD$ is cyclic as well, we must have that $\angle TID = \alpha$ .
This implies that $\angle TFD = \frac{1}{2}\alpha$ .

As well we have that $AFIE$ is cyclic we have that $\angle FIE = 180 - \alpha$ , this implies that $\angle FTE = 90 - \frac{1}{2}\alpha$

Since we have that $\angle FTE = \angle FTP$ and since $\angle TFD = \angle TFP$ by angle calculation we have that $\angle TPF= 90$ .
Thus our claim holds true.
$\color{red}\rule{25cm}{0.5pt}$

Now our final lemma is going to kill the problem
$\color{red}\rule{25cm}{0.5pt}$
Lemma: $\angle BIZ = 90$
Proof:
Let's call an inversion around the incircle with $\varphi$ .
Let $B'$ and $Z'$ be the inverse images of $B$ and $Z$ under $\varphi$ , respectively.
Since $B'$ is the midpoint of $FD$ and $Z'$ is the midpoint of $TE$ , this implies that $\angle IB'P=\angle IZ'P = 90$ , which implies that $IB'PZ'$ is cyclic.
Because of the second lemma we have that $\angle B'PZ' = 90$ , this implies that $\angle B'IZ' = 90$ , this implies that $\angle BIZ =90$
$\color{red}\rule{25cm}{0.5pt}$

Now let's use these lemmas to solve the problem.

Since we have that $IP \perp BC$ and $O,I,P$ are colinear points we must have that $Q$ is the Miquel point of $ACYX$ .
Because of our third lemma we have that $\angle BIZ = 90$ , this implies a direct similarity between $ZIQ$ and $BIZ$ , this gives us:
$\frac{ZI}{ZQ}=\frac{ZB}{ZI}$ thus we have that $ZI^2 = ZQ.ZB$ .

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#3 h Jan 7, 2021, 12:17 AM

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or just use brianchon to get the very very very wellknown configuration.

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Krits

#4 h Jan 22, 2021, 12:35 PM • 3 Y

Y by Mango247, Mango247, Mango247

Denote $\ell(G)$ the polar of point $G$ wrt incircle.Let $\ell(A)\cap\ell(Y)=\{J\}$ then $P\in\ell(J)\to J\in\ell(P)=\overleftrightarrow{BZ}$ ,let $\ell(C)\cap \ell(X)=\{T\}$ then $P\in\ell(T)\to T\in\ell(P)=\overleftrightarrow{BZ}.$
Invert about incircle we see that $Q\equiv P^{*}$ and $\overleftrightarrow{IQ}\perp \overleftrightarrow{BZ}$ so its enough to show that $\angle BIZ=90^{\circ}$ which is true because $\measuredangle (\ell(B),\ell(Z))=90^{\circ}$ from angle chasing $\to IZ^{*}PB^{*}$ is a rectangle.

This post has been edited 2 times. Last edited by Krits, Jan 22, 2021, 12:38 PM

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