Hard excircle geo

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Hard excircle geo

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Source: MEMO 2023 T6

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#1 h Aug 25, 2023, 7:47 AM • 1 Y

Y by v4913

Let $ABC$ be an acute triangle with $AB < AC$ . Let $J$ be the center of the $A$ -excircle of $ABC$ . Let $D$ be the projection of $J$ on line $BC$ . The internal bisectors of angles $BDJ$ and $JDC$ intersectlines $BJ$ and $JC$ at $X$ and $Y$ , respectively. Segments $XY$ and $JD$ intersect at $P$ . Let $Q$ be the projection of $A$ on line $BC$ . Prove that the internal angle bisector of $QAP$ is perpendicular to line $XY$ .

Proposed by Dominik Burek, Poland

This post has been edited 4 times. Last edited by a_507_bc, Aug 26, 2023, 3:26 PM

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#2 h Aug 25, 2023, 1:07 PM

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Bump this beauty

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CyclicISLscelesTrapezoid

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CyclicISLscelesTrapezoid

#3 h Aug 25, 2023, 1:59 PM • 3 Y

Y by v4913, VicKmath7, Newmaths

Since $\overline{AQ} \parallel \overline{DJ}$ , it suffices to prove that $\overline{XY}$ bisects $\angle APJ$ . Let $\omega_X$ be the circle centered at $X$ tangent to $\overline{AB}$ , $\overline{BC}$ , and $\overline{DJ}$ , and let $\omega_Y$ be the circle centered at $Y$ tangent to $\overline{AC}$ , $\overline{BC}$ , and $\overline{DJ}$ . The following claim finishes.

Claim: $\overline{AP}$ is tangent to $\omega_X$ and $\omega_Y$ . (This is a special case of the main claim to ISL 2009 G8.)

Proof: Let the tangent to $\omega_X$ passing through $A$ other than $\overline{AB}$ intersect $\overline{DJ}$ at $P'$ . By (excircle) Pitot, $AC-CD=AB-BD=AP'-P'D$ , so by the converse of Pitot, $ACDP$ is tangential, which means $\overline{AP'}$ is tangent to $\omega_Y$ . Thus, $P'$ lies on $\overline{XY}$ , so $P'=P$ . $\square$

This post has been edited 3 times. Last edited by CyclicISLscelesTrapezoid, Sep 12, 2023, 1:00 AM

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#4 h Aug 25, 2023, 3:13 PM

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The problem is equivalent to show that the intersection of $AP$ and $BC$ lies on $(DXY)$ (or actually just on the perpendicular bisector of $XY$ is sufficient). Is there any solution based on this approach, which I find much more motivated than the above solution (it's beautiful, but still hard to think of)?

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#5 h Aug 26, 2023, 3:05 PM

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This problem was proposed by Burii.

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#6 h Aug 28, 2023, 9:07 AM

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does anyone have a solution based on the approach of a_507_bc ?

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#7 h Aug 28, 2023, 10:28 AM

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Oops...seems I solved the incircle version of the problem...although they're not different at all. Rename $J$ to be $I$ and the rest are defined similarly , with incircle though .We shall prove that the intersection of $AP$ and $BC$ lies on $(DXY)$ which finishes the problem after an easy angle chasing.
Lemma(Serbia 2018). Let $Z,T$ be on $BI,CI$ st $\angle ZDT = 90$ , Then $\angle ZAT = \angle BAI$ .
Proof. Trivial by moving points.
Now , consider a point $Q$ on $ID$ st $ABDP$ is tangential with incircle $C(X,dist(X,BC))$ . Then $AQ-QD = AB-BD = AC - CD$ Which means $ACDQ$ is tangential with incircle $C(Y,dist(Y,BC))$ . So since $Q$ is the intersection of the common tangents of $C(X,dist(X,BC))$ and $C(Y,dist(Y,BC))$ , It would lie on $XY$ so $Q \equiv P$
Now , Let $R = AP \cap BC$ . By the lemma , since $\angle XAY = \angle BAI$ we get that $\angle XDY =90$ . But $X,Y$ are the incenters of $\triangle ABR, \triangle ACR$ respectively. So $\angle XRY =90$ and $R \in (XDY)$ so we're done.

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hakN

#8 h Aug 28, 2023, 11:05 AM

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It's also easy to coordbash this:

Set $D=(0,0), B=(-b,0), C=(c,0)$ and $J=(0,-1)$ with $b,c>0$ .

Then we have $\ell_{BJ}: y=\frac{-1}{b}x-1$ and $\ell_{CJ}: y=\frac{1}{c}x-1$ and thus $X=\left(\frac{-b}{b+1},\frac{-b}{b+1}\right)$ and $Y=\left(\frac{c}{c+1},\frac{-c}{c+1}\right)$ .

Thus we have $\ell_{XY}: y=\frac{b-c}{2bc+b+c}x - \frac{2bc}{2bc+b+c}$ . Also by tangent double angle we have $\ell_{AB}: y=\frac{2b}{1-b^2}(x+b)$ and $\ell_{AC}: y=\frac{2c}{c^2-1}(x-c)$ , thus $A=\left(\frac{b-c}{bc-1},\frac{-2bc}{bc-1}\right)$ .

We need to show the reflection of $A$ over $XY$ lies on $JD$ , or the foot from $A$ to $XY$ has $x$ -coordinate $\frac{b-c}{2(bc-1)}$ .

The line from $A$ perpendicular to $XY$ has equation: $y=\frac{2bc+b+c}{c-b}x+\frac{b+c}{bc-1}$ , intersecting this with $\ell_{XY}$ , we have the foot from $A$ to $XY$ is $\left(\frac{b-c}{2(bc-1)},\frac{b+c-2bc}{2(bc-1)}\right)$ , so we're done.

This post has been edited 2 times. Last edited by hakN, Aug 28, 2023, 11:06 AM

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#9 h Aug 29, 2023, 1:34 PM • 1 Y

Y by Gms68bx

Let the $A$ -excircle be tangent to sides $AB$ and $AC$ at points $E$ and $F$ respectively, furthermore let us introduce the point $Z$ on the ray $DJ$ with $DZ=AE=AF.$

$AQ\parallel{DJ},$ so we have to prove that the internal bisector of $\angle{APD}$ (which is parallel to the bisector of $\angle{QAP}$ ) is perpendicular to $XY,$ which is equivivalent to proving that $XY$ is the external bisector of $\angle{APD}.$

$D$ and $E$ are symmetric w.r.t. $BJ,$ so we have $\angle{ZDX}=\angle{XDB}=\angle{BEX}=\angle{AEX}=45°.$ We also now that $XD=XE$ (by symmetry), and $DZ=AE$ (by definition), hence $ZDX\triangle\simeq{AEX\triangle},$ from which we obtain $AX=ZX.$ Similarly $AY=ZY,$ hence $Z$ is the reflection of $A$ w.r.t. line $XY.$ This shows that lines $AP$ and $DPZ$ are symmetric w.r.t. $XY,$ hence $XY$ is one of their bisectors. Since $A$ and $D$ lie on the same side of $XY,$ it immediatly follows that it is the external bisector of $\angle{APD},$ yielding the desired result.

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lksb

#10 h Apr 13, 2025, 11:22 PM

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hakN wrote:

It's also easy to coordbash this:

Set $D=(0,0), B=(-b,0), C=(c,0)$ and $J=(0,-1)$ with $b,c>0$ .

Then we have $\ell_{BJ}: y=\frac{-1}{b}x-1$ and $\ell_{CJ}: y=\frac{1}{c}x-1$ and thus $X=\left(\frac{-b}{b+1},\frac{-b}{b+1}\right)$ and $Y=\left(\frac{c}{c+1},\frac{-c}{c+1}\right)$ .

Thus we have $\ell_{XY}: y=\frac{b-c}{2bc+b+c}x - \frac{2bc}{2bc+b+c}$ . Also by tangent double angle we have $\ell_{AB}: y=\frac{2b}{1-b^2}(x+b)$ and $\ell_{AC}: y=\frac{2c}{c^2-1}(x-c)$ , thus $A=\left(\frac{b-c}{bc-1},\frac{-2bc}{bc-1}\right)$ .

We need to show the reflection of $A$ over $XY$ lies on $JD$ , or the foot from $A$ to $XY$ has $x$ -coordinate $\frac{b-c}{2(bc-1)}$ .

The line from $A$ perpendicular to $XY$ has equation: $y=\frac{2bc+b+c}{c-b}x+\frac{b+c}{bc-1}$ , intersecting this with $\ell_{XY}$ , we have the foot from $A$ to $XY$ is $\left(\frac{b-c}{2(bc-1)},\frac{b+c-2bc}{2(bc-1)}\right)$ , so we're done.

Doesn't this solve only for the incenter version of this problem?

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