A difficult problem [tangent circles in right triangles]

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ThAzN1

867 posts

ThAzN1

#1 h Oct 17, 2004, 4:13 AM • 7 Y

Y by Davi-8191, nguyendangkhoa17112003, Adventure10, jhu08, Mango247, Rounak_iitr, and 1 other user

Let $ABC$ be a triangle such that $\angle A=90^{\circ }$ and $\angle B<\angle C$ . The tangent at $A$ to the circumcircle $\omega$ of triangle $ABC$ meets the line $BC$ at $D$ . Let $E$ be the reflection of $A$ in the line $BC$ , let $X$ be the foot of the perpendicular from $A$ to $BE$ , and let $Y$ be the midpoint of the segment $AX$ . Let the line $BY$ intersect the circle $\omega$ again at $Z$ .

Prove that the line $BD$ is tangent to the circumcircle of triangle $ADZ$ .

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grobber

7849 posts

grobber

#2 h Oct 17, 2004, 6:38 PM • 4 Y

Y by Adventure10, jhu08, Mango247, and 1 other user

It's still nice to have a solution posted

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Let $F$ be the point on the circle $(ABC)$ s.t. $BF\|AY$ . We see that the four lines $BF,BZ,BA,BE$ for a harmonic quadruple, so their intersections with the circle (other than $B$ ) for a harmonic quadrilateral, i.e. the quadrilateral $AFEZ$ is harmonic. This means that the tangents from $A,E$ to the circle meet on the diagonal $ZF$ , so $D,Z,F$ are collinear. We will keep this in mind in what follows.

Since $\angle ZAB=\angle BFD$ and $\angle AZB=\angle ACB=\angle FBD$ , we find the triangles $ZAB,BFD$ to be similar. At the same time, the triangles $ZCD,BFD$ are similar, so $ZAB,ZCD$ are similar. What we have to prove is now reduced to a simple angle chase: $\angle CDZ=\angle ZBA=\angle ZAD$ , and that's it.

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ThAzN1

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ThAzN1

#3 h Oct 17, 2004, 7:32 PM • 3 Y

Y by Adventure10, jhu08, Mango247

Cool, thanks.

Btw, where can I find more information on harmonic quadrilaterals? I tried searching on google, but that didn't bring up much.

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grobber

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grobber

#4 h Oct 17, 2004, 7:50 PM • 3 Y

Y by Adventure10, jhu08, Mango247

Sorry.. I could find next to nothing. I remember seeing long thread on Math Forum once, but couldn't find it. I'll keep looking for info.

[Edit: try looking for stuff in this thread. It's not about harminic quadrilaterals per se, but I think there's quite a bit of useful information on the subject in those messages]

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al.M.V.

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al.M.V.

#5 h Aug 9, 2005, 6:56 PM • 3 Y

Y by Adventure10, jhu08, Mango247

It's somewhere in the 2004 or 2005 issues of http://www.cms.math.ca/Competitions/MOCP/

[Moderator edit: Indeed, it's Olymon 2004 problem #300, solved at http://www.cms.math.ca/Competitions/MOCP/2004/sol_mar.html and at http://www.kalva.demon.co.uk/short/soln/sh98g8.html .]

I remember I had an algebraic solution---coordinates! lol

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Amir.S

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Amir.S

#6 h Aug 20, 2007, 7:31 PM • 5 Y

Y by zuss77, Adventure10, jhu08, Mango247, and 1 other user

not important which one of B and C are bigger in my solution B is bigger one.
let $AE\cap BC\equiv H\ ,AZ\cap BC\equiv M$ we have $\triangle ABX\sim\triangle ADH$ and also $\angle XBY=\angle ZAH$ hence AZ bisect DH and cause $(DH,BC)=-1$ hence $MD^{2}=MB.MC=MZ.MA$

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Joao Pedro Santos

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Joao Pedro Santos

#7 h Jan 31, 2010, 12:08 AM • 3 Y

Y by jhu08, Adventure10, Mango247

Let $O$ be the midpoint of $[BC]$ and let $F$ be the orthogonal projection of $A$ in $BD$ . So we have $O\hat{F}A = O\hat{A}D = 90^{\circ}$ , which means $D$ and $F$ are inverses with respect to the circumcircle of $[ABC]$ . So invert through the circumcircle of $[ABC]$ . Line $BD$ is sent to itself and the circumcircle of $[ADZ]$ is sent to the circumcircle of $[AFZ]$ . Therefore it suffices to prove line $BD$ is tangent to the circumcircle of $[AFZ]$ . So we just have to prove $AZ\perp FZ$ . Now use complex numbers. Let $a,b,c,e,f,x,y,z$ be the complex numbers corresponding to $A,B,C,E,F,X,Y,Z$ , respectively. Suppose $b = 1$ and $c = - 1$ . So $a,e,z$ are on the unit circle and $f$ is on the real axis. Therefore we have:
$e = \bar{a} = \frac {1}{a}$
$f = \frac {1}{2}(b + c + a - bc\bar{a}) = \frac {a^2 + 1}{2}$
$\bar{f} = f = \frac {a^2 + 1}{2}$
$x = \frac {1}{2}(b + e + a - be\bar{a}) = \frac {a^3 + a^2 + a - 1}{2a^2}$
$y = \frac {a + x}{2} = \frac {3a^3 + a^2 + a - 1}{4a^2}$
$\bar{y} = \frac {3 + a + a^2 - a^3}{4a}$
Since $z$ belongs to line $by$ : $\frac {y - b}{\bar{y} - \bar{b}} = \frac {z - b}{\bar{z} - \bar{b}}\Leftrightarrow\frac {y - b}{\bar{y} - \bar{b}} = - bz\Leftrightarrow z = - \frac {y - 1}{\bar{y} - 1}$ , which means $z = \frac {3a^2 + 1}{a(a^2 + 3)}$ and $\bar{z} = \frac {a(a^2 + 3)}{3a^2 + 1}$ .
Now we want to prove $az\perp fz$ : $\frac {z - f}{\bar{z} - \bar{f}} = - \frac {z - a}{\bar{z} - \bar{a}}\Leftrightarrow\frac {z - f}{\bar{z} - \bar{f}} = - ( - az)\Leftrightarrow f - z = a(fz - 1)$ , and it's trivial to check the last equality.

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jayme

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jayme

#8 h Feb 10, 2010, 12:28 PM • 3 Y

Y by jhu08, Adventure10, Mango247

Dear Mathlinkers,
for the beauty of this nice result, can we imagine a proof without calculation ?
I think so.
Sincerely
Jean-Louis

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Raúl

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Raúl

#9 h Aug 28, 2010, 8:57 PM • 3 Y

Y by jhu08, Adventure10, Mango247

jayme wrote:

Dear Mathlinkers,
for the beauty of this nice result, can we imagine a proof without calculation ?
I think so.
Sincerely
Jean-Louis

I think that bouth solutions are interesting, in fact we must lear to solve problems by bouth ways

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jayme

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jayme

#10 h Aug 29, 2010, 10:54 AM • 2 Y

Y by jhu08, Adventure10

Dear Raül and Mathlinkers,
you are right... all solutions of a problem is interesting for all members of our community... after each one of us has his proper sensibility. Do you have find yours?
Sincerely
Jean-Louis

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Raúl

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Raúl

#11 h Aug 29, 2010, 1:29 PM • 3 Y

Y by jhu08, Adventure10, Mango247

I found one mine yes

this problem was used for one friendly competition of geometry (in witch each participant propose 2 problems) and when I find out that it is a G8... xD

My solution is the same as one presented here, is like that:

We create the following points: $Q=AE\cap BC$ , $P=A\cap BC$ , $E'=DZ\cap C(ABC)/\{Z\}$ and $U=AE'\cap BC$ (we will find out that this point doesn't exist, by now assume that it exists and we goal one contradiction) where $C(ABC)$ is the circuncircle of $ABC$
We have the similarity of the triangles $[AXB]\sim [DQA]$ and $[AYB]\sim [DZA]$ so, since $\overline{AX}=2\overline{AY}$ we can conclude $\overline{DQ}=2\overline{DP}$ and so $P$ is the midpoint of $[DQ]$ .

By one well known fact (we know that $(A, E, Z, E')$ is one harmonic conjugate, just take the pencil from $A$ and intersect with $BC$ ) we get the points $(D, Q, P, U)$ are in harmonic division, but since $\overline{DP}=\overline{PQ}$ we find that the point $U$ must be in the infinity, and $AE'\parallel BC$ , and now just angle chase with inscribed arcs:
$\angle DAZ+\angle ZAC=\angle DBC=\angle CBA=\angle E'CB=\angle E'DB+\angle ZE'C=\angle ZDC+\angle ZAC\Leftrightarrow \angle DAZ=\angle ZDC$

witch is our goal

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skytin

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skytin

#12 h Sep 5, 2010, 11:05 AM • 2 Y

Y by jhu08, Adventure10

let DB intersect AE at point H and let M is midpoint of DH , let MA intersect BY at point Z' easy to see that trangle ABX ~ DAH and AM and BY are their medians so angle AZ'B = 180 - BAD so Z' is on circle arround ABC so Z' =Z and M is on radical line of point D and circle arround ABC so MZ*MA = MD*MD so done

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alkjash

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alkjash

#13 h Nov 8, 2010, 2:41 AM • 3 Y

Y by jhu08, Adventure10, Mango247

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jayme

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jayme

#14 h Nov 9, 2010, 4:49 AM • 3 Y

Y by jhu08, Adventure10, Mango247

Dear Mathlinkers,
a some up of my synthetic proof.
1. T is the second point of intersection of DZ with the circumcircle (omega).
2. If we prove that AT // BC we are done.
Who want to continue?
Sincerely
Jean-Louis

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sankha012

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sankha012

#15 h Mar 29, 2011, 6:40 AM • 3 Y

Y by jhu08, Adventure10, Mango247

Here's another inversive proof
Invert with center $D$ and radius $DA$ .This takes $C$ to $B$ , $B$ to $C$ and leaves $A,E$ Invariant.Let $Z'$ be the intersection of $\omega$ with $DZ$ .This $Z'$ is the inverse of $Z$ ,So the line $AZ'$ is the inverse of the circumcircle of $\bigtriangleup ADZ$ .So it suffices to prove that $AZ'||CB$ .
Taking $\omega$ as the unit circle,This is same as $z'=-\frac{b^2}{a}$ .
Some calculations(not too long...finding out the value of $x=\frac{a^2b+ab^2+a^3-b^3}{2a^2}$ ,then of $y=\frac{a^2b+ab^2+3a^3-b^3}{4a^2}$ ,then of $z=\frac{b^2(3a^2+b^2)}{a(3b^2+a^2)}$ and finally of $z'$ ) prove this result

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Fakesolver19

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Fakesolver19

#38 h Sep 8, 2021, 3:32 PM • 1 Y

Y by jhu08

Let $BC \cap AE=F$
It's trivial to see that $E \in \odot(ABC)$ as $A$ is the orthocentre.
From angle chase we get, $\measuredangle ABF=\measuredangle EBF=\measuredangle XAF=\measuredangle EAC=\measuredangle CEA \Rightarrow EC||AX.$
Notice that $DE$ is tangent to $\odot(ABC)$
Now draw a pont $K$ such that $AX||BK||EC$
Let $AX \cap BK=P_{\infty}$
$-1=(P_{\infty},Y;A,X)\stackrel{B}{=}(K,Z;A,E) \Rightarrow AKEZ \text {is a Harmonic quadrilateral}$ It suffices to show that $DKFE$ is cyclic.
This is just simple angle chasing,
$\measuredangle KBE=\measuredangle KZE=\measuredangle DZE=\measuredangle CFE=90^\circ \Rightarrow$ $DKFE$ is cyclic
As $AD$ is tangent to $\odot(ABC)$ ,
$\measuredangle DAZ= \measuredangle ZEC= \measuredangle EDZ$ which implies $BD$ is tangent to $\odot(DAZ)$ where the last equality came from cyclic quadrilateral.

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tigerzhang

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tigerzhang

#39 h Oct 22, 2021, 3:37 AM • 1 Y

Y by RP3.1415

Let $\overline{BC}$ and $\overline{AE}$ intersect at $M$ . Since $\overline{BE} \parallel \overline{MY}$ , $AYMZ$ is cyclic by Reim's. Notice that $\angle MAC=\angle AEC=\angle CAD$ , so $C$ lies on the angle bisector of $\angle MAD$ . Since $\angle BAC=90^\circ$ , $B$ lies on the external angle bisector of $\angle MAD$ . Since $Z$ lies on the circumcircle of $\triangle AMY$ and on the Apollonian circle of $\overline{BC}$ passing through $A$ , $Z$ is the $A$ -humpty point of $\triangle AMD$ . Thus, the circumcircle of $\triangle ADZ$ is tangent to $\overline{BD}$ .

$\mathcal{YUH}$

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Knty2006

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Knty2006

#41 h Apr 3, 2022, 11:57 AM

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Let the line $AE$ intersect $BC$ at $J$ .

Note that $Y$ and $J$ are the midpoints of $AX$ and $AE$ respectively, hence $YJ//XE$
This implies that $\angle AYJ= 90$

Also, since $\angle BJA= \angle AYJ= 90$ it means that $\angle YJB= \angle YAJ$ Hence, $BC$ tangent to $AYJ$

$\angle AZB= \angle AEX= \angle AJY$ , so $A,Y,J,Z$ concyclic.

Also, note $\angle ZED= \angle EAZ= \angle ZJD$ which implies that $J,Z,D,E$ concyclic

Therefore, $\angle ZDC=\angle ZEA=\angle ZBA= \angle ZAD$

Hence, $BC$ is tangent to $ZAD$

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Reason: oops

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awesomeming327.

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awesomeming327.

#42 h Jul 25, 2022, 7:12 PM

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Diagram
Note that $\angle BAX=90^\circ-\angle ABX=90^\circ-2\angle ABC$ and that $\angle ADC=\angle ACB-\angle CAD=\angle ACB-\angle BAC=90^\circ-2\angle ABC$ so $\angle BAX=\angle ADC.$

$~$
Let $F,G$ be intersections of $BC$ with $AE$ and $AZ$ respectively. $AF\perp BD$ so $\triangle BAX\sim \triangle ADF.$ Since $\angle FAG=\angle EAZ=\angle EBZ=\angle XBY$ so $G$ is midpoint of $FD.$ Since $GC\cdot GB=GZ\cdot GA$ It remains to show that $GC\cdot GB=GF^2.$

$~$
Note that
$\begin{align*} &= FB\cdot FC \\ &= FA^2 \\ &= AD^2 - FD^2 \\ &= DC\cdot DB - FD^2 \\ &= (DF+FB)(DF-FC) - FD^2 \\ &= DF(FB-FC) - FB\cdot FC \\ \end{align*}$ so $FB\cdot FC = \tfrac12 DF(FB-FC)=GF(FB-FC)$ and $GC\cdot GB = (GF+FB)(GF-FC)=GF^2+GF(FB-FC)-FB\cdot FC=GF^2$ as desired.

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JAnatolGT_00

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JAnatolGT_00

#43 h Jul 26, 2022, 6:48 AM

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By symmetry $DE$ is tangent to $\omega$ at $E.$ Constructing $E-\text{antipode}$ $F$ we get $AF\parallel BC$ and $BF\parallel AX\implies (AEZF)_{\omega}\stackrel{B}{=} (AXY\infty_{AX})=-1\implies D\in FZ.$ Finally $\measuredangle DAZ$ $=$ $\measuredangle AFD$ $=$ $\measuredangle BDZ$ yields the conclusion.

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ike.chen

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ike.chen

#44 h Jul 29, 2022, 4:30 AM • 1 Y

Y by Mango247

Let $A_1$ lie on $(ABC)$ such that $AA_1 \parallel BC$ , $M = AE \cap BC$ , and $T = AZ \cap BD$ .

Because $AA_1 \parallel BC \perp AE$ we know $A_1E$ is a diameter of $(ABC)$ . Thus, $BA_1 \perp BE \perp AX$ implies $BA_1 \parallel AX$ , which means $-1 = (A, X; Y, \infty_{AX}) \overset{B}{=} (A, E; Z, A_1) \overset{A}{=} (D, M; T, \infty_{BC})$ so $T$ is the midpoint of $DM$ .

It's clear that $ABEC$ is a cyclic kite, so $-1 = (A, E; B, C) \overset{A}{=} (D, M; B, C).$ It follows that $B$ and $C$ are inverses wrt $(DM)$ , yielding $TD^2 = TB \cdot TC = | Pow_{(ABC)}(T) | = TA \cdot TZ$ which finishes. $\blacksquare$

Remarks: It's very easy to reverse engineer the fact that $T$ is the midpoint of $DM$ , as $-1 = (D, M; B, C)$ is well-known and commonly seen. At this point, the definition of $D$ motivates us to use projective methods to show the aforementioned result, and the rest follows naturally.

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signifance

140 posts

signifance

#45 h Oct 9, 2023, 6:55 PM

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remark that is the same length as my sol

Let W be the antipode of E, $V=AE\cap BC$ , and $M=AZ\cap BC$ . Then $WBE=90=AXE\Rightarrow AX\parallel BW\implies(DMVP_{\infty})\stackrel{A}{=}(AZEW)\stackrel{B}{=}(AYXP_{\infty})=-1,$ so M is the midpoint of DV. It follows that $M(DBVC)=-1\implies MD^2=MD\cdot MV=MB\cdot MC=MZ\cdot MA$ , so $MD\equiv BC$ is tangent to (ADZ), as needed.

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math_comb01

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math_comb01

#46 h Dec 25, 2023, 12:27 PM

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How is this G8?
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -15.98006838884981, xmax = 9.683036625251171, ymin = -10.19069829689637, ymax = 5.434389088786898; /* image dimensions */ /* draw figures */ draw((-1.0388665184850865,4.327988392401811)--(-5.374150879830733,-2.3104157859087104), linewidth(0.4)); draw((xmin, -0.6530612244897959*xmin + 3.649544951758489)--(xmax, -0.6530612244897959*xmax + 3.649544951758489), linewidth(0.4)); /* line */ draw((-5.374150879830733,-2.3104157859087104)--(9.338330097801927,-2.4489563366019524), linewidth(0.4)); draw(circle((1.982089608985597,-2.379686061255332), 7.356566624475532), linewidth(0.4)); draw((xmin, 0.4503730985071294*xmin + 4.795865925267252)--(xmax, 0.4503730985071294*xmax + 4.795865925267252), linewidth(0.4)); /* line */ draw((xmin, -0.00941653218813773*xmin-2.3610216506525457)--(xmax, -0.00941653218813773*xmax-2.3610216506525457), linewidth(0.4)); /* line */ draw((-15.56556976958695,-2.2144479618905266)--(-1.1646456874358049,-9.029282250092894), linewidth(0.4)); draw((xmin, -1.59611787065206*xmin-10.888194044787035)--(xmax, -1.59611787065206*xmax-10.888194044787035), linewidth(0.4)); /* line */ draw((xmin, 0.6265201451516054*xmin + 4.97885919435623)--(xmax, 0.6265201451516054*xmax + 4.97885919435623), linewidth(0.4)); /* line */ draw((-4.088852374348884,2.417110811275681)--(-5.374150879830733,-2.3104157859087104), linewidth(0.4)); draw(circle((-15.386643015476231,16.786894627489083), 19.00218500547524), linewidth(0.4)); draw((xmin, 0.6265201451516054*xmin + 1.0565980033894682)--(xmax, 0.6265201451516054*xmax + 1.0565980033894682), linewidth(0.4)); /* line */ draw((-4.396583103819279,1.285229616255652)--(-1.1646456874358049,-9.029282250092894), linewidth(0.4)); draw((-1.1646456874358049,-9.029282250092894)--(5.128824905406998,4.2699101275822295), linewidth(0.4)); draw((5.128824905406998,4.2699101275822295)--(-1.0388665184850865,4.327988392401811), linewidth(0.4)); draw((-1.0388665184850865,4.327988392401811)--(-4.396583103819279,1.285229616255652), linewidth(0.4)); draw((-15.56556976958695,-2.2144479618905266)--(5.128824905406998,4.2699101275822295), linewidth(0.4)); /* dots and labels */ dot((-1.0388665184850865,4.327988392401811),dotstyle); label("$A$", (-0.9485480942903854,4.553784452888563), NE * labelscalefactor); dot((-5.374150879830733,-2.3104157859087104),dotstyle); label("$B$", (-5.283832455636032,-2.084619725421958), NE * labelscalefactor); dot((9.338330097801927,-2.4489563366019524),dotstyle); label("$C$", (9.438070688100227,-2.2200973617140094), NE * labelscalefactor); dot((-15.56556976958695,-2.2144479618905266),linewidth(4pt) + dotstyle); label("$D$", (-15.467234783588566,-2.0394605133246073), NE * labelscalefactor); dot((-1.1646456874358049,-9.029282250092894),dotstyle); label("$E$", (-1.084025730582437,-8.813342327927181), NE * labelscalefactor); dot((-7.138838230212681,0.5062332301495509),linewidth(4pt) + dotstyle); label("$X$", (-7.045041727432701,0.692671818565097), NE * labelscalefactor); dot((-4.088852374348884,2.417110811275681),linewidth(4pt) + dotstyle); label("$Y$", (-3.9967949108615435,2.5893587266538174), NE * labelscalefactor); dot((-4.396583103819279,1.285229616255652),linewidth(4pt) + dotstyle); label("$Z$", (-4.312909395542997,1.4603784242200553), NE * labelscalefactor); dot((5.128824905406998,4.2699101275822295),linewidth(4pt) + dotstyle); label("$F$", (5.215684356997956,4.440886422645187), NE * labelscalefactor); dot((1.9820896089855973,-2.3796860612553314),linewidth(4pt) + dotstyle); label("$O$", (2.077119116232097,-2.1975177556653342), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Let $F$ be the antipode of $E$ , so $AF \parallel BC$
Let $P_\infty$ denote the point at infinity along $AX$
$-1=(AX;YP_\infty)=(AE;ZF)$ , $\therefore D-Z-F$ , so $\measuredangle ZAD = \measuredangle AFZ = \measuredangle ZDB$ , hence we're done.

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rstenetbg

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rstenetbg

#47 h Mar 21, 2024, 7:41 PM

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Let $L=AE\cap BC$ and let $R=AZ\cap BC$ .

Claim 1: $\triangle BXA \sim \triangle ALD$ .
Proof: Since $\angle BXA = \angle ALD=90^{\circ}$ , it is enough to show that $\angle XBA=\angle LAD$ . However, $\angle LAD = \angle EAC+\angle CAD = \angle EBC + \angle CBA = \angle XBA$ , so we are done.

Now back to the problem.
Note that $\angle RAD = \angle ZAD = \angle YBA$ . Since $BY$ is the median in $\triangle BXA$ , it follows that $AR$ is the median in $\triangle LAD.$

From PoP, we have $RZ\cdot RA = RC\cdot RB$ . It is well-known that $(D,L;C,B)=-1$ and since $R$ is the midpoint of $LD,$ we obtain that $RC\cdot RB=RD^2=RZ\cdot RA$ and we are done.

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EthanWYX2009

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EthanWYX2009

#48 h Jul 30, 2024, 1:50 PM

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Solution

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kamatadu

477 posts

kamatadu

#49 h Oct 30, 2024, 10:39 AM

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Diagram.

Note that $DA$ is tangent to the circumcircle of $\odot(ABC)$ . Now since D lies on the diameter $BC$ and $E$ is the reflection of $A$ over $BC$ , we get that $DE$ is tangent to $\odot(ABC)$ due to symmetry.

Let the line parallel to $BC$ through $A$ intersect $\odot(ABC)$ again at $T$ .

Now note that $\measuredangle EAT = 90^{\circ}$ as $AT \parallel BC$ and $AE \perp BC$ . This gives us that,
$90^{\circ}=\measuredangle EAT=\measuredangle EBT \implies TB\perp XE \implies XA \parallel BT$ where the last implication follows due to the fact that $XA\perp XE$ .

Now note that $D$ is basically the intersection of tangents to $\odot(ABC)$ at $A$ and $E$ , which implies that $TD$ is the $T$ -symmedian of $\triangle TAE$ .

Thus we have,
$\begin{align*} (T,TD\cap \odot(ABC);A,E)=-1&=(\infty_{XA},Y;A,X) \\ &\overset{B}{=}(B\infty_{XA}\cap \odot(ABC),A,E)\\ &= (T,Z;A,E) .\end{align*}$
This implies that $TD\cap \odot(ABC)\equiv Z$ which gives us that $\overline{T-Z-D}$ are collinear.

Now note,
$\begin{align*} \measuredangle (AD, BD) = \measuredangle ADB &= \measuredangle ABD+\measuredangle DAB \\ &= \measuredangle ABC+\measuredangle ACB\\ &= \measuredangle ABC-\measuredangle BCA\\ &= \measuredangle ABC-\measuredangle TBC\\ &= \measuredangle ABT\\ &= \measuredangle AZT\\ &= \measuredangle AZD .\end{align*}$
This finally gives that $BD$ is tangent to $\odot(AZD)$ and we are done.

This post has been edited 2 times. Last edited by kamatadu, Oct 30, 2024, 10:41 AM

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bin_sherlo

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bin_sherlo

#50 h Nov 13, 2024, 12:38 PM

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Consider $\sqrt{bc}$ inversion.
New Problem Statement: $ABC$ is a triangle with $\measuredangle CAB=90$ and $AB<AC$ . Let $E$ be the midpoint of $BC$ and $D$ is on $(ABC)$ such that $AD\parallel BC$ . $X$ is taken on $(AEC)$ with $\measuredangle ACX=90$ . Tangent to $(ABC)$ at $D$ intersects $BC$ at $T$ and $Y$ is the reflection of $A$ over $X$ . Prove that $A,C,T,Y$ are concyclic.
Let $A'$ be the antipode of $A$ on $(ABC)$ . Note that $A',C,X$ are collinear.
$\measuredangle TEA'=2\measuredangle CAE=2\measuredangle ECA=2\measuredangle DAC=\measuredangle DEC=\measuredangle DET$ And $T$ lies on $EC$ hence $TA'$ is tangent to $(ABC)$ . Since $\measuredangle XEA=90$ and $AE=EA'$ , we observe that $XY=XA=XA'$ thus, $A',T,Y$ are collinear.
$180-\measuredangle ACT=\measuredangle ECA=\measuredangle CAE=\measuredangle CAA'=\measuredangle CA'T=\measuredangle XA'Y=\measuredangle A'YX=\measuredangle TYA$ Which completes the proof as desired. $\blacksquare$

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Aiden-1089

278 posts

Aiden-1089

#51 h Dec 13, 2024, 4:26 PM

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Let $A'$ be the reflection of $A$ across the perpendicular bisector of $BC$ .
Now we have $A'B // AX$ , so $-1=(A,X;Y,\infty_{AX}) \stackrel{B}{=} (A,E;Z,A') \implies A',Z,D$ collinear.

It suffices to show that $\angle BDZ = \angle DAZ$ , or $\angle BDA' = \angle ABY$ .
Let $O$ be the centre of $\omega$ , and let $P$ be the point such that $APDO$ is an isosceles trapezium with $AP//OD$ .
First note that $\angle OPD = \angle OAD = 90^\circ = \angle AXB$ .
Next, note that $PD//OE$ , so $\angle ODP = \angle DOE = \frac{1}{2} \angle AOE = \angle ABX$ . It follows that $\Delta ABX \sim \Delta ODP$ .
Now, it is not hard to see that $A'P=OD$ (by considering $A'A=2d(O,AE)$ ), and since $A'P//OD$ , we have that $A'PDO$ is a parallelogram.
So $DA'$ passes through the midpoint $M$ of $OP$ . We also have $\Delta ABY \sim \Delta ODM$ , so $\angle ABY = \angle ODM = \angle BDA'$ as desired. $\square$

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L13832

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L13832

#52 h Jan 1, 2025, 5:33 AM

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Let $BC\cap AE=F$ and let line through $B$ parallel to $AX, CE$ meet $\odot(ABC)$ at $E$ .
Observe that
$(A,X;Y,P_{\infty})\stackrel{B}{=}(A,E;Z,E')\implies \overline{D-Z-E'}$ Now it is enough to show $\odot(DZFE)$ is cyclic
$\begin{align*} \angle DZE=\angle E'BE=\angle DFE \end{align*}$ as then we also have
$\angle ZDF=\angle ZEA=\angle ZE'A=\angle ZAD$ giving $\odot(ADZ)$ tangent to $BD$ .
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Jupiterballs

40 posts

Jupiterballs

#53 h Mar 9, 2025, 12:45 PM

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@4th above
Why did you make the diagram on the table

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