A difficult problem [tangent circles in right triangles]

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A difficult problem [tangent circles in right triangles]

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Source: IMO ShortList 1998, geometry problem 8; Yugoslav TST 1999

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#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 $\overline{BD}$ .

$\mathcal{YUH}$

This post has been edited 2 times. Last edited by tigerzhang, Oct 23, 2021, 3:45 AM

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#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 $ZAD$

This post has been edited 2 times. Last edited by Knty2006, Apr 3, 2022, 12:20 PM
Reason: oops

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awesomeming327.

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

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How is this G8?
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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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#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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#48 h Jul 30, 2024, 1:50 PM

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Solution

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

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Diagram.

:rotfl:

https://i.imgur.com/XX1s1YK.png

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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#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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#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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#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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#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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#59 h Yesterday at 1:17 AM

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ThAzN1 wrote:

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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Complex Bash!
let b=-1 and c=1thus we will have x=(1/2)*(a+(1/a)+(1/a^2)-1) thus y=(1/4)*(3a+(1/a)+(1/a^2)-1) and z=((1+(1/4)*(3a+(1/a)+(1/a^2)-1))/((1+(1/4)*((3/a)+a+a^2-1))) and notice D is on tangent of E and e is conjugate of a thus d=2a/(a^2+1) now FTSOC we would assume there exist k != d such that AZKD is cyclic and K is on line BC (K is real) thus we have ((k-z)(k-d))/((a-z)(a-d)) is a real number and therefore equal to its conjugate thus we have (k-z)(a-d)=(kz-1)(ad-1) notice following equation is at most degree of 1 (for variable k)and therefore has at most 1 solution we will prove d is a solution and contradiction is trivial by putting k=d in equation we will get equation is true ifoif z=(3a^4-2a^2-1)/(a^5+2a^3-3a) so we need to prove (3a^4-2a^2-1)(1+(1/4)*(3/a+a+a^2-1))=(a^5+2a^3-3a)(1+(1/4)*(3a+1/a+1/(a^2)-1)) with is trivial

This post has been edited 2 times. Last edited by Autistic_Turk, Yesterday at 1:19 AM
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