APMO 2017: (ADZ) passes through M

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159 posts

#1 h May 14, 2017, 3:20 PM • 9 Y

Y by Davi-8191, artsolver, LoveIsABeautifulPain, me9hanics, AlastorMoody, MatBoy-123, Adventure10, farhad.fritl, ehuseyinyigit

Let $ABC$ be a triangle with $AB < AC$ . Let $D$ be the intersection point of the internal bisector of angle $BAC$ and the circumcircle of $ABC$ . Let $Z$ be the intersection point of the perpendicular bisector of $AC$ with the external bisector of angle $\angle{BAC}$ . Prove that the midpoint of the segment $AB$ lies on the circumcircle of triangle $ADZ$ .

Olimpiada de Matemáticas, Nicaragua

This post has been edited 1 time. Last edited by MellowMelon, May 17, 2017, 5:50 PM
Reason: add proposer

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jlammy

1099 posts

jlammy

#2 h May 14, 2017, 3:38 PM • 14 Y

Y by Plasma_Vortex, tapir1729, lifeisgood03, Plops, Tuleuchina, lahmacun, Physicsknight, k12byda5h, sabkx, lanjecrure, Adventure10, solidgreen, Funcshun840, ehuseyinyigit

Let $M$ be the midpoint of $\overline{AB}$ and $\ell$ the perpendicular bisector of $\overline{AD}$ . Then $OM,OZ$ are symmetric in $\ell$ , so $OM$ passes through $Z'$ , the reflection of $Z$ in $\ell$ . Now $AZZ'D$ is a rectangle and $\angle AMZ'=90^{\circ}$ , so we're done.

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anantmudgal09

1980 posts

anantmudgal09

#3 h May 14, 2017, 4:18 PM • 8 Y

Y by Ankoganit, me9hanics, tsbktl, Wizard_32, Aniruddha07, A-Thought-Of-God, Adventure10, Mango247

Apply inversion at $A$ . We obtain the equivalent problem:

inverted APMO 2017/2 wrote:

In triangle $ABC$ with $AB>AC$ , let $M$ be a point on $AB$ such that $BA=BM$ and $M \ne A$ , $D$ be the foot of the $A$ angle bisector, $Z$ be a point on the external bisector of angle $A$ such that $CA=CZ$ . Prove that $M, D, Z$ are collinear.

(Proof) To see this, let $X=MD \cap AC$ and $Y=MZ \cap AC$ . Note that $AB \parallel CZ$ and $CZ=CA$ so $\frac{AY}{YC}=\frac{AM}{CZ}=\frac{2AB}{AC}.$ Finally, by Menelaus' Theorem in $\triangle ABC$ for transversal $MDX,$ $\frac{AX}{XC} \cdot \frac{CD}{DB} \cdot \frac{BM}{MA}=1 \Longrightarrow \frac{AX}{XC}=\frac{2AB}{AC}.$ The last result is due to the Angle-Bisector Theorem. Evidently, $X \equiv Y$ so we are done! $\blacksquare$

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tenplusten

1000 posts

tenplusten

#4 h May 14, 2017, 5:27 PM • 5 Y

Y by FabrizioFelen, Elnuramrv, Adventure10, Mango247, Ulapu_Kanama

Let $(AZD)\cap AB =K$ .By angle chasing we get triangles $ZKD$ and $ZNC$ are similar.So there exist a spiral similarity centered at Z that takes KD to NC .Then by lemma if $KN\cap CD=O$ $OZKD$ is cyclic.Again angle chasing gives $KN\parallel BC$ .So done!!!(here N is midpoint of AC).

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FabrizioFelen

241 posts

FabrizioFelen

#5 h May 14, 2017, 5:30 PM • 7 Y

Y by tenplusten, DashTheSup, Spectralon, Aurn0b, Adventure10, Mango247, Warideeb

My solution in the contest

:
Let $O$ the circumcenter of $\odot(ABC)$ , $M$ the midpoint of $AB$ , let $E$ the midpoint of the arc $\overarc {BAC}$ and $N$ the midpoint of $EZ$ , so from $AEDB$ is cyclic we get: $\measuredangle ABD=\measuredangle ZED=\measuredangle ZEO...(1)$ , so from $DA\perp AZ$ we get $\measuredangle BAD$ $=$ $\measuredangle DAC$ $=$ $\measuredangle AZO$ $=$ $\measuredangle EZO..(2)$

Combining $(1)$ and $(2)$ we get $\triangle BAD\cup \{M\}\sim \triangle EOZ\cup \{N\}$ $\Longrightarrow$ $\measuredangle BMD=\measuredangle ENO$ , so from $NO\parallel ZD$ we get $\measuredangle BMD$ $=$ $\measuredangle ENO$ $=$ $\measuredangle EZD=\measuredangle AZD$ hence $AZDM$ is cyclic.

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EulerMacaroni

851 posts

EulerMacaroni

#6 h May 14, 2017, 6:23 PM • 2 Y

Y by IFA, Adventure10

Solution 1: Define $\{A,M\}\equiv \odot(ADZ)\cap AB, \{A,X\}\equiv \odot(ADZ)\cap AC$ . Notice that $D$ is the spiral center sending segment $\overline{CX}$ to $\overline{BM}$ , and since $DB=DC$ , this implies that $BM=CX$ . It is also easy to check that $BA$ is tangent to $\odot(AZC)$ , which quickly implies that $Z$ is the spiral center sending segment $\overline{AM}$ to $\overline{CX}$ , and since $ZA=ZC$ , this implies that $AM=CX$ . Therefore, $AM=BM$ , and the conclusion follows.

Solution 2: Invert about $A$ with power $r^2=\tfrac{1}{2}AB\cdot AC$ , composed with a reflection about the $A$ -angle bisector. The problem now becomes to show that if $Z$ is the projection of $C$ onto the external angle bisector of $\angle BAC$ , and if the internal angle bisector of $\angle BAC$ intersects $\overline{BC}$ at $X$ , then $BZ$ bisects $\overline{AX}$ at a point $D$ . It is well-known that $Z$ is collinear with the midpoints $M_B, M_A$ of $\overline{AC}, \overline{BC}$ , respectively, so we have $\frac{ZM_B}{ZM_A}=\frac{M_BC}{M_BC+M_BM_A}=\frac{AC}{AC+AB}=\frac{\tfrac{BC\cdot AC}{2(AC+AB)}}{\tfrac{BC}{2}}=\frac{DM_B}{BM_A}$ so $\triangle ZDM_B\sim\triangle ZBM_A$ , as desired.

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trumpeter

3332 posts

trumpeter

#7 h May 14, 2017, 7:11 PM • 2 Y

Y by Adventure10, Mango247

Obligatory Barybash

After I finished this, I realized that complex would have probably been faster, but this was only 10 minutes so I didn't mind

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ABCDE

1963 posts

ABCDE

#8 h May 14, 2017, 8:10 PM • 2 Y

Y by SHARKYBOY, Adventure10

Let $A'$ be the reflection of $A$ across $Z$ and $B'$ be the reflection of $B$ across $D$ . It is easy to see that $CB'B$ and $CAA'$ are similar right triangles so $AB'\perp A'B$ , which means that $\angle DMZ=90^\circ$ as $AB'\parallel DM,A'B\parallel MZ$ , immediately implying the conclusion.

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atmchallenge

980 posts

atmchallenge

#9 h May 14, 2017, 9:06 PM • 2 Y

Y by Adventure10, Mango247

Obligatory Complex Bash :)

Did I mess up in not explicitly incorporating $AB < AC$ ?

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v_Enhance

6877 posts

v_Enhance

#10 h May 14, 2017, 10:07 PM • 11 Y

Y by thunderz28, Wizard_32, HolyMath, v4913, math31415926535, aansc1729, sabkx, Adventure10, Mango247, JordanLv111, Rounak_iitr

Redefine $Z$ to be the point on $(AMD)$ such that $\overline{AZ}$ is an external bisector; hence $\measuredangle DMZ = \measuredangle DAZ = 90^{\circ}$ . Then let $N$ be the midpoint of $\overline{AC}$ . As $\measuredangle AMN = \measuredangle ABC = \measuredangle ADC$ , we find $T = \overline{MN} \cap \overline{CD}$ lies on $(AMD)$ .

$[asy] pair A = dir(110); pair B = dir(200); pair C = dir(340); draw(unitcircle, lightred); draw(A--B--C--cycle, lightred); pair D = dir(270); pair M = midpoint(A--B); pair Z = extension(origin, A+C, A, dir(90)); pair N = midpoint(A--C); draw(circumcircle(A, D, Z), dashed+heavycyan); draw(A--Z, lightred); draw(A--D, lightred); draw(D--M--Z--cycle, heavycyan); pair T = extension(M, N, D, C); draw(M--T--D, dashed+cyan); pair L = dir(90); draw(circumcircle(Z, T, C), dotted); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$M$", M, dir(M)); dot("$Z$", Z, dir(Z)); dot("$N$", N, dir(225)); dot("$T$", T, dir(T)); dot("$L$", L, dir(L)); [/asy]$

Let $L$ be the $D$ -antipode. We have $\triangle ZMD \sim \triangle LBD$ also by spiral similarity, so $\frac{ZM}{ZD} = \frac{LB}{LD} = \cos A/2 = \frac{MN}{DC}$ plus $\measuredangle ZMN = \measuredangle ZMT = \measuredangle ZDT = \measuredangle ZDC$ . So we obtain the direct similarity $\triangle ZMN \sim \triangle ZDC$ , hence $\triangle ZMD \sim \triangle ZNC$ , so $\overline{ZN} \perp \overline{AC}$ .

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djmathman

7938 posts

djmathman

#11 h May 15, 2017, 2:08 AM • 5 Y

Y by USHdog, amar_04, sabkx, Adventure10, Math_Kobekeye

Man, my oly geo is pretty rusty

Let $M$ , $N$ , and $E$ be the midpoints of $\overline{AB}$ , $\overline{AC}$ , and $\overline{BC}$ respectively, noting that $DE\perp BC$ . Remark that $\triangle CED\sim\triangle ZNC$ , so $\dfrac{DE}{EM}=\dfrac{DE}{CN}=\dfrac{DC}{CZ}.$ Furthermore, $\angle DEM=90^\circ+\angle MEB = 90^\circ + \angle ACB = (\angle ZCN+\angle DCE)+\angle ACB=\angle ZCD.$ Hence $\triangle DEM\sim\triangle DCZ$ , so by spiral similarity $\triangle DEC\sim\triangle DMZ\quad\implies\quad \angle DMZ=\angle DEC = 90^\circ.$ Combining this with $\angle DAZ=90^\circ$ establishes the desired cyclicity.

$[asy] size(250); defaultpen(linewidth(0.8)); pair A = dir(120), B = dir(215), C = dir(325), D = dir(270), M = (A+B)/2, N = (A+C)/2, E = (B+C)/2, P = dir(90), Z = extension(A,P,origin,N); draw(A--B--C--A--Z--C^^Z--N^^B--D--C^^unitcircle); draw(circumcircle(D,A,Z),linetype("4 4")); draw(D--M--E--cycle,linetype("8 8")); label("$A$",A,dir(origin--A)); label("$B$",B,dir(origin--B)); label("$C$",C,dir(origin--C)); label("$D$",D,dir(origin--D)); dot(M^^N^^E); label("$M$",M,dir(origin--M)); label("$Z$",Z,dir(N--Z)); label("$E$",E,NE); label("$N$",N,dir(Z--N)); draw(rightanglemark(Z,N,C,3)^^rightanglemark(D,E,C,3)); [/asy]$

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jayme

9795 posts

jayme

#12 h May 15, 2017, 10:09 AM • 3 Y

Y by Ankoganit, Plops, Adventure10

Dear Mathlinkers,
an outline of my proof

1. the circle with diameter CZ intersect (AZD) in T
2. TN is parallel to BC (angle chasing)
3. Reim's theorem for (O) and (AZD) and we are done...

Sincerely
Jean-Louis

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MarkBcc168

1595 posts

MarkBcc168

#13 h May 15, 2017, 10:10 AM • 3 Y

Y by sa2001, Adventure10, GeoKing

Took me 20 minutes during contest

Bary-bash :)

Remarks

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andrei.pantea

172 posts

andrei.pantea

#14 h May 15, 2017, 10:56 AM • 2 Y

Y by Adventure10, Mango247

ABCDE wrote:

Let $A'$ be the reflection of $A$ across $Z$ and $B'$ be the reflection of $B$ across $D$ . It is easy to see that $CB'B$ and $CAA'$ are similar right triangles so $AB'\perp A'B$ , which means that $\angle DMZ=90^\circ$ as $AB'\parallel DM,A'B\parallel MZ$ , immediately implying the conclusion.

Could you please explain more, why is $AB'\perp A'B$ ? Thank you!

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ABCDE

1963 posts

ABCDE

#15 h May 16, 2017, 12:08 AM • 2 Y

Y by Adventure10, Mango247

$CAB'$ and $CA'B$ are spirally similar with an angle of $90^\circ$ .

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Knty2006

50 posts

Knty2006

#68 h Mar 14, 2023, 8:55 AM

Y by

Let $E$ be the midpoint of $AC$ , and define $M$ to be the intersection of $(ZAD)$ with line $AB$

Claim: $Z$ is the spiral center sending $CE$ to $MD$

Proof:

Note that $\angle{ZMD}=\angle{ZAD}=90=\angle{ZEC}$

$\angle{CZE}=\angle{EZA}=\angle{CAD}=\angle{DAM}=\angle{DZM}$

So, $\triangle{CZE} \sim \triangle{DZM}$

Using our claim, we have that $Z$ is the spiral center sending $CD$ to $EM$

Which implies $\angle{ZEM}=\angle{ZCD}=\angle{ZCA}+\angle{ACB}+\angle{BCD}=90+\angle{ACB}$

$90+\angle{ACB}=\angle{ZEM}=90+\angle{AEM}$

This implies that $EM$ parallel to $CB$ , since $E$ is the midpoint of $AC$ it means that $M$ must be the midpoint of $AB$

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eibc

600 posts

eibc

#69 h Aug 21, 2023, 7:31 PM

Y by

this is pretty contrived but whatever

Let $E$ be the second intersection of $AZ$ with $(ABC)$ and $M$ be the midpoint of $\overline{AB}$ . Since $\overline{EA} \perp \overline{AD}$ , $E$ must be the midpoint of arc $BAC$ of $(ABC)$ . A quick angle chase now gives $\angle BEC = \angle BAC = \angle AZC$ , and since $\triangle EBC$ and $\triangle ZAC$ are both isosceles, there must a spiral similarity at $C$ taking $\triangle EBC$ to $\triangle ZAC$ .

In particular, this spiral similarity takes $\overline{BE}$ to $\overline{AZ}$ , so there also exists a spiral similarity $\tau$ at $C$ taking $\overline{BA}$ to $\overline{EZ}$ . So, $\tau(M) = M'$ is just the midpoint of $\overline{EZ}$ .

Claim: $\tau(D)$ is the center of $(ABC)$ .

Proof: Let $O$ be the center of $(ABC)$ . Then $O$ is the midpoint of diameter $\overline{ED}$ , and $O$ lies on the perpendicular bisector of $\overline{EC}$ . However, since $\angle DEC = \angle DBC$ and $\triangle ABC \sim \triangle ZEC$ (so lines $BD$ and $ED$ "correspond" to each other), we find that $\tau(D)$ must lie on $\overline{ED}$ . But $\tau(D)$ is also the intersection of the internal bisector of $\angle EZC$ with $(EZC)$ and hence lies on the perpendicular bisector of $\overline{EC}$ , so indeed $\tau(D) = O$ .

To finish, since $\overline{OM'} \parallel \overline{DZ}$ , we get
$\measuredangle AMD = \measuredangle ZM'O = \measuredangle M'ZD = \measuredangle AZD,$ so $M$ indeed lies on $(AZD)$ .

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shendrew7

796 posts

shendrew7

#70 h Oct 1, 2023, 9:31 PM

Y by

Let $M$ be the midpoint of $AB$ and $E$ , $F$ be the midpoints of arcs $\overarc{CA}$ , $\overarc{AB}$ .

$\textcolor{blue}{\textbf{Claim 1:}}$ The perpendicular bisector of $AD$ is parallel to $EF$ .
$\begin{align*} \angle (AC, EF) &= \frac 12 \left(\overarc{AF} + \overarc{DE}\right) \\ &= \frac 12 \left(\angle A + \angle B + \angle C\right) \\ &= 90.~{\color{blue} \Box} \end{align*}$
$\textcolor{blue}{\textbf{Claim 2:}}$ $OE$ and $OF$ are symmetric about the perpendicular bisector of $AD$ .

First we note $O$ lies on the perpendicular bisector. Since $\triangle OEF$ is isosceles, $\angle OEF = \angle OFE$ , and we have the conclusion by alternate interior angles. ${\color{blue} \Box}$

$\textcolor{blue}{\textbf{Claim 3:}}$ Pentagon $MYZAD$ is cyclic.

We know $AD$ and $AZ$ are the internal and external bisectors of $\angle BAC$ , respectively, so $\angle ZAD = 90$ . As a result, we can see each half of $YZAD$ is a rectangle, so $YZAD$ itself is a (cyclic) rectangle with diameter $AY$ . We also have

$\angle AMY = \angle AMO = 90,$
so $M$ also lies on $(ADZ)$ . $\blacksquare$

$[asy] size(250); pair A, B, C, O, M, X, D, E, F, Z, Y; A = dir(120); B = dir(210); C = dir(330); O = (0, 0); M = .5A + .5B; X = dir(90); D = dir(270); E = dir(45); F = dir(165); Z = extension(A, X, O, E); Y = D + Z - A; draw(A--B--C--cycle^^circumcircle(A, B, C)); draw(Y--F^^O--Z); draw(D--A--Z--Y--cycle^^E--F); draw(dir(195)--foot(O, Y, Z), red); draw(circumcircle(A, Y, Z), blue+dashed); dot("$A$", A, NW); dot("$B$", B, SW); dot("$C$", C, SE); dot("$D$", D, S); dot("$E$", E, 2dir(0)); dot("$F$", F, W); dot("$M$", M, dir(210)); dot("$Y$", Y, SE); dot("$Z$", Z, NE); dot("$O$", O, S); [/asy]$

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IAmTheHazard

5001 posts

IAmTheHazard

#71 h Oct 20, 2023, 5:20 PM • 1 Y

Y by centslordm

Apply $\sqrt{bc}$ inversion at $A$ , which yields the following problem.

Restated problem wrote:

Let $ABC$ be a triangle. Let $E$ be the intersection of the internal $\angle A$ -bisector and $\overline{BC}$ , $A'$ be the reflection of $A$ over $C$ , and $X$ be the point on the external $\angle A$ -bisector such that $\overline{BX} \parallel \overline{AC}$ . Prove that $X,E,A'$ collinear.

The reason this is the inverted problem statement is because $Z$ gets sent to the point $X$ on the external $\angle A$ -bisector such that $\measuredangle ACZ=\measuredangle AXB$ , so $\triangle ACZ \sim \triangle AXB$ . But since $\angle ZAC=90^\circ-\angle A/2$ , $\angle ABX=\angle AZC=\angle A$ , hence $\overline{BX} \parallel \overline{AC}$ .

It then suffices to prove that $\frac{EB}{EX}=\frac{EC}{EA'} \iff \frac{EB}{AB}=\frac{EC}{AC}$ . This is just the angle bisector theorem. $\blacksquare$

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Shreyasharma

682 posts

Shreyasharma

#72 h Feb 21, 2024, 7:32 AM • 1 Y

Y by ATGY

We barybash.

Set $\triangle ABC$ as reference and note that it suffices to show that $Z \in (AMD)$ . Note that $D$ can be parametrized as $(t: b: c)$ . Plugging this into circle equation we have,
$\begin{align*} a^2bc + b^2tc + c^2tb &= 0\\ \iff t &= -\frac{a^2bc}{b^2c + bc^2}\\ \iff t&= -\frac{a^2}{b + c} \end{align*}$ So we have $D = (-a^2 : b(b+c) : c(b+c))$ . Now we move onto computing $Z$ . Note that it can be parametrized as $(t: -b : c)$ . This should satisfy the equation of the perpendicular bisector of $\overline{AC}$ given by,
$\begin{align*} 0 = b^2(z - x) + y(c^2 - a^2) \end{align*}$ Plugging in we have,
$\begin{align*} 0 &= b^2(c - t) + (-b)(c^2 - a^2)\\ \frac{(c^2 - a^2)}{b} &=c - t\\ t &= \frac{bc - c^2 + a^2}{b} \end{align*}$ Thus we have $Z = (a^2 + bc - c^2, -b^2, bc)$ . Now consider the equation of the circle $(AMD)$ . It is given for constants $v$ and $w$ by,
$\begin{align*} -a^2yz - b^2xz - c^2xy + (vy + wz)(x+y+z) = 0 \end{align*}$ Plugging in the coordinates of $M = (1 : 1 : 0)$ we find that,
$\begin{align*} -c^2 + 2v &= 0\\ v &= \frac{c^2}{2} \end{align*}$ Then plugging in the coordinates of $D$ we find that,
$\begin{align*} -a^2bc(b+c)^2 + a^2b^2c(b+c) + a^2bc^2(b+c) + \left(\frac{bc^2(b+c)}{2} + wc(b+c) \right)(-a^2 + (b+c)^2) &= 0\\ -a^2bc(b+c) + a^2b^2c + a^2bc^2 + \left( \frac{bc^2}{2} + wc \right)(b + c - a)(a + b + c) &= 0\\ \left(\frac{bc^2}{2} + wc \right)(b + c - a)(a + b +c) &= 0\\ \frac{bc^2}{2} + wc = 0 \end{align*}$ so we find $w = \frac{-bc}{2}$ . Then our circle has equation,
$\begin{align*} -a^2yz - b^2xz - c^2xy + \left(\frac{c^2}{2}y - \frac{bc}{2}z\right)(x+y+z) = 0 \end{align*}$ It is easy to verify $Z$ satisfies this so we're done.

This post has been edited 1 time. Last edited by Shreyasharma, Feb 21, 2024, 7:33 AM

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john0512

4189 posts

john0512

#73 h Apr 21, 2024, 3:29 AM

Y by

We $\sqrt{bc}$ invert. The new problem becomes:

Quote:

In triangle $\triangle ABC$ , $M$ is the reflection of $A$ across $B$ , and $D$ is the foot of the internal bisector of $\angle A$ to $BC$ , and $Z$ is the point on the external bisector such that $AC=ZC$ . Prove that $M,D,Z$ are collinear.

Let $T$ be the intersection of $AZ$ and $BC$ . We will Menalaus on $\triangle ABT$ . We have by ratio lemma that $\frac{ZA}{ZT}=\frac{\sin\alpha}{\sin\beta}\cdot\frac{AC}{CT}=\frac{\sin\alpha}{\sin\beta}\cdot\frac{\sin\angle ATC}{\sin90-\alpha/2},$ $\frac{MB}{MA}=\frac{1}{2},$ and $\frac{DT}{DB}=\frac{1}{\sin\alpha/2}\cdot\frac{AT}{AB}=\frac{1}{\sin\alpha/2}\cdot\frac{\sin\beta}{\sin \angle ATC}.$
Thus, $\frac{ZA}{ZT}\cdot\frac{MB}{MA}\cdot\frac{DT}{DB}=\frac{\sin\alpha}{2\sin(90-\alpha/2)\sin\alpha/2}=1,$ as desired.

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N3bula

277 posts

N3bula

#74 h Sep 20, 2024, 12:12 PM

Y by

Let $K$ denote the midarc of $BAC$ , and define $Z$ as $(AMD)\cup AK$ , let $T$ be $MN \cup DC$ , $\angle DTM =\angle DCB=\angle DBC=\angle MAD$ , thus $T$ lies on $(AMD)$ , by spiral similarity
we have that $\triangle ZMD ~ \triangle KBD$ , thus $\frac{ZM}{ZD}=\frac{KB}{KD}$ , let $R$ denote $KD\cup BC$ , we have that $RK$ is the perpendicular bisector of $BC$ , and thus $MN=BR$ ,
we also have that $\angle KBD=90$ so $\triangle KBR ~ \triangle KBD$ , so $\frac{KB}{KD}=\frac{BR}{BD}=\frac{MN}{BD}$ , finally we have $\angle ZMN=\angle ZDC$ , thus $\triangle ZMN ~ \triangle ZDC$ ,
so by spiral similarity $\triangle ZNC ~ \triangle ZMD$ which suffices.

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Likeminded2017

391 posts

Likeminded2017

#75 h Oct 7, 2024, 4:25 AM

Y by

Used Evan's diagram for this problem, but with a different solution:

Let $D$ be the midpoint of arc $BC$ not containing $A,$ let $M$ be the intersection of $(ADZ)$ and $AB,$ and let $N$ be the midpoint of $AC.$ Observe $\angle ZMD=\angle ZAD=90^\circ$ and $\angle ZNC=90^\circ.$ Next,
$\angle CZN=\angle ZAN=\angle CAD=\angle MAD=\angle MZD.$ Thus $\triangle ZMD \sim \triangle ZNC.$ Then there exists a unique point $T=MN \cap CD$ that is on $(ZMD)$ and $(ZNC)$ by the fundamental lemma of spiral similarity. Now,
$\angle DTM=\angle DAM=\angle DAB=\angle DCB$ so $TM \parallel BC$ and $M$ must be the midpoint of $AB.$

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ihatemath123

3447 posts

ihatemath123

#76 h Nov 30, 2024, 9:11 PM

Y by

Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AC}$ , respectively. Now, invert at $A$ and refer to the images of points (other than $A$ ) by their original names. Let $Y$ be the foot from $M$ to the external bisector of $A$ so that $YMNZ$ is a trapezoid. Let $D'$ be the reflection of $A$ across $D$ . It's a well known property of trapezoids that, since $YA:AZ = YM : ZN$ and $D$ is the midpoint of $AD'$ , $D$ is the intersection of the diagonals $MZ$ and $YN$ . In particular, $M$ , $D$ and $Z$ are collinear, as desireed.

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andrewthenerd

17 posts

andrewthenerd

#77 h Dec 9, 2024, 4:18 AM

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Perform $\sqrt{\frac{bc}{2}}$ -inversion, we have
- $E \rightarrow C$ ( $E$ is midpoint of $AB$ ), $H \rightarrow B$
- $D \in (ABC) \implies D \rightarrow I$ where $I$ is intersection of $AF$ and $EH$
- $G$ goes to feet of perpendicular from $B$ to $AZ$
- $A \rightarrow A$

Suffice to prove that $G,I,C$ collinear, as it is equivalent to $(AEDZ)$ cyclic. Let $K = AC \cap GB$ , and redefine $G$ to be the intersection of parallel to $AF$ through $B$ and $CJ$ , we wish to prove that $AG \perp BG$ . Indeed, $\frac{AI}{IF} = \frac{KG}{GB}$ so $G$ is midpoint of $KB$ and $\angle BKC = \angle FAC = \angle BAF = \angle KBA$ so we are done. $\blacksquare$

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ehuseyinyigit

837 posts

ehuseyinyigit

#78 h Dec 14, 2024, 10:22 PM

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I don't have inversion skills yet, so lets prove classically.

Let $J$ and $H$ be midpoints of sides $AB$ and $AC$ , respectively. Let $JH$ meet $DC$ at $K$ . Since $\angle AJH=\angle ABC=\angle ADC$ , we have points $A$ , $D$ , $K$ and $J$ are concyclic. On the other hand by using the concyclic property we reached $\angle BAD=\angle HKC=\angle HZC\angle DAC=\angle HZC$ which implies points $C$ , $H$ , $K$ and $Z$ are concylic. We complete the proof as
$\angle ZKC=\pi-\angle ZHC=90^{\circ}=\angle DBZ$ implying $Z\in (AJDK)$ as desired.

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mananaban

36 posts

mananaban

#79 h Dec 27, 2024, 10:07 PM

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spiral spiral yes yes

Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$ , respectively. Note that the problem statement is true iff $\triangle ZMD \sim \triangle ZNC$ , a spiral similarity centered at $Z$ . Thus we complete the spiral similarity by considering $MN \cap CD = K$ . $ZNCK$ should be cyclic, so we let $\odot ZNC \cap MN = K'$ and prove $K=K'$ . We have
$\begin{align*} \angle NZK' &= 180 - \angle ZNK' - \angle ZKN \\ &= 180 - (90 - \angle C + \angle ZCN) \\ &= 90 + \angle C - \angle CAZ \\ &= \angle C +\frac{\angle A}{2} \\ \angle NZK' &= \angle ACD. \end{align*}$ Thus $DCK'$ is collinear and $K=K'$ . Then $AZKD$ is cyclic since $\angle ZKD = 90$ , so it is sufficient to prove that $AMZK$ is cyclic. This is evident since
$\angle ZKM = \angle ZCN = \angle ZAN = 180 - \angle MAZ.$ $\blacksquare$

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Mathgloggers

87 posts

Mathgloggers

#80 h Apr 10, 2025, 8:10 AM

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Let $M ,K$ be the midpoints of $AB,BC$

Take the homothety factor , $h(k)=2$ at $A$ ,consider the quadilateral $AZ'D'B$ . So if we can prove that this is cyclic so our initial condition would hold.
$\angle D'AZ'=90=\angle D'BA$ ,but we have $MD \parallel BD'$ so it remains to prove that $Z'B \perp MD$

CLAIM: $DKM \sim BCZ'$ by $90^{.}$
$\angle DKM=90+C=\angle BCZ'=\angle BCA+ \angle ACZ'=90+C$ ,

Also we want to have $\frac{KM}{KD}=\frac{CZ'}{CB} \implies \frac{KM}{CZ'}=\frac{KD}{CB}=\frac{KD}{2KC}=\frac{CA}{2CM}$

Now notice that ,
$DK \perp BC$ and $MK \parallel AC \perp CM$ hence we should also have
$BM \perp GD$ as required.

This post has been edited 2 times. Last edited by Mathgloggers, Apr 10, 2025, 11:20 AM
Reason: m

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Ilikeminecraft

652 posts

Ilikeminecraft

#81 h May 11, 2025, 9:34 PM

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Let $M$ denote the second intersection of $AB$ with $(ADZ).$ Let $N$ denote the antipode of $D$ in $(ABC).$ Let $N'$ denote the reflection of $N$ across $Z.$ Let $E$ denote the second intersection of $ZC$ with $(ABC).$

Claim: $DEN'$ are collinear
Proof: First note that $\angle NED = 90.$ Furthermore, $Z$ must be the circumcenter of $NN'E$ since $ZE = ZN = ZN'.$ However, $Z$ lies on $NN',$ implying that $\angle NEN' = 90.$

Claim: $\angle NN'D = \angle MAD$
Proof: Observe that $\angle NN'D = 90 - \angle ADE = 90 - \angle ADN - \angle NDE = \angle DNE - \angle CAE = \angle DAE = \angle MAD$ as desired.

Finally, this implies spiral sim centered at $D,$ which implies our problem.

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Ihatecombin

61 posts

Ihatecombin

#82 h Yesterday at 3:34 PM

Y by

Trig bash strikes again. Define $\alpha = \angle CAB$ , $\beta = \angle ABC$ , $\gamma = \angle BCA$ .
Define $M$ as the midpoint of $AB$ and define $X$ as the center of $\triangle AMD$ , and define $Z'$ as the reflection of $D$ over $X$ . Clearly $Z'$ lies on $(AMD)$ , thus it suffices to show that $Z' = Z$ . Notice that since $\angle ZAD = 90$ , it follows that $2ZA \cos(90 - \frac{\alpha}{2}) = AC$ . Thus
$ZA = \frac{AC}{2\cos(90 - \frac{\alpha}{2})} = \frac{AC}{2\sin(\frac{\alpha}{2})}$ Since $\angle ZAD = 90$ , it follows that $ZA^2 + AD^2 = ZD^2$ . Since we wish to show that $Z' = Z$ , it suffices to show that $Z'D = ZD$ (notice that $\angle Z'AD = \angle ZAD = 90$ ).
We know that $Z'D = 2DX = \frac{MD}{\sin(\frac{\alpha}{2})}$ . Thus it suffices to show that
$\left(\frac{MD}{\sin(\frac{\alpha}{2})}\right)^2 = \left(\frac{AC}{2\sin(\frac{\alpha}{2})}\right)^2 + AD^2$ However
$\left(\frac{MD}{\sin(\frac{\alpha}{2})}\right)^2 = \left(\frac{AC}{2\sin(\frac{\alpha}{2})}\right)^2 + AD^2 \iff AC^2 = (2MD)^2 - \left(2\sin(\frac{\alpha}{2})AD\right)^2$ We can define the altitude from $D$ to $AB$ as $X$ . It will suffice to show that
$2MX = AC$ Which is not that difficult

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