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Source: 2024 USAMO Problem 5, JMO Problem 6

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DottedCaculator

7303 posts

DottedCaculator

#65 h Mar 23, 2024, 2:54 PM • 5 Y

Y by IAmTheHazard, scannose, ESAOPS, crazyeyemoody907, Alex-131

crazyeyemoody907 wrote:

(imagine bashing it)

also crazyeyemoody907 wrote:

The second is lengths- with the direction of line $D_1D_2$ down, we need to show
$\text{dist}(C,\overline{D_1D_2})=b\sin C$ using the usual $a=BC$ and $A=\angle BAC$ shorthand (angles undirected for $A,B,C$ ). Indeed:
$\text{dist}(C,\overline{D_1D_2})\overset{s^{-1}}= \frac{CC'}{CB}\text{dist}(C,\ell) =\frac{b\tan A}a \text{dist}(A,\text{foot}(B,\overline{AC}))=\frac{b\sin A}{a\cos A}c\cos A =bc\frac{\sin A}a=b\sin C.$ (last line by law of sines)

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#66 h Mar 23, 2024, 3:20 PM • 1 Y

Y by R8kt

Let $BD$ intersect $\odot(ADC)$ again at $X$ . $\angle BAX = \angle BAC + \angle CAX = \angle BAC+\angle CDX = 90^\circ.$ Moreover, $\angle BXC = \angle DAC = \angle BCA.$ Thus, letting $D'$ be the a point on $AC$ such that $BD'=BC$ , we have $B,C,D',X$ are concyclic on a circle $\omega$ .

Let $Y$ be the intersection of $BA$ and $\omega$ . The inversion at $B$ with radius $BC$ sends $A'\to X$ and $A\to Y$ . That is, $\angle YA'B = \angle BAX = 90^\circ$ . Let $H$ be the orthocenter of $\triangle BXY.$ Hence, $H'$ , the reflection of $H$ over $BX$ , lies on $\omega$ , the circumcircle of $\triangle BXY$ .

Let $Z$ be the reflection of $B$ across $E$ . Let $P_B,P_Z$ be the projection of $B,Z$ on $AC$ . Since the midpoint of $BZ$ lies on the bisector of $AC$ , $AP_B = CP_Z$ . Also, cyclic $HA'BA$ implies $\triangle BAP_B\sim BA'H$ . Now, $\frac{ZP_Z}{P_ZC} = \frac{BP_B \cdot A'Z/BA'}{AP_B}=\frac{BP_B}{AP_B}\cdot \frac{A'Z}{BA'} = \frac{BA'}{A'H}\cdot \frac{A'Z}{B'A} = \frac{A'Z}{A'H'}$ That is, $\triangle ZP_ZC \sim \triangle ZA'H'$ , and $Z,C,A',H$ are concyclic.

Let $ZC$ intersect $\omega$ again at $C'$ . By Reim's Theorem, $YC' \parallel A'Z$ . That is $\angle BCZ = \angle BXC' = \angle XBY = \angle ABA'$ . Note that $CZ \parallel ME$ , so $\angle BCE = \angle BCZ = \angle ABA' = \angle ABE$ , which means $AB$ is tangent to $BEM$ as desired.

Remark Replacing $D$ with $X$ in this problem seems to be a good idea, but it is not.

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624 posts

#67 h Mar 24, 2024, 3:03 AM

Y by

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -5.251487363189673, xmax = -0.47214371509479935, ymin = 0.9997359177069542, ymax = 5.337909382900762; /* image dimensions */ pair E2 = (-3.1073786161585977,3.426981657246437), E1 = (-2.3139719839488744,4.239503969108282), D2 = (-2.865570819495604,1.7542575305395893), D1 = (-2.9208621797355425,3.9871396950503657), T = (-2.9119472921690517,3.6271215410484516), C = (-2.0399616132538543,3.881136226214614), B = (-3.1731868306193642,3.8822150681750345), C1 = (-1.6179607179072883,3.4690634692226383), B1 = (-3.597099177310678,2.286482284501728), A = (-2.94,4.76); draw(D1--C--D2--cycle, linewidth(2) + lightblue); draw(E1--C--E2--cycle, linewidth(2) + lightgreen); draw(E1--C--E2, linewidth(1)); draw(D1--C--D2, linewidth(1)); /* draw figures */ draw(circle((-2.6075299476089824,2.8777728768621826), 1.1527670297967343), linewidth(1)); draw(circle((-2.6067169714891376,3.7317302440423794), 0.5861175510059863), linewidth(1)); draw((-2.94,4.76)--(-3.597099177310678,2.286482284501728), linewidth(1)); draw((-2.94,4.76)--(-1.6179607179072883,3.4690634692226383), linewidth(1)); draw((-3.597099177310678,2.286482284501728)--(-1.6179607179072883,3.4690634692226383), linewidth(1)); draw((-3.597099177310678,2.286482284501728)--(-2.0399616132538543,3.881136226214614), linewidth(1)); draw((-3.1731868306193642,3.8822150681750345)--(-1.6179607179072883,3.4690634692226383), linewidth(1)); draw((-3.1731868306193642,3.8822150681750345)--(-2.3139719839488744,4.239503969108282), linewidth(1)); draw((-3.1731868306193642,3.8822150681750345)--(-2.865570819495604,1.7542575305395893), linewidth(1)); draw((-2.94,4.76)--(-2.865570819495604,1.7542575305395893), linewidth(1)); draw((-2.3139719839488744,4.239503969108282)--(-3.1073786161585977,3.426981657246437), linewidth(1) + linetype("2 2")); draw((-3.1731868306193642,3.8822150681750345)--(-2.9119472921690517,3.6271215410484516), linewidth(1) + linetype("2 2")); /* dots and labels */ dot((-2.94,4.76),dotstyle); label("$A$", (-2.9169618120048693,4.823210836182852), NE * labelscalefactor); dot((-3.597099177310678,2.286482284501728),dotstyle); label("$B'$", (-3.5725897227050636,2.3477559210157644), SW * labelscalefactor * 4); dot((-1.6179607179072883,3.4690634692226383),dotstyle); label("$C'$", (-1.5934512633016735,3.5303371057366753), NE * labelscalefactor * 0.5); dot((-3.1731868306193642,3.8822150681750345),linewidth(4pt) + dotstyle); label("$B$", (-3.1498016307582093,3.928615743077915), NW * labelscalefactor); dot((-2.0399616132538543,3.881136226214614),linewidth(4pt) + dotstyle); label("$C$", (-2.0162393552485276,3.928615743077915), NE * labelscalefactor); dot((-2.9119472921690517,3.6271215410484516),linewidth(4pt) + dotstyle); label("$T$", (-2.886324993747851,3.6773938333703637), dir(60) * labelscalefactor); dot((-2.9208621797355425,3.9871396950503657),linewidth(4pt) + dotstyle); label("$D_{1}$", (-2.898579721050658,4.038908288803181), NE * labelscalefactor); dot((-2.865570819495604,1.7542575305395893),linewidth(4pt) + dotstyle); label("$D_{2}$", (-2.843433448188025,1.8024205560408368), dir(20) * labelscalefactor); dot((-2.3139719839488744,4.239503969108282),linewidth(4pt) + dotstyle); label("$E_{1}$", (-2.2919707195616934,4.290130198510732), NE * labelscalefactor * 0.5); dot((-3.1073786161585977,3.426981657246437),linewidth(4pt) + dotstyle); label("$E_{2}$", (-3.2600941764834754,3.3710256508001795), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Interpret the $\pi/2+\angle A$ condition by constructing $B' \in AB, C' \in AC$ such that $B,C$ are the feet from $B',C'$ as shown; then the condition is that $D \in (B'C')$ and $\angle DAC = \angle C$ . Naturally, we ignore the condition that $D$ is inside $\triangle ABC$ , so we actually have two points $D_1,D_2$ as shown.

Construct the point $T$ such that $ABTC$ is an isosceles trapezoid as shown. It's clear that $T$ is collinear with $AD_1D_2$ . Draw the circle $\omega$ through $B,C$ tangent to $AB$ . Let $BD_1, BD_2$ intersect $\omega$ again at $E_1,E_2$ . By a homothety of ratio $\tfrac12$ at $B$ , it suffices to show that $\overline{E_1E_2T}$ are collinear and perpendicular to $AC$ . (The midpoint of $BE_1$ will be the point $E$ in the original problem).

Indeed, note $C$ is the spiral center sending $(B,E_1,E_2) \mapsto (B',D_1,D_2)$ because this is how spiral similarity works. The rotation angle of this spiral similarity is $\angle BCB' = \pi/2 - \angle C$ . Hence $\angle(E_1E_2,D_1D_2) = \pi/2-\angle C \implies E_1E_2 \perp AC$ .

It remains to show $T \in E_1E_2$ , which follows from letting $T' \equiv E_1E_2 \cap D_1D_2$ so that
$\angle AT'C = \angle D_1T'C = \pi - \angle D_1 E_1 C = \angle ABC$ hence $ABCT'$ is cyclic $\implies T=T'$ .

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7578 posts

#68 h Mar 24, 2024, 3:29 AM • 1 Y

Y by Danielzh

well i didn't solve but here's how I would motivate the easiest solution

If $D'=AD\cap BC$ , we get the most goofy isosceles triangle ever. Let's draw in the perpendicular bisector of $AC$ and rotate $AC$ to be horizontal.

The tangency is annoying. Let $E'$ be the point where $BE'$ has midpoint $E$ . In particular we notice two points that have the same $x$ -coordinate as $E'$ . Let $P$ be the point such that $BPCA$ is an isosceles trapezoid; let $Q$ be the antipode of $B$ in $(ABC)$ .

Notice that $P\in AD$ . That's a pretty nice green flag. Can we do anything else? As it turns out, we know the angle $\angle E'PC$ . We also know the angle $\angle E'DC$ (this is pretty straightforward arguments of lines intuition). Quick check reveals the stunning cyclic quadrilateral, and we are home free. $\blacksquare$

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52 posts

#69 h Mar 24, 2024, 2:41 PM • 1 Y

Y by Danielzh

Finally $B'$ worked...

Let $\omega$ denote the circumcircle of triangle $ABC$ and let $O$ be its center. We also let $B'$ and $C'$ be the points on lines $(AC)$ and $(AB)$ respectively such that $(BB')\perp (AB)$ and $(CC')\perp (AC)$ , and let $\gamma$ be the circle with diameter $B'C'$ , passing through $B$ and $C$ . By the angle condition, $\gamma$ also passes through $D$ . Finally, let $A'=(BB')\cap (CC')$ be the antipode of $A$ in $\omega$ , let $P$ be on $(BB')$ such that $PB=PC$ , and let $J=(AD)\cap (BC)$ , so that $JA=JC$ .
The proof completely relies on this point $X$ :
Claim : Circles $\gamma$ , $(BMP)$ and $(OCA')$ all intersect at a point $X$ .
Proof : First note that $\angle POC=\angle A=180-\angle PA'C$ , so $P$ lies on $(OCA')$ . We now define $X\ne P$ to be the point where $(BMP)$ and $(OCA'P)$ meet.
Angle chasing gives $\angle BXC=\angle BXP+\angle PXC=\angle BMP+\angle POC=90+\angle A$ . $\square$

Now, note that $\angle OBC=\angle BB'C=90-\angle A$ , so $OB$ and $OC$ are tangent to $\gamma$ . In particular, the inversion w.r.t $\omega$ fixes $\gamma$ . Since it also fixes $A'$ and $C$ , it maps $X$ to the second intersection of $(CA')$ with $\gamma$ , i.e. $C'$ .
Inverting back gives that $X$ lies on $(AOB)$ , given that $C'$ lies on $(AB)$ . Furthermore, since $\angle BJA=2\angle ACB=\angle AOB$ , we have that points $A$ , $B$ , $J$ , $X$ and $O$ are all concyclic.
Therefore, we have $\angle EJX=\angle OJX=\angle OBX=180-\angle BDX=\angle EDX$ , where we also used the fact that $OB$ is tangent to $\gamma$ . This tells us that points $E$ , $J$ , $D$ and $X$ are concyclic.

Finally, we have $\angle BEX=\angle DEX=180-\angle DJX=\angle AOX=\angle XPA'$ , so that $E$ lies on $(BPMX)$ .
Circle $(BEM)$ thus has diameter $(BP)$ , which is perpendicular to $(AB)$ . This concludes the proof.
$\blacksquare$

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486 posts

#70 h Mar 30, 2024, 8:58 PM

Y by

Bro im such a clown ? (didn't finish writeup oops :wallbash_red:

:wallbash_red:

:blush:

:oops_sign:

:noo:

:noo:

:rotfl: )

Solution

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#71 h Apr 11, 2024, 5:11 PM

Y by

khina wrote:

clean

How is $B’E’ \perp AC$ ?

This post has been edited 1 time. Last edited by NuMBeRaToRiC, Apr 11, 2024, 5:12 PM

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#74 h Apr 22, 2024, 9:24 AM

Y by

Wow.

Define $B'$ and $D'$ be the reflections of $B$ and $D$ about $\overline{AC}$ . Redefine $E$ as the point on the perpendicular bisector of $\overline{AC}$ such that $(BEM)$ is tangent to $\overline{AB}$ . Then we may redefine $D = \overline{AB'} \cap \overline{BE}$ . It suffices then to show that $\angle BDC = 90 + \angle BAC$ .

Note that,
$\begin{align*} \angle BDC &= \angle BDD' + \angle CDD'\\ &= (180 - \angle BB'D') + \angle DCA\\ &= 180 - (\angle BB'C - \angle EB'C) + \angle DCA\\ &= 180 - (180 - \angle BAC) + \angle EBA + \angle DCA\\ &= \angle BAC + \angle EMB + \angle DCA\\ \end{align*}$ Thus the problem reduces to showing $\angle EMB + \angle DCA = 90$ . Let $F$ be on $\overline{BE}$ so that $\angle BCF = \angle BME$ . Then if $N$ is the midpoint of $\overline{AC}$ it follows that $\triangle BCF \sim \triangle ACG$ as
$\begin{align*} \angle ACG = \angle ABG = \angle AME = \angle ACF \end{align*}$ and we have,
$\begin{align*} \angle CAG = \angle CBG = \angle CBF \end{align*}$ However then it follows that $B'DFC$ is cyclic as $\angle DFC = \angle AGC = 180 - \angle DB'C$ Now we will let $K = \overline{AD'} \cap \overline{CD}$ and angle chase to find,
$\begin{align*} \angle D'ME = \angle EBA = \angle EB'C &= (90 - \angle EB'B) + \angle FB'C\\ &= 90 + \angle FDC - \angle EDD' \\ &= 90 - \angle DCA \end{align*}$ as desired. $\square$

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CyclicISLscelesTrapezoid

371 posts

CyclicISLscelesTrapezoid

#75 h May 15, 2024, 7:11 PM • 1 Y

Y by crazyeyemoody907

Here's my in-contest solution. Configuration issues are ignored, and I give no guarantee that this solution is without typoes.

Let $O$ be the circumcenter of $ABC$ , and let $K=\overline{AD} \cap \overline{BC}$ . Notice that $\angle CAD=\angle ACB$ implies $KA=KC$ , so $E$ lies on $\overline{OK}$ . Notice that $\angle AKB=\angle DAC+\angle DCA=2\angle ACB=\angle AOB$ , so $AOKB$ is cyclic.

Now, we invert about $B$ , denoting images with $*$ . Notice the following:

$M^*$ is the reflection of $B$ over $C^*$ .
$(ABC)$ maps to $\overline{A^*C^*}$ , so $O^*$ is the reflection of $B$ over $\overline{A^*C^*}$ .
$K^*=\overline{BC^*} \cap \overline{A^*O^*}$ , so $\overline{A^*C^*}$ bisects $\angle BA^*K^*$ .
$D^*$ satisfies $\angle BC^*D^*=\angle BC^*A^*+90^\circ$ , so $D^*$ lies on the line through $C^*$ perpendicular to $\overline{A^*C^*}$ . It also lies on the inverse of $\overline{AK}$ , which is $(BA^*K^*)$ .
$E^*$ is the second intersection of $\overline{BD^*}$ and $(BO^*K^*)$ .
It suffices to show the inverse of the circle through $B$ and $M$ tangent to $\overline{AB}$ passes through $E^*$ . This inverse must be parallel to $\overline{A^*B}$ , so it is the line through $M^*$ parallel to $\overline{A^*B}$ , or the reflection of $\overline{A^*B}$ over $C^*$ .

Notice that the line perpendicular to $\overline{A^*C^*}$ at $C^*$ intersects $(BA^*K^*)$ at two points, one of which is $D^*$ . Let $D^*=D_1$ , and let the other section be $D_2$ . Let $E^*=E$ , and let $\overline{BD_2}$ intersect $(BO^*K^*)$ again at $E_2$ . We will prove simultaneously that $E_1$ and $E_2$ lie on the reflection of $\overline{A^*B}$ over $C^*$ .

Notice that $\angle K^*D_1D_2=\angle K^*BD_2=\angle K^*BE_2=\angle K^*E_1E_2$ and $\angle K^*D_2D_1=\angle K^*BD_1=\angle K^*BE_1=\angle K^*E_2E_1$ , so $K^*D_1D_2 \sim K^*E_1E_2$ .

Claim: $\overline{E_1E_2} \parallel \overline{A^*B}$ .
Proof: This spiral similarity at $K^*$ mapping $D_1$ to $E_1$ maps $D_2$ to $E_2$ , as shown above. This spiral similarity has an angle of rotation of
$\angle D_1K^*E_1=\angle BE_1K^*-\angle BD_1K^*=\angle BO^*K^*-\angle BA^*K^*=180^\circ-\angle BO^*A^*-\angle BA^*O^*=\angle A^*BO^*.$ Notice that a rotation counterclockwise of $\angle A^*BO^*$ maps $\overline{BO^*}$ to $\overline{BA^*}$ . Since $\overline{BO^*}$ and $\overline{D_1D_2}$ are perpendicular to $\overline{A^*C^*}$ , we have $\overline{BO^*} \parallel \overline{D_1D_2}$ . Thus, a rotation of $\angle A^*BO^*$ must map $\overline{D_1D_2}$ to a line parallel to $\overline{BA^*}$ , as desired. $\square$

Let the spiral similarity mapping $D_1$ to $E_1$ at $K^*$ map $C^*$ to $L$ . We have
$\angle K^*D_1E_1=\angle BA^*K^*$ $\angle K^*E_1D_1=180^\circ-\angle BE_1K^*=180^\circ-\angle BO^*K^*=\angle A^*O^*B$ $\angle D_1K^*E_1=\angle A^*BO^*=\angle A^*O^*B,$ where the first equality of the last equation has been shown above. Thus, the corresponding angles of $K^*C^*L$ must have these measures too. In particular, $\angle K^*C^*L=\angle BA^*K^*$ and $\angle K^*LC^*=\angle LK^*C^*$ , which implies $C^*L=C^*K^*$ . Notice that since $C^*$ lies on $\overline{D_1D_2}$ , we know that $L$ lies on $\overline{E_1E_2}$ .

Claim: $L$ lies on the reflection of $\overline{A^*B}$ over $C^*$ .
Proof: Let $\overline{C^*L}$ and $\overline{A^*B}$ intersect at $L'$ . We have (in directed angles modulo $180^\circ$ )
$\measuredangle K^*C^*L=\measuredangle K^*A^*B \implies \measuredangle K^*C^*L'=\measuredangle K^*A^*B=\measuredangle K^*A^*L',$ so $A^*L'C^*K^*$ is cyclic. Furthermore, $\overline{A^*C^*}$ bisecting $\angle L'A^*K^*$ implies $C^*L'=C^*K^*$ . Thus, $C^*L'=C^*K^*=C^*L$ , so $L'$ is the reflection of $L$ over $C^*$ . Therefore, $L$ lies on the reflection of $A^*B$ over $C^*$ . $\square$

By the previous two claims, $E_1$ and $E_2$ lie on the reflection of $\overline{A^*B}$ over $C^*$ , as desired. $\square$

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5123 posts

Zhaom

#76 h May 20, 2024, 1:22 AM • 2 Y

Y by OronSH, peace09

Without loss of generality assume that $BC\ge{}AB$ . Let $X$ be the point on $(ABC)$ such that $(A,X;B,C)=-1$ .

Claim $1$ . We have that $(BMX)$ is tangent to $\overline{AB}$ .

Proof. Since $\overline{XM}$ and $\overline{XA}$ are isogonal with respect to $\angle{}BXC$ , we see that
$\angle{}BXM=\angle{}CXA=\angle{}ABC,$ proving the claim.

Now, perform force overlaid inversion at $A$ swapping $B$ and $C$ . We get the following problem.

Let $D$ be the point in the exterior of $\triangle{}ABC$ on the circle with diameter $\overline{BC}$ such that $\angle{}BCA=\angle{}DAB$ . Let $(CAD)$ intersect the circle centered at $B$ through $A$ at $E$ other than $A$ . Then, let $M$ be the midpoint of $\overline{BC}$ . Prove that $(CME)$ is tangent to $\overline{CA}$ .

Claim $2$ . The circle centered at $B$ through $A$ swaps with the $B$ -Apollonius circle of $\triangle{}ABC$ under force overlaid inversion at $C$ swapping $A$ and $B$ .

Proof. If $P$ is an intersection of the circle centered at $B$ through $A$ with $\overline{BC}$ , then $\overline{AP}$ is parallel to the internal or external angle bisector of $\angle{}ABC$ , and both can occur. If $P'$ is the image of $P$ under force overlaid inversion at $C$ swapping $A$ and $B$ , then $\overline{AP}\parallel\overline{BP'}$ , so $P'$ is the intersection of the internal or external angle bisector of $\angle{}ABC$ with $\overline{CA}$ , and both can occur. The image of the circle centered at $B$ through $A$ under force overlaid inversion at $C$ swapping $A$ and $B$ is the circle through the intersections of the internal and external angle bisector of $\angle{}ABC$ with $\overline{CA}$ and the point $B$ , which is the $B$ -Apollonius circle of $\triangle{}ABC$ .

Claim $3$ . Let $Z$ be the point on $\overline{BC}$ so that $\angle{}BCA=\angle{}ZAB$ . The image of $Z$ under force overlaid inversion at $C$ swapping $A$ and $B$ is the center of the $B$ -Apollonius circle of $\triangle{}ABC$ .

Proof. We see that $\angle{}CAZ=\angle{}CAB-\angle{}BCA$ , meaning that $\overline{AZ}$ is parallel to the tangent to $(ABC)$ at $B$ , proving the claim.

Now, we will perform force overlaid inversion at $C$ swapping $A$ and $B$ . We get the following problem.

Let $\omega$ with center $Z$ be the $B$ -Apollonius circle of $\triangle{}ABC$ . The perpendicular from $A$ to $\overline{CA}$ intersects $(BCZ)$ at a point $D$ , and the intersection of $\overline{BD}$ with $\omega$ other than $B$ is $E$ . Also, let $M$ be the reflection of $C$ over $A$ . Prove that $\overline{EM}\parallel\overline{BC}$ for some choice of $D$ .

We will prove that it is true for all choices of $D$ . Let those choices be $D$ and $D'$ and let the corresponding choices of $E$ be $E$ and $E'$ , respectively.

Claim $4$ . We have that $\overline{AB},(BCZ),$ and $\omega$ concur at two points $B$ and $J$ .

Proof. Note that $\overline{AB}$ and $(BCZ)$ swap under inversion at $\omega$ , proving the claim.

Claim $5$ . There exists a fixed spiral similarity $\psi$ centered at $J$ sending a point $P$ on $(BCZ)$ to the intersection $Q$ of $\overline{BP}$ and $\omega$ other than $B$ .

Proof. Fix one such pair $(P,Q)$ to be $\left(P_1,Q_1\right)$ . Then, note that $J$ is the Miquel point of $PQQ_1P_1$ , and therefore the fixed spiral similarity centered at $J$ sending $P_1$ to $Q_1$ sends $P$ and $Q$ .

Now, if $\psi$ sends the perpendicular from $A$ to $\overline{CA}$ to the line through $M$ parallel to $\overline{CA}$ , then we are done.

Claim $6$ . We have that $\psi$ rotates an angle of $90^\circ-\angle{}BCA$ .

Proof. The angle between the tangents to $\omega$ and $(BCZ)$ at $B$ is the obtuse angle between the tangent to $(BCZ)$ at $B$ and $\overline{BZ}$ minus $90^\circ$ , which is $180^\circ-\angle{}BCA-90^\circ=90^\circ-\angle{}BCA$ .

The angle between the perpendicular from $A$ to $\overline{CA}$ and the line through $M$ parallel to $\overline{CA}$ is also $90^\circ-\angle{}BCA$ , so it suffices that $\psi(A)$ is on the line through $M$ parallel to $\overline{BC}$ .

Claim $7$ . We have $A\psi(A)=AJ$ .

Proof. Note that $\psi(Z)$ is on $\omega$ , so $Z\psi(Z)=ZJ$ , implying the claim.

Claim $8$ . We have that $\psi(A)$ is a $2\angle{}BCA$ rotation of $J$ around $A$ .

Proof. This follows from claim $7$ since
$\angle{}JA\psi(A)=180^\circ-2\angle{}AJ\psi(A)=180^\circ-2\left(90^\circ-\angle{}BCA\right)=2\angle{}BCA$ by claim $6$ .

Reflecting $\psi(A)$ around $A$ , it suffices that a $180^\circ-2\angle{}BCA$ rotation of $J$ around $A$ is on $\overline{BC}$ , so it suffices that a $180^\circ-2\angle{}BCA$ rotation of $\overline{BC}$ around $J$ goes through $\overline{BC}$ . However, this line is $\overline{JC}$ since $\overline{CA}$ bisects $\angle{}BCJ$ since $ZB=ZJ$ and therefore $\angle{}BCJ=2\angle{}BCA$ , so we are done.

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#77 h Jun 9, 2024, 11:39 PM

Y by

If we have a point $D$ selected outside acute $\triangle ABC$ so that $\angle DAC = \angle ACB$ and $\angle BDC = 90^{\circ} - \angle BAC$ , and everything else defined as in the original statement, the problem is also true. The proof is similar to the original problem.

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mcmp

#78 h Oct 28, 2024, 9:35 AM • 1 Y

Y by Rounak_iitr

I do not know what got to me when I solved this problem, nor do I know how I even thought of it. Nevertheless, it is time for me to present:

The most American-as-that can-possibly-could solution to this one absolute wild beast of a geo.

USAMO 2024/5 wrote:

Point $D$ is selected inside acute $\triangle ABC$ so that $\angle DAC = \angle ACB$ and $\angle BDC = 90^{\circ} + \angle BAC$ . Point $E$ is chosen on ray $BD$ so that $AE = EC$ . Let $M$ be the midpoint of $BC$ .

Show that line $AB$ is tangent to the circumcircle of triangle $BEM$ .

Proposed by Anton Trygub

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draw((4.914428342473825,-8.709098349811534)--(-0.0421506280239867,9.999911165833284), linewidth(0.5) + ccqqqq); draw(circle((-9.254715806054826,-16.70455790202032), 16.269358295667054), linewidth(0.5) + qqwuqq); draw((-9.989787996616482,-0.45181388054956395)--(-25.52392957423316,-16.77313426899128), linewidth(0.5) + ffvvqq); draw((-25.52392957423316,-16.77313426899128)--(4.914428342473825,-8.709098349811534), linewidth(0.5) + ffvvqq); draw((-9.989787996616482,-0.45181388054956395)--(7.014497962123509,-16.635981535049357), linewidth(0.5) + ffvvqq); draw((4.914428342473825,-8.709098349811534)--(7.014497962123509,-16.635981535049357), linewidth(0.5) + ffvvqq); draw((-0.0421506280239867,9.999911165833284)--(0.0421506280239867,-9.999911165833284), linewidth(0.5) + ffvvqq); draw((-25.52392957423316,-16.77313426899128)--(7.014497962123509,-16.635981535049357), linewidth(0.5) + ffvvqq); draw((-9.989787996616482,-0.45181388054956395)--(0,0), linewidth(0.5) + ffvvqq); draw((0,0)--(4.914428342473825,-8.709098349811534), linewidth(0.5) + ffvvqq); draw(circle((16.269213768178343,0.06857636697096868), 19.096911251634147), linewidth(0.5) + qqwuqq); draw((-2.7362438814359997,-1.7981316663075286)--(0.0421506280239867,-9.999911165833284), linewidth(0.5) + ffvvqq); draw((-25.52392957423316,-16.77313426899128)--(-2.7362438814359997,-1.7981316663075286), linewidth(0.5) + ffvvqq); draw(circle((-6.801332963624456,-3.486489428701771), 4.401761145265004), linewidth(0.5) + qqwuqq); draw((-2.7362438814359997,-1.7981316663075286)--(0,0), linewidth(0.5) + ffvvqq); draw((-7.3098051055596365,-1.936587088394866)--(0,0), linewidth(0.5) + ffvvqq); draw((-7.3098051055596365,-1.936587088394866)--(-0.0421506280239867,9.999911165833284), linewidth(0.5) + ffvvqq); draw((-2.7362438814359997,-1.7981316663075286)--(-2.537679827071328,-4.5804561151805485), linewidth(0.5) + ffvvqq); draw((-2.7362438814359997,-1.7981316663075286)--(-9.989787996616482,-0.45181388054956395), linewidth(0.5) + ffvvqq); draw((-2.7362438814359997,-1.7981316663075286)--(-0.0421506280239867,9.999911165833284), linewidth(0.5) + ffvvqq); draw(circle((-26.729721794923844,9.887420378018781), 26.687808245820563), linewidth(0.5) + qqwuqq); draw(circle((-5.004844411628021,-2.467897198388346), 2.3654036090170742), linewidth(0.5) + qqwuqq); draw((-9.989787996616482,-0.45181388054956395)--(-3.2652805802903995,-0.865070972544453), linewidth(0.5) + ffvvqq); draw((-7.3098051055596365,-1.936587088394866)--(-2.7362438814359997,-1.7981316663075286), linewidth(0.5) + ffvvqq); /* dots and labels */ dot((0,0),dotstyle); label("$O$", (0.16378305715615946,0.41247255747594136), NE * labelscalefactor); dot((-0.0421506280239867,9.999911165833284),dotstyle); label("$A$", (0.12064408040383652,10.420715164014931), NE * labelscalefactor); dot((-9.989787996616482,-0.45181388054956395),dotstyle); label("$B$", (-9.80132057263044,-0.018917210047290897), NE * labelscalefactor); dot((4.914428342473825,-8.709098349811534),dotstyle); label("$C$", (5.081626406920974,-8.258461769741027), NE * labelscalefactor); dot((7.014497962123509,-16.635981535049357),dotstyle); label("$X$", (7.195436267784798,-16.1960334921685), NE * labelscalefactor); dot((-25.52392957423316,-16.77313426899128),dotstyle); label("$Y$", (-25.331352203466697,-16.325450422425472), NE * labelscalefactor); dot((-7.3098051055596365,-1.936587088394866),dotstyle); label("$F$", (-7.126704013986417,-1.4856424196262805), NE * labelscalefactor); dot((-2.537679827071328,-4.5804561151805485),dotstyle); label("$M$", (-2.381416571230894,-4.1602589782703205), NE * labelscalefactor); dot((0.0421506280239867,-9.999911165833284),dotstyle); label("$G$", (0.20692203390848238,-9.552631072310724), NE * labelscalefactor); dot((-6.535054525477722,-0.6641258046072265),dotstyle); label("$D$", (-6.350202432444604,-0.23461209380890705), NE * labelscalefactor); dot((-3.2652805802903995,-0.865070972544453),dotstyle); label("$E$", (-3.0716401992680606,-0.4503069775705232), NE * labelscalefactor); dot((-2.7362438814359997,-1.7981316663075286),dotstyle); label("$H$", (-2.5539724782401856,-1.3562254893693109), NE * labelscalefactor); 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We are first and foremost going to spam point constructions; let:

$X$ the intersection of the perpendicular line to $AB$ at $B$ at $AC$
$Y$ the intersection of the perpendicular line to $AC$ at $C$ at $AB$
$O$ the circumcentre of $\triangle ABC$
$G=\overline{BX}\cap\overline{CY}$ , i.e. the point diametrically opposite $A$ on $(ABC)$
$F$ the intersection of the perpendicular bisector of $AC$ with $BC$
$H=\overline{YO}\cap(BCDXY)\neq Y$

To deal with the condition in the most blatantly, direct, possible way, we introduce the circle $(BCXY)$ . By some angle chasing we can immediately see that $D\in(BCXY)$ ; this is going to be crucial.

We can now also say that $D\in\overline{AF}\cap(BCXY)$ (since there's actually two intersections; the other one's outside). We now claim that we just magically got an orthocentre configuration to spawn, specifically a humpty point configuration has spawned in front of us very very blatantly. Let's walk through it however:

$X$ is the orthocentre (!) of $\triangle AGY$ (very obvious).
$H$ is the $Y$ -humpty point (!!) of $\triangle AGY$ (not that obvious but it follows from definition of the humpty point)

I actually now claim the following, absolutely magical facts:

$ABFHO$ cyclic (!)
$DEFH$ cyclic (!!)
$BEHM$ cyclic and $(BEHM)$ tangent to $AB$ (!!!)

Let's pull this off shall we?

It's quite clear that $\overline{FO}$ is the perpendicular bisector of $AC$ and $F-E-O$ . We use this to prove our first claim. To show $ABFO$ cyclic it suffices to note that $\measuredangle AFB=\measuredangle ACF+\measuredangle FAC=2\measuredangle ACB=\measuredangle AOB$ . To show that $H$ lies on $(ABFO)$ , note $\measuredangle BHO=\measuredangle BHY=\measuredangle BXY=\measuredangle BAG=\measuredangle BAO$ as desired.

Now to prove our second fact, let $E'=(DFH)\cap\overline{BD}\neq D$ ; I claim $E'=E$ and it suffices to show that $E'\in\overline{FO}$ . But it's well known that $(AYH)$ tangent to $\overline{AG}$ , so $\measuredangle E'FH=\measuredangle E'DH=\measuredangle BDH=\measuredangle BYH=\measuredangle GAH=\measuredangle OAH=\measuredangle OFH$ , thus done.

We now finally move on to the third and final fact; we first show that $(BHM)$ tangent to $AB$ . But by some really well known properties of symmedians $\measuredangle BHM=\measuredangle YHX=\measuredangle YBC=\measuredangle ABC$ , so we are done by alternate segment. So it remains to show that $E\in(BHM)$ . We however now know the tangency holds so $\measuredangle BEH=\measuredangle DEH=\measuredangle DFH=\measuredangle AFH=\measuredangle ABH=\measuredangle BMH$ , thus done. Hence $(BEMH)$ tangent to $AB$ , completing the proof.

Some motivational remarks

oops looks like someone beat me to it… :noo:

:noo:

almost

This post has been edited 4 times. Last edited by mcmp, Oct 28, 2024, 9:40 AM

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#79 h Jan 18, 2025, 9:52 AM • 2 Y

Y by SomeonesPenguin, Kaus_sgr

Solved with SomeonesPenguin

Setup
Clearly $ABFC$ is an isoscelles trapezoid. Then $\overline{N-E-O}$ collinear. Therefore by homothety at $B$ with ratio $2$ , $\overline{F-K-B’}$ are collinear. Therefore $\angle B’FC=90^{\circ}-A=\angle KDC$ so $KCFD$ is cyclic. The following angle chase finishes the problem:
$\angle ABC=\angle AFC=180^{\circ}-\angle BKC=180^{\circ}-\angle BEM$

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283 posts

#80 h Mar 14, 2025, 7:15 AM

Y by

solved with the hint to complete an isosceles trapezoid

Complete isosceles trapezoid $ABKC.$ That is, draw $BK\parallel AC$ so that $ABKC$ is cyclic.
Let $F = BD\cap (DKC).$ Let $L = AK\cap BC.$
Observe that $\angle CDF = 90 - \angle BAC = 90 - \angle KCA,$ and so $KF\parallel LE.$
Furthermore, we have that $2AE = AF$ since $KF\perp BK.$ Thus, $EM\parallel FC.$ Hence, $\angle BEM = \angle BFC = 180-\angle AKC = 180 - \angle ABC$ which finishes.

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#81 h Saturday at 5:38 AM

Y by

very nice

The key is to orient $\overline{AC}$ as parallel to the $x$ -axis; then, constructing $D$ and $E$ feels way more natural by drawing in the perpendicular bisector of $\overline{AC}$ . Moreover, the point at which $\overline{AD}$ intersects $\overline{BC}$ lies on the perpendicular bisector as well. Hence, we extend $\overline{AD}$ to a point $F$ such that $\overline{BF} \parallel \overline{AC}$ .

An immediate result of this is that $ABFC$ is an isosceles trapezoid. However, we still do not have any useful information from point $M$ , so we reflect $B$ over point $E$ to point $G$ , so that $\triangle BEM \sim \triangle BGC$ with a scale factor of $2$ . Notice that the perpendicular bisector of $\overline{AC}$ is also the perpendicular bisector of $\overline{BF}$ , implying that $EF = EG$ and $\overline{BF} \perp \overline{FG}$ .

Now, we can angle chase that

$\begin{align*} \angle GDC &= 180^\circ - \angle BDC \\ &= 180^\circ - (90^\circ - \angle BAC) \\ &= (180^\circ - \angle BAC) - 90^\circ \\ &= \angle BFC - 90^\circ = \angle GFC. \end{align*}$
Therefore, points $C$ , $D$ , $F$ , and $G$ are concyclic. Finally, we have

$\angle ABM = \angle AFC = 180^\circ - \angle DGC = 180^\circ - \angle BEM,$
which shows the desired tangency. $\blacksquare$

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