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AMC and other contests, summer programs, etc.
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Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

Are you ready to level up with Olympiad training? Registration is open with early bird pricing available for our WOOT programs: MathWOOT (Levels 1 and 2), CodeWOOT, PhysicsWOOT, and ChemWOOT. What is WOOT? WOOT stands for Worldwide Online Olympiad Training and is a 7-month high school math Olympiad preparation and testing program that brings together many of the best students from around the world to learn Olympiad problem solving skills. Classes begin in September!

Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
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0 replies
jlacosta
Mar 2, 2025
0 replies
2025 ROSS Program
scls140511   12
N 3 minutes ago by Alex-131
Since the application has ended, are we now free to discuss the problems and stats? How do you think this year's problems are?
12 replies
scls140511
Today at 2:36 AM
Alex-131
3 minutes ago
is this really supposed to be #13???
hgmium   2
N 3 minutes ago by Rice_Farmer
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_13

I managed to do all the other geo problems for that year besides this one
misplaced?
2 replies
hgmium
14 minutes ago
Rice_Farmer
3 minutes ago
MathPath 2025 form.
BraveCobra22aops   0
30 minutes ago
I created a form for people going to MathPath 2025: https://artofproblemsolving.com/community/c136h3528968_mathpath_2025.
0 replies
BraveCobra22aops
30 minutes ago
0 replies
AMC- IMO preparation
asyaela.   9
N an hour ago by Schintalpati
I'm a ninth grader, and I recently attempted the AMC 12, getting 18 questions correct and leaving 7 empty. I started working on Olympiad math in November and currently dedicate about two hours per day to preparation. I'm feeling a bit demotivated, but if it's possible for me to reach IMO level, I'd be willing to put in more time. How realistic is it for me to get there, and how much study would it typically take?
9 replies
+1 w
asyaela.
4 hours ago
Schintalpati
an hour ago
No more topics!
Geo equals ABsurdly proBEMatic
ihatemath123   73
N Yesterday at 5:38 AM by joshualiu315
Source: 2024 USAMO Problem 5, JMO Problem 6
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
73 replies
ihatemath123
Mar 21, 2024
joshualiu315
Yesterday at 5:38 AM
Geo equals ABsurdly proBEMatic
G H J
G H BBookmark kLocked kLocked NReply
Source: 2024 USAMO Problem 5, JMO Problem 6
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DottedCaculator
7303 posts
#65 • 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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k12byda5h
104 posts
#66 • 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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VulcanForge
624 posts
#67
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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asdf334
7578 posts
#68 • 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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Pitchu-25
52 posts
#69 • 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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Mr.Sharkman
486 posts
#70
Y by
Bro im such a clown ? (didn't finish writeup oops :wallbash_red: :blush: :oops_sign: :noo: :noo: :rotfl: )

Solution
This post has been edited 3 times. Last edited by Mr.Sharkman, Dec 17, 2024, 2:05 AM
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NuMBeRaToRiC
14 posts
#71
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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Shreyasharma
666 posts
#74
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. $\square$
This post has been edited 1 time. Last edited by Shreyasharma, Apr 22, 2024, 9:27 AM
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CyclicISLscelesTrapezoid
371 posts
#75 • 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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Zhaom
5123 posts
#76 • 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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TestX01
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#77
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
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#78 • 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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[/asy]

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:

almost :P
This post has been edited 4 times. Last edited by mcmp, Oct 28, 2024, 9:40 AM
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cosdealfa
27 posts
#79 • 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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Ilikeminecraft
283 posts
#80
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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joshualiu315
2513 posts
#81
Y by
very nice :P


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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