i spent 4 hours on this problem to no avail

We perform a $\sqrt{bc}$ inversion composed with a reflection about the angle bisector. The angle conditions tell us $D$ is sent onto the semicircle with diameter $BC$. $E$ is sent to the intersection of the circle with center $B$ and radius $AB$. $M$ is mapped to the symmedian chord intersection point on $(ABC)$. Note that if we define $J$ as the midpoint of $BC$, $(BJM)$ is the desired circle tangent to $AB$. At this point we are in familiar territory with nine-point circles, Humpty-Dumpty points, etc, so our problem can be reformulated into a bashable question: Restatement wrote: Let $ABC$ be a triangle, with $M$, $N$ as the midpoints opposite $B$, $C$, and $X$ as the Dumpty point. Let $K$ be the point on $(ABC)$ such that $\angle KAN = \angle C$, and let $L$ be the midpoint of $MN$. Define $E = (AB) \cap (MLX)$ and $D = AK \cap (AME)$. Show that $D \in (MN)$.

The hardest angle chasing problem to have ever existed

khina wrote:

if we got a claim but couldnt finish angle chasing would we get a point for cyclic

Out of all the possible things you can get in this weird config they decide to ask this, :skull: (at first i complicated this problem at it took me around 1h 30 mins, which is why i said this as during that time i found a LOT of cool facts in this weird config). Let $AD$ meet $BC, (ABC)$ at $J,B'$ and relfect $B$ over $E$ (call it $B_1$), let $\angle DBC=x$ and $\angle DCB=y$. $$\angle DCB'=y+\angle BCB'=y+A-C=90-x-C=\angle EJC=\angle BEJ=\angle DB_1B' \implies DB_1CB' \; \text{cyclic}$$Now $\angle DB_1C=180-\angle DB'C=180-B$ hence $(BB_1C)$ is tangent to $AB$ and by homothety at $B$ with scale factor $0.5$ we are done. Note: $A>C$ holds in the angle chasing in case someone wonders about config issues, this is because if $C \ge A$ then $D$ has to lie outside $ABC$ which cannot happen.

This problem seems very misleading on what to try... The angle condition simply begs for a perpendicular $BQ\perp AB$ where $Q$ is on $AC$, thus giving $B,D,C,Q$ cyclic, yet this approach is ultimately futile and I got sucked down a rabbit hole trying to prove a collinearity that was false, and my two (!) different diagrams both didn't catch the error.

Nice problem! Let $P$ be the point on $BC$ such that $AP=PC$. Let $X$ be the point on $\odot(BDC)$ such that $\odot(BMX)$ is tangent to $AB$. The solution splits into two claims. The first doesn't do anything with point $D$ and $E$. Claim. We have $\angle CXP = 90^\circ + \angle C$. Proof. Let $Q$ be the point on $AC$ such that $BQ=QC$. Since $\angle QBC = \angle C = \angle PAC$, it follows that $ABPQ$ is cyclic. Now, notice that $$\left. \setlength{\arraycolsep}{2pt} \begin{array}{rcl} \angle CXM &=& \angle BXC - \angle BXM \\[4pt] &=& (90^\circ + \angle A) - (180^\circ - \angle B) \\[4pt] &=& 90^\circ - \angle C = \angle CQM \end{array} \right\} \implies C,Q,X,M\text{ are concyclic.} $$$$\left. \setlength{\arraycolsep}{2pt} \begin{array}{rcl} \angle BXQ &=& 360^\circ - \angle BXM -\angle MXQ \\[4pt] &=& 360^\circ - (180^\circ - \angle B) - (180^\circ - \angle C) \\[4pt] &=& \angle BAQ \end{array}\right\} \implies A,B,P,Q,X\text{ are concyclic.} $$Hence, we have that \begin{align*} \angle CXP &= \angle CXB - \angle BXP \\ &= (90^\circ + \angle A) - (\angle A - \angle C) = 90^\circ + \angle C. \qquad\blacksquare \end{align*} We now finish the problem by angle chasing. First, note that \begin{align*} \angle EPX &= (90^\circ - \angle C) - \angle XPC \tag{$PE\perp AC$} \\ &= \angle XCP \tag{the first claim} \\ &= 180^\circ - \angle BDX = \angle EDX, \end{align*}so $D, E, X, P$ are concyclic. Hence, we have that \begin{align*} \angle BEX &= \angle 180^\circ - \angle DPX \\ &= 2\angle C - \angle XPC \\ &= 90^\circ + \angle C - \angle XCP = \angle XMB \end{align*}(where in the last step, we used $\angle MXC$ $= \angle BXC - \angle BXM$ $= (90^\circ + \angle A) - (180^\circ - \angle B) $ $= 90^\circ - \angle C$). Thus, $BEXM$ is cyclic, and we are done. Remark. The motivation for point $X$ is as follows: let $AD$ intersect $\odot(BDC)$ again at $D_1$ and let $E_1$ be on $BD_1$ such that $AE_1=E_1C$ (i.e., the other points if the problem has no configuration issues). Then, point $X$ is Miquel point of $DD_1EE_1$. However, proving this is not sufficient to show the problem, since we only get that $BEE_1X$ is cyclic, and not $M$. Doing a little bit more work leads to the proof of the second half.

Apparently, $e = \frac{a(b+c)}{a+b}$ when you toss onto the complex plane. Source: a friend EDIT: apparently that was false, source: below

ihatemath123 wrote: Apparently, $e = \frac{a(b+c)}{a+b}$ when you toss onto the complex plane. Source: a friend I.... don't think this is true? That would imply that $E$ lies on line $BC$. Here's my complex bash solution, which took roughly half an hour: Let $\triangle ABC$ be inscribed in the unit circle, so that \begin{align*} |a|=|b|=|c|&=1 \\ m &= \frac{b+c}2 \end{align*}Let ${AD}, {BD}$ meet the unit circle again at $X, Y$ respectively. Then since $\angle DAC = \angle ACB$ as directed angles, we have $\frac{x}c = \frac{a}b$, so $$x = \frac{ac}b$$Then $$d = \frac{ax(b+y) - by(a+x)}{ax-by} = \frac{a^2bc+a^2cy-ab^2y-abcy}{a^2c-b^2y}$$Now $\angle BDC = 90^{\circ} + \angle BAC$, so we have $$\frac{\left(\frac{b-d}{c-d}\right)}{\left(\frac{b-a}{c-a}\right)} \in i\mathbb{R}$$$$\frac{(a-c)(b-d)}{(a-b)(c-d)} \in i\mathbb{R}$$$$\frac{y(a-b)(a-c)(b^2-ac)}{(a-b)(a^2c^2-b^2cy-a^2bc-a^2cy+ab^2y+abcy)} \in i\mathbb{R}$$$$\frac{y(a-c)(b^2-ac)}{a^2c^2-b^2cy-a^2bc-a^2cy+ab^2y+abcy} = -\frac{(a-c)(b^2-ac)}{b^2y-a^2c-bcy-b^2c+ac^2+abc}$$$$a^2c^2-b^2cy-a^2bc-a^2cy+ab^2y+abcy = -b^2y^2+a^2cy+bcy^2+b^2cy-ac^2y-abcy$$$$\boxed{a^2bc-a^2c^2+2a^2cy-ab^2y-ac^2y-2abcy-b^2y^2+bcy^2+2b^2cy = 0}$$We will use this equation later. First, we find the coordinate of $E$: $$\overline{e} = \frac{e}{ac} = \frac{b+y-e}{by} \implies e = \frac{ac(b+y)}{ac+by}$$Now we want to show that $AB$ is tangent to the circumcircle of $\triangle BEM$. Thus we want to show that $$\frac{(a-b)(e-m)}{(b-e)(b-m)} \in \mathbb{R}$$Now $$e-m = \frac{abc+2acy-ac^2-b^2y-bcy}{2(ac+by)}$$Thus $$\frac{(a-b)(e-m)}{(b-e)(b-m)} = \frac{(a-b)(abc+2acy-ac^2-b^2y-bcy)}{y(b-c)(b^2-ac)}$$and conjugating, we have $$\overline{\left[\frac{(a-b)(e-m)}{(b-e)(b-m)}\right]} = \frac{(a-b)(abc+ac^2+b^2y-2b^2c-bcy)}{a(b-c)(b^2-ac)}$$Subtracting, we have $$\frac{(a-b)(e-m)}{(b-e)(b-m)} - \overline{\left[\frac{(a-b)(e-m)}{(b-e)(b-m)}\right]} \\ = \frac{(a-b)(a^2bc+2a^2cy-a^2c^2-ab^2y-2abcy-ac^2y-b^2y^2+2b^2cy+bcy^2)}{ay(b-c)(b^2-ac)}$$The big factor in the numerator is zero due to the boxed equation, so we are done. $\blacksquare$

The angle condition actually isn't scary, there is a natural way to eliminate it ... It's just that the most natural way to eliminate it turns out to be a trap that gets you stuck for 3 hours.

I saw bdc = a + 90 and immediately thought isogonal conjugates, since then we can get a point on the circumference on the circle with diameter bc, from there I concluded with angle chasing and angle ceva. A friend of mine claims to have done it with trig bash.

Define $N, P, Q, R$ to be the midpoint of $AC$, $CD\cap NE$, $BC\cap AD$, and $AB\cap NE$ respectively. Clearly, $Q$ lies on line $NE$. Notice that since $\angle ARP=90^\circ - \angle BAC$ and $\angle ADP=90^\circ +\angle BAC$, quadrilateral $RBDP$ is cyclic. So we have $\angle ABE=\angle DPR=\angle CPN=\angle APN$, so quadrilateral $ABPE$ is also cyclic. We now use the brilliant method of Cartesian coordinates. Let $A=(-1,0)$, $N=(0,0)$, $C=(1,0)$, $P=(0,p)$, $Q=(0,q)$, $R=(0,r)$ for $p,q,r>0$. We have $D=AQ\cap CP$, which are the lines \begin{align*} y&=qx+q\\ y&=-px+p\\ D&= \left(\frac{p-q}{p+q},\frac{2pq}{p+q}\right). \end{align*}Similarly, $B=CQ\cap AR$ satisfies \[ B=\left(\frac{q-r}{q+r},\frac{2qr}{q+r}\right).\]We note that $E$ is simply the point on line $BD$ satisfying $x=0$. Therfore, \begin{align*} E_y&=\frac{\frac{2qr}{q+r}-\frac{2pq}{p+q}}{\frac{q-r}{q+r}-\frac{p-q}{p+q}}\left(-\frac{p-q}{p+q}\right)+\frac{2pq}{p+q}\\ &= -\frac{q^2(r-p}{q^2-pr}\left(-\frac{p-q}{p+q}\right)+\frac{2pq}{p+q}\\ &=\frac{q^r-q^2rp+q^2p^2+q^3p-2p^2qr}{(q^2-pr)(p+q)}\\ &= \frac{pq^2-2pqr+qr^2}{q^2-pr}. \end{align*}Now, applying Power of a Point at $R$ on circle $(ABPE)$, we obtain $RB\cdot RA=RP\cdot RE$. Note that \begin{align*} RB\cdot RA&= r(r-B_y)\frac{1+r^2}{r^2}\\ &=\frac{(r-q)(1+r^2)}{q+r}, \end{align*}while \begin{align*} RP\cdot RE&=(r-p)\left(r- \frac{pq^2-2pqr+qr^2}{q^2-pr}\right)\\ &= \frac{(r-p)(2pqr-pr^2-q^2p)}{q^2-pr}\\ &= \frac{p(r-p)(r-q)^2}{pr-q^2}. \end{align*}Therefore, \begin{align*} \frac{1+r^2}{q+r}&=\frac{p(r-p)(r-q)}{pr-q^2}\\ pr-q^2+pr^3-q^2r^2 &= pr^2q+pr^3-p^2rq-p^2r^2-pq^2r-pqr^2+p^2q^2+p^2qr\\ pr-q^2&=p^2q^2+q^2r^2-p^2r^2-pq^2r. \end{align*}It suffices to show that \[ \angle ABE=\angle APN=\angle BME,\]where \[ M=\left(\frac{q}{q+r},\frac{qr}{q+r}\right).\]Create line $MT$ through $M$ parallel to $AC$, where $T$ is closer to $A$ than $C$. We need \[ \angle APN=\angle QMT+\angle TME\implies \angle APN-\angle TME=\angle QMT.\]Obviously, $\tan(\angle APN)=1/p$ while $\tan(\angle QMT)=q$. So \[\tan(\angle APN-\angle QMT)=\frac{1/pq}{1+q/p}=\frac{1-pq}{p+q}.\]Meanwhile, \begin{align*} \tan (\angle TME)&=\frac{q+r}q \left(\frac{qr}{q+r}-\frac{pq^2-2pqr+q^2r}{q^2-pr}\right)\\ &= \frac{q^2r-pr^2-pqr+2pqr+2pr^2-q^2r-qr^2}{q^2-pr}\\ &= \frac{pqr+pr^2-pq^2-qr^2}{q^2-pr}. \end{align*}Setting these two expressions equal, we get \begin{align*} \frac{pqr+pr^2-pq^2-qr^2}{q^2-pr}&=\frac{1-pq}{p+q}\\ p^2qr+p^2r^2-p^2q^2-pqr^2+pq^2r+pqr^2-pq^3-q^2r^2 &=(q^2-pr)-pq(q^2-pr). \end{align*}To homogenize, we can substitute the first $q^2-pr$ with $p^2r^2+pq^2r-p^2q^2-q^2r^2$ to get \[ pq^2r+p^2r^2-q^2r^2-p^2q-q^3p+qp^2r= pq^2r+p^2r^2-q^2r^2-p^2q-q^3p+qp^2r,\]which is equal.

I hope you enjoyed my problem! I made it by a quite long sequence of adding and erasing lots of extra points; the actual clean solution below is found by my friend Mykhailo Shtandenko, whom I thank a lot! Same as #2 but will post anyways. Let $F$ be symmetric to $B$ w.r.t. $E$, and let $B'$ be such that $ABB'C$ is isosceles trapezoid. Since $E$ is on perpendicular bisector of $BB'$, $B'F \perp AC$. Then $\angle FB'C = 90^\circ - \angle B'CA = 90^\circ - \angle A = 180^\circ - \angle BDC = \angle FDC$, so points $F, D, B', C$ lie on one circle. Since $EM$ is midline in $\triangle BFC$, we get $\angle BEM = \angle BFC = 180^\circ - \angle AB'C = 180^\circ - \angle B$, and tangency follows.

hukilau17 wrote: ihatemath123 wrote: Apparently, $e = \frac{a(b+c)}{a+b}$ when you toss onto the complex plane. Source: a friend I.... don't think this is true? That would imply that $E$ lies on line $BC$. Here's my complex bash solution, which took roughly half an hour: Let $\triangle ABC$ be inscribed in the unit circle, so that \begin{align*} |a|=|b|=|c|&=1 \\ m &= \frac{b+c}2 \end{align*}Let ${AD}, {BD}$ meet the unit circle again at $X, Y$ respectively. Then since $\angle DAC = \angle ACB$ as directed angles, we have $\frac{x}c ... My new method for becoming a better complex basher is reading all of your AoPS posts

And now, an absolute disaster of a solution that I found by divine intervention in contest. I do not have ANY idea how I came up with this. Let $X$ denote $AD \cap BC$ and observe that $X$ is collinear with $E$ and the circumcenter of $ABC$ by angle conditions. Secondly the angle condition gives $(BDC)$ is orthogonal to $(ABC)$; this is just angle chasing. Also note that $BAOX$ is cyclic, again this is easy to show. Perform an inversion at $B$. Then $O$ gets mapped to the reflection of $B^*$ over $A^*C^*$, the perpendicular bisector of $AC$ is mapped to a circle orthogonal to $(A^*B^*C^*)$ passing through $O^*$, so it is the $B^*$-apollonian circle. This meets line $B^*C^*$ at $X^*$, which should also lie on $A^*O^*$. $M^*$ maps to the reflection of $B^*$ over $C^*$. Now $D*$ is one of the intersections of $(B^*A^*X^*)$ with the line perpendicular to $A^*C^*$ at $C^*$ (orthogonality). It turns out it does not matter which one. Then $E*$ is the intersection of $B^*D^*$ with the $B^*$-apollonian circle. So the problem rewrites thus: Quote: Let $ABC$ be a triangle, and let $O$ be the reflection of $B$ over $AC$. Let $M$ be the reflection of $B$ over $C$. $AO$ meets $BC$ at $X$, and $D$ is on $(BAX)$ such that $CD \perp AC$. $BC$ meets the circumcircle of $(BOX)$ again at $E$. Prove that $EM \parallel AB$. We define $T = \overline{AX} \cap\overline{CD}$. Now since $ATC$ is right, it follows from reflection that the line through $C$ and the midpoint of $AT$ is parallel to $AB$. Therefore it is clear that $TM$ is parallel to $AB$. Now, in directed angles, \[ \measuredangle XED = \measuredangle XEB = \measuredangle XOB = \measuredangle XTC = \measuredangle XTD \]which implies that $XTED$ is cyclic, \[ \measuredangle TEB = \measuredangle TED = \measuredangle TXD = \measuredangle AXD = \measuredangle ABD = \measuredangle ABE \]which implies $TE \parallel AB$. Therefore $EM \parallel AB$. This finishes the problem.

I hate myself

Batsuh wrote: I hate myself me too

Let $O$ be the circumcenter, $G$ be the point on $\overline{BO}$ such that $ACB \sim OMG$, and $E$ be the point such that $ACBD \sim OMGE$. We will show that $E$ is the same as the $E$ in the problem and $(BME)$ is tangent to $\overline{AB}$. We have $\angle OBC = 90^\circ - \angle A = 180^\circ - \angle BDC$, so $\overline{OB}$ is tangent to $(BDC)$ which gives $\angle BCD = \angle OBD$. Also $\angle GEM = 90^\circ + \angle A = 180^\circ - \angle OBM$, so $BGEM$ is cyclic. Now we prove the following: $B$, $D$, $E$ collinear: follows from $\angle GBD = \angle BCD = \angle GME = \angle GBE$. $(BMEG)$ tangent to $\overline{BC}$: follows from $\angle BMG = 90^\circ - \angle OMG = 90^\circ - \angle C = \angle ABG$. $AE = CE$: we have $\angle MOE = \angle CAD = \angle C$, so $\overline{OE} \perp \overline{AC}$.

wow beautiful nonstandard problem! (imagine bashing it)

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)

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.

[asy][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][/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 $\implies T=T'$.

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. $\blacksquare$

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. $\blacksquare$

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

khina wrote:

How is $B’E’ \perp AC$?

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. $\square$

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. $\square$

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.

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.

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

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.

oops looks like someone beat me to it… almost

Solved with SomeonesPenguin

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: $$\angle ABC=\angle AFC=180^{\circ}-\angle BKC=180^{\circ}-\angle BEM$$