Let $ABC$ be a triangle. The circle $\omega_A$ through $A$ is tangent to line $BC$ at $B$. The circle $\omega_C$ through $C$ is tangent to line $AB$ at $B$. Let $\omega_A$ and $\omega_C$ meet again at $D$. Let $M$ be the midpoint of line segment $[BC]$, and let $E$ be the intersection of lines $MD$ and $AC$. Show that $E$ lies on $\omega_A$.
Problem
Source: BxMO 2020, Problem 3
Tags: geometry, BxMO, Benelux
02.05.2020 23:54
Note that $D$ is the $B-dumpty$ point of $\triangle ABC$ and it's well-known that $OD\perp DB$ where $O$ is the circumcenter of $\triangle ABC$, let $F=BD\cap (ABC)$, then $D$ is the midpoint of $BF$ and $DM\| FC$. The conclusion now follows by Reim's.
03.05.2020 00:12
Let $E'$ be the point in which $AC$ cuts $\omega_A$. Note that $\angle CBD= \angle BE'D$ and $\angle DCB = \angle DBA = \angle DE'C$ Hence, $BC$ is tangent to the circumcircles of $\triangle E'BD$ and $\triangle DE'C$. Then, $DE'$ is their radical axis, so it cuts $BC$ in $M$. Finally, this proves $E'=E$, so we are done.
05.05.2020 10:06
Let $E'$ be intersection of $\omega_A$ and $AC$. Claim: Line $E'D$ passes though midpoint of $BC$. Proof: Let $O$ be circumcircle of $\omega_B$. Note that $OB \perp AB$. Therefore $\angle CBO = 90 -\angle B = \angle BCO \implies \angle BOC=2\angle B \implies \angle BDC = 180 - \angle B$. Let $\angle CBD = \alpha$, then $ABD = \angle B -\alpha = \angle DEC$. Note that $\angle CBD = 180-(180 -\angle B) - \alpha = \angle B - \alpha = \angle DEC$. If $E'D$ intersects $BC$ at $M'$, then it follows that $M'C$ is tangent to $\odot(EDC)$, therefore $M'C^2 = M'D \cdot M'E' =BM'^2$, hence $M'$ is midpoint of $BC$. In fact $M=M'$. From claim it follows that $E = E'$, which finishes problem.
30.06.2020 02:14
Consider a $\sqrt{BA.BC}$ inversion followed by a reflection over $B$- anglebisecter.Obviously $D$ is the $B$-dumpty point in $ABC$ 1) $D$ goes to the reflection of $B$ over midpoint of $AC$.Call it $D'$ 2) $M$ goes to the reflection of $B$ in $A$.Call it $M'$. 3)$E$ goes to the point where $(BAC)$ and $(BM'D')$ meet.Call it $E'$ 4) $A$ goes to $C$ Now proving $EADB$ is concyclic is equivalent to prove $D',C,E'$ are collinear. Now $B,E'$ are the 2 common point in $(BAC)$ and $(BD'M')$,$B,A,M'$ are collinear, and $M'D'||AC$.So by Reim's theorem,$E',C,D'$ are collinear.Inverting back we get the comclusion$\blacksquare$
09.06.2021 15:06
A quick recap of Evan Chen's Lemma in Chapter 2 simplifies the problem and is essentially equivalent.
16.04.2024 18:02
Denote N the midpoint of AB so MN$\|$AC. We have $\angle BAD$ = $\angle DBC$ , $\angle DBA$ = $\angle DCB$ so $\triangle$ ABD and $\triangle$ BCD are similar so $\triangle$ BND and $\triangle$ CMD are similar so$\angle DNB$ = $\angle DMC$ so BMDN is cyclic. We have $\angle ACM$ = $\angle BMN$. it suffices to show that 180 - $\angle DMC$ - $\angle ACM$ = $\angle DBN$ $\iff$ $\angle DBN$ + $\angle DMC$ + $\angle BMN$ = 180 which is clear since $\angle DBN$ = $\angle DMN$.
16.04.2024 18:54
Radin.A wrote: Denote N the midpoint of AB so MN$\|$AC. We have $\angle BAM$ = $\angle DBC$ , $\angle DBA$ = $\angle DCB$ so $\triangle$ ABD and $\triangle$ BCD are similar so $\triangle$ BND and $\triangle$ CMD are similar so$\angle DNB$ = $\angle DMC$ so BMDN is cyclic. We have $\angle ACM$ = $\angle BMN$. it suffices to show that 180 - $\angle DMC$ - $\angle ACM$ = $\angle DBN$ $\iff$ $\angle DBN$ + $\angle DMC$ + $\angle BMN$ = 180 which is clear since $\angle DBN$ = $\angle DMN$. Sorry, I think you mistyped $\angle BAM$ = $\angle DBC$?..I think it's supposed to be $\angle BAD$ = $\angle DBC$?
05.12.2024 00:32
Why such an easy problem? Solved in 2 minutes... Define $E'=AC\cap \omega_{A}$. All we have to prove is that $E'-D-M$ are collinear. By the tangencies we have: $\angle DCB= \angle DBA= \angle DE'A$. $\newline$ $\Longrightarrow$ Thus, the circle $(E'DC)$ is tangent to $BC$ at $C$. $\newline$ $\Longrightarrow$Thus, it implies that $D$ is the Humpty point of triangle $\Delta E'BC$. This means that $E'-D-M$ are collinear. $\Longrightarrow$ $E'\equiv E$. Which is what we wanted. $\blacksquare$