Point $H$ is the orthocenter of the scalene triangle $ABC.$ A line, which passes through point $H$, intersect the sides $AB$ and $AC$ at points $D$ and $E$, respectively, such that $AD=AE.$ Let $M$ be the midpoint of side $BC.$ Line $MH$ intersects the circumscribed circle of triangle $ABC$ at point $K$, which is on the smaller arc $AB$. Prove that Nibab can draw a circle through $A, D, E$ and $K.$
Problem
Source: Moldova TST 2019
Tags: geometry
10.03.2019 18:55
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10.03.2019 19:30
Let $A’$ be the antipode of $A$ w.r.t $(ABC)$. Note that $K,H,M,A’$ are collinear. Note that it is enough to prove that the antipode of $A$ w.r.t $(ADE)$ lies on $HM$. Let $\Delta PQR$ be the orthocentre triangle of $\Delta ABC$ and $X = HM \cap A-$anglebisector. Note that $\frac{HX}{XA’} = \frac{AH}{AA’} = \frac{HQ}{HC} = \frac{QE}{EC}$ by repeated use of angle bisector theorem. Also, $HQ \parallel CA’$. Thus $XE \perp AC$. Similarly (or by Simson’s line), we have that $XD \perp AB$. Thus, the claim holds true. $\blacksquare$.
10.03.2019 20:10
Another way to show this is to use the fact that $H$ is the $K$-Humpty point of $\triangle KBC$. Then, by the spiral similarity lemma, we just need to show that $$\frac{BD}{CE}=\frac{KB}{KC}=\frac{HB}{HC} \Leftrightarrow \frac{BD}{BH}=\frac{CE}{CH} \Leftrightarrow \angle BHD=\angle CHE$$where the last statement follows by using sine law in $\triangle BHD$ and $\triangle CHE$, and because $\angle BDH=\angle CEH$. Thus, it suffices to show that $DE$ is the external angle bisector of $\angle BHC$; or equivalently that the internal angle bisector of $\angle BHC$ and $\angle BAC$ are parallel to each other. But this is a well known property of an orthocentric system. Hence, done. $\blacksquare$
10.03.2019 21:03
Let, $MH\cap AO=R$ which lies on $(ABC)$ It's enough to prove that $KA$ is the external angle bisector of $\angle{DKE}$ By Reim's theorem, $DE||BP||QC$ where $P=KE\cap(ABC)$ and $Q=KD\cap(ABC)$ Easy angle chasing gives that, $CA$ is the internal angle bisector of $\angle{QCP}$ which implies that arc$AQ=$arc$PA$ implying arc$RQ=$arc$RP$ So we are done...
10.03.2019 21:06
This literally took me like 2.5 hours, so here we go. Also note that i think i set $D$ as $E$ and $E$ as $F$, so this might be kinda confusing. First, let $Q$ be the foot of the perpendicular from $B$ to $AC$ and let $R$ be the equivalent for $C$. It’s well-known (or just look at the solutions above) that $APRHQ$ is cyclic. Now, perform a $\sqrt{AR \cdot AB}$ inversion around $A$. Note that $R$ gets sent to $B$, $Q$ gets sent to $C$, and vice versa. If we let $Y$ be the point on $BC$ such that $Y$ bisects $\angle{BAC}$, we have that the line $EF$ is sent to the circle with diameter $Y$. Thus, if $E’$ and $F’$ denote the images of $E$ and $F$ under the inversion respectively, we have that they are the foots of $Y$ onto $AB$ and $AC$. We also have that $H’$ is sent to the foot of the altitude from $A$ to $BC$. Finally, $P$ gets sent to the intersection of $QR$ and $BC$. Thus, it suffices to prove that the image of $P$ lies on the line $E’F’$. Now, with our previous definition of $P$ we had that $(P, H ; R, Q) = -1$. Since inversion preserves cross-ratio, if we define $P’$ as the intersection of $E’F’$ and $BC$, it suffices to prove that $(P’, H’ ; B, C) = -1$. Now, let $B’$ be the reflection of $B$ over $AY$. We have that: \begin{align*} \frac{P’B}{P’C} &= \frac{E’B’}{E’C} \\ &= \frac{F’B}{E’C} \end{align*} Now, since $AF’H’Y’E’$ is cyclic, we have that: \begin{align*} \frac{BF’}{BH’} &= \frac{BY’}{BA} \\ &= \frac{CY’}{CA} \\ &= \frac{CE’}{CH’} \\ \frac{P’B}{P’C} &= \frac{F’B}{E’C} \\ &= \frac{H’B}{H’C} \\ (P’, H’ ; B, C) &= -1 \end{align*} and thus we are done.
10.03.2019 23:34
This is the necessary and sufficient condition to prove ISL 2005 G5...
10.03.2019 23:46
No it's very different, read the title.
11.03.2019 00:01
It’s different, but each problem implies directly in the other.
11.03.2019 05:29
Based on khina's solution. With same inversion, let $X$ to be the intersection of $RQ$ with $BC$ (or the image of $P$). By brocards, since $RC$ and $BQ$ intersect at $H$, we get $AH$ is the polar of $X$ with respect to $(BRQC)$. Since $AH$ intersects $BC$ and $H'$, we find that $(X,H';C,B)=-1$. Let $X'$ be the intersection of $E'F'$ with $BC$. We claim that $F'C$ and $BE'$ intersect on $AH$. Notice that this suffices to prove the problem as we may use brocards again to prove $(X',H';C,B)=-1$. To prove this, we use Ceva's theorem w.r.t triangle ABC. Since $\frac{E'C}{AE'}=\frac{E'Y\cot(C)}{E'Y\cot{A/2}}=\frac{\cot{C}}{\cot{A/2}}$, $\frac{AF'}{F'B}=\frac{F'Y\cot{A/2}}{F'Y\cot(B)}=\frac{\cot{A/2}}{\cot{B}}$, $\frac{H'B}{CH'}=\frac{AH'cot(B)}{AH'cot{C}}=\frac{cot{B}}{cot{C}}$, we get that $\frac{E'C}{AE'}\frac{AF'}{F'B}\frac{H'B}{CH'}=\frac{\cot{C}}{\cot{A/2}}\frac{\cot{A/2}}{\cot{B}}\frac{cot{B}}{cot{C}}=1$. This proves the above claim. Extension: I seek to prove that $X$, the intersection of $HM$ and the A-bisector is on (ADEP). By the same inversion, it suffices to prove that $X$ gets sent to $N$, the midpoint of $E'F'$. $M$ gets sent to $M'$, the intersection of $(ARQ)$ and $AM$ (since M was on $BC$ and line $BC$ gets send to $(ARQ)$). We see that, by phantom points, it suffices to prove that $M'NP'A$ concyclic, as that would imply that, reversing the inversion, $N'$ is both on the angle bisector of A and line $MP$, which is infact point $X$. With the above proof, $P'$ is on $E'F'$, so <$ANP'$=<$ANE'=90$. It suffices to prove that <$AM'P'=90$ or $M'HP'$ collinear as <$AM'H=90$ since $AH$ is diameter of $(ARQ)$. This is true since $<MH'A=90=<MPA$, so $AMPH'$ concyclic. Under inversion, this means that $M'P'H$ is collinear. This completes the proof.
06.01.2020 20:50
Let $X$,$Y$,$Z$ be the foots of the altitudes from $A$,$B$,$C$,respectively,and redefine $K$ as the intersection of circles $(ABC)$, $(ADE)$ , $K \neq A$ Claim:The circles $(ABC)$,$(AYZ)$,$(ADE)$ are coaxial
Now,if $O$ and $O'$ are the centers of $(ABC)$,$(AYZ)$ and $A’$ is the antipode of $A$ w.r.t $(ABC)$ note that $ OO'|| A'H$ so $ OO'|| MH$ by the fact that $A',M,H$ are collinear.Also $KA \perp KA'$.But,by the claim $KA$ is the radical axis of $(ABC)$,$(AYZ)$ so $KA \perp OO'$,therefore $KA' || OO'$.Therefore $K \in MH$, so the problem is solved. PS:My first post
06.01.2020 21:21
Problem wrote: Prove that Nibab can draw a circle I hope Nihab has a steady hand :0
06.01.2020 23:49
rocketscience wrote: Problem wrote: Prove that Nibab can draw a circle I hope Nihab has a steady hand :0 Nibab*, not Nihab
02.11.2024 08:46
Let $ \overarc{ADE} \cap \overarc{ABC} : K' $ Let the foot of altitute from $ B $ and $ C$ $Y$ and $Z$. Then, does $\bar{AK'}, \bar{YS}, \bar{BC} $ concurrent?