Let $K$ be a point inside an acute triangle $ ABC $, such that $ BC $ is a common tangent of the circumcircles of $ AKB $ and $ AKC$. Let $ D $ be the intersection of the lines $ CK $ and $ AB $, and let $ E $ be the intersection of the lines $ BK $ and $ AC $ . Let $ F $ be the intersection of the line $BC$ and the perpendicular bisector of the segment $DE$. The circumcircle of $ABC$ and the circle $k$ with centre $ F$ and radius $FD$ intersect at points $P$ and $Q$. Prove that the segment $PQ$ is a diameter of $k$.
Problem
Source: Middle European Mathematical Olympiad 2013 T-6
Tags: geometry, circumcircle, trigonometry, geometric transformation, reflection, perpendicular bisector, power of a point
Luis González
18.05.2014 17:43
$AK$ is common radical axis of $\odot(AKB),\odot(AKC)$ cutting its common tangent $\overline{BC}$ at its midpoint $M$ $\Longrightarrow$ $AK$ is the A-median of $\triangle ABC$ $\Longrightarrow$ $DE \parallel BC$ for any $K$ on $AM,$ thus $\widehat{DEK}=\widehat{CBK}=\widehat{DAK}$ $\Longrightarrow$ $ADKE$ is cyclic. Hence, by Miquel's theorem $\odot(BDK)$ and $\odot(CEK)$ meet again at $X \in BC.$ Hence $\widehat{KDX}=\widehat{KBC}=\widehat{KAD}$ $\Longrightarrow$ $\widehat{EDX}=\widehat{KAD}+\widehat{KDE}=\widehat{BAC}$ and similarly $\widehat{DEX}=\widehat{BAC},$ thus $\triangle XED$ is X-isosceles $\Longrightarrow$ $X \equiv F.$ Since $\widehat{DFB}=\widehat{EFC}$ and $\widehat{BDF}=\widehat{BKF}=\widehat{ECF},$ then $\triangle DBF \sim \triangle CEF$ $\Longrightarrow$ $\tfrac{FD}{FC}=\tfrac{FB}{FE}$ $\Longrightarrow$ $FD^2=FB \cdot FC$ $\Longrightarrow$ power of $F$ WRT $\odot(ABC)$ equals $-FD^2$ $\Longrightarrow$ $\odot(ABC)$ bisects $\odot(F,FD),$ i.e. $PQ$ is a diameter of $\odot(F,FD).$
Mikasa
10.06.2014 09:28
Luis González wrote: $AK$ is common radical axis of $\odot(AKB),\odot(AKC)$ cutting its common tangent $\overline{BC}$ at its midpoint $M$ $\Longrightarrow$ $AK$ is the A-median of $\triangle ABC$ $\Longrightarrow$ $DE \parallel BC$ for any $K$ on $AM,$ thus $\widehat{DEK}=\widehat{CBK}=\widehat{DAK}$ $\Longrightarrow$ $ADKE$ is cyclic. I did the same thing at the start of my proof. Now wlog assume that $AB>AC$. Now, let $AF'$ be the $A$-symmedian of $\triangle ABC$. Then it is a well known fact that $\frac{BF'}{F'C}=\frac{AB^2}{AC^2}$. Now, applying sine law in $\triangle ABK$ we get that $KB^2=AB^2\cdot \frac{\sin \angle BAK}{\sin \angle AKB}$. Similarly, $KC^2=AC^2\cdot \frac{\sin \angle CAK}{\sin \angle AKC}$. So, $\frac{KB^2}{KC^2}=\frac{AB^2}{AC^2}\cdot r$, where $r=\frac{\sin \angle BAK\cdot \sin \angle AKC}{\sin \angle AKB\cdot \sin \angle CAK}$. Now we know that, $\angle BAK=\angle KBC,\angle CAK=\angle KCB$ and $\angle AKC=180^{\circ}-\angle CKM, \angle AKB=180^{\circ}-\angle BKM$. Also, $\frac{\sin \angle KBC}{\sin \angle KCB}=\frac{KC}{KB}$, so we get that, $r=\frac{KC\cdot \sin \angle CKM}{KB\cdot \sin \angle BKM}$ We introduce a lemma here: Lemma: In $\triangle XYZ$, the point $W$ lies on $YZ$ (Possibly on the extension). Then, $\frac{YW}{WZ}=\frac{XY\cdot \sin \angle YXW}{XZ\cdot \sin \angle ZXW}$. Proof: Applying sine law in $\triangle YXW, \triangle ZXW$ finishes it. Now by this lemma, we get that $r=\frac{CM}{MB}=1$. So, $\frac{KB^2}{KC^2}=\frac{AB^2}{AC^2}=\frac{BF'}{F'C}$. Thus $KF'$ is the $K$-symmedian of $\triangle KBC$. Now $\angle CKF=\angle BKM=\angle EKA=\angle EDA=\angle DBF$ i.e. $F'\in \odot BKD$. Similarly $F'\in \odot CKE$. Now, $\angle DEK=\angle KBC, \angle F'EK=\angle KCB$. So, $\angle F'ED=\angle KBC+\angle KCB=\angle BKD=\angle BAC$. Similarly, $\angle F'DE=\angle BAC$. Thus we have that $F\equiv F'$. Also, $FD,FE$ are tangents to $\odot ADKE$. Some simple angle chasing now provides that $\angle BDF=\angle ACB, \angle CDF=\angle KBC$. So by applying sine law in $\triangle DBF, \triangle DCF$ we get that $\frac{DF}{FB}=\frac{\sin \angle ABC}{\sin \angle ACB}=\frac{AC}{AB}$ and that $\frac{FC}{DF}=\frac{\sin \angle KBC}{\sin \angle KCB}=\frac{KC}{KB}$. But we proved before that $\frac{AC^2}{AB^2}=\frac{KC^2}{KB^2}$. So we get that $\frac{DF}{FB}=\frac{FC}{DF}\Rightarrow FD^2=FP^2=BF\cdot FC$. Since $F$ is the center of $\odot DEP$ we have that $F$ is on the radical axis, $PQ$, of $\odot ABC, \odot DEP$. Thus $PQ$ is a diameter of $k$.
Mikesar
10.06.2014 16:06
Luis González wrote: Hence, by Miquel's theorem $\odot(BDK)$ and $\odot(CEK)$ meet again at $X \in BC.$ Hence $\widehat{KDX}=\widehat{KBC}=\widehat{KAD}$ $\Longrightarrow$ $\widehat{EDX}=\widehat{KAD}+\widehat{KDE}=\widehat{BAC}$ and similarly $\widehat{DEX}=\widehat{BAC},$ thus $\triangle XED$ is X-isosceles $\Longrightarrow$ $X \equiv F.$ We can also avoid angle chasing using the well known fact that if $O$ is the circumcenter of $ADKE$ then $OX\perp BC.$ But $OF\perp BC$ since $O,F$ belong to the perpendicular bisector of $DE$ and $DE\parallel BC.$ The rest of my proof is the same.
saturzo
12.06.2014 14:23
Let $AK$, the radical axis of $\odot AKB$ and $\odot AKC$, intersect $BC$ and $DE$ at $M$ and $M'$, respectively; so $MB^2=MC^2 \implies M$ is the midpoint of $BC \implies \pi - \angle DKE = \angle MBK + \angle MCK = \angle BAK + \angle KAC = \angle DAE.$ $\therefore K \in \odot ADE\implies \angle EDC = \angle CAK = \angle BCD \implies DE \parallel BC$. So, $M'$ is the midpoint of $DE$. Now note that, in $\odot ADE, BC$ is the polar of the pole $M'$ and $M'F \perp BC [\because M'$ is the midpoint of $DE]$. So, $F$ is the inverted point of $M'$ w.r.t. $\odot ADE$. And if $O$ is the center of $\odot ADE$, then $OM' \perp DE$ as $M'$ is the midpoint of chord the $DE$. So, $FD$ and $FE$ touch $\odot ADE$, simple angle chasing implies $\angle BDE = \angle ECF, \angle DBF = \angle CEF \implies \triangle DBF \sim \triangle CEF.$ So, $FB.FC = FD.FE = FP^2$. So if $FP$ intersects $\odot ABC$ again at $Q'$, then, $FQ' = FP \implies Q' \in k \implies Q' \equiv Q \implies PQ$ is the diameter of $k$ - done!
Particle
20.06.2014 10:13
The statement is equivalent to proving $DF^2=BF\cdot FC$. So I'll prove only that.
Obviously line $AK$ is the median of $ABC$. Now applying Ceva's theorem for $D,E, AK\cap BC$, we get $DE||BC$. Simple angle-chasing shows $A,D,E,K$ are concyclic. Now suppose circle $KBD$ intersects $BC$ at $F'$. From Miquel's theorem $F'CEK$ is also concyclic. So $\angle F'BK=\angle F'DK\implies \angle F'DE=\angle A$. Similarly $F'ED=\angle A$. Thus $F=F'$. Now notice that $FD$ and $FE$ are tangents to $\bigcirc ADE$. So from a well-known lemma $AF$ is a symmedian of $ADE$ and therefore of $ABC$, too. Suppose $AF$ meets circles $ABC$ and $ADE$ at $J,H$, resp. $\angle JCB=\angle JAB=\angle KAE=\angle KCB$. Similarly $\angle JBC=\angle KBC$ So $J$ is the reflection of $K$ across $BC$ yielding $FJ=FK$. Notice that in circle $ADE$, $\widehat {DH}=\widehat {EK}$ implying $H$ is the reflection of $K$ across the perpendicular bisector of $DE$. So $FH=FK=FJ$. Hence $BF\cdot FC=AF\cdot FJ=AF\cdot FH=DF^2$. Therefore $PQ$ is the diameter of circle $(F,DF)$.
JANMATH111
13.07.2018 17:31
Particle wrote:
The statement is equivalent to proving $DF^2=BF\cdot FC$. So I'll prove only that.
Obviously line $AK$ is the median of $ABC$. Now applying Ceva's theorem for $D,E, AK\cap BC$, we get $DE||BC$. Simple angle-chasing shows $A,D,E,K$ are concyclic. Now suppose circle $KBD$ intersects $BC$ at $F'$. From Miquel's theorem $F'CEK$ is also concyclic. So $\angle F'BK=\angle F'DK\implies \angle F'DE=\angle A$. Similarly $F'ED=\angle A$. Thus $F=F'$. Now notice that $FD$ and $FE$ are tangents to $\bigcirc ADE$. So from a well-known lemma $AF$ is a symmedian of $ADE$ and therefore of $ABC$, too. Suppose $AF$ meets circles $ABC$ and $ADE$ at $J,H$, resp. $\angle JCB=\angle JAB=\angle KAE=\angle KCB$. Similarly $\angle JBC=\angle KBC$ So $J$ is the reflection of $K$ across $BC$ yielding $FJ=FK$. Notice that in circle $ADE$, $\widehat {DH}=\widehat {EK}$ implying $H$ is the reflection of $K$ across the perpendicular bisector of $DE$. So $FH=FK=FJ$. Hence $BF\cdot FC=AF\cdot FJ=AF\cdot FH=DF^2$. Therefore $PQ$ is the diameter of circle $(F,DF)$.
beautiful! Well done!