In an acute triangle $ABC$, the circle with diameter $AC$ intersects $AB$ and $AC$ at $K$ and $L$ different from $A$ and $C$ respectively. The circumcircle of $ABC$ intersects the line $CK$ at the point $F$ different from $C$ and the line $AL$ at the point $D$ different from $A$. A point $E$ is choosen on the smaller arc of $AC$ of the circumcircle of $ABC$ . Let $N$ be the intersection of the lines $BE$ and $AC$ . If $AF^{2}+BD^{2}+CE^{2}=AE^{2}+CD^{2}+BF^{2}$ prove that $\angle KNB= \angle BNL$ .
Problem
Source: Turkey NMO 2007 Problem 1
Tags: geometry, circumcircle, geometry unsolved
oneplusone
29.09.2011 14:29
By Carnot's theorem, $AD,CF$ and the perpendicular from $E$ onto $AC$ are concurrent. Thus $AD,BE,CF$ are altitudes and the rest is easy.
efoski1687
02.10.2011 10:32
Posted also there: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=336562&hilit=Turkey+2007
MatjazL
16.10.2011 22:04
There is typo in the statement. It should be "intersects $AB$ and $BC$". Because $L$ and $K$ lie on a circle with $AC$ as its diameter we know that $CF \perp AB$ and $AD \perp CB$. We can now apply Pythagoras's theorem several times to make the condition from the problem nicer: $ AF^{2}+BD^{2}+CE^{2}=AE^{2}+CD^{2}+BF^{2} $ $ AK^{2}+KF^2+BL^{2}+DL^2+CE^{2}=AE^{2}+CL^{2}+DL^2+BK^{2}+KF^2 $ $ AK^{2}+BL^{2}+CE^{2}=AE^{2}+CL^{2}+BK^{2} $ $ CA^2-CK^2+AB^2-AL^2+CE^{2}=AE^{2}+CA^2-AL^2+BC^2-KC^2 $ $ AB^{2}+CE^{2}=AE^{2}+CB^{2} $ $ CE^{2}-CB^2=AE^{2}-AB^2 $ Let's draw the perpendicular from $B$ to $AC$ and let their intersection be $N'$ and let $E'$ be the intersection of this line with the circumcirlce. We can change our condition one last time : $ CE^{2}-CB^2=AE^{2}-AB^2 $ $ CE^{2}-CN'^2-BN'^2=AE^{2}-AN'^2-BN'^2 $ $ CE^{2}-CN'^2=AE^{2}-AN'^2$ We can easily check that if $E=E'$ the above condition is satisfied and one can check with a simple calculation that there is no other point $E$ that satisfies. Therefore $E=E'$. Since $AL,BN,CK$ are the altitudes of the triangle, the triangle $LNK$ is an orthic triangle of the triangle $ABC$ and it is well known that the altitudes bisect the angles of the orthic triangle, therefore $\angle KNB=\angle BNL$ and we're done.
ehuseyinyigit
17.09.2024 20:06
Here is an elemantary solution just by using some orthocenter properties. It is clear that $AL$ and $CK$ are altitudes. Let $H$ be the orthocenter. It is well-known that reflection of $H$ wrt triangle's sides are on circumcircle (Since $\angle FAB=\angle FCB=\angle BAL$). This implies $$AF=AH\quad ,\quad BD=BH=BF \quad \text{and} \quad CD=CH$$ Claim: $BN\perp AC$ According to the given condition $$AF^2+BD^2+CE^2=AH^2+BF^2+CE^2=AE^2+CD^2+BF^2$$$$\Longleftrightarrow AH^2+CE^2=AE^2+CD^2$$Observe that all the lengths above are segments of quadrilateral $AHCE$ and by perpendicularity condition, we have $HE\perp AC\Longleftrightarrow BN\perp AC$. Thus, $N$ is the foot of altitude drawn from point $B$. Orthocenter $H$ is incenter of the orthic triangle $\triangle KLN$, and the result follows as $\angle KNB=\angle BNL$ directly. Remark: I think that the problem can be generalized via Blanchet's Theorem.