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Source: Balkan MO 2014 G-5

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dizzy

#1 h Jun 10, 2015, 2:27 PM • 3 Y

Y by Adventure10, Mango247, lian_the_noob12

Let $ABCD$ be a trapezium inscribed in a circle $k$ with diameter $AB$ . A circle with center $B$ and radius $BE$ ,where $E$ is the intersection point of the diagonals $AC$ and $BD$ meets $k$ at points $K$ and $L$ . If the line ,perpendicular to $BD$ at $E$ ,intersects $CD$ at $M$ ,prove that $KM\perp DL$ .

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tranquanghuy7198

253 posts

tranquanghuy7198

#2 h Jun 10, 2015, 4:09 PM • 1 Y

Y by Adventure10

My solution:
$O$ is the midpoint of $AB$ and also the center of the circle $(ABCD)$
$EM\cap{AB} = N$
It’s easy to see that $\angle{MNB} = \angle{CBN}$
$\Rightarrow CMNB$ is an isoceles trapezium.
Moreover: $BO.BN = BE^2 = BK^2$
$\Rightarrow \frac{BO}{BK} = \frac{BK}{BN}$
$\Rightarrow \triangle{BOK} \sim \triangle{BKN}$
$\Rightarrow KN = KB (\because OK = OB)$
$\Rightarrow KL$ is the perpendicular bisector of $BN$
$\Rightarrow M = R_{KL}(C)$
$\Rightarrow M$ is the orthocenter of $\triangle{DKL}$ and the conclusion follows.
Q.E.D

This post has been edited 1 time. Last edited by tranquanghuy7198, Jun 11, 2015, 2:38 AM

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#3 h Jun 10, 2015, 5:23 PM • 2 Y

Y by Adventure10, Mango247

Why do we have $KN=KB$ ? I don´t think that that´s obvious.
Actually, $BO \cdot BN=BK^2$ holds because the triangles $\triangle BEO$ and $\triangle BEN$ are similar ( $\angle BOE=\angle NEB=90^\circ$ ), and therefore $BK^2=BE^2=BO \cdot BN$ . From this it follows that $KN=KB$ .

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1699 posts

#4 h Jun 10, 2015, 5:38 PM • 2 Y

Y by Adventure10, Mango247

Another more ugly solution:
Again we show that $M$ is the orthocenter of $\triangle DKL$ . Because of $KL \perp DM$ , it suffices to prove that $DM=AB \cos \angle LDK$ , because $AB$ is two times the circumradius of $\triangle DKL$ .
Because of $ME \parallel AD$ and $AB \parallel CD$ , we have $DM = \frac{AE \cdot CD}{AC}=\frac{AB \cdot CD}{AB + CD}$ , therefore the claim is equivalent to $\cos \angle LDK = \frac{CD}{AB+CD}$ .
We have $\angle LDK = 180^\circ - \angle KBL=180^\circ -2\angle KBO$ . Because of $OK=OB$ , we get $\cos \angle KBO=\frac{BK}{2BO}=\frac{BE}{AB}$ , and we have $BE=\frac{BD \cdot AB}{AB+CD}$ . This implies
$\cos \angle LDK = 1 - 2 \cos^2 \angle KBO=1-2\left( \frac{BE}{AB} \right) ^2 = \frac{(AB+CD)^2-2BD^2}{(AB+CD)^2}$ ,
and this should be equal to $\frac{CD}{AB+CD}$ , which is equivalent to $AB^2+AB \cdot CD = 2BD^2$ .
But, by ptolemy, $AB \cdot CD = BD^2-AD^2$ and by pythagoras $AB^2=AD^2+BD^2$ . This implies the desired statement.

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41 posts

#5 h Jun 10, 2015, 7:02 PM • 3 Y

Y by sydneymark, Adventure10, Mango247

My solution.
Let $O$ be the center of the circle k. Then $EO$ $\perp$ $BA$ $\implies$ $BCEO$ cyclic.
Observe that the centers of the circles $\odot$ $BCEO$ and $\odot$ $BCDA$ lies on the line $BE$ $\implies$ $ME$ is the radical axis of these two circles.
Hence by applying radical axis theorem to circles $\odot$ $BCEO$ , $\odot$ $BCDDA$ and $\odot$ $KEL$ $\implies$ $BC,LK,EM$ are concurrent. Assume that they meets at $P$ .
Since $PE$ $\parallel$ $DE$ $\implies$ $\angle$ $PMC$ $=$ $\angle$ $DAB$ $=$ $\angle$ $ABC$ $=$ $\angle$ $MCP$ by using this together with the fact $PL$ $\parallel$ $MC$ $\implies$ $PK$ is the perpendicular bisector of $CM$ . Rest is easy angle chasing
If $M'$ $=$ $KM$ $\cup$ $DL$ $\implies$ $\angle$ $EMM'$ $=$ $\angle$ $KMP$ $=$ $KCP$ $=$ $\angle$ $BLK$ $=$ $\angle$ $BKL$ $=$ $\angle$ $BDL$
Hence $MEM'D$ are cyclic and we are done.

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111 posts

#6 h Aug 2, 2019, 7:02 AM • 3 Y

Y by top1csp2020, Adventure10, Mango247

Notice EM//BC,AB//MC so KL bisect CM=F so FM.FD=FK.FL but KL perp CD so done

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1900 posts

#7 h Aug 2, 2019, 4:22 PM • 3 Y

Y by RudraRockstar, Adventure10, Mango247

If $KM\perp DL$ then why not $LM\perp DK$ since assumptions are the same for both points? That's why we will prove both which means $M$ is the orthocenter of triangle $DLK$ .
Complex coordinates: $a=-1,b=1,|c|=1,d=-\frac1c$ Then $e=\frac{ab(c+d)-cd(a+b)}{ab-cd}=\frac{c-1}{c+1}$ $C,M,D$ are collinear:
$\frac{c-m}{c-d}=\overline{\left(\frac{c-m}{c-d}\right)}$ $\overline{m}=m+\frac{1}{c}-c$ $EM\perp BE$ yields
$\frac{b-e}{e-m}=-\overline{\left(\frac{b-e}{e-m}\right)}$ as we know relationship between $m$ and $\overline m$ we deduce
$m=\frac{2c^3-c^2-1}{c(c+1)^2}$ $k,l$ are all solutions $x$ of two equations:
$|x|=1\wedge |x-1|=\left| 1-\frac{c-1}{c+1}\right|$ Squaring both sides of the second gives
$0=x^2-x\cdot\frac{2(c+1)^2-4c}{(c+1)^2}+1$ By Viete formulas
$k+l+d=k+l-\frac{1}{c}=\frac{2(c+1)^2-4c}{(c+1)^2}-\frac{1}{c}=\frac{2c^3-c^2-1}{c(c+1)^2}=m$ Because $|k|=|l|=|d|=1$ (they all lie on circle $k$ ) we have that $M$ is orthocenter of triangle $KLD$ .

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2125 posts

#8 h Aug 11, 2019, 9:26 PM • 11 Y

Y by Naruto.D.Luffy, e_plus_pi, Delta0001, Physicsknight, MathPassionForever, Pluto1708, Varuneshwara, hansu, amar_04, Adventure10, Mango247

Balkan MO 2014 G5 wrote:

Let $ABCD$ be a trapezium inscribed in a circle $k$ with diameter $AB$ . A circle with center $B$ and radius $BE$ ,where $E$ is the intersection point of the diagonals $AC$ and $BD$ meets $k$ at points $K$ and $L$ . If the line ,perpendicular to $BD$ at $E$ ,intersects $CD$ at $M$ . Prove that $KM\perp DL$ .

Solution:
Note that Proving the following Lemma directly implies the question:

Equivalent Lemma wrote:

Let $\Delta ABC$ be a right triangle at $B$ . Let $M,M'$ be midpoints of arc $BC$ and $BAC$ . Let $AM$ meet $BM'$ at $E$ . Let $\omega$ be circle $(M,ME)$ . Let $\omega$ meet $\odot (ABC)$ at $L$ and $N$ . Let tangent to $\omega$ at $E$ meet $AB$ at $K$ . Prove, $B$ is reflection of $K$ over $LN$

Solution #1: (AlastorMoody, MathPassionForever & Naruto.D.Luffy)

$E$ is the incenter WRT $\Delta ALN$ and also $EO||LN||BC$ . Then, using the following lemma solves the problem:

Lemma#: In $\Delta ABC$ , If $OI || BC$ and $H$ is orthocenter WRT $\Delta ABC$ . Then, $AI \perp IH$
Proof: Let $D$ be $A-$ intouch point and $D'$ be reflection of $D$ over $I$ . Let $M$ be midpoint of $BC$ . Let $T_A$ be $A-$ extouch point and $H'$ be reflection of $H$ over $BC$ . $\angle D'OI$ $=$ $\angle ODC$ $=$ $\angle OT_AD$ $=$ $180^{\circ}-\angle IOT_A$ $\implies$ $A-D'-O-T_A$ . Now, $\angle MOA'$ $=$ $\angle MOD$ $=$ $\angle HAA'$ $=$ $\angle HOM$ $\implies$ $H-D-O$ $\implies$ $HD||AO$ . Now, $HD=H'D=AD'=D'I=DI$ $\implies$ $D$ is center of $\odot (IHH')$
$\angle AHI=180^{\circ}-\angle H'HD-\angle DHI=90^{\circ}-\angle H'AA'+\angle AH'I=90^{\circ}-\angle HAI \implies AI \perp HI \qquad \blacksquare$

Solution #2: (e_plus_pi)

Let $F$ = $MB \cap EE$ . Also let $O$ be the circumcenter of $\odot(ABC)$ . Consider the following claim:
Claim: $MOEB$ is cyclic.
Proof: Firstly, $\angle MOE = 90^{\circ}$ . Next, note that $ABMM^{\prime}$ is a isosceles trapezoid which means that $EM = EM^{\prime} \implies EO \perp MO$ . Hence the claim.

So by radical axis theorem $F \in LN$ . Consider the inversion about $\omega$ . Clearly, $\odot(ABC) \mapsto LN$ and so $F \mapsto B$ . Also, $\overline{EE} \mapsto \odot (MOEB)$ . Thus we have, $\angle BFK = \angle MFE = \angle MEB = \angle MOB = \angle A$ And,
$\angle FBK = 90^{\circ} - \angle KBE = \angle ABM^{\prime} = \frac{1}{2} \cdot \angle (90^{\circ} + \angle C)$ These two imply the desired result $\qquad \blacksquare$

Solution #3: (Naruto.D.Luffy)

Let $KE \cap LN=I$ . Applying Radical Axes Theorem on $\odot (ABC)$ , $\odot (MOEB)$ and $\odot (LEN)$ $\implies$ $B-M-I$ Now, $EA=EB$ and $\Delta MEI$ $\sim$ $\Delta EBI$ . Hence,
$\angle BIK=\angle MIE=\angle BEM=2\angle BAE=\angle BAC$ And, $\angle BKI=\angle EKA =90^{\circ}-\frac{A}{2} \qquad \blacksquare$

Solution #4: (Pluto1708)

Redefine $K$ as reflection of $B$ over $LN$ . Let $U=AM\cap BC$ .Note that it suffices to show $\odot{KEBU}$ concyclic.Equivalently since $EA=EB=EU=\dfrac{c}{2\cos \dfrac{A}{2}} \Rightarrow AU.AE=2AE^2=\dfrac{c^2}{2\cos^2 \dfrac{A}{2}}=\dfrac{c^2}{1-\cos A}=\dfrac{bc^2}{b-c}$ .Because of POP it suffices to show $AK.AB=\dfrac{bc^2}{b-c} \Rightarrow AK=\dfrac{bc}{b-c} \implies BK=\dfrac{c^2}{b-c} \implies BT=\dfrac{c^2}{2(b-c)}$ .By Los $\dfrac{BN}{\sin{\dfrac{A}{2}+\angle NAC}}=2R=\dfrac{BM}{\sin{\dfrac{A}{2}}}=\dfrac{c}{\sin C}=\dfrac{MN}{\sin{\dfrac{A}{2}+\angle NAC}}=\dfrac{ME}{\sin{\dfrac{A}{2}+\angle NAC}}$ .
Recall that $ME=\dfrac{R}{\cos \dfrac{A}{2}}$ .Hence $BN=\dfrac{R\sin{A+\angle NAC}}{\cos{\dfrac{A}{2}}(\sin{\dfrac{A}{2}}+\angle NAC})$ .Now $BT=BN\sin NAC=\dfrac{R\sin{A+\angle NAC}\sin NAC}{\cos{\dfrac{A}{2}}\sin{\dfrac{A}{2}}+\angle NAC}$ .Now note that $2R=\dfrac{ME}{\sin{\dfrac{A}{2}+\angle NAC}} \implies \sin{\dfrac{A}{2}+\angle NAC}=\dfrac{1}{2\cos \dfrac{A}{2}}$ .Thus $BT=\dfrac{R\sin{((\dfrac{A}{2}+\angle NAC)+\dfrac{A}{2})}\sin{((\dfrac{A}{2}+\angle NAC)-\dfrac{A}{2}})}{\cos \dfrac{A}{2} \sin{(\dfrac{A}{2}+\angle NAC})}=2R(\dfrac{1}{4\cos^2\dfrac{A}{2}}-\sin^2\dfrac{A}{2})$ .This after some easy simplifications gives $\dfrac{c^2}{b-c}$ as desired. $\qquad \blacksquare$

This post has been edited 5 times. Last edited by AlastorMoody, Aug 12, 2019, 9:20 AM

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lian_the_noob12

173 posts

lian_the_noob12

#9 h Aug 4, 2024, 11:22 PM

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$\color{green} \textbf{Construction :}$ Let $EM \cap AB \equiv Z, ZM \cap BC \equiv Y, KL \cap AB \equiv P, KL \cap CD \equiv Q$ $AC \cap \omega_2 \equiv R, KM \cap DL \equiv S, LD \cap AB \equiv T$
$\color{blue} \textbf{Claim 1 :}$ $BCMZ$ is a trapezoid
$\textbf{Proof :}$ $KL$ is the radical axis of the circles, So $KL \perp AB \equiv CD$
We have $\angle ECB=\angle ACB=90°$ as $AB$ is a diameter, $ER$ is a chord of $\omega_2$ and $B$ is the center $\implies \angle EBC=\frac{1}{2} \angle EBR=\angle MEC$ Again $\angle EBC= \angle DBC=\angle DAC$ So, $\angle DAC =\angle MCC \implies AD \parallel ME\equiv MZ \implies AD=MZ=BC \square$
$\color{blue} \textbf{Claim 2 :}$ $KL$ passes through $Y$
$\textbf{Proof :}$
$\triangle EYC \sim \triangle EBY \implies YE^{2}=YC.YB \implies Pow_{\omega_2} Y =Pow_{\omega_1} Y$ Hence the radical axis $KL$ passes through $Y\square$

As $BCMZ$ is a trapezoid $YBZ$ is isosceles and $YD$ is the perpendicular bisector of $BZ$
Hence, $\angle SKD=\angle MKQ=\angle QKC=\angle LKC=\angle LDC=\angle LTD$ So, $KSTD$ cyclic $\implies \angle KST=180-\angle KDT=90 \blacksquare$

$\color{black} \textbf{Remark :}$ Really great problem!

This post has been edited 2 times. Last edited by lian_the_noob12, Aug 4, 2024, 11:31 PM

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