Trapezium inscribed in a circle

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

shivangjindal

676 posts

shivangjindal

#1 h May 4, 2014, 12:59 PM • 3 Y

Y by ImSh95, Adventure10, Rounak_iitr

Let $ABCD$ be a trapezium inscribed in a circle $\Gamma$ with diameter $AB$ . Let $E$ be the intersection point of the diagonals $AC$ and $BD$ . The circle with center $B$ and radius $BE$ meets $\Gamma$ at the points $K$ and $L$ (where $K$ is on the same side of $AB$ as $C$ ). The line perpendicular to $BD$ at $E$ intersects $CD$ at $M$ . Prove that $KM$ is perpendicular to $DL$ .

Greece - Silouanos Brazitikos

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

sedrikktl

102 posts

sedrikktl

#2 h May 4, 2014, 3:22 PM • 4 Y

Y by ImSh95, Adventure10, Mango247, and 1 other user

Is that true?
I mean seems like LM is perpendicular to DK

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Aiscrim

409 posts

Aiscrim

#3 h May 4, 2014, 3:26 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

Quite an easy and computable problem

Solution

This post has been edited 1 time. Last edited by Aiscrim, May 6, 2014, 3:39 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

mathuz

1512 posts

mathuz

#4 h May 4, 2014, 5:04 PM • 9 Y

Y by Sx763_, B101099, Davrbek, thczarif, ImSh95, Real_Math, Adventure10, Mango247, and 1 other user

It's equivalent to prove that $KM=KC$ .
Let $O$ is center of $(ABCD)$ . We have $OECB$ is cyclic and tangents to $ME$ . By radical lines theorem for $(O),(B),(OECB)$ we get that $ME,BC,KL$ are concurrent at one point $P$ . Since $ME\parallel AD$ we have $\angle PMC=\angle PCM$ and $PM=PC$ . $KL\perp AB$ it gives $KL\perp MC$ and $PK\perp MC$ . So $KM=KC$ .

Hence the point $M$ is orthocenter of the triangle $KDL$ . :wink:

This post has been edited 1 time. Last edited by mathuz, Jan 27, 2022, 4:53 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

SamISI1

46 posts

SamISI1

#5 h May 4, 2014, 5:51 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

Assume that $<LAB=x$ and $<BDC=y$ It is enough to prove that $M$ is orthocentere of the triangle $\triangle{KDL}$ . In other words we have to prove that $DM=2Rcosx$ . From base we clearly know that $PE$ parallel to $AD$ . From theorem sinus we have $DE=DM cosy$ ; $\frac{AD} {sin2y}=\frac{DE} {cos2y}$ . From general theorem sinus we have that
$2R=\frac{AD} {siny}$ . Then $\frac{DM} {cos2x}=\frac{DE} {cos2x cosy}=\frac{AD cos2y} {sin2y cosy cos2x}=\frac{AD} {siny}$ . To be brief we have to prove that $sin^2 xcos^2 y=\frac{1} {4}$ . It is clearly shown that $ADMP$ is parallelogram.. $DM=AP$ . $AB=AP+PB=2R$ . From theorem sinus $BE=2R sinx$ . Then $DM+\frac{2R sinx} {cosy}=2R$ . we know $DM=\frac{DE} {cosy}$ as well as $DE=2R sinx cos2y$ . Then we have $cos2y sinx+sinx=cosy$ . then $2sinx cos^2 y=cosy$ . Then we have $sinx cos y=\frac{1} {2}$ . Problem has been proven.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

mathworld1

29 posts

mathworld1

#6 h May 4, 2014, 7:41 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

This problem was submitted by Greece. Does anyone know who proposed it?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

mathuz

1512 posts

mathuz

#7 h May 4, 2014, 8:34 PM • 2 Y

Y by ImSh95, Adventure10

i don't understand! Why there written 'where $K$ is on the same side of $AB$ as $C$ '?
Of course, $M$ is orthocenter of the triangle $DKL$ , so $LM\perp DK$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

NickNafplio

422 posts

NickNafplio

#8 h May 6, 2014, 7:02 PM • 3 Y

Y by ImSh95, Adventure10, and 1 other user

mathworld1 wrote:

This problem was submitted by Greece. Does anyone know who proposed it?

Yes, It was proposed by the Greek team leader: Silouanos Brazitikos (Silouan in this forum)

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

cretanman

430 posts

cretanman

#9 h May 6, 2014, 7:52 PM • 4 Y

Y by reshadqedim, ImSh95, Adventure10, Mango247

There is an analytic geometry solution using very simple calculations. You can see it here http://www.mathematica.gr/forum/viewtopic.php?p=206825#p206825 (I am sure that you can understand it).

Alexandros

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

StanleyST

46 posts

StanleyST

#10 h May 7, 2014, 6:14 AM • 4 Y

Y by Chevrolet23, ImSh95, Adventure10, and 1 other user

Let $X=ME \cap BC$ . We know that $\angle XEC=\angle XBE \Rightarrow \Delta EBC$ and $\Delta XEC$ are equivalent triangles. Thus $XE^2=XC*XB$ so $X$ lie on the radical axis $\Rightarrow M$ and $C$ are symetric wrt. radical axis, so by previous observation, $M$ is the orthocenter of triangle $DKL$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

NewAlbionAcademy

910 posts

NewAlbionAcademy

#11 h May 11, 2014, 11:14 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

$ME$ meets $BC$ on the radical axis, and the rest follows. This implies that $M$ is the orthocenter of $\triangle DKL$ .

Note also that $E$ is the incenter of $\triangle DKL$ . In fact, this is the exact same configuration as in this problem.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

polya78

105 posts

polya78

#12 h May 15, 2014, 2:56 PM • 3 Y

Y by ImSh95, Adventure10, Mango247

Let $\Omega$ be the circle with center $B$ , $O$ be the circumcenter of $\triangle ABC$ , and let $ME,KL$ intersect $AB$ at $P,S$ respectively.

Then $A,S$ are inverses with respect to $\Omega$ , as are $P,O$ . So $AP= 2* OS$ , which means that $DM=AP=2*OS$ , which implies that $M$ is the orthocenter of $\triangle DKL$ .

Attachments:

balkan.pdf (397kb)

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

jlammy

1099 posts

jlammy

#13 h May 25, 2014, 5:55 PM • 3 Y

Y by ImSh95, Adventure10, and 1 other user

My solution is similar to mathuz and StanleyST.

My solution

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

sayantanchakraborty

505 posts

sayantanchakraborty

#14 h Aug 9, 2014, 5:30 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

Let $ME \cap BC=X$ .Then it is easy to see that $XE^2=XC \times XB \Rightarrow X$ lies on the radical axis of the two circles,or in other words $X,K,L$ are collinear.From this it immediately follows that $MK=MC$ (note that $ABCD$ is an isosceles trapezoid).Then $\angle{DLK}=\angle{DCK}=\angle{MCK}=\angle{CMK}=180-\angle{CMJ} \Rightarrow LCMJ$ is cyclic.Thus $\angle{KJL}=\angle{MJL}=\angle{MCL}=90^{\circ}$ or in other words $M$ is the orthocenter of $\triangle{LKD}$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

buzzychaoz

178 posts

buzzychaoz

#15 h Aug 29, 2015, 9:37 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

Note $B$ is the midpoint of arc $\overarc{KL}$ not containing $D$ , and since $BE=BK=BL$ we see that $E$ is the incenter of $\triangle DKL$ . Let the incircle of $\triangle DKL$ touch $KL$ at $F$ , let $G$ be the reflection of $F$ across $E$ , let $O$ be the midpoint of $AB$ , and $M'$ the orthocenter of $\triangle DKL$ .

From IO//BC implies AO//HK, we see that $D,G,O$ are collinear and $DGFM'$ is a parallelogram. From $GE//OB$ and $OD=OB$ , $\Rightarrow GD=GE$ . Let $N$ be the midpoint of $DM'$ , it follows that $DGEN$ is a rhombus so $DE\perp GN\Rightarrow DE\perp EM'$ so $M'\equiv M$ , giving $M$ is the orthocenter of $\triangle DKL$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

EulerMacaroni

851 posts

EulerMacaroni

#16 h Oct 13, 2015, 4:01 AM • 2 Y

Y by ImSh95, Adventure10

Applying the radical axis theorem on $\Gamma$ , the circle through $B, C$ and $E$ , and the circle centered at $B$ with radius $BE$ , lines $KL$ , $BC$ , and $EM$ are concurrent at a point $F$ . Remark that $BE\perp EM$ and $BE \perp AD$ ; hence $AD \parallel EM$ . Now, let $G \equiv ME \cap AB$ ; clearly $MGCB$ is an isosceles trapezoid; hence, line $ME$ is the reflection of line $BC$ about line $KL$ . But $MC \perp KL$ , which implies that $C$ is the reflection of $M$ over line $KL$ . However, $DM\perp KL$ , and the reflection of $M$ , $C$ , lies on the circumcircle of $DKL$ , whence $M$ is the orthocenter of $\triangle DKL$ , as desired $.\:\blacksquare\:$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Satvikgupta

17 posts

Satvikgupta

#17 h Nov 8, 2015, 6:16 AM • 3 Y

Y by ImSh95, Adventure10, Mango247

The problem can also be solved using coordinate geometry. Take the center of ABC as origin and AB as X-axis. then simply write the equations of the given lines and their intersection points.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

kapilpavase

595 posts

kapilpavase

#18 h Feb 3, 2016, 3:33 PM • 5 Y

Y by anantmudgal09, Farruxjon_fallen, ImSh95, Adventure10, Mango247

An interesting overkill....
Observe that $E$ is the incentre & $C$ is the $D$ mixtilinear point of $DLK$ .Since $ME\perp DE$ it is well known that $EM,LK,BC$ are concurrent...and its easy from here as shown by posts above

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Tumon2001

449 posts

Tumon2001

#19 h Dec 11, 2017, 8:19 PM • 3 Y

Y by Farruxjon_fallen, ImSh95, Adventure10

This has also got a solution using inversion. Here it is:

Solution: Let $BC\cap KL=N$ .

Claim: $C$ and $M$ are reflections of each other in $KL$ .

Proof of the claim: Consider the inversion $\Psi$ around $\Gamma$ . It is easy to see that $\odot (BCE)$ and $\Gamma$ are tangent to each other at $E$ . Now, $\Psi (NE)=\odot (BCE)\implies NE$ is tangent to $\odot (BCE)$ at $E\implies M$ lies on $NE$ . As $ME\parallel AD$ , so, $NM=NC$ . The claim now follows immediately as $KL\perp CD$ .

Main problem: It is easy to deduce from our claim using angle chase that $M$ is the orthocenter of $\Delta DKL\implies MK\perp DL$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

anantmudgal09

1979 posts

anantmudgal09

#20 h Jan 31, 2018, 7:13 PM • 3 Y

Y by Farruxjon_fallen, ImSh95, Adventure10

shivangjindal wrote:

Let $ABCD$ be a trapezium inscribed in a circle $\Gamma$ with diameter $AB$ . Let $E$ be the intersection point of the diagonals $AC$ and $BD$ . The circle with center $B$ and radius $BE$ meets $\Gamma$ at the points $K$ and $L$ (where $K$ is on the same side of $AB$ as $C$ ). The line perpendicular to $BD$ at $E$ intersects $CD$ at $M$ . Prove that $KM$ is perpendicular to $DL$ .

Greece - Silouanos Brazitikos

Let $X$ be a point on $\overline{AB}$ with $XB=XE$ . Then $\triangle BEA \sim \triangle BXE$ hence $BE^2=BX \cdot BA$ , proving $X=\overline{KL} \cap \overline{AB}$ . Now reflect $E, M$ in $X$ to obtain $E^{*}, M^{*}$ . Then $\angle DBE^{*}=90^{\circ}$ and $\overline{E^{*}M^{*}} \parallel \overline{EM}$ hence $\overline{DM^{*}}$ is a diameter in $\Gamma$ . Thus, $M$ is the orthocenter of $\triangle DKL$ and the claim follows. $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Farruxjon_fallen

2 posts

Farruxjon_fallen

#21 h Dec 21, 2021, 9:52 AM • 1 Y

Y by ImSh95

sedrikktl wrote:

Is that true?
I mean seems like LM is perpendicular to DK

It's also true btw

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Chevrolet23

11 posts

Chevrolet23

#22 h May 1, 2022, 5:57 PM • 1 Y

Y by ImSh95

All are good solution

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Ibrahim_K

62 posts

Ibrahim_K

#23 h Jan 12, 2023, 4:03 PM

Y by

We use complex numbers.Set $(ABC)$ unit circle...so that
$a=-1,b=1,d=-\frac{1}{c},l=\frac{1}{k}$ Let us firstly note that
$BE=BK \iff (k-1)(\bar{k}-1)=(e-1)(\bar{e}-1) \iff k+\frac{1}{k}=2-\frac{4c}{(c+1)^2}(\spadesuit)$ Since $M$ lies on $DC$
$m+\bar{m}dc=d+c \iff \bar{m}=\frac{1}{c}+m-c \ \ \ \ (1)$ On the other hand $ME \bot BE$
$\frac{m-e}{1-e}+\frac{\bar{m}-\bar{e}}{1-\bar{e}}=0 \iff \bar{m}=\frac{(m-e)(1-\bar{e})}{e-1} \ \ \ \ (2)$ Equalizing $1$ and $2$ yields
$m=\frac{(c-1)(2c^2+c+1)}{(c+1)^2}$
Now we are going to show $M$ is orthocenter of $\triangle DKL$ which suffices to show that $m=d+k+l$
$m=d+k+l \iff \frac{(c-1)(2c^2+c+1)}{(c+1)^2}=k+\frac{1}{k}-\frac{1}{c} \implies \text{ using } \spadesuit \text { gives LHS=RHS }$
So we are done!

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

trinhquockhanh

522 posts

trinhquockhanh

#24 h Aug 5, 2023, 12:37 AM

Y by

$\text{nice one, a solution without using trigonometric ratio:}$

https://i.ibb.co/x13SMBf/2014-BMO-P3.png

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Tellocan

35 posts

Tellocan

#25 h Jan 26, 2024, 4:09 PM • 3 Y

Y by Mirhabib, ismayilzadei1387, bmoimo

Similar to some solutions above. It's obvious that $ME$ is tangent to $(KEL)$ .Let $KM$ intersect $DL$ at $J$ .
First, observe that $ME$ is tangent to $(CEB)$ . This is doable through easy angle-chasing.(More precisely, let $\angle EBK$ be $2x$ and $\angle KEC$ be $y$ . Then, $\angle KEM$ is $x$ and so $\angle CEB$ is $90-x-y$ . Therefore, $\angle CBE$ is $x+y$ and so $\angle CBE$ is equal to $\angle CEM$ , as desired.)
Now, use the radical axis theorem to $(ABCD)$ , $(ECB)$ and $(KEL)$ .Since $ME$ is tangent to both of the latter $2$ circles, their radical axis is $ME$ . $KL$ intersects $CM$ at $X$ .
As a result, $ME$ , $KL$ and $BC$ are concurrent. Let the concurrency point be $P$ . Note that $ME$ and $AD$ are perpendicular to $BD$ , so $ME$ is parallel to $AD$ . Therefore, $\angle EMC$ is equal to $\angle MCB$ , or equivalently, $\triangle PMC$ is isosceles. Since $KL$ is perpendicular to $AB$ , we get that $PK$ is the perpendicular bisector of $MC$ . In other words, $KM$ = $KC$ .
Finally, we get that $\angle DMJ$ = $\angle KMC$ = $\angle KCM$ = $\angle KLD$ , and so $JMXD$ is cyclic. We are done. :-D

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

SimplisticFormulas

94 posts

SimplisticFormulas

#27 h Jan 16, 2025, 6:55 AM

Y by

aah found a slick soln

Let $EM \cap KL=T$ and $KL \cap BD=R$
CLAIM 1: $T-C-B$
PROOF: Indeed, note that $TE$ is tangent to $(KEL)$
$\implies TE^2=TK \cdot TL$ . Since $AB$ is a diameter, $AD \perp MD$
$\implies ME \parallel AD$
$\implies \angle CEB=\angle CBD=\angle CAD =\angle CET$
$\implies TE$ is tangent to $(CEB)$
$\implies TE^2= TC \cdot TB= TK \cdot TL$ . Since $B,C,K,L$ are concyclic , we get $T-C-B$ .

CLAIM 2: $T,C,R,D$ are concyclic
PROOF: Consider an inversion around $(B,BE)=(KEL)$ . Note that $(ABCD) \mapsto KL$ , so $C \mapsto T$ and $R \mapsto D$ , so $T,C,R,D$ are concyclic.

In particular, $C$ is reflection of $M$ in $KL$ since $M$ is the orthocentre of $\triangle TDR$ . This implies that $M$ is also the orthocentre of $\triangle KDL \implies KM \perp DL$ as required. $\blacksquare$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Nari_Tom

106 posts

Nari_Tom

#28 h Jan 23, 2025, 1:26 PM

Y by

By using radical axis theorem on the circles $(ABC)$ , $(CEB)$ , $KEL$ we deduce that $EM$ , $BC$ , $KL$ are concurrent. Since $DA$ and $EM$ are parallel $GM=GC$ . So $M$ and $C$ are symetryc about $KL$ we know that $M$ is the orthocenter of $\triangle DKL$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

andrewthenerd

17 posts

andrewthenerd

#29 h Apr 2, 2025, 5:32 AM

Y by

Solution. Let $EM \cap BC=G$ , then $\angle GMC = \frac{\pi}{2}- \angle EDC = \frac{\pi}{2} - \angle CAB = \angle ABC = MCG$ , so $G$ lies on the perpendicular bisector of $MC$ , Furthermore, $CE\perp BC$ and $BE \perp GE \implies GE^2 = GC\cdot GB$ so $G$ lies on the radical axis of both circles, hence $G,K,L$ collinear. Note that the radical axis must be a line perpendicular to $AB\parallel DC$ , so the radical axis is precisely the perpendicular bisector of $MC$ , so $M,C$ are reflections across $KL$ , hence $M$ is the orthocenter of $\triangle DKL$ and we are done.

Z K Y