CWMO 2010, Day 1, Problem 2

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CWMO 2010, Day 1, Problem 2

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analytic geometry

perpendicular bisector

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#1 h Oct 30, 2010, 1:05 PM • 2 Y

Y by Adventure10, Mango247

$AB$ is a diameter of a circle with center $O$ . Let $C$ and $D$ be two different points on the circle on the same side of $AB$ , and the lines tangent to the circle at points $C$ and $D$ meet at $E$ . Segments $AD$ and $BC$ meet at $F$ . Lines $EF$ and $AB$ meet at $M$ . Prove that $E,C,M$ and $D$ are concyclic.

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lssl

#2 h Oct 30, 2010, 1:31 PM • 2 Y

Y by Adventure10, Mango247

This problem is famous , it is from Russia , in that problem , you have to prove $EF$ is perpendicular to $AB$ and this is just the key step to solve this CWMO problem.
Let $AC$ and $BD$ meet at point $L$ , easy to get : $\angle ALB=\frac{\pi}{2}-\angle CAD$ , $\angle CED=\pi-2\angle EDC =\pi-2\angle CAD=2\angle CLD$ , Notice that $EC=DE$ , hence $E$ is the circumcentre of triangle $LCD$ .because $\angle FCL=\angle FDL$ ,hence $L,E,F$ is collinear , Obviously $F$ is the orthocentre of triangle $LAB$ ,hence $EF$ is perpendicular to $AB$ .
Join $CM,DM$ , $C,A,M,F$ ; $M,F,D,B$ are both concyclic , $\Longrightarrow$ $\angle EDC=\angle CAF=\angle CMF$ , Hence $E,C,M,D$ are concyclic .
QED.

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1896 posts

#3 h Oct 31, 2010, 3:43 AM • 2 Y

Y by Adventure10, Mango247

Solution 1
Note that $\angle OCA= \angle OAC = \angle BCE = \alpha$ and
$\angle ODB= \angle OBD= \angle ADE = \beta$
Note that since $CE=DE$ , $E$ lies on the perpendicular bisector of $CD$ .
Also,
$\angle CFD= \angle FCA + \angle CAF= 90+ \frac{1}{2} \angle COD= 90+ \frac{1}{2} (180- \angle CED)= \newline 180- \frac{1}{2} \angle CED$
Thus $E$ is the circumcenter of $\triangle CFD$ , so $\angle EFC= \angle ECF= \alpha$ and thus $\angle CEF= \angle COA$ . Thus $CEOM$ is cyclic, similarly $DEMO$ is cyclic, so $CEDOM$ is cyclic.

In fact, here is a more interesting and quicker solution:
Solution 2
Rotate $\triangle DOB$ about $O$ until $B$ coincides with $A$ , let $D$ rotate to $D'$ . Note that we have $CO=OD'$ and $CE=ED$ . Also,
$\angle COD'=180- \angle COD =\angle CED$ . Next,
$\angle ACO=\angle CAO= \angle ECF$ and $\angle AD'O= \angle ODB= \angle OBD= \angle EDF$ . Thus, $ECFD \sim OCAD'$ . It follows that $\angle CEF= \angle COA$ , so $CEOM$ is cyclic, similarly $DEMO$ is cyclic, so $CEDOM$ is cyclic.

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#4 h Nov 1, 2010, 6:17 PM • 4 Y

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

Just apply Pascal theorem to $ACCBDD$

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#5 h Nov 20, 2010, 12:14 PM • 2 Y

Y by Adventure10, Mango247

mahanmath wrote:

Just apply Pascal theorem to $ACCBDD$

Brilliant idea!

I actually went to this year's CWMO, and to be honest, almost the entire Hong Kong team used coordinate geometry to solve both geometry problems - i guess that's our style.

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1343 posts

jgnr

#6 h Nov 20, 2010, 12:44 PM • 2 Y

Y by Adventure10, Mango247

Andy Loo wrote:

mahanmath wrote:

Just apply Pascal theorem to $ACCBDD$

Brilliant idea!

I actually went to this year's CWMO, and to be honest, almost the entire Hong Kong team used coordinate geometry to solve both geometry problems - i guess that's our style.

Wow, that's quite amazing.

I also used Pascal to solve this. This is my friend's solution:

$\angle CFD=\frac12(\widehat{CD}+\widehat{AB})=\frac12\widehat{CD}+90^{\circ}$ and $\angle CED_{maj}=360^{\circ}-\frac12(\widehat{CD}_{maj}-\widehat{CD})=360^{\circ}-\frac12(360^{\circ}-2\widehat{CD})=180^{\circ}+\widehat{CD}$ . Thus $\angle CED_{maj}=2\angle CFD$ , therefore $C,F,D$ lie on a circle centered at $E$ , which gives $EC=EF=ED$ . So $\angle BFM=\angle CFE=\angle FCE=\frac12\widehat{BC}=90^{\circ}-\frac12\widehat{AC}=90^{\circ}-\angle FBM$ , thus $\angle FMB=90^{\circ}$ . So $\angle EMO=\angle EDO=\angle ECO=90^{\circ}$ , which implies that $E,D,M,O,C$ are cyclic.

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jayme

#7 h Nov 20, 2010, 1:23 PM • 2 Y

Y by Adventure10, Mango247

Dear mathlinkers,
yes, the idea of using Pascal is good in order to have a synthetic proof.
After, we can can show thet this circle goes through the midpoint of AB...
Sincerely
Jean-Louis

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vslmat

#8 h Sep 19, 2012, 5:14 PM • 1 Y

Y by Adventure10

Let $CD$ cut $AB$ at $K$ , $AC$ cut $BD$ at $G$ .
$GF$ is $k$ , the polar of $K$
$CD$ is $e$ , the polar of $E$ . As $K$ is on $e$ , $E$ must be on $k$ , the polar of $K$ . Thus $G, E, F$ are all on $k$ , they are collinear and $GM \perp KO$ .
It follows that $E, C, M, O, D$ are concyclic.

Attachments:

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#9 h Sep 21, 2012, 9:27 PM • 2 Y

Y by Adventure10, Mango247

From figure above, $\angle AFC=\angle AGB\implies \angle AGB=\frac{\angle COA+\angle BOD}{2}$ $=90^\circ-\frac{\angle COD}{2}=\angle CDO=\angle DCO$ , so $CO, DO$ are tangent to the circle $(CDG)$ , hence $E$ is its circumcenter, $\iff GE\bot AB$ .

Best regards,
sunken rock

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jred

#10 h May 29, 2014, 3:56 AM • 2 Y

Y by Adventure10, Mango247

There's another simple solution. Suppose $AC$ and $BD$ meet at $G$ (still using the figure above), we easliy get $F$ is the orthocenter of $\triangle ABG$ . Let $E'$ be the midpoint of $GF$ , then we have $\angle GCE'=\angle CGE'=\angle ABF=\angle BCO$ , hence $\angle E'CO=\angle GCF=90^\circ$ , which means $E'C$ is tangent to circle $O$ at $C$ . Similarly, we have $E'D$ is tangent to circle $O$ at $D$ , so $E'$ is $E$ , thus $EM\bot AB$ . Finally, we have $E, C, M, D$ are on the nine-point-circle of $\triangle ABG$ .

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sayantanchakraborty

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sayantanchakraborty

#11 h Sep 27, 2014, 3:13 PM • 1 Y

Y by Adventure10

Solution:

Extend $CD$ to meet $AB$ at $X$ .Note that $CD$ is the polar of $E$ and it passes through $X$ .So the polar of $X$ must pass through $E$ .Also it is well known that the polar of $X$ passes through $F$ .So $EF$ is the polar of $X$ which means that $EF \perp OX \implies EF \perp AB$ .Now the rest is easy:Note that $MFDB$ concyclic(why?) which means $\angle{ECD}=\angle{CBD}=\angle{FBD}=\angle{FMD}=\angle{EMD}$ and everything is oK.

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#12 h Jun 3, 2018, 6:35 AM • 2 Y

Y by Adventure10, Mango247

Just a question, do u still need to deal with the config issue where F could be inside or outside the semicircle, or it's ok to just deal with the inside case?

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#13 h Nov 19, 2019, 11:49 AM • 3 Y

Y by RudraRockstar, RAMUGAUSS, Adventure10

Let $AB \cap CD=P$ and $AC \cap DB=Q$ . Then by Brocard's theorem $QF$ is the polar of $P$ , and hence $QF \perp AB$ . Now $CD$ is the polar of $E \implies P$ lies on the polar of $E$ . By La hire's theorem the polar of $P$ which is $QF$ passes through $E$ . So $Q,E,F,M$ collinear. Hence $\angle EMA =90$ . So $\angle CME=\angle CAD=\angle ECD \implies C,E,D,M$ is concyclic.

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#14 h Feb 13, 2021, 1:50 PM

Y by

https://i.imgur.com/VrXKcrx.png

We claim that $E$ is the circumcenter of triangle $CDF$ , indeed $\angle DEC=2 \angle DFC \iff 180^{\circ}-\angle EDC-\angle ECD=360^{\circ}-2\angle A-2\angle B\iff \angle A+\angle B-\angle CBD=\angle A+\angle DBA=90^{\circ}$ which is true as $AB$ is the diameter of the circle with center $O$ . So, we now prove that quadrilateral $OCEM$ is cyclic, $\angle COM=2\angle B=2\angle CDF=\angle CEF \implies \textrm{OCEM is cyclic. $\square$}$ Then, we show that quadrilateral $OCED$ is cyclic, $\angle CED=180^{\circ}-\angle EDC-\angle ECD=180^{\circ}-2\angle CBD=180^{\circ}-\angle COD \implies \textrm{OCED is cyclic. $\square$}$ Therefore, since quadrilaterals $OCEM$ and $OCED$ are cyclic, we have quadrilateral $ECMD$ is cyclic. $\blacksquare$

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