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Source: IMO Shortlist 2011, G3

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1347 posts

WakeUp

#1 h Jul 13, 2012, 11:38 AM • 7 Y

Y by Davi-8191, Amir Hossein, mathematicsy, Adventure10, Mango247, Rounak_iitr, and 1 other user

Let $ABCD$ be a convex quadrilateral whose sides $AD$ and $BC$ are not parallel. Suppose that the circles with diameters $AB$ and $CD$ meet at points $E$ and $F$ inside the quadrilateral. Let $\omega_E$ be the circle through the feet of the perpendiculars from $E$ to the lines $AB,BC$ and $CD$ . Let $\omega_F$ be the circle through the feet of the perpendiculars from $F$ to the lines $CD,DA$ and $AB$ . Prove that the midpoint of the segment $EF$ lies on the line through the two intersections of $\omega_E$ and $\omega_F$ .

Proposed by Carlos Yuzo Shine, Brazil

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222 posts

#2 h Jul 13, 2012, 6:55 PM • 1 Y

Y by Adventure10

This problem was posted in april I posted a solution then with this picture, but now looking at my posts it seems gone.
NOPE, found it: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=476973 (seems someone from russia posted it)?

Attachments:

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154 posts

msecco

#3 h Jul 25, 2012, 11:22 AM • 2 Y

Y by Adventure10, Mango247

This problem is due to Carlos Yuzo Shine, from Brazil.

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22 posts

Orin

#4 h Jul 25, 2012, 7:24 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

I have some changes in the notations.Let the circles with diameters $AD,BC$ meet at $X,Y$ inside the quadrilateral $ABCD$ .
So $\angle AXD=\angle AYD=\angle BXD=\angle BYD=90^{\circ}$ .
Let $X_1,X_2,X_3,X_4$ be the foot of perpendiculars from $X$ to $BC,CD,DA,AB$ respectivley.
Let $Y_1,Y_2,Y_3,Y_4$ be the foot of perpendiculars from $Y$ to $BC,CD,DA,AB$ respectively.
Observation:From the perpendiculars we get $(X,X_3,A,X_4);(X,X_4,B,X_1);(X,X_1C,X_2);(X,D,X_2,X_3)$ are cyclic.The similar cases with $Y$ .
Note that these properties are independent of the positions of the foot of perpendiculars meaning the properties are valid even if some the foots lie on the extensions of sides $AB,BC,CD,DA$ .
We'll be using these properties from now on.
Claim1: $X_1X_2X_3X_4,Y_1Y_2Y_3Y_4$ are cyclic.
Proof: $\angle X_4X_1X_2+\angle X_4X_3X_2=\angle X_4X_1X+\angle XX_1X_2+\angle X_4X_3X+\angle XX_3X_2$
$=\angle ABX+\angle XCD+\angle BAX+\angle CDX$
$=360^{\circ}-(\angle XBC+\angle XCB)-(\angle XAD+\angle XDA)$
$=360^{\circ}-90^{\circ}-90^{\circ}=180^{\circ}$ .
So $X_1X_2X_3X_4$ is cyclic.Let the circle be $\omega_X$ .
Similarly $Y_1Y_2Y_3Y_4$ is cyclic.Let the circle be $\omega_Y$ .
Let $P,Q=\omega_X \cap \omega_Y$ .
Let $X_2'=XX_2 \cap AB,Y_2'=YY_2 \cap AB,X_4'=XX_4 \cap CD,Y_4'=YY_4 \cap CD$
Claim: $X_2',X_4' \in \omega_X,Y_2',Y_4' \in \omega_Y$ .
$\angle X_4X_2'X_2=\angle X_2'BX+\angle X_2'XB=\angle X_4BX+90^{\circ}-\angle CXX_2$
$=\angle X_4X_1X+\angle XCX_2=\angle X_4X_1X+\angle XX_1X_2=X_4X_1X_2$ .
So, $X_2' \in \omega_X$ .The others can be proved similarly.
Further observation: $\angle X_2'Y_4Y_4'=\angle X_2'X_2Y_4'=90^{\circ}$ ,so $X_2',Y_4,X_2,Y_4'$ are cyclic.
Similarly $X_4,Y_2',X_4',Y_2$ are cyclic.
Let $M=Y_4Y_4' \cap X_2X_2',N=X_4X_4' \cap Y_2Y_2'$
So from intersecting chord theorem $MY_4.MY_4'=MX_2.MX_2'$
But $LHS=$ power of $M$ wrt $\omega_Y$ and $RHS=$ power of $M$ wrt $\omega_X$
$\Rightarrow M \in PQ$ .
Similarly $N \in PQ$ .
$MY \bot AB,NX \bot AB \rightarrow MY \parallel NX;MX \bot CD,NY \bot CD \rightarrow MX \parallel NY$
So $MXNY$ is a parallelogram of which $MN,XY$ are diagonals.
Hence $MN$ bisects $XY$ meaning the midpoint of $XY$ lies on $PQ$ .

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49 posts

#5 h Aug 13, 2012, 9:00 AM • 2 Y

Y by Adventure10, Mango247

WakeUp wrote:

...Prove that the midpoint of the segment $EF$ lies on the line through of $\omega_E$ and $\omega_F$ ...

Typo. It should be like "on the line through the two intersections of ~", according to the official version. [WakeUp: edited]

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236 posts

#6 h Sep 11, 2012, 7:54 PM • 3 Y

Y by Adventure10, Mango247, khina

msecco wrote:

This problem is due to Carlos Yuzo Shine, from Brazil.

I do have to point out that, even though I wrote the problem, the first person to solve it was Thiago Costa Leite Santos, a former IMO contestant (IMO 2004).

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29 posts

#7 h Sep 11, 2012, 9:32 PM • 2 Y

Y by Adventure10, Mango247

cyshine wrote:

msecco wrote:

This problem is due to Carlos Yuzo Shine, from Brazil.

I do have to point out that, even though I wrote the problem, the first person to solve it was Thiago Costa Leite Santos, a former IMO contestant (IMO 2004).

Why didn't you solve it yourself?

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4378 posts

#8 h Sep 12, 2012, 6:08 AM • 1 Y

Y by Adventure10

@mathworld1: because it is a proposed problem!

Best regards

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2 posts

0091p

#9 h Feb 27, 2013, 4:17 PM • 1 Y

Y by Adventure10

this is the shortlist G3

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105 posts

#10 h May 25, 2013, 5:58 PM • 6 Y

Y by hongduc_cqt, mad, myh2910, Adventure10, Mango247, Math_legendno12

Let $G$ be the midpoint of $EF$ , $M$ the midpoint of $AB$ , $R,S,T$ the feet of the perpendiculars from $E$ , and $R',S',T'$ the corresponding feet from $F$ , and let $RG$ intersect $w_E$ again at $U$ .

Then $MG \perp EF$ , so $R,E,G,M$ are concyclic, as are $B,S,E,R$ , so we have $\angle RGE =\angle RME = 2 \angle RAE = 2 \angle BER =2 \angle BSR$ . Similarly, $\angle TGE = 2 \angle TSC$ , so $\angle RGT =2*( \angle BSR + \angle TSC) = 2*(180-\angle RST) =2 *\angle RUT$ , which means that $\triangle TGU$ is isosceles at vertex $G$ , or $GT=GU$ . Then the power of $G$ with respect to $w_E$ is thus equal to $RG * GT$ .

It is easily seen that $G$ lies on the perpendicular bisector of $RR'$ , so $GR=GR'$ , and similarly $GT=GT'$ . So it follows that $RG*GT =R'G*GT'$ and $G$ lies on the radical axis of $w_E,w_F$ .

Attachments:

france.pdf (493kb)

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1514 posts

ksun48

#11 h Jul 4, 2013, 4:55 PM • 5 Y

Y by Binomial-theorem, jam10307, Phie11, Adventure10, Mango247

Let the feet of the perpendiculars from $E$ to $AB, BC, CD, DA$ , be $X, Y, Z, W$ , respectively. Then $\begin{align*} \angle XYZ &= \angle XYE + \angle EYZ \\ &= \angle XBE + \angle ECZ \\ &= \frac \pi 2 - \angle EAB + \frac \pi 2 - \angle EDC \\ &= \pi - \angle EAX - \angle EDZ \\ &= \pi - \angle XWE - \angle XWZ \\ &= \pi - \angle XWZ \end{align*}$ so $WXYZ$ is cyclic, which means that $W$ , $X$ , $Y$ , and $Z$ are on circle $\omega_E$ .
Now, let $EW$ intersect $BC$ at $P$ , and $EY$ intersect $AD$ at $Q$ . We chase some angles to get that $\begin{align*} \angle EPY &= \frac \pi 2 - \angle YEP \\ &= \angle WEY - \frac \pi 2 \\ &= \angle WEA + \angle AEB + \angle BEY - \frac \pi 2 \\ &= \angle WEA + \angle BEY \\ &= \angle WXA + \angle BXY \\ &= \pi - \angle WXY \end{align*}$ Thus $P$ is on $\omega_E$ . Similarly $Q$ is on $\omega_E$ , so $\omega_E$ is the circumcircle of $W$ , $Y$ , $Q$ , and $P$ .
Similarly, if $U$ and $V$ are the feet of the perpendiculars from $F$ to $AD$ and $BC$ , and $UF \cap BC = R$ , $VF \cap AD = S$ , then $\omega_F$ is the circumcircle of $UVRS$ .
(now we can basically ignore points $A,B,C,D,X,Z$ )

Let $EQ \cap UF = M_1$ and $EP \cap VF = M_2$ . Since $\angle QUR = \frac \pi 2 = \angle QYR$ , $QURY$ is cyclic, so $M_1Q \cdot M_1Y = M_1U \cdot M_1 R$ . Thus $M_1$ lies on the radical axis of $\omega_E$ and $\omega_F$ . Similarly, $WPVS$ is cyclic, so $M_2P \cdot M_2W =M_2V \cdot M_2 S$ , so $M_2$ is also on the radical axis of $\omega_E$ and $\omega_F$ .

Since $QY \perp BC$ and $FU \perp BC$ , $QY \parallel FU$ , so $M_1E \parallel M_2F$ . Similarly, $M_1F \parallel M_2E$ , so $M_1EM_2F$ is a parallelogram. If $M$ is the midpoint of $EF$ , $M_1, M, M_2$ are collinear, so $M$ is on the radical axis of $\omega_E$ and $\omega_F$ .

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366 posts

#12 h May 5, 2015, 7:36 PM • 2 Y

Y by Adventure10, Mango247

First notice that the intersections of the perpendiculars from E to BC and AD and the lines BC and AD form a cyclic quadrilateral, the circumcircle is w_E. Same for F.

The four lines: the perpendiculars from E and F to BC and AD, form a parallelogram with vertices EFXY. M is the midpoint of XY. Notice X and Y both have the same power of a point to w_E and w_F (which is not hard) to finish.

This post has been edited 1 time. Last edited by JuanOrtiz, May 5, 2015, 7:37 PM

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2079 posts

pi37

#13 h May 21, 2015, 12:08 AM • 2 Y

Y by Adventure10, Mango247

Let $E_1,E_2,E_3,E_4$ be the projections of $E$ onto $AB,BC,CD,DA$ . We claim $E_1,E_2,E_3,E_4$ are concyclic. Indeed,
$\angle E_1E_2E_3+\angle E_3E_4E_1 = \angle EBE_1+\angle EAE_1+\angle ECE_3+\angle EDE_3 = \pi$
So this is true for $E$ and similarly $F$ . Now note
$\angle E_2E_3E_4=\angle E_2CE+\angle E_4DE=90-\angle(BC,AD)$
So if $E_4E$ meets $BC$ at $W$ , $E_2E$ meets $AD$ at $X$ , circle $(E_1E_2E_3E_4)$ passes through $W$ and $X$ . Similarly, let $F_4F$ meet $BC$ at $Y$ , $F_2F$ meet $AD$ at $Z$ . Then the other circle has diameter $YZ$ .

Let $WEE_4$ meet $F_2FZ$ at $G$ and $YFF_4$ meet $E_2EX$ at $H$ . Clearly, $EGFH$ is a parallelogram, so $GH$ passes through the midpoint of $EF$ . We claim that both $G,H$ lie on the radical axis of the two circles, so $GH$ is their radical axis (in the case that $G=H$ , both points are the midpoint of $EF$ , so regardless the claim still holds).

But note that $WE_4ZF_2$ is cyclic, so
$GW\cdot GE_4=GZ\cdot GF_2$
and $G$ has equal powers with respect to $\omega_E$ and $\omega_F$ , so $H$ does similarly, as desired.

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563 posts

AMN300

#14 h Feb 7, 2016, 9:48 PM • 1 Y

Y by Adventure10

Let the projections of $E$ onto $AB, BC, CD, DA$ respectively be $E_a, E_b, E_c, E_d$ . Similarly define $F_a, F_b, F_c, F_d$ . By cyclic quads we have
$\angle E_aE_bE_c+\angle E_aE_dE_c = \angle E_aBE+\angle E_aAE+\angle E_cCE+\angle E_cDE=90^{\circ}+90^{\circ}=180^{\circ}$ Thus $E_aE_bE_cE_d$ and likewise $F_aF_bF_cF_d$ are cyclic quadrilaterals.

Let $E_bE \cap AD \equiv P_d$ . Denote $P_b$ analogously. We will prove $E_aE_bE_cE_dP_d$ is cyclic. By definition $E_b - E - P_d$ collinear, hence
$90-\angle E_bCE+\angle CED + \angle P_dED = 180-\angle E_bCE+\angle P_dED = 180 \implies \angle P_dED=\angle E_bCE$
Then by lots of cyclic quads
$\angle E_bE_cE_d = \angle E_bE_cE + \angle E_dE_cE = \angle E_bCE + \angle E_dDE = \angle P_dED + \angle P_dDE = \angle E_dP_dE = \angle E_dP_dE_b$ so $E_aE_bE_cP_dE_d$ is cyclic.

Now let $E_bE \cap F_dF \equiv P$ . By projections we have $\angle CE_bP_d = \angle DF_dP_b = 90^{\circ}$ so $E_bP_bP_dF_d$ is cyclic. By PoP, this means $PE_b \cdot PP_d = PF_d \cdot PP_b$ .

But $P_d \in \omega_E$ from earlier and $P_b \in \omega_F$ can be proved exactly the same way. Set $N, O \equiv \omega_E \cap \omega_F$ . So
$PE_b \cdot PP_d = PF_d \cdot PP_b \implies p(P, \omega_E) = p(P, \omega_F) \implies P \in NO$
Similarly one can show $S \in NO$ where $S = F_bF \cap E_dE$ . By perpendicularity $SEPF$ is a parallelogram so the conclusion follows.

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1979 posts

#16 h Jun 5, 2018, 10:47 PM • 3 Y

Y by e_plus_pi, Adventure10, Mango247

WakeUp wrote:

Let $ABCD$ be a convex quadrilateral whose sides $AD$ and $BC$ are not parallel. Suppose that the circles with diameters $AB$ and $CD$ meet at points $E$ and $F$ inside the quadrilateral. Let $\omega_E$ be the circle through the feet of the perpendiculars from $E$ to the lines $AB,BC$ and $CD$ . Let $\omega_F$ be the circle through the feet of the perpendiculars from $F$ to the lines $CD,DA$ and $AB$ . Prove that the midpoint of the segment $EF$ lies on the line through the two intersections of $\omega_E$ and $\omega_F$ .

Proposed by Carlos Yuzo Shine, Brazil

Let $P,Q,R,S$ be projections of $E$ on lines $AD, BC, AB, CD$ respectively. Let $X,Y,Z,T$ be projections of $F$ on lines $AD, BC, AB, CD$ respectively. Let $O=AD \cap BC$ ; $K=EQ \cap AD$ ; $L=EP \cap BC$ ; $U=FY \cap AD$ ; $V=FX \cap BC$ ; $G=EP \cap FY$ and $H=EQ \cap FX$ . Then $EGFH$ is a parallelogram. Now we conclude by showing line $HG$ coincides with the radical axis of $\omega_E, \omega_F$ .

Claim. $P,R,Q,L,S,K$ lie on $\omega_E$ .

(Proof) Note that $\angle RPS+\angle RQS=\angle RAE+\angle SDE+\angle RBE+\angle SCE=360^{\circ}-\angle AEB-\angle CED=180^{\circ}$ hence $PRQS$ is cyclic. Now $\angle PKQ=90^{\circ}-\angle AOB=\angle PDE+\angle PCE=\angle PSQ$ hence $K \in \omega_E$ . Likewise $L \in \omega_E$ . $\blacksquare$

Analogously, $X,Z,Y,V,T,U$ lie on $\omega_F$ . Now $G$ is the orthocenter of $\triangle OUL$ hence $p(G, \omega_E)=GP \cdot GL=GY \cdot GU=p(G, \omega_F)$ so $G$ indeed lies on their radical axis. Thus, $GH$ coincides with radical axis of $\omega_E, \omega_F$ and the result follows.

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1614 posts

yayups

#21 h Dec 13, 2019, 10:18 PM • 2 Y

Y by Pathological, Adventure10

All angles in this solution are directed mod $\pi$ .

Claim: The points $E$ and $F$ have pedal circles with respect to $ABCD$ . Said differently, $\omega_E$ also passes through the foot from $E$ to $DA$ , and $\omega_F$ also passes through the foot from $F$ to $BC$ .

Proof: We see that $\angle AEC=\angle AEB+\angle BEC=\angle CED+\angle BEC=\angle BED.$ It is well known that this implies that $E$ has an isogonal conjugate in $ABCD$ .

Now, let $W$ , $X$ , $Y$ , $Z$ be the feet from $E$ to $CD$ , $AB$ , $AD$ , $BC$ , and let $W_1$ , $X_1$ , $Y_1$ , $Z_1$ be the corresponding feet from $E^*$ , the isogonal conjugate of $E$ . We then have that the following pedal circles exist: $(WXYW_1X_1Y_1),(WXZW_1X_1Z_1),(WYZW_1Y_1Z_1),(XYZX_1Y_1Z_1),$ which imply that $(WXYZW_1X_1Y_1Z_1)$ is cyclic, as desired. $\blacksquare$

The problem now reduces to the following if we let $V=AD\cap BC$ .

New Problem wrote:

Let $VAB$ be a triangle, and let $E$ and $F$ be arbitrary points on the circle $(AB)$ . Let $\omega_E$ and $\omega_F$ denote the pedal circles of $E$ and $F$ respectively. Show that the midpoint $S$ of $EF$ lies on the radical axis of $\omega_E$ and $\omega_F$ .

$[asy] unitsize(1.5inches); pair V=dir(115); pair A=dir(210); pair B=dir(-30); pair E=(A+B)/2+abs(A-B)/2 * dir(130); pair F=(A+B)/2+abs(A-B)/2 * dir(100); pair S=(E+F)/2; pair X=foot(E,A,B); pair XX=foot(F,A,B); pair Y=foot(E,V,A); pair YY=foot(F,V,A); pair Z=foot(E,V,B); pair ZZ=foot(F,V,B); pair L=2*foot(circumcenter(X,Y,Z),X,S)-X; pair LL=2*foot(circumcenter(XX,YY,ZZ),XX,S)-XX; draw(A--B--V--cycle); draw(circle((A+B)/2,abs(A-B)/2)); draw(E--F); draw(X--L); draw(XX--LL); draw(E--X); draw(E--Y); draw(F--XX); draw(F--YY); draw(circumcircle(X,Y,Z)); draw(circumcircle(XX,YY,ZZ)); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$V$",V,dir(V)); dot("$E$",E,dir(90)); dot("$F$",F,dir(-40)); dot("$S$",S,dir(-20)); dot("$X$",X,dir(X)); dot("$X'$",XX,dir(XX)); dot("$Y$",Y,dir(Y)); dot("$Z$",Z,dir(80)); dot("$Z'$",ZZ,dir(210)); dot("$Y'$",YY,dir(YY)); dot("$L$",L,dir(L)); dot("$L'$",LL,dir(90)); [/asy]$
Let $X$ , $Y$ , $Z$ denote the feet from $E$ to $AB$ , $VA$ , and $VC$ , and define $X'$ , $Y'$ , $Z'$ similarly for $F$ . Let $L$ be the second intersection of $XS$ with $\omega_E$ , and let $L'$ be the second intersection of $X'S$ with $\omega_F$ . It suffices to show that $SX\cdot SL=SX'\cdot SL'$ . But we already know $SX=SX'$ , so it suffices to show that $SL=SL'$ .

To do this, we use moving points. Animate $V$ on the line $\ell$ (which is line $AV$ ). The only objects defined that move are $Z$ , $Z'$ , $\omega_E$ , $\omega_F$ , $L$ , and $L'$ . We have the following key claim.

Claim: The map $V\mapsto L$ (and $V\mapsto L'$ ) is projective.

Proof: Note that the center of $\omega_E$ varies on the perpendicular bisector of $XY$ (which is fixed). Note that $Z$ varies projectively on $(BE)$ and $X$ is a fixed point on $(BE)$ . Thus, the perpendicular bisector of $XZ$ varies projetively on the pencil of lines through the midpoint of $BE$ . Thus, the center of $\omega_E$ , which is the intersection of projective perpendicular bisector of $XZ$ and fixed perpendicular bisector of $XY$ , must also vary projectively. Call it $O_E$ .

Thus, the foot from $O_E$ to fixed $XS$ is projective, so the reflection of $X$ in the previously mentioned point is also projective, so $L$ is projective, as desired. $\blacksquare$

Thus, it suffices to solve the problem for three values of $V$ on line $\ell$ (say by constructing $L'$ instead as the intersection of $X'S$ and the line through $L$ perpendicular to the angle bisector of $\angle XSX'$ , which is clearly projective). First, we take $V$ to be the foot from $B$ to $\ell$ . In this case we see that $\omega_E$ is the Simson line of $E$ with respect to $AVD$ , so $L$ is the infinity point of line $XS$ , and similarly $L'$ is the infinity point of $X'S$ , so the problem works.

We'll now solve the problem in the case that $BV$ passes through $F$ . $[asy] unitsize(1.5inches); pair V=dir(115); pair A=dir(210); pair B=dir(-30); pair E=(A+B)/2+abs(A-B)/2 * dir(130); pair F=foot(A,B,V); pair S=(E+F)/2; pair X=foot(E,A,B); pair XX=foot(F,A,B); pair Y=foot(E,V,A); pair YY=foot(F,V,A); pair Z=foot(E,V,B); pair ZZ=foot(F,V,B); pair L=2*foot(circumcenter(X,Y,Z),X,S)-X; pair LL=2*foot(circumcenter(XX,YY,ZZ),XX,S)-XX; pair P=extension(A,E,B,F); draw(A--B--V--cycle); draw(circle((A+B)/2,abs(A-B)/2)); draw(E--F); draw(X--L); draw(XX--LL); draw(E--X); draw(E--Y); draw(F--XX); draw(F--YY); draw(circumcircle(X,Y,Z)); draw(circumcircle(XX,YY,ZZ)); draw(A--P); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$V$",V,dir(V)); dot("$E$",E,dir(90)); dot("$F$",F,dir(70)); dot("$S$",S,dir(-20)); dot("$X$",X,dir(X)); dot("$X'$",XX,dir(XX)); dot("$Y$",Y,dir(Y)); //dot("$Z$",Z,dir(80)); //dot("$Z'$",ZZ,dir(210)); dot("$Y'$",YY,dir(YY)); dot("$L$",L,dir(80)); dot("$L'$",LL,dir(100)); dot("$P$",P,dir(70)); [/asy]$
Let $P=AE\cap BV$ . The key claims in this case are the following.

Claim: In this case, we have $L=Z$ .

Proof: Clearly $Z$ lies on $\omega_E$ , so it suffices to show that $X,S,Z$ collinear. This is because $XZ$ is the simson line of $E$ with respect to $\triangle FAB$ , so it bisects $FE$ , since $F$ is the orthocenter of $\triangle FAB$ . $\blacksquare$

Claim: In this case, we have that $L'$ is the foot from $F$ to $AP$ .

Proof: We see that $\omega_F$ is $(AF)$ , so it suffices to show that $X',S,Q$ collinear where $Q$ is the foot from $F$ to $AP$ . This follows exactly as in the previous claim, since $X'Q$ is the simson line of $F$ with respect to $\triangle AEB$ . $\blacksquare$

Now, we have a triangle $PEF$ with feet of altitudes from $E$ and $F$ called $L$ and $L'$ , and we have $S$ , the midpoint of $EF$ . It is well known that $SL=SL'$ here, so we're done.

The case where $BV$ passes through $E$ is similar, so we have our three cases, thus solving the problem.

This post has been edited 2 times. Last edited by yayups, Dec 13, 2019, 10:49 PM

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#22 h Feb 26, 2020, 10:56 AM • 6 Y

Y by GeoMetrix, AlastorMoody, amar_04, BinomialMoriarty, Limerent, karitoshi

Solution(with GeoMetrix and amar_04)
Let $E_1,E_2,E_3,E_4$ be the perpendiculars from $E$ onto $AB,BC,CD,DA$ respectively. $F_1,F_2,F_3,F_4$ are defined similarly.
$O,P,Q$ are the midpoints of $EF,AB,CD$ respectively.

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.137383023245283, xmax = 8.381300997964356, ymin = -4.063105688248307, ymax = 6.34758768501158; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); /* draw figures */ draw((-5.732170768152634,5.979167929957042)--(5.191536445768684,4.798953366127438), linewidth(0.6) + rvwvcq); draw((5.191536445768684,4.798953366127438)--(4.140088404886246,-1.8935968623465549), linewidth(0.6) + rvwvcq); draw((4.140088404886246,-1.8935968623465549)--(-4.288185573615843,-3.3455965378508874), linewidth(0.6) + rvwvcq); draw((-4.288185573615843,-3.3455965378508874)--(-5.732170768152634,5.979167929957042), linewidth(0.6) + rvwvcq); draw(shift((-0.2703171611919748,5.38906064804224))*xscale(5.493639178908462)*yscale(5.493639178908462)*arc((0,0),1,173.83359279591065,353.83359279591065), linewidth(0.4) + linetype("4 4") + wrwrwr); draw(shift((-0.07404858436479866,-2.6195967000987213))*xscale(4.276216356615977)*yscale(4.276216356615977)*arc((0,0),1,9.774808122102176,189.7748081221022), linewidth(0.4) + linetype("4 4") + wrwrwr); draw((-2.916802125592067,0.57489522075972)--(-2.3003874165870393,-3.003144182085526), linewidth(0.4)); draw((-2.916802125592067,0.57489522075972)--(-2.3721375834721763,5.616144642912431), linewidth(0.4)); draw((-2.916802125592067,0.57489522075972)--(-4.848959620871087,0.2756912219532954), linewidth(0.4)); draw((2.608811142991501,0.7103117088398352)--(4.502441398643733,0.412808701093976), linewidth(0.4)); draw((2.608811142991501,0.7103117088398352)--(3.0886125119121783,-2.0747422142054552), linewidth(0.4)); draw(circle((-0.11554187045567219,1.3249863351272293), 4.848325879566928), linewidth(1) + dtsfsf); draw(circle((-0.2531693316995378,1.4701575054345088), 4.87173684763416), linewidth(1) + sexdts); draw((-2.916802125592067,0.57489522075972)--(2.608811142991501,0.7103117088398352), linewidth(0.4)); draw((-2.916802125592067,0.57489522075972)--(4.348577310512546,-0.5665484312966041), linewidth(0.4)); draw((-0.2703171611919748,5.38906064804224)--(-0.07404858436479866,-2.6195967000987213), linewidth(0.4)); draw((-4.740037111644647,-0.42769323629595274)--(2.608811142991501,0.7103117088398352), linewidth(0.4)); draw((-3.295036384637054,-2.3352156359078293)--(3.7089615197203,4.304824863136913), linewidth(0.4) + linetype("4 4")); draw((-4.327920873244868,4.140404645170897)--(3.0886125119121783,-2.0747422142054552), linewidth(0.4)); draw((2.608866910716772,5.335455213893016)--(-2.3003874165870393,-3.003144182085526), linewidth(0.4)); draw((-2.3721375834721763,5.616144642912431)--(-0.153995491300283,0.6426034647997776), linewidth(0.4)); draw((-2.3721375834721763,5.616144642912431)--(2.608866910716772,5.335455213893016), linewidth(0.4)); /* dots and labels */ dot((-5.732170768152634,5.979167929957042),linewidth(3pt) + dotstyle); label("$A$", (-5.6767249502030515,6.060396143680272), NE * labelscalefactor); dot((5.191536445768684,4.798953366127438),linewidth(3pt) + dotstyle); label("$B$", (5.250913197453122,4.882910824221912), NE * labelscalefactor); dot((4.140088404886246,-1.8935968623465549),linewidth(3pt) + dotstyle); label("$C$", (4.274461956926683,-2.182001092528246), NE * labelscalefactor); dot((-4.288185573615843,-3.3455965378508874),linewidth(3pt) + dotstyle); label("$D$", (-4.55667793901096,-3.53180133678539), NE * labelscalefactor); dot((-2.916802125592067,0.57489522075972),linewidth(3pt) + dotstyle); label("$E$", (-3.1063606552878675,0.7042738978513918), NE * labelscalefactor); dot((2.608811142991501,0.7103117088398352),linewidth(3pt) + dotstyle); label("$F$", (2.6661893254713727,0.790431360250784), NE * labelscalefactor); dot((-2.3721375834721763,5.616144642912431),linewidth(3pt) + dotstyle); label("$E_1$", (-2.129909414761429,5.241900250886046), NE * labelscalefactor); dot((-4.848959620871087,0.2756912219532954),linewidth(3pt) + dotstyle); label("$E_4$", (-4.815150326209135,0.5463185501191729), NE * labelscalefactor); dot((-2.3003874165870393,-3.003144182085526),linewidth(3pt) + dotstyle); label("$E_3$", (-2.545303070759908,-3.3594864119866057), NE * labelscalefactor); dot((4.502441398643733,0.412808701093976),linewidth(3pt) + dotstyle); label("$F_2$", (4.2170236486604225,0.6468355895851303), NE * labelscalefactor); dot((3.0886125119121783,-2.0747422142054552),linewidth(3pt) + dotstyle); label("$F_3$", (2.5943914401385464,-1.9953265906628963), NE * labelscalefactor); dot((3.0752577236838756,5.027599458522165),linewidth(3pt) + dotstyle); label("$F_1$", (2.781065942003895,4.59675336182252), NE * labelscalefactor); label("$\omega_{E}$", (-1.5555263320988177,5.644485448215834), NE * labelscalefactor,dtsfsf); label("$\omega_{F}$", (-3.608945852617652,5.371653483951092), NE * labelscalefactor,sexdts); dot((-0.153995491300283,0.6426034647997776),linewidth(3pt) + dotstyle); label("$O$", (-0.5359963603726834,0.6611951666516958), NE * labelscalefactor); dot((-0.2703171611919748,5.38906064804224),linewidth(3pt) + dotstyle); label("$P$", (-0.20572608784168206,5.098304480220393), NE * labelscalefactor); dot((-0.07404858436479866,-2.6195967000987213),linewidth(3pt) + dotstyle); label("$Q$", (-0.019051585976333496,-2.5266309421258146), NE * labelscalefactor); dot((4.348577310512546,-0.5665484312966041),linewidth(3pt) + dotstyle); label("$E_2$", (4.5042151899917275,-0.7029646546720137), NE * labelscalefactor); dot((-4.740037111644647,-0.42769323629595274),linewidth(3pt) + dotstyle); label("$F_4$", (-5.416701444607006,-0.5019305757400987), NE * labelscalefactor); dot((-3.295036384637054,-2.3352156359078293),linewidth(3pt) + dotstyle); label("$G$", (-3.608945852617652,-2.54099051919238), NE * labelscalefactor); dot((3.7089615197203,4.304824863136913),linewidth(3pt) + dotstyle); label("$H$", (3.771876759596899,4.39468520395869), NE * labelscalefactor); dot((2.608866910716772,5.335455213893016),linewidth(3pt) + dotstyle); label("$K$", (2.6661893254713727,5.428574752751397), NE * labelscalefactor); dot((-4.327920873244868,4.140404645170897),linewidth(3pt) + dotstyle); label("$L$", (-4.628475824343787,4.17929154796021), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Claim 1: $OE_i=OF_i$ for $1 \leq i \leq 4$
Proof:
Let the projection from $O$ to $E_iF_i$ be $M_i$ . Clearly, $M_i$ is the midpoint of $E_iF_i$ since $O$ is the midpoint of $EF$ . Therefore, $O$ lies on the perpendicular bisector of $E_iF_i$ .

Claim 2:The quadrilaterals $OEE_3Q$ and $OEE_1P$ are cyclic.
Proof:
$EF$ is the common chord of circles $\odot(AB)$ and $\odot(CD)$ with centres $P,Q$ . So, $PQ$ is the perpendicular bisector of $EF$ and passes through $O$ . So, we have $\angle EOP=\angle EE_1P=90^{\circ}$ and $\angle EOQ=\angle EE_3Q=90^{\circ}$

Let $OE_3 \cap \omega_E=K$ and $OF_3 \cap \omega_F=L$

Main proof:
We claim that $OE_1=OK$ and $OF_1=OL$ .

$\begin{align} \angle OKE_1=\angle E_3KE_1=\angle E_3E_2E_1 \\ \angle OE_1K=\angle E_1OE_3-\angle OKE_1=\angle E_1OE_3-\angle E_3E_2E_1 \end{align}$ Now we compute $\angle E_1OE_3$
$\angle E_1OE_3=\angle E_1OE+\angle EOE_3=\angle E_1PE+\angle EQE_3=2\angle E_1BE+2\angle ECE_3=2(\angle E_1E_2E+\angle EE_2E_3)=2 \angle E_3E_2E_1$ Substituting this in $(2)$ we get $\angle OE_1K=\angle E_3E_2E_1=\angle OKE_1$
So, $OK=OE_1$ and similarly $OL=OF_1$ . From Claim 1, $OF_1=OE_1$ and $OF_3=OE_3$ . This gives us $OK=OE_1=OF_1=OL$ .

Finally, we show that the power of point $O$ wrt circles $\omega_E$ and $\omega_F$ is equal.
$\text{Pow}_{\omega_E}(O)=OE_3 \cdot OK=OF_3 \cdot OL=\text{Pow}_{\omega_F}(O)$

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691 posts

#23 h Jun 1, 2020, 6:22 PM • 1 Y

Y by Han1728

Oh my god. What an annoying geo problem. The observation was terrific, at least for me.

Denote $E_1, E_2, E_3, E_4$ as the projection of $E$ wrt $AB, BC, CD, DA$ . Do it similarly to $F$ .
$\textbf{Claim.}$ $E_1 E_2 E_3 E_4$ is cyclic.
$\textit{Proof.}$ Basically
$\angle E_3 E_2 E_1 = \angle E_3 E_2 E + \angle E E_2 E_1 = \angle E_3 CE + \angle EBE_1 = \angle E_3 E D + \angle E_1 EA = \angle E_3 E_4 D + \angle A E_4 E_1 = 180^{\circ} - \angle E_3 E_4 E_1$
Similarly, $F_1 F_2 F_3 F_4$ is cyclic. Now, let $EE_4 \cap BC = G$ and $EE_2 \cap AD = H$ . Then, $(E_1 E_2 E_3 E_4 GH)$ is cyclic. Furthermore, $GH$ is the diameter. Denote $I,J$ similarly as $G,H$ , then the other circle has $IJ$ as its diameter. Let $IF_2 \cap GE_4 = K$ and $HE_2 \cap F_4 J = L$ .
$\textbf{Claim 02.}$ $K,L$ lies on the radical axis of $\omega_E$ and $\omega_F$ .
$\textit{Proof.}$
To prove this, notice that $\angle IE_4 G = \angle IF_2 G$ . So, $KF_2 \cdot KI = KE_4 \cdot KG$ , which means $K$ lie on radical axis of $(\omega_E, \omega_F)$ . Similarly, $L$ satisfies that too.

To finish this, note that $EKLF$ is a parallelogram. Therefore, $KL$ passes the midpoint of $EF$ , done.

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2 posts

#24 h Oct 4, 2020, 8:00 PM

Y by

Please can you explain why GR= GR’

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2 posts

#25 h Oct 4, 2020, 8:01 PM

Y by

polya78 wrote:

Let $G$ be the midpoint of $EF$ , $M$ the midpoint of $AB$ , $R,S,T$ the feet of the perpendiculars from $E$ , and $R',S',T'$ the corresponding feet from $F$ , and let $RG$ intersect $w_E$ again at $U$ .

Then $MG \perp EF$ , so $R,E,G,M$ are concyclic, as are $B,S,E,R$ , so we have $\angle RGE =\angle RME = 2 \angle RAE = 2 \angle BER =2 \angle BSR$ . Similarly, $\angle TGE = 2 \angle TSC$ , so $\angle RGT =2*( \angle BSR + \angle TSC) = 2*(180-\angle RST) =2 *\angle RUT$ , which means that $\triangle TGU$ is isosceles at vertex $G$ , or $GT=GU$ . Then the power of $G$ with respect to $w_E$ is thus equal to $RG * GT$ .

It is easily seen that $G$ lies on the perpendicular bisector of $RR'$ , so $GR=GR'$ , and similarly $GT=GT'$ . So it follows that $RG*GT =R'G*GT'$ and $G$ lies on the radical axis of $w_E,w_F$ .

Please can you explain why is GR=GR’

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597 posts

#26 h Dec 30, 2020, 6:35 AM • 4 Y

Y by Aryan-23, LinuxAvocado, maikewl_, CT17

fantastic problem!

First, we define a crud ton of points. All perpendiculars will be to $AB$ , $BC$ , $CD$ , $DA$ , respectively: from $E$ we have $P$ , $Q$ , $R$ , $S$ , and from $F$ we have $W$ , $X$ , $Y$ , $Z$ . The desired midpoint is $M$ , and $w_e \cap w_f = N_1, N_2$ . Additionally, let $X'$ be the second intersect of $\overline{AD}$ with $w_f$ , and $Z'$ the second intersect of $\overline{BC}$ with $w_f$ .

Observe that $PQRS$ and $WXYZ$ cyclic is well known, in fact after taking a homothety of scale factor $2$ it becomes a USAMO 1996 problem.

$\textbf{Claim:}$ We have $Z'$ , $F$ , $Z$ collinear, as well as $X$ , $F$ , $X'$ collinear.

We will use a lot of "obvious" cyclic quadrilaterals freely in the angle chase without mentioning what quadrilateral it is, because typing is sad. We have $\begin{align*} \measuredangle WXF = \measuredangle WBF = \measuredangle ABF = 90^\circ - \measuredangle FAB = \measuredangle WFA. \end{align*}$ However, we also have $\begin{align*} \measuredangle WFA = \measuredangle WZA = \measuredangle WZX' = \measuredangle WXX', \end{align*}$ proving the collinearity for $X$ , $F$ , and $X'$ . By symmetry the $Z$ 's follow. Additionally, for the other circle, we get the same thing for $S$ and $Q$ , so define $S'$ and $Q'$ analogously. $\square$

Even more points. I'm running out of alphabet letters here. Let $K = \overline{ZZ'} \cap \overline{QQ'}$ and $L = \overline{XX'} \cap \overline{SS'}$ .

$\textbf{Claim:}$ $KFLE$ is a parallelogram.

We have $\overline{EQ} \perp \overline{BC}$ by construction, similarly $\overline{FX} \perp \overline{BC}$ by construction. Thus $\overline{EK} \parallel \overline{LF}$ , and a similar result holds for the others. $\square$

Now we know that $\overline{KL}$ passes through $M$ . It suffices for $K$ and $L$ to lie on the common radical axis; however, note that from the perpendicularities we get $QQ'ZZ'$ cyclic, at which point $KZ \cdot KZ' = KQ \cdot KQ'$ , thus $K$ lies on the radical axis. By an identical argument $L$ lies on the radical axis, thus done.

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656 posts

#27 h May 16, 2021, 5:04 PM • 1 Y

Y by HoripodoKrishno

Can someone please tell the motivation behind any of the solutions, like, I cannot get how could anyone think of such things.

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336 posts

#28 h May 17, 2021, 11:13 AM

Y by

WLOG assume $E$ is closer to $AB$ than $F$ . Let the feet of perpendiculars from $E$ to $AB,BC,CD$ be $I,J,K$ and from $F$ to $CD,DA,AB$ be $M,N,L$ . Let $Q,W$ be $EK \cap FL, FM \cap EI$ respectively. Let $R$ be the midpoint of $EF$ . Let $P,O$ be $\omega_E \cap \omega_F$ .

Claim 1: $EK$ intersects $AB$ on $\omega_E$ and similar for other perpendiculars.
Proof:It is equivalent to proving $SIJK$ is cyclic. Note that $\angle BEJ=\angle ECJ$ because $\angle BEC =90^{\circ}$ Now observe that $\begin{align*} \angle BIJ=\angle BEJ &=\angle ECJ \\&= \angle EKJ \end{align*}$ which implies our result. $\square$

Claim 2: $R$ lies on $WQ$
Proof: Note that $EQ \parallel WF$ and $WE \parallel FQ$ . Hence $WEQF$ is a parallelogram, so $WQ$ bisects $EF$ . $\square$

Claim 3: $W,Q$ lie on $OP$ .
Proof:Consider the points as mentioned in the diagram. Note that $STKL$ is cyclic because $\angle SKT= 90^{\circ} =\angle TLS$ . So $\begin{align*} Pow_{\omega_E}(Q) &=-QS\cdot QK \\ &=-QL \cdot QT \\ &=Pow_{\omega_F}(Q) \end{align*}$ Hence $Q$ lies on the radical axis of $\omega_E,\omega_F$ i.e. $OP$ . By a similar argument $W$ lies on $OP$ . $\square$

Now, by claim 3, $WQ$ is the same as $OP$ , combining this with claim 2, we are done. $\blacksquare$
Diagram(the diagram got messed up on addition of $R$ , so I did not show it here):

Attachments:

This post has been edited 3 times. Last edited by Wizard0001, May 17, 2021, 11:16 AM

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465 posts

#29 h Nov 1, 2022, 9:49 AM • 1 Y

Y by HoripodoKrishno

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -13, xmax = 8, ymin = -6, ymax = 10; /* image dimensions */ /* draw figures */ draw((-7.41,4.51)--(-9,-4), linewidth(0.4)); draw((-9,-4)--(2.23,-4.09), linewidth(0.4)); draw((2.23,-4.09)--(-0.71,8.73), linewidth(0.4)); draw((-0.71,8.73)--(-7.41,4.51), linewidth(0.4)); draw(circle((-8.205,0.255), 4.328631423440901), linewidth(0.4) + linetype("4 4")); draw(circle((0.76,2.32), 6.576397189951349), linewidth(0.4) + linetype("4 4")); draw((-7.487217852530451,4.096714512557141)--(-5.657940781114573,3.7549341901656894), linewidth(0.4) + blue); draw((-5.657940781114573,3.7549341901656894)--(0.1267025219638667,5.081521655926268), linewidth(0.4) + blue); draw(circle((-3.1873616620351193,0.7783507203910318), 5.431418002336044), linewidth(0.4)); 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$P_1, P_2, P_3, P_4$ be the foot of perpendiculars from $E$ onto $AB, BC, CD, DA$ , $P_5 = EP_2 \cap AD$ , $P_6 = EP_4 \cap BC$ .

$Q_1, Q_2, Q_3, Q_4$ be the foot of perpendiculars from $F$ onto $AB, BC, CD, DA$ , $Q_5 = FQ_2 \cap AD$ , $Q_6 = FQ_4 \cap BC$ .

$X = Q_2Q_5 \cap P_4P_6$ , $Y = Q_4Q_6 \cap P_2P_5$ .

Firstly,
$\begin{align*} \measuredangle P_3P_4P_1 &= \measuredangle P_3P_4E + \measuredangle EP_4P_1\\ &= \measuredangle P_3DE + \measuredangle EAP_1\\ &= \measuredangle P_3EC + \measuredangle BEP_1\\ &= \measuredangle P_3P_2C + \measuredangle BP_2P_1\\ &= \measuredangle P_3P_2P_1\\ &\implies P_1P_2P_3P_4 \text{ is cyclic.} \end{align*}$
Now,
$\begin{align*} \measuredangle P_4P_3P_2 &= \measuredangle P_4P_3E + \measuredangle EP_3P_2\\ &= \measuredangle P_4DE + \measuredangle ECP_2\\ &= 90^{\circ} + \measuredangle P_4ED + \measuredangle ECP_6\\ &= \measuredangle DEC + \measuredangle P_4ED + \measuredangle ECP_6\\ &= \measuredangle P_4EC + \measuredangle ECP_6\\ &= \measuredangle P_6EC + \measuredangle ECP_6\\ &= \measuredangle EP_6P_2\\ &= \measuredangle P_4P_6P_2\\ &\implies P_4P_3P_6P_2 \text{ is cyclic.} \end{align*}$
Thus we conclude that $P_1P_2P_3P_4P_5P_6$ is cyclic. Similarly $Q_1Q_2Q_3Q_4Q_5Q_6$ is cyclic.

Let $\omega_1 = \odot P_1P_2P_3P_4P_5P_6$ , $\omega_2 = \odot Q_1Q_2Q_3Q_4Q_5Q_6$ .

Finally, $Pow_{\omega_1}(X) = XP_4 \cdot XP_6 = Pow_{\odot P_4Q_2P_6Q_5}(X) = XQ_5 \cdot XQ_2 = Pow_{\omega_2}(X)$ $\implies X \text{ lies on the radical axis of } \omega_1 \text{ and } \omega_2.$
Similarly, $Y$ also lies on the radical axis of $\omega_1$ and $\omega_2$ .

To finish, notice that $EXFY$ is a parallelogram, and thus $XY$ bisects $EF$ and we are done. $\blacksquare$

This post has been edited 4 times. Last edited by kamatadu, Nov 1, 2022, 2:53 PM
Reason: asdf

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awesomeming327.

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awesomeming327.

#30 h Apr 13, 2023, 7:59 PM

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This was positively brutal for a G3 :skull:

Diagram

Let $E_1$ , $E_2$ , $E_3$ , and $E_4$ be the feet of the altitudes from $E$ to $DA$ , $AB$ , $BC$ , and $CD$ , respectively. Let $F_1$ , $F_2$ , $F_3$ , $F_4$ be the feet of the altitudes from $F$ to $BC$ , $AB$ , $DA$ , and $CD$ , respectively. Let $FF_1$ and $EE_3$ intersect $AD$ at $F_5$ and $E_6$ respectively. Let $FF_3$ and $EE_1$ intersect $AD$ at $F_6$ and $E_5$ respectively. Let $M$ and $N$ be the intersection points of $\omega_E$ and $\omega_F$ . Let $P$ be the intersection of $F_1F_5$ and $E_1E_5$ , and let $Q$ be the intersection of $E_3E_6$ and $F_3F_6$ .

Note that $FP\parallel FF_1\parallel EE_3\parallel EQ$ and similarly $EP\parallel FQ$ so $FPEQ$ is a parallelogram. Thus, $PQ$ bisects $EF$ . It suffices to show that $P$ and $Q$ are both on $MN$ .

Note that $A,F_2,F,F_3$ , $B,F_1,F,F_2$ , $C,F_1,F,F_4$ , and $D,F_4,F,F_3$ are all concyclic by perpendiculars. Thus,
$\begin{align*} \angle F_1F_2F_3+\angle F_4F_3F_1 &= \angle FF_2F_3 + \angle FF_2F_1 + \angle FF_4F_3+\angle FF_4F_1 \\ &= \angle FAD+\angle FBC + \angle FDA+\angle FCB \\ &= 180^\circ - \angle AFD + 180^\circ - \angle BFC \\ &= \angle AFB+\angle CFD \\ &= 90^\circ+90^\circ=180^\circ \end{align*}$ So $F_1F_2F_3F_4$ and similarly $E_1E_2E_3E_4$ are cyclic. Now, note that
$\begin{align*} \angle FF_5F_3 &= 90^\circ - \angle F_5FF_3 \\ &= \angle F_3FF_1 - 90^\circ \\ &= \angle F_3FF_2 + \angle F_2FF_1 - 90^\circ \\ &= 180^\circ - \angle DAB + 180^\circ - \angle ABC - 90^\circ \\ &= 270^\circ - \angle FAD - \angle FBC - (\angle AFB + \angle FBA) \\ &= 180^\circ - \angle FF_2F_3 - \angle FF_2F_1 \\ &= 180^\circ - \angle F_3F_2F_1 \end{align*}$ so $F_5$ lies on $\omega_F$ . Similarly, $F_6$ also lies on $\omega_F$ , and $E_5$ and $E_6$ lie on $\omega_E$ . Now, since $E_1E_3E_5E_6$ is cyclic, and $F_1F_5\parallel E_3E_6$ , we have $E_1F_1E_5F_5$ is cyclic. By radical center theorem on $(MNE_1E_5)$ , $(MNF_1F_5)$ , and $(E_1F_1E_5F_5)$ , $P$ lies on $MN$ . Similarly, $Q$ lies on $MN$ . We are done.

Motivation

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#31 h Aug 4, 2023, 8:45 PM

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A few remarks: the most difficult part of the problem is making sense of the diagram. It's physically impossible to draw the circumcircles too far apart, so it's difficult to distinguish main claims just by staring at what's going on. It helps immensely to draw two separate diagrams (one for $E$ and one for $F$ ).

We spend most of the solution defining points. Let $E_{BC}$ be the perpendicular from $E$ to $\overline{BC}$ , and define all $7$ other points similarly. Let $I$ and $K$ be the second intersections of $\omega_E$ with $\overline{BC}$ and $\overline{AD}$ , respectively, and define $J$ and $L$ similarly for $\omega_F$ . Let $\overline{PQ}$ be the endpoints of the radical axis, $O$ to be the midpoint of $\overline{EF}$ , $X = \overline{E_{BC}E} \cap \overline{F_{DA}F}$ , and $Y = \overline{E_{DA}E} \cap \overline{F_{BC}F}$ .

Note first that $\omega_E$ contains all four feet from $E$ ; this is by USAMO 1993/2. A similar result holds for $\omega_F$ .

Claim. [Main Claim] $I$ lies on $\overline{EE_{DA}}$ , and the three other cyclic permutations follow similarly.

Proof. Angle chase. Redefine $I$ as $\overline{EE_{DA}} \cap \overline{BC}$ instead, and show it lies on the circle. This is fairly easy:
$\begin{align*} \angle E_{BC}IE &= \angle E_{BC}EE_{CD} + \angle E_{CD}EE_{DA} - 90^\circ \\ &= 270^\circ - \angle E_{BC}CE - \angle ECE_{CD} - \angle EDE_{CD} - \angle EDE_{DA} \\ &= 180^\circ - \angle E_{BC}E_{CD}E_{DA}. \ \blacksquare \end{align*}$
Note that this implies $\angle LE_{BC}K = 90^\circ$ and similar permutations.

Now I claim $\overline{XY} = \overline{PQ}$ . Suffices to show $Y$ lies on the radical axis, but this is true by radical center theorem as $(LE_{BC}KF_{DA})$ is cyclic by the previous observation. Then note $EYFX$ is a parallelogram, so the midpoint $O$ of $\overline{EF}$ lies on $\overline{XY}$ or the radical axis.

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#32 h Aug 31, 2023, 11:27 PM

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Sketch because it took so long.
We'll look at E first.
Define some feet of altitudes JKLM from E and NPQR from F onto AB,BC,CD,DA and STUV the extensions of EJ,EL,FN,FQ onto the opposite sides. By easy angle chasing similar to USAMO 1993/2 we get that JKLMST is cyclic; do the same for F. Again by a good diagram you see that $X=LT\cap NU,Y=JS\cap QV$ forms a parallelogram with the points E and F. It follows that since XNT\sim XLU and similarly for Y, we have pow_{w_e}X=XN*XU=XT*XL=pow_{w_f}X and similarly for Y XY is radax, so the midpoint of XY which is also midpoint of EF lies on this radax, as desired. $\blacksquare$

Remark. The extension is pretty well motivated when you draw the circle and it looks like they're collinear; nothing else (at least for my incompetency

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also even tho i posted an hour ago, this was a long in the making

This post has been edited 1 time. Last edited by huashiliao2020, Aug 31, 2023, 11:37 PM

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#33 h Sep 30, 2023, 4:45 AM

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Cute, at least until the write up.

Begin with a bunch of point definitions. Let the feet of the altitudes from $E$ to $AB$ , $BC$ , $CD$ and $DA$ be $W_e$ , $X_e$ , $Y_e$ and $Z_e$ . Also let $M$ be the midpoint of $EF$

Claim: $W_e$ , $X_e$ , $Y_e$ and $Z_e$ are concylic.
Proof: This is just a simple angle chase. Note that
$\begin{align*} \angle W_eZ_eY_e &= 180 - \angle W_eZ_eD - \angle Y_eZ_eB\\ &= 180 - \angle W_eED - \angle Y_eEC\\ &= 180 - \angle EAW_e - \angle EBY_e\\ &= 180 - \angle W_eX_eY_e \end{align*}$
We can find a similar conclusion for $F$ . Now extend $X_eE$ past $E$ to meet $AD$ at a point $P_e$ .

Claim: $P_e \in (W_eX_eY_eZ_e)$ .
Proof: Note that,
$\begin{align*} \angle EP_eZ_e &= 90 - \angle P_eEZ_e\\ &= \angle X_eEY_e + \angle Y_eEZ_e - 90\\ &= \angle X_eEC + \angle Z_eED\\ &= \angle 180 - X_eY_eZ_e \end{align*}$ as desired.

Now define $Q_e$ to be the intersection of $Z_eE$ and $BC$ and define $P_f$ and $Q_f$ , noting that similar results hold. Finally let $R$ be the intersection of $P_eE$ and $Q_fF$ and let $S$ be the intersection of $P_fF$ and $Q_eE$ . Then we can easily find that $ERSF$ is a parallelogram from which we get that $R$ , $M$ , $S$ collinear. Also noting that $R$ and $S$ lie on the radical axis $PQ$ we are done.

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#34 h Oct 7, 2023, 8:24 PM • 1 Y

Y by OronSH

Let the feet from $E$ to $AB$ , $BC$ , $CD$ , $DA$ be $E_1$ through $E_4$ in order, and the feet from $F$ to $CD$ , $DA$ , $AB$ , $BC$ be $F_i$ .

Claim: $F_4$ and $E_4$ lie on $\omega_E$ and $\omega_F$ .
Proof. This is just angle chasing. From the cyclic quads formed by feet we have $\measuredangle E_2E_1E_4 =\measuredangle E_2E_1E+\measuredangle EE_1E_4=\measuredangle CBE+\measuredangle EAD= \measuredangle ECB+\measuredangle ADE=\measuredangle E_2E_3E_4.$ Some steps use the fact that $\angle BEC+\angle AED=180^{\circ}$ which is true from $\angle BEA=\angle CED=90^{\circ}.$ $\blacksquare$

Claim: We have $O=E_4E\cap BC$ , $E_2E\cap DA$ lie on $\omega_E$ and $R=FF_4\cap DA$ and $FF_2\cap BC$ lie on $\omega_F$ .
Proof. We show $O$ lies on $\omega_E$ and the rest follow by symmetry. To do so, observe that $\begin{align*} & \measuredangle E_2E_1E_4=\measuredangle E_2E_1E+\measuredangle EE_1E_4=\measuredangle E_2BE+\measuredangle EAE_4 \\ & =90^{\circ} + \measuredangle E_2EB+\measuredangle BEA=\measuredangle =90+\measuredangle E_2EO \\ & =\measuredangle E_2OE=\measuredangle E_2OE_4 \end{align*}$ as desired. During this proof we make use of the many right angels, and some cycic quads formed by them yet again. $\blacksquare$

Now define $P=EE_4\cap FF_4$ and $Q=EE_2\cap FF_2$ . First, note that $EP\perp AD$ and $QF\perp AD$ so $QF\parallel EP$ and $QE\parallel PF$ so $EPFQ$ is a parallelogram. To finish, we present the final claim.

Claim: Line $PQ$ is the radical axis of $\omega_E$ and $\omega_F$ .
Proof. We prove that $P$ lies on the radical axis and it follows that $Q$ does as well by symmetry. To do so, note that by right angles $OF_4E_4R$ is cyclic and hence $OP\cdot PE_4=PR\cdot PF_4$ . But we have that $\operatorname{pow}_{\omega_E}(P)=PO\cdot PE_4 =PR\cdot PF_4=\operatorname{pow}_{\omega_{F}}(P)$ Implying the result. $\blacksquare$

Then, we have that the diagonals of $EPFQ$ bisect each other so $PQ$ , the radical axis of $\omega_E$ and $\omega_F$ bisects $EF$ as desired.

This post has been edited 1 time. Last edited by GrantStar, Oct 7, 2023, 8:25 PM

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#35 h Dec 31, 2023, 5:39 AM

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Define the feet to $AB$ , $BC$ , $CD$ , and $DA$ from $E$ be $E_{1,2,3,4}$ and $F$ be $F_{1,2,3,4}$ , respectively. These induce several cyclic quadrilaterals.

Note that $E_1E_2E_3E_4$ is cyclic, as
$\angle E_4E_1E_2 + \angle E_2E_3E_4 = 360 - \angle BEC + \angle DEC = 180,$
and analogously for $F$ . Letting $E_2E \cap AD = E_2'$ , and similarily for other even indexed points, we have that this point also lies on $\omega_C$ , as
$\angle E_2E_2'E_4 = \angle E_2EE_3 + \angle E_3EE_4 - 90 = 180 - \angle E_2CE - \angle E_4DE = 180 - \angle E_2E_3E_4,$
and analogously for the other altitudes from $E$ , $F$ to $BC$ and $DA$ . Consider points $K = EE_2 \cap FF_4$ and $L = EE_4 \cap FF_2$ :

Cyclic quadrilateral $E_2E_2'F_4F_4'$ tells us $K$ lies on the radical axis of $\omega_E$ and $\omega_F$ , and similarily for $L$ .
$EKFL$ is a parallelogram, so $LK$ bisects $EF$ , and hence the midpoint of $EF$ lies on the radical axis, as desired. $\blacksquare$

This post has been edited 1 time. Last edited by shendrew7, Dec 31, 2023, 5:41 AM

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Ywgh1

#36 h Aug 18, 2024, 6:02 PM

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2011 G3

Here is a Sketch of the solution.

Let $P_1,P_2,P_3,P_4$ denote the perpendiculars from $E$ to $AB,BC,CD,AD$ . Define $Q_1,Q_2,Q_3,Q_4$ similarly wrt to $F$ .

Let $M$ and $N$ denote midpoints of $AB$ and $CD$ ,let $R$ denote the midpoint of $EF$ , we show that $R$ lies on radicle axis of these two circles.

Let $RP_3 \cap \omega_E=K$ and $RQ_3 \cap \omega_F=L$

Claim: $RP_i=RQ_i$ for $i=1,2,3,4$ .

After that we can easily show that $RMEP_1$ and $RNFQ_1$ are cyclic.

Now for the main claim.

Claim: $RK=RP_1$ and $RL=RQ_1$ .

After proving the claim above, we would be done by POP.

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