See http://www.mathlinks.ro/Forum/viewtopic.php?t=4344 . darij

Let the reflection of $E$ across $AB,BC,CD,DA$ are $P,Q,R,S$ and let $EP,EQ,ER,ES$ intersect $AB,BC,CD,DA$ at $K,L,M,N$.clearly $AKEN$ is cyclic,and $EK=KP$,$EN=NS$,so $KN\parallel PS$,hence $\angle KAE=\angle KNE=\angle PSE$,similarly $\angle EDM=\angle RSE$,$\angle KBE=\angle PQE,\angle ECM=\angle RQE$,but $\angle EAK+\angle EBK+\angle ECM+\angle EDM=180$ because $\angle AEB=90$,hence $\angle PSR+\angle PQR=180$.

homothety(dilation)centred at E with ratio $-2$ brings the reflections back to the feet of perpendiculars simplifying the solution

Let's say that the reflections intersects the sides $BA$,$AD$,$DC$ and $CB$ at the points $A"$,$B"$,$C"$ and $D"$, respectively. And let's name the points of the reflections $A'$,$B'$,$C'$ and $D'$ across the sides $BA$,$AD$,$DC$ and $CB$ , respectively. Since the quadrilaterals $A'B'C'D'$ and $A"B"C"D"$ have parallel sides, they are similar. So, we must to show that the quadrilateral $A"B"C"D"$ is cyclic: from the cyclic quadrilaterals $AA"EB"$,$DB"EC"$,$C"ED"C$ and $D"BA"E$ we have $\angle B"AE=\angle B"A"E $, $\angle B"DE=\angle B"C"E$,$\angle D"C"E=\angle D"CE$ and $\angle D"BE=\angle D"A"E$, respectively. And since $\angle B"AE+\angle B"DE=90^\circ$ and $\angle D"BE +\angle D"CE=90^\circ$. From equal angles we get $\angle B"A"E+\angle D"A"E+\angle D"C"E+\angle B"C"E=180^\circ$, as desired.

aayush-srivastava wrote: homothety(dilation)centred at E with ratio $-2$ brings the reflections back to the feet of perpendiculars simplifying the solution Shouldn't the ratio be $\frac{1}{2}$? Can anyone explain reason of ratio being $-2$?

Let the reflections of $E$ over $AB,BC,CD,DA$ respectively be $P,S,R,Q$ respectively. Let $\omega_A,\omega_B,\omega_C,\omega_D$ be the circles centered at $A,B,C,D$ with radius $EA,EB,EC,ED$. We see that, $P=\omega_A\cap\omega_B, Q=\omega_A\cap\omega_D, R=\omega_C,\omega_D$ and $S=\omega_A\cap\omega_C$. Now By an inversion $\Psi$ centered at $E$ with an arbitary radius, (let the circle be $\omega$) we see that $\omega_A\mapsto$ Radical Axis of $\omega$ and $\omega_A$. Similarly for others. Let $\omega_A,\omega_B,\omega_C,\omega_D\mapsto \omega_A',\omega_B',\omega_C',\omega_D'$ respectively . Now as $P=\omega\cap\omega_A$. Hence, $P\mapsto\omega_A'\cap\omega_B'$. Similarly for others. If $P,Q,R,S\mapsto P',Q',R',S'$ then $P'Q'R'S'$ is a rectangle which is actually a cyclic quadrilateral. Now again inverting back we get that $PQRS$ is a cyclic quadrilateral. Hence, proved. $\blacksquare$.

Take a homothety centered at $E$ with scale factor $\frac{1}{2}$, which brings the reflection of $E$ across each side of $ABCD$ to the projection from $E$ onto that side. Because dilations preserve circles, it suffices to prove that these four projections are concyclic, so let the projection from $E$ onto $AD, AB, BC$, and $CD$ be $W, X, Y$, and $Z$, respectively. Because $\angle{EXB}=\angle{EYB}=90$, we have that $EXBY$ is cyclic and thus $\angle{EXY}=\angle{EBY}$, and by right angles, $\angle{EBY}=\angle{CEY}$. Because $\angle{EZC}=\angle{EYC}=90$, we have that $EYCZ$ is cyclic as well, so thus $\angle{CEY}=\angle{YZC}$, and therefore $\angle{EXY}=\angle{YZC}$. Now we do basically the same angle chase on the "other side": we know $\angle{AWE}=\angle{AXE}=90$, so $\angle{WXE}=\angle{EAW}$, and by right angles, $\angle{EAW}=\angle{WED}$. Because $\angle{DWE}=\angle{EZD}=90$, we have that $WEZD$ is cyclic as well, so thus $\angle{WED}=\angle{WZD}$, and therefore $\angle{WXE}=\angle{WZD}$. Hence, $\angle{WXY}=\angle{WXE}+\angle{EXY}=\angle{WZD}+\angle{YZC}=180-\angle{WZY}$, so $\angle{WXY}+\angle{WZY}=180$, so thus $WXYZ$ is cyclic, and we are done.

Redacted.....

let E1 E2 E3 E4 be the reflectons E. It is req to prove that E1 E2 E3 E4 are cocyclic define the intersections of E and E1 E2 E3 E4 with the sides of the quadrilateral as P1 P2 P3 P4 consider a circular inversion centered at E with arbitary radius. it is obvious that P1^ P2^ P3^ P4^ are cocyclic. So P1 P2 P3 P4 must also be concyclic. moreover we notice a homothety centered at E which maps P1 P2 P3 P4 to E1 E2 E3 E4 resp. So E1 E2 E3 E4 must be concyclic as well (as desired)

let $E,F,G,H$ be foot of altitudes from $E$ to sides. obviously $EFGH$ is cyclic. with scaling with center of $E$ and $k=2$ we are done.

Let $X,Y,Z,W$ denote the foot of the perpendicular from $AB,AD,DC,BC$ respectively. Let the reflection of $E$ over $AB,AD,DC,CB$ be $X’,Y’,Z’,W’$. $X’Y’Z’W’$ are $XYZW$ merely dilations of each other, so by homothety, it suffices to prove that $XWZY$ is cyclic. Noting that $AXEY, BWEX,WCZE, YEZD$ are all cyclic, $$\angle XWE + \angle XYE = \angle XBE + \angle XAE = 90$$$$\angle EYZ + \angle EWZ = \angle EDZ + \angle ECZ = 90$$Hence, $\angle XWE + \angle XYE + \angle EYZ + \angle EWZ = \angle XYZ + \angle XWZ = 180$, which implies that $XYZW$ is cyclic, as desired.

$\angle AEB + \angle CED = \pi \implies E$ has isogonal conjugate ($E'$) in $ABCD$ and it's well-known that $E'$ is a center of circle in question.

MithsApprentice wrote: Let $\, ABCD \,$ be a convex quadrilateral such that diagonals $\, AC \,$ and $\, BD \,$ intersect at right angles, and let $\, E \,$ be their intersection. Prove that the reflections of $\, E \,$ across $\, AB, \, BC, \, CD, \, DA \,$ are concyclic.

Let $E_1$ be the reflection of $E$ across $AB$, $E_2$ be the reflection of $E$ across $BC$, $E_3$ be the reflection of $E$ across $CD$, and $E_4$ be the reflection of $E$ across $DA$. Consider a homothety of ratio 1/2 centered at $E$. This sends the points $E_1, E_2, E_3, E_4$ to points on $AB, BC, CD, DA$ respectively. Call the points after the homothety $E_1', E_2', E_3', E_4'$, respectively. Since $E_1, E_2, E_3, E_4$ were originally reflections, $EE_1', EE_2', EE_3', EE_4'$ are perpendicular to $AB, BC, CD, DA$ respectively. Since circles are preserved in homotheties, if we show that $E_1', E_2', E_3', E_4'$ are concyclic, then $E_1, E_2, E_3, E_4$ are concyclic. Now we angle chase. Since $E_1'BE_2'E$ is cyclic ($\angle BE_2'E=\angle BE_1'E=90^{\circ}$) and $AE_1'EE_4'$ is cyclic, $\angle BE_2'E_1'=\angle E_1'EB=\angle BAE=\angle E_1'E_4'E.$ Let these angles have measure $m^{\circ}$ Since $\angle EE_3'D=\angle EE_4'E=\angle EE_2'C=90^{\circ}, $ $EE_3'DE_4'$ and $EE_3'CE_2'$ are cyclic. Hence, $\angle EE_4'E_3'=\angle EDE_3'=\angle CEE_3'=\angle CE_2'E_3'.$ Let these angles have measure $n^{\circ}.$ Since $\angle E_1'E_2'E_3'=180-m-n$ and $\angle E_1'E_4'E_3'=m+n$, $\angle E_1'E_2'E_3'$ and $\angle E_1'E_4'E_3'$ are supplementary, and $E_1', E_2', E_3', E_4'$ are concyclic. Therefore, $E_1, E_2, E_3, E_4$ are concyclic.

oops i needed hints for this(looking at the first problem(2003 IMO SL G4) which was similar to this of the first result of this and looking at the hints section of that for P3 too), but after some work, I figured it out

@below oops my sol is also inversion but way longer bc i used different circles

Solution from EGMO forum This problem gets destroyed pretty fast if you know inversions well

Let $W,X,Y,Z$ be the foot of the altitudes from $E$ to $AD, AB, BC, CD$, respectively. Let $W', X', Y', Z'$ be the reflections of $E$ cross $AD, AB, BC, CD$ respectively. We have $E, W, W'$ are collinear, and $EW = \dfrac{EW'}{2}$; the same applies to $EXX', EYY', EZZ'$. Claim: $WXYZ$ is a cyclic quadrilateral. Proof. This is simple angle chasing combined with similar triangles $\Delta AED \sim \Delta EWD$ and $\Delta BEC \sim \Delta EYC$. We have, \begin{align*} \angle WXY &= \angle WXE + \angle YXE \\ &= \angle WAE + \angle YBE \\ &= \angle WED + \angle YEC \\ &= \angle WZD + \angle YZC \\ &= 180^{\circ} - \angle WXY \end{align*}and the claim follows. $\blacksquare$ Then, a homothety centered at $E$ with factor $2$ sends $W,X,Y,Z$ to their reflections $W',X',Y',Z'$, and the proof is complete since the homothety preserves concyclic points.

Beautiful problem. Let the reflections about $AB, BC, CD, DA$ respectively be $W, X, Y, Z$. Invert about E, and let the inversion take object $P$ to $P'$. Then $W', X', Y', Z'$ are the midpoints of $A'B', B'C', C'D', D'A'$ (i. e. the circumcentres of the triangles $A'B'E, B'C'E, C'D'E, D'A'E$. But now taking a homothety of scale factor two about $E$ takes $W', X', Y', Z'$ to the vertices of a rectangle, which is cyclic. Therefore $W'X'Y'Z'$ is cyclic, and so is $WXYZ$.$\square$ (Shown below is the inverted diagram)

If we take a homothety with scale factor $\dfrac{1}{2}$ centered at $E$, the reflections of $E$ across the sides would map to the foot of the perpendicular from $E$ to each of the sides of $ABCD$. Let the points $P, Q, R, S$ be the feet of the perpendiculars from $E$ to sides $AB, BC, CD, DA$. If we prove that $PQRS$ is cyclic, then the reflections of $E$ across the sides of the quadrilateral would also be cyclic because homothety preserves circles. We have $\angle ESP = \angle EAP$ since $ESAP$ is cyclic, $\angle EAB = 90^{\circ} - \angle PBE = \angle PEB$, and $\angle PEB = \angle PQB$ since $EPBQ$ is cyclic. So, we have $\angle ESP = \angle EAP = \angle PEB = \angle PQB$. Similarly, we have $\angle RSQ = \angle EDC = \angle REC = \angle RQC$. We have $\angle RSP = \angle ESP + \angle RSQ$ and $\angle PQR = 180^{\circ} - (\angle PQB + \angle RQC) = 180^{\circ} - (\angle ESP + \angle RSQ)$, so $\angle RSP = 180^{\circ} - \angle PQR$, hence $PQRS$ is cyclic.$\blacksquare$

We define $W$, $X$, $Y$ and $Z$ to be reflections over $AB$, $BC$, $CD$ and $DA$ respectively and $P$, $Q$, $R$ and $S$ to be the intersection of $EW$, $EX$, $EY$ and $EZ$ with $AB$, $BC$, $CD$ and $DA$ respectively. Lemma Quadrilateral $PQRS$ is cyclic. We know that $\angle ASE$ = $\angle APE = 90^{\circ}$ and $\angle BQE$ = $\angle BPE = 90^{\circ}$. So, quadrilaterals $APES$ and $PBQE$ are cyclic and since $\Delta AEB$ is right angled at $E$ we get, \[\angle ESP = \angle EAP = \angle EAB = 90^{\circ} - \angle ABE = 90^{\circ} - \angle PBE = 90^{\circ} - \angle PQE\] Similarly $\angle RSE = 90^{\circ} - \angle EQR$. So we get, \begin{align*} \angle ESP + \angle RSE &= 180^{\circ} - \angle PQE - \angle EQR\\ \Rightarrow \angle PSR &= 180^{\circ} - \angle PQR \end{align*}Hence, quadrilateral $PQRS$ is cyclic. Now, we take a homothety $h$ at $E$ with a scale factor of 2. So, $h(P) = W, h(Q) = X, h(R) = Y \text{ and } h(S) = Z$. Now, from the above lemma, since $P$, $Q$, $R$ and $S$ are concyclic and since homothety preserves circles, we can say that $W$, $X$, $Y$ and $Z$ are concyclic as well.

Let the reflections of \(E\) across \(AB, BC, CD, DA\) be \(P, Q, R, S\) respectively, and let \(EP, EQ, ER, ES\) intersect \(AB, BC, CD, DA\) at \(K, L, M, N\) respectively. Using midlines, we find \(LK = \frac{PQ}{2}\), \(KN = \frac{PS}{2}\), \(NM = \frac{SR}{2}\), \(ML = \frac{RQ}{2}\). From similar triangles, we get \(\angle LKN = \angle QPS\), \(\angle KNM = \angle PSR\), \(\angle NML = \angle SRQ\), \(\angle MLK = \angle RQP\), proving \(KNML \sim PSRQ\) with ratio \(\frac{1}{2}\). To show \(KNML\) is cyclic, we need \(\angle LKN + \angle NML = 180^\circ\), which follows from \(\angle BKL + \angle AKN + \angle LMC + \angle NMD = 180^\circ\). Since quadrilaterals \(ANEK, KELB, LEMC, MEND\) are cyclic, we have \(\angle BEC + \angle AED = 180^\circ\) due to the perpendicular diagonals, thus we are done.

Let the reflections of $E$ across $AB$, $BC$, $CD$, and $DA$ be $W$, $X$, $Y$, and $Z$, respectively. Note $\angle AZD + \angle BXC = 180^{\circ}$. Then, angle chasing: \begin{align*} \measuredangle AZW + \measuredangle YZD = \dfrac{180 - \measuredangle WAZ}{2} + \dfrac{180 - \measuredangle ZDY}{2} = \dfrac{180 - 2 \cdot \measuredangle BAD}{2} + \dfrac{180 - 2 \cdot \measuredangle ADC}{2} = 180 - \measuredangle BAD - \measuredangle ADC \\ \measuredangle BXW + \measuredangle CXY = \dfrac{180 - \measuredangle WBX}{2} + \dfrac{180 - \measuredangle XCY}{2} = \dfrac{180 - 2 \cdot \measuredangle ABC}{2} + \dfrac{180 - 2 \cdot \measuredangle BCD}{2} = 180 - \measuredangle ABC - \measuredangle BCD \end{align*}Summing implies \begin{align*} \measuredangle AZW + \measuredangle YZD + \measuredangle BXW + \measuredangle CXY = \left(180 - \measuredangle BAD - \measuredangle ADC\right) + \left(180 - \measuredangle ABC - \measuredangle BCD\right) = 0. \end{align*}Therefore, \begin{align*} \measuredangle WZY + \measuredangle WXY = \measuredangle AZD + \measuredangle BXC - \left(\measuredangle AZW + \measuredangle YZD + \measuredangle BXW + \measuredangle CXY\right) = 180^{\circ}. \end{align*}Hence, $WXYZ$ is cyclic, as desired. OoPsOoPs.

Let $W, X, Y, Z$ be the reflections of $E$ across $\overline{AB}, \overline{BC}, \overline{CD}, \overline{DA}$, respectively. Let $F, G, H, I$ be the midpoints of $\overline{WE}$, $\overline{XE}$, $\overline{YE}$, $\overline{ZE}$, respectively (note that $F$ lies on $\overline{AB}$, and so on). Now, take a homothety centered at $E$ with scale factor $\frac 12$. This sends $WXYZ$ to $FGHI$. It remains to show that $FGHI$ is cyclic. Fortunately, this is quite easy to see by straight angle chasing. Note that $AFEI$, $BGEF$, $CHEG$, $DIEH$ are cyclic. Thus, we have: $$\angle DAE = \angle IAE =\angle EFI$$$$\angle CBE = \angle GBE = \angle EFG$$$$\angle BCE = \angle GCE = \angle EHG$$$$\angle ADE = \angle IDE = \angle EHI$$However, $\angle ADE + \angle DAE + \angle CBE + \angle BCE = 180^\circ$, so $$\angle IFG + \angle GHI = (\angle EFI + \angle EFG) + (\angle EHG + \angle EHI) = 180^\circ$$. Therefore, $FGHI$ is cyclic, so $WXYZ$ is also cyclic. $\blacksquare$

Let use call the reflection of $E$ across sides $AB, BC, CD, DE$ as $P, Q, R, S$ respectively. Now we take a homothety with scale factor $\frac{1}{2}$ and $E$ as center of homothety. Since $P, Q, R, S$ are reflections of E , they get reflected onto sides $AB, BC, CD, DA$ , let us call these points $P', Q', R', S'$. Now if we prove that the new quadrilateral formed $P'Q'R'S'$ is cyclic, this is equivalent to $PQRS$ being cyclic. Now, take a look at quadrilateral $AP'ES'$.since $EP \perp AB$. $$\angle AP'E = 90^{\circ} =\angle AS'E$$ Thus quadrilateral $AP'ES'$ is cyclic. Similarly the quadrilaterals $BP'EQ', CR'EQ'and DR'ES'$ are also cyclic. Now, using these cyclic quadrialterals we have $$\angle S'P'E + \angle S'R'E = \angle S'AE + \angle S'DE = 180^{\circ} - \angle AED = 180^{\circ} - 90^{\circ} = 90^{\circ} $$ similarly$$\angle Q'P'E + \angle Q'R'E = \angle Q'BE + \angle Q'CE = 180^{\circ} - \angle BEC = 180^{\circ} - 90^{\circ} = 90^{\circ}$$ thus $$\angle S'P'Q' + S'R'Q' = \angle S'P'E + \angle S'R'E + \angle Q'P'E + \angle Q'R'E = 90^{\circ} + 90^{\circ} = 180^{\circ}$$ Hence the quadrilateral $P'Q'R'S'$ is cyclic , therefore points $P, Q, R, S$ are also concyclic.

Funny generalization: $E$ is any point in any $ABCD$ such $\measuredangle AEB+\measuredangle CED=180^\circ$. Prove the result.

TestX01 wrote: Funny generalization: $E$ is any point in any $ABCD$ such $\measuredangle AEB+\measuredangle CED=180^\circ$. Prove the result. invert across the miquel point then reflect across angle bisector of AMC (clawson-schmidt)

Let $P,Q,R,S$ be the perpendiculars drawn form $E$ to sides $AB,BC,CD$ and $DA$ respectively. Let $E_{P},E_{Q},E_{R},E_{S}$ be the reflection points. Take a homothety $H$ with center $E$ and ratio 2. then, $$H(P) = E_{P}, H(Q) = E_{Q}, H(R) = E_{R}, H(S) = E_{S} $$Let $M_{P},M_{Q},M_{R},M_{S}$ denote the midpoints of sides $AB,BC,CD,DA$. Let $M$ be the midpoint of $EC$.Then from the $Nine\text{ }Point\text{ }Circle$, we have that $$QM_{Q}ME \text{ and } RM_{R}ME \text{ is cyclic}$$From $PoP, \text{ } QRM_{R}M_{Q} \text{is cyclic}$ Hence $PQRS$ is cyclic which implies that the reflections are cyclic as well

Let $P, Q, R, S$ be the feet of the altitudes from $E$ to $AB, BC, CD, DA$ respectively. Claim: $PQRS$ is cyclic. All angles are directed. Clearly, $$\measuredangle{SPQ} = \measuredangle{SPE} + \measuredangle{EPQ},$$so $$\measuredangle{SPQ} = \measuredangle{SAE} + \measuredangle{EBQ} = \measuredangle{DAC} + \measuredangle{DBC}$$(since points E, P, A, and S are concyclic, as are points P, B, Q, and E). Similarly, $\measuredangle{SRQ}=\measuredangle{ADB} + \measuredangle{ACB}.$ We want to prove that $$\measuredangle{DAC} + \measuredangle{DBC} = \measuredangle{ADB} + \measuredangle{ACB},$$or $$\measuredangle{DAC} + \measuredangle{DBC} - \measuredangle{ADB} - \measuredangle{ACB} = \measuredangle{DAC} + \measuredangle{DBC} + \measuredangle{BDA} + \measuredangle{BCA} = 0.$$Since $\measuredangle{DAC}+\measuredangle{BDA} = 90^\circ$ and $\measuredangle{DBC}+\measuredangle{BCA} = 90^\circ$, we are done as $90^\circ + 90^\circ = 180^\circ = 0$. Now, we finish the problem. Take a homothety at $E$ with scale factor $k = 2$. It will bring $PQRS$ to the quadrilateral in the problem. Since $PQRS$ is cyclic, we are done.

@above orz.

peace09 wrote: @above orz. nou jason admits