Lots of Cyclic Quads

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1354 posts

#1 h Apr 19, 2018, 11:00 PM • 19 Y

Y by anantmudgal09, W.R.O.N.G, Math-Ninja, tenplusten, Ultroid999OCPN, samuel, ramirahma, blacksheep2003, mathlogician, Muaaz.SY, icematrix2, OlympusHero, centslordm, megarnie, jhu08, HWenslawski, Adventure10, Mango247, Rounak_iitr

In convex cyclic quadrilateral $ABCD$ , we know that lines $AC$ and $BD$ intersect at $E$ , lines $AB$ and $CD$ intersect at $F$ , and lines $BC$ and $DA$ intersect at $G$ . Suppose that the circumcircle of $\triangle ABE$ intersects line $CB$ at $B$ and $P$ , and the circumcircle of $\triangle ADE$ intersects line $CD$ at $D$ and $Q$ , where $C,B,P,G$ and $C,Q,D,F$ are collinear in that order. Prove that if lines $FP$ and $GQ$ intersect at $M$ , then $\angle MAC = 90^\circ$ .

Proposed by Kada Williams

This post has been edited 2 times. Last edited by djmathman, Jun 22, 2020, 5:49 AM

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wu2481632

4233 posts

wu2481632

#2 h Apr 19, 2018, 11:00 PM • 32 Y

Y by Einstein314, alifenix-, whatshisbucket, Vfire, PythoN-, samuel, budu, InCtrl, ThisIsASentence, MathCube1717, Mudkipswims42, Math-Ninja, hwl0304, Ultroid999OCPN, luka2002, Wictro, Benq, rzlng, joey8189681, qweDota, sriraamster, enzoP14, mathlogician, Muaaz.SY, OlympusHero, centslordm, jhu08, megarnie, icematrix2, rg_ryse, Adventure10, Mango247

Solution. We invert about $C$ swapping $E$ and $A$ .

Under this inversion, we must have $A \mapsto E$ , $B \mapsto P$ , and $D\mapsto Q$ ; therefore as $ABCD$ is a cyclic quadrilateral, $P, E,$ and $Q$ are collinear.

Next, let $G'$ be the image of $G$ under this inversion. I claim that $EB = EG'$ . Note that $\angle{EBC} = \angle{EAC} = 180^{\circ} - \angle{EAG} = \angle{EG'G}$ , so the claim follows. Similarly if $F'$ is the image of $F$ , then $ED = EF'$ .

Then the image of line $GQ$ is the circumcircle of $\triangle{G'DC}$ , and similarly the image of line $FP$ is the circumcircle of $\triangle{F'BC}$ . Suppose these circles meet at $X$ , which is the image of $M$ under the inversion. Then we want to show that $\angle{CXE} = 90^{\circ}$ .

The good thing about this inversion is that it completely eliminates the necessity of points $A, P, Q, G$ and $F$ ; we may restate the problem as follows:

Inverted USAMO 5 wrote:

Suppose that $CDB$ is a triangle, and let $E$ be some point on side $BD$ . Let $U$ and $V$ be the feet of the perpendiculars from $E$ to $CD$ and $CB$ , respectively, and let $F'$ and $G'$ be the reflections of $D$ and $B$ over $U$ and $V$ , respectively. Let the circumcircles of $\triangle{G'DC}$ and $\triangle{F'BC}$ meet at $X \neq C$ . Show that $\angle{CXE} = 90^{\circ}$ .

Equivalently, it suffices to demonstrate that $CUXEV$ is a cyclic quadrilateral; we know that $CUEV$ is cyclic with diameter $CE$ , so we want to show that $\angle{UXV} = 180^{\circ} - \angle{C}$ . We claim that $X$ is the center of the spiral similarity mapping $DG'$ to $F'B$ . Observe that $\angle{XDG'} = \angle{XCB} = \angle{XF'B}$ ; similarly, $\angle{XG'D} = \angle{XCD} = \angle{XBF'}$ . Thus this claim is proven.

Therefore, by the Gliding Principle and the principle that spiral similarities come in pairs, we observe that $X$ maps $U$ , the midpoint of $DF'$ , to $V$ , the midpoint of $G'B$ . Therefore $\angle{UXV} = \angle{F'XB} = 180^{\circ} - \angle{C}$ , and we may conclude.

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Plasma_Vortex

268 posts

Plasma_Vortex

#3 h Apr 19, 2018, 11:01 PM • 9 Y

Y by GuamUp, ramirahma, Muaaz.SY, jhu08, megarnie, icematrix2, Adventure10, Mango247, Rounak_iitr

Since $\measuredangle APC = \measuredangle APB = \measuredangle AEB = \measuredangle AED = \measuredangle AQD = \measuredangle AQC,$ $APCQ$ is cyclic. Also, $\measuredangle ADQ = \measuredangle ADC = \measuredangle ABC = \measuredangle ABP,$ so $\triangle ADQ \sim \triangle ABP$ , and $A$ is the Miquel point of quadrilateral $BDQP$ . Note that $BDPQ$ is cyclic because $CD\cdot CQ = CA\cdot CE = CB \cdot CP.$ Also $P$ , $E$ , and $Q$ are collinear by the spiral similarity lemma.

Now let $X = BQ \cap DP$ , and by Pappus' theorem on hexagon $BFPDGQ$ , we get that $A$ , $M$ , and $X$ are collinear. From Brocard's theorem on cyclic quadrilateral $BDQP$ , $EC$ is the polar of $X$ w.r.t. $(BDQP)$ , and from well-known properties of the Miquel point configuration (namely that the Miquel point and the intersection of the diagonals are inverses w.r.t. the circumcircle), $XA \perp AC$ , so $MA \perp AC$ .

This post has been edited 10 times. Last edited by Plasma_Vortex, Apr 4, 2019, 1:15 PM

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v_Enhance

6866 posts

v_Enhance

#4 h Apr 19, 2018, 11:01 PM • 22 Y

Y by Anar24, InCtrl, ramirahma, sansae, Omeredip, Lemon293, v4913, HamstPan38825, math31415926535, microsoft_office_word, jhu08, icematrix2, Mop2018, Quidditch, sabkx, Adventure10, Rounak_iitr, Mango247, bhan2025, k12byda5h, Sedro, ehuseyinyigit

This is indeed a Miquel point problem, but the main idea is to focus on the self-intersecting cyclic quadrilateral $PBQD$ as the key player, rather than on the given $ABCD$ .

Indeed, we will prove that $A$ is its Miquel point; this follows from the following two claims.

Claim: The self-intersecting quadrilateral $PQDB$ is cyclic.

Proof. By power of a point from $C$ : $CQ \cdot CD = CA \cdot CE = CB \cdot CP$ . $\blacksquare$

Claim: Point $E$ lies on line $PQ$ .

Proof. $\measuredangle AEP = \measuredangle ABP = \measuredangle ABC = \measuredangle ADC = \measuredangle ADQ = \measuredangle AEQ$ . $\blacksquare$

$[asy]size(10cm); pair P = dir(240); pair B = dir(300); pair Q = dir(20); pair D = dir(50); pair E = extension(Q, P, B, D); pair C = extension(D, Q, B, P); pair H = extension(D, P, B, Q); pair A = foot(H, E, C); draw(unitcircle, lightblue); pair G = extension(D, A, B, C); pair F = extension(B, A, C, D); draw(circumcircle(P, A, E), heavygreen); draw(circumcircle(Q, A, E), heavygreen); draw(circumcircle(A, B, C), heavygreen+dotted); draw(B--F--C, grey); draw(D--G--C, grey); draw(P--H--B, blue); draw(P--B--Q--D--cycle, blue); draw(A--C, grey); draw(P--Q, blue); draw(B--D, blue); pair M = extension(P, F, G, Q); draw(F--P, heavycyan); draw(G--Q, heavycyan); dot("$P$", P, dir(P)); dot("$B$", B, dir(B)); dot("$Q$", Q, dir(Q)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$C$", C, dir(C)); dot("$H$", H, dir(H)); dot("$A$", A, dir(E-C)); dot("$G$", G, dir(G)); dot("$F$", F, dir(F)); dot("$M$", M, dir(M)); /* TSQ Source: P = dir 240 B = dir 300 Q = dir 20 D = dir 50 E = extension Q P B D C = extension D Q B P H = extension D P B Q A = foot H E C RE-C unitcircle 0.1 lightcyan / lightblue G = extension D A B C F = extension B A C D circumcircle P A E 0.1 green / heavygreen circumcircle Q A E 0.1 green / heavygreen circumcircle A B C 0.05 yellow / heavygreen dotted B--F--C grey D--G--C grey P--H--B blue P--B--Q--D--cycle blue A--C grey P--Q blue B--D blue M = extension P F G Q F--P heavycyan G--Q heavycyan */ [/asy]$

To finish, let $H = \overline{PD} \cap \overline{BQ}$ . By properties of the Miquel point, we have $A$ is the foot from $H$ to $\overline{CE}$ . But also, points $M$ , $A$ , $H$ are collinear by Pappus theorem on $\overline{BPG}$ and $\overline{DQF}$ , as desired.

This post has been edited 1 time. Last edited by v_Enhance, Apr 19, 2018, 11:03 PM
Reason: enlarge diagram

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shawnee03

113 posts

shawnee03

#5 h Apr 19, 2018, 11:02 PM • 5 Y

Y by GuamUp, jhu08, icematrix2, Adventure10, Mango247

Here's an incomplete solution I submitted; how many points do you think I would get?
First, note that cyclic quadrilaterals give us
$\angle AQC =180^{\circ}-\angle AQD = 180^{\circ} - \angle AED = \angle AEB$ ,
$\angle ACQ = \angle ABE$ ,
so $\Delta AQC \sim \Delta AEB$ . Similarly, $\Delta APC \sim \Delta AED$ .
Also,
$\angle BAC = \angle BDC = \angle EAD$ ,
so $\overline{AC}$ bisects $\angle BAD$ . In order to show that $\angle MAC = 90^{\circ}$ , it suffices to show that the pencil of lines $(\overline{AM},\overline{AC};\overline{AB},\overline{AQ})$ is harmonic.
Define $S=\overline{FT}\cap\overline{AC}$ and $T=\overline{FS}\cap\overline{BC}$ . Now,
$(\overline{AM},\overline{AC};\overline{AB},\overline{AQ})\stackrel{A}{=} (M, S; \overline{AB}\cap\overline{GQ}, Q) \stackrel{F}{=} (P, T; B, C)$ ,
so we only need to show $\frac{TB}{TC}=\frac{PB}{PC}$ . Menelaus's Theorem on $\Delta ABC$ with collinear points $F, T, S$ gives
$\frac{BT}{TC}\cdot\frac{CS}{SA}\cdot\frac{AF}{FB}=1$ ,
while Menelaus on $\Delta ACD$ with collinear points $Q, S, G$ gives
$\frac{AS}{SC}\cdot\frac{CQ}{QD}\cdot\frac{DG}{GA}=1$ ,
so
$\frac{TB}{TC}=\frac{FB}{AF}\cdot\frac{QD}{CQ}\cdot\frac{GA}{DG}$ ,
and it suffices to show
$\frac{PB}{PC}=\frac{FB}{AF}\cdot\frac{QD}{CQ}\cdot\frac{GA}{DG}$
From our similar triangles, we see
$CP=\frac{DE\cdot AC}{DA}$ and $CQ=\frac{BE\cdot CA}{BA}$ . Substituting these, as well as $PB=PC-BC$ and $QD=CD-QC$ , we need to show an equation in terms of just the sides of complete cyclic quadrilateral $ABCD$ , which should be easy to show with power of a point, Ptolemy's Theorem, etc.

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mathmaster2012

636 posts

mathmaster2012

#6 h Apr 19, 2018, 11:08 PM • 4 Y

Y by kingofgeedorah, icematrix2, Adventure10, Mango247

Outline

This post has been edited 5 times. Last edited by mathmaster2012, Oct 8, 2019, 12:08 AM

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whatshisbucket

975 posts

whatshisbucket

#7 h Apr 19, 2018, 11:10 PM • 43 Y

Y by 62861, wu2481632, leequack, mathwizard888, shawnee03, brainiac1, hwl0304, MathCube1717, InCtrl, Mudkipswims42, thecoolchicken, Ultroid999OCPN, TheUltimate123, xilias, Kagebaka, MathbugAOPS, coolak, sa2001, pad, yayups, e_plus_pi, joey8189681, Wizard_32, Aryan-23, OlympusHero, Toinfinity, tigerzhang, ike.chen, icematrix2, Mogmog8, rayfish, SHZhang, lrjr24, IAmTheHazard, r00tsOfUnity, mathlove_13520, Adventure10, Mango247, Scilyse, vrondoS, balllightning37, ehuseyinyigit, aidan0626

whatshisbucket wrote:

Me trying to do a geometry problem is like trying to eat soup with a toothpick.

Some things never change.

And I thought the only things certain in life were death and taxes.

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yrnsmurf

20654 posts

yrnsmurf

#8 h Apr 19, 2018, 11:12 PM • 3 Y

Y by icematrix2, Adventure10, Mango247

My solution:
After proving E lies on PQ, define F' as intersection of line QA and circle PBEA and define G' as intersection of line PA and circle QDEA.
Then define M' as the intersection of circles F'AD and G'AB.
Then do root bc inversion with B and Q and find out that it inverts to the same thing, so F' switches with F, G' switches with G, M' switches with M, E and C switch, B and Q switch, D and P switch

then angle chase with DAG' to get that MAM' is a straight line.

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AllenWang314

661 posts

AllenWang314

#9 h Apr 19, 2018, 11:17 PM • 3 Y

Y by icematrix2, Adventure10, Mango247

I only got the first two lemmas of Evan's proof and finished sketchily . Would that be 1,2,5, or 6 pts edit or 0pts

This post has been edited 1 time. Last edited by AllenWang314, Apr 19, 2018, 11:18 PM
Reason: 0

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lucasxia01

908 posts

lucasxia01

#10 h Apr 19, 2018, 11:18 PM • 3 Y

Y by icematrix2, Adventure10, Mango247

Did anyone successfully bash this one?

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hwl0304

1840 posts

hwl0304

#11 h Apr 19, 2018, 11:19 PM • 4 Y

Y by icematrix2, Adventure10, Mango247, ehuseyinyigit

lucasxia01 wrote:

Did anyone successfully bash this one?

spent 3 hours trying to?

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brainiac1

1909 posts

brainiac1

#12 h Apr 19, 2018, 11:20 PM • 4 Y

Y by icematrix2, Adventure10, Mango247, ehuseyinyigit

lucasxia01 wrote:

Did anyone successfully bash this one?

>4 pages of bary bash wasted

:cry:
Getting line CQ was untenable, at least for me.

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mathisawesome2169

1823 posts

mathisawesome2169

#13 h Apr 19, 2018, 11:21 PM • 4 Y

Y by AlastorMoody, icematrix2, Adventure10, Mango247

lucasxia01 wrote:

Did anyone successfully bash this one?

I considered complex bashing this one but I don't know how to complex bash

and I doubt that's efficient anyway. other bash methods are probably even worse

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cpma213

560 posts

cpma213

#14 h Apr 19, 2018, 11:22 PM • 3 Y

Y by icematrix2, Adventure10, Mango247

AllenWang314 wrote:

I only got the first two lemmas of Evan's proof and finished sketchily . Would that be 1,2,5, or 6 pts edit or 0pts

I would guess 0-1, unless "finished sketchily" meant a lot more nontrivial observations than the two lemmas. I personally found the two lemmas within the last 15 minutes

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wu2481632

4233 posts

wu2481632

#15 h Apr 19, 2018, 11:22 PM • 5 Y

Y by samuel, icematrix2, Adventure10, Mango247, ehuseyinyigit

My solution was motivated by the fact that $FP$ and $GQ$ are actually quite difficult to draw on paper and pencil without making the diagram really ugly. I figured inverting would probably clean some things up.

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trk08

614 posts

trk08

#97 h Jan 6, 2024, 3:51 AM • 1 Y

Y by radian_51

Claim:
$P,E,Q$ are collinear
Proof:
We can say
$\begin{align*} \angle PEQ&=\angle AEQ+\angle AEP\\ &=\angle ABP+180-\angle ADQ\\ &=180\\ \end{align*}$ $\square$

Claim:
$A$ is the Miquel Point of quadrilateral $PBDQ$
Proof:
Let $A'$ be the Miquel Point. Then:
$A'=(DQE)\cap (PBE)\cap (BCD)=A$ $\square$

Claim:
$PBDQ$ is cyclic
Proof:
We can say:
$\angle DQP=180-\angle DAE=180-\angle PBC=\angle PBD$ $\square$

Let $R=BP\cap PD$ . As $PBDQ$ is cyclic, $\angle RAE=90$ , by the special properties of Miquel Points in cyclic quadrilaterals. It remains to show that $R,A,M$ are collinear, which is true by Pappus' Theorem on $BRG$ and $DQF$ $\blacksquare$

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Shreyasharma

666 posts

Shreyasharma

#98 h Jan 6, 2024, 7:41 PM • 1 Y

Y by radian_51

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.898319437038159, xmax = 1.721309433790064, ymin = -1.2091941798378278, ymax = 5.085859335855192; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen qqwwzz = rgb(0,0.4,0.6); pen cczzqq = rgb(0.8,0.6,0); /* draw figures */ draw((-2.344645147638352,1.1130030809515516)--(-2.093371596036182,-0.40327742089290003), linewidth(0.7)); draw((-2.093371596036182,-0.40327742089290003)--(-0.20726904334706847,-0.3300674029964366), linewidth(0.7) + qqwuqq); draw(circle((-1.1849783579010602,0.5262192470965725), 1.299670085403655), linewidth(0.7) + dotted); draw((-0.20726904334706847,-0.3300674029964366)--(-1.6730242915905422,1.7307745581978542), linewidth(0.7) + qqwuqq); 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/* dots and labels */ dot((-2.344645147638352,1.1130030809515516),dotstyle); label("$A$", (-2.54087741336256,1.148247842321295), NE * labelscalefactor); dot((-2.093371596036182,-0.40327742089290003),dotstyle); label("$B$", (-2.1394289528091064,-0.6027507610939955), NE * labelscalefactor); dot((-0.20726904334706847,-0.3300674029964366),dotstyle); label("$C$", (-0.14072810664935878,-0.48317036866563423), NE * labelscalefactor); dot((-1.6730242915905422,1.7307745581978542),dotstyle); label("$D$", (-1.5586099034977268,1.7034425214529723), NE * labelscalefactor); dot((-1.8592579243618086,0.7852891265429016),linewidth(4pt) + dotstyle); label("$E$", (-2.0881802131639846,0.7126335556179787), NE * labelscalefactor); dot((-2.682141858241821,3.149587028851064),linewidth(4pt) + dotstyle); label("$F$", (-2.745872371943047,3.2579876230216693), NE * labelscalefactor); dot((-4.076793811892611,-0.48026495039497513),linewidth(4pt) + dotstyle); label("$G$", (-4.206461451829016,-0.6625409573081762), NE * labelscalefactor); dot((-2.9286091162282704,-0.43569758474553844),linewidth(4pt) + dotstyle); label("$P$", (-2.9679502437385743,-0.6539995007061504), NE * labelscalefactor); dot((-1.3853858810925042,1.326356899304287),linewidth(4pt) + dotstyle); label("$Q$", (-1.302366205272118,1.3019940611577594), NE * labelscalefactor); dot((-2.8762750934106447,0.3255896073486023),linewidth(4pt) + dotstyle); label("$M$", (-2.8712869380182497,0.1916047028944045), NE * labelscalefactor); dot((-0.1301321950380226,4.39298636671745),linewidth(4pt) + dotstyle); label("$T$", (-0.06385499718167618,4.402542807693127), NE * labelscalefactor); dot((-2.119250331019013,0.9608255552249374),linewidth(4pt) + dotstyle); label("$S$", (-2.0283900169113425,0.9774187102807788), NE * labelscalefactor); dot((-1.661509412919258,0.6517772754772615),linewidth(4pt) + dotstyle); label("$R$", (-1.541526990282686,0.6613848160058239), NE * labelscalefactor); dot((-3.40268969572076,-0.45409926055024313),linewidth(4pt) + dotstyle); label("$X$", (-3.454813270367231,-0.6625409573081762), NE * labelscalefactor); 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Claim: The points $Q$ , $E$ and $P$ are collinear.
Proof. Note that $\angle BPE = \angle BAE = \angle BAC = \angle BDC.$ Also we have $\angle PBE = 180 - \angle EBC = 180 - \angle DBC = 180 - \angle DAC = 180 - \angle DAE = DQE$ However then $\angle PEB = \angle DEQ$ , hence proven. $\blacksquare$

Then it follows that $A$ is the Miquel Point of complete quadrilateral $QEBC$ . As a result $\overline{AC}$ is the angle bisector of $\angle BAQ$ and of $\angle PAD$ . Now let $\ell$ denote the external bisector of $\angle PAQ$ . Clearly it suffices to show that $M \in \ell$ .

Claim: The lines $BQ$ , $DP$ and $\ell$ are concurrent.
Proof. Let $T = \ell \cap \overline{DP}$ and define $T' = \overline{BQ} \cap \overline{DP}$ . Also define $R = \overline{AC} \cap \overline{BQ}$ . Note that $(PD,ST) = -1$ from right angles and bisectors. However we also have $(PD, ST') = -1$ from Ceva-Menalus configuration. Thus $T \equiv T'$ . $\blacksquare$

However now by Pappus Theorem on $\overline{G PB}$ and $\overline{FQ D}$ we find that $M = \overline{FP} \cap \overline{GQ}$ , $T = \overline{PD} \cap \overline{BQ}$ and $A = \overline{FB} \cap \overline{DG}$ are collinear, hence $M \in \ell$ .

Remark Wow. I was doing this in spinvert so I already knew to utilize Miquel points, and from the defintion of $P$ and $Q$ it was obvious that $A$ is our desired Miquel point. From there I saw angle bisectors centered at $A$ , so I leaped to projective. To force the right angles and bisectors lemma I extended $\overline{AM}$ , $\overline{BQ}$ and $\overline{DP}$ . However here I got stuck. I had two harmonic bundles and no clue how to use them to show that $M$ lies on $\ell$ . Usually this would be a job for La Hire's but there were no useful circles present. Then I was hinted to use Pappus, which prior to this problem I didn't know. From there it was basically instakill.

This post has been edited 1 time. Last edited by Shreyasharma, Jan 6, 2024, 8:45 PM

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IAmTheHazard

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IAmTheHazard

#99 h Feb 25, 2024, 6:09 PM • 1 Y

Y by radian_51

By inversion about $E$ fixing $(ABCD)$ —or just angle chase—we have $P,E,Q$ collinear. Now, $\measuredangle QAE=\measuredangle QDE=\measuredangle CDB=\measuredangle CAB=\measuredangle CAD$ , so $\overline{AC}$ is the bisector of $\angle QAB$ .

Let $\overline{BQ} \cap \overline{DP}=X$ . Since $\overline{BD},\overline{PQ},\overline{CA}$ concur in $\triangle CBQ$ , by Ceva-Menelaus we have $(B,Q;\overline{AC} \cap \overline{BQ},X)=-1$ , hence by a well-known fact $\overline{AX}$ is the other bisector of $\angle QAB$ . It therefore suffices to show that $X,A,M$ are collinear, but this follows by Pappus on $FBQGDP$ since $\overline{BPG}$ and $\overline{DQF}$ are lines. $\blacksquare$

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IAmTheHazard

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IAmTheHazard

#100 h Feb 25, 2024, 6:35 PM • 1 Y

Y by radian_51

Alternatively, after finding that $\overline{AC}$ is the bisector of $\angle QAB$ (and similarly $\angle PAD$ ) we can employ a coordinate bash. This has the funny effect of revealing that $ABCD$ cyclic only matters in having $\overline{AP}$ and $\overline{AQ}$ be reflections of $\overline{AD}$ and $\overline{AB}$ over $\overline{AC}$ —if we redefine $P$ and $Q$ in this manner, the perpendicularity is true for any quadrilateral $ABCD$ .

Let $A=(0,0)$ and $C=(1,0)$ . Suppose lines $\overline{AB}$ and $\overline{AD}$ have equations $y=bx$ and $y=dx$ respectively and let $B=(p,bp)$ and $D=(q,dq)$ . We compute $F$ as the intersection of $y=bx$ and $\overline{CD}$ , which has equation $y=\tfrac{dq}{q-1}(x-1)$ :
$bx=\frac{dq}{q-1}(x-1) \implies (bq-dq-b)x=-dq \implies x=\frac{dq}{dq-bq+b} \implies y=\frac{bdq}{dq-bq+b} \implies F=\left(\frac{dq}{dq-bq+b},\frac{bdq}{dq-bq+b}\right).$ On the other hand, $P$ is the intersection of $\overline{CB}$ , which has equation $y=\tfrac{bp}{p-1}(x-1)$ , with $y=-dx$ :
$-dx=\frac{bp}{p-1}(x-1) \implies (-bp-dp+d)x=-bp \implies x=\frac{bp}{bp+dp-d} \implies y=-\frac{bdp}{bp+dp-d} \implies P=\left(\frac{bp}{bp+dp-d},-\frac{bdp}{bp+dp-d}\right).$ We will intersect $\overline{FP}$ with the $y$ -axis. First, we calculate the slope of the line as
$\frac{\frac{bdq}{dq-bq+b}+\frac{bdp}{bp+dp-d}}{\frac{dq}{dq-bq+b}-\frac{bp}{bp+dp-d}}=bd\cdot\frac{q(bp+dp-d)+p(dq-bq+b)}{dq(bp+dp-d)-bp(dq-bq+b)}=bd\cdot \frac{2dpq+bp-dq}{b^2pq+d^2pq-d^2q-b^2p}.$ By point-slope, the $x$ -intercept of $\overline{FP}$ is now
$\begin{align*} \frac{bdq}{dq-bq+b}-bd\cdot \frac{2dpq+bp-dq}{b^2pq+d^2pq-d^2q-b^2p}\cdot \frac{dq}{dq-bq+b}&=\frac{bdq}{dq-bq+b}\left(1-\frac{d(2dpq+bp-dq)}{b^2pq+d^2pq-d^2q-b^2p}\right)\\ &=\frac{bdq}{dq-bq+b}\left(\frac{b^2pq-d^2pq-b^2p-bdp}{b^2pq+d^2pq-d^2q-b^2p}\right)\\ &=\frac{bdpq}{dq-bq+b}\cdot \frac{(b+d)((b-d)q-b)}{b^2pq+d^2pq-d^2q-b^2p}\\ &=-\frac{bdpq}{b^2pq+d^2pq-d^2q-b^2p}, \end{align*}$ which is symmetric in $(b,p)$ and $(d,q)$ , hence equal to the $x$ -intercept of $\overline{GQ}$ , hence $M$ lies on the $y$ -axis as desired. $\blacksquare$

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Markas

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Markas

#101 h May 5, 2024, 8:23 PM • 2 Y

Y by Rounak_iitr, radian_51

We will prove that P, E, Q lie on one line. Now $\angle QEA = 180 -\angle QDA = \angle ABC = 180 - \angle PBA = 180 - \angle PEA$ $\Rightarrow$ $\angle QEA + \angle PEA = 180$ $\Rightarrow$ P, E, Q lie on one line. Denote X lies on CG and Y lies on CF and $XY \perp AC$ , $\angle PAC = 180 - \angle PBE = \angle CBD = \angle CAD$ $\Rightarrow$ AC is angle bisector of $\angle PAD$ $\Rightarrow$ AX is bisector of $\angle RAG$ . So by lemma 9.18 (G, P; X, C) = -1. We have $\angle QAC = \angle QDE = \angle CAB$ $\Rightarrow$ AC is bisector of $\angle BAQ$ $\Rightarrow$ AY is bisector of $\angle FAQ$ . From lemma 9.18 (Q, F; Y, C) = -1. So (Q, F; Y, C) = (G, P; X, C) = -1. From Prism lemma we get that GQ, FP and XY concur at M, M lies on XY and $XY \perp AC$ $\Rightarrow$ $\angle MAC = 90^{\circ}$ .

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YaoAOPS

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YaoAOPS

#102 h Aug 1, 2024, 2:37 PM • 4 Y

Y by bjump, Rounak_iitr, golue3120, radian_51

Am I the only clown to find an arbitrary circumconic?

Let $T$ be the Miquel point of $ABCD$ .

Claim: $E$ lies on $PQ$ .
Proof. Note that $\measuredangle PEA = \measuredangle PBA = \measuredangle CBA = \measuredangle CDA = \measuredangle QEA$ . $\blacksquare$

Claim: $N \coloneq PQ \cap GF$ satisfies $\measuredangle NAC = 90^\circ$
Proof. It remains to show that $NATE$ is cyclic. To do this, note that $\measuredangle NEA = \measuredangle PEA = \measuredangle PBA = \measuredangle GBA = \measuredangle GTA = \measuredangle NTA.$ $\blacksquare$
Let $H = AC \cap FG$ . Then it's well known that $(A, C; E, H) = -1$ .
Let $\mathcal{C}$ be a circumconic of $PGFQ$ which intersects $AC$ at $X, Y$ . As such, by DIT on $PGFQ$ and $CA$ there exists an involution swapping $(C, C), (E, H), (PF \cap AC, GQ \cap AC), (X, Y)$ Then this involution must consist of inversion about $(AC)$ or harmonic conjugation wrt $A, C$ .
This means that $A$ lies on the polar of $C$ wrt $\mathcal{C}$ . Since by Brokard's $NM$ is the polar of $A$ , we get that $N, C, M$ concur which finishes.

Remark: This can also be proved by homographying $ABCD$ into a rectangle and generalizing to $E \in PQ$ .

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bjump

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bjump

#103 h Aug 4, 2024, 11:33 PM • 1 Y

Y by radian_51

Solved with Arch Hints :oops:

Begin by observing $P$ , $E$ , $Q$ collinear by the spiral similarity $\triangle PAB \sim \triangle QAD$ . Note that $\overline{CA}$ bisects $\angle BAQ$ and $\angle PAD$ , since $\angle BAC = \angle BDC = \angle EDQ = \angle EAQ$ , and $\angle PAB = \angle QAD$ . Now let the external angle bisectors of $\angle PAD$ and $\angle BAQ$ meet $\overline{BC}$ , and $\overline{CD}$ at $X$ , and $Y$ respectively. By right angles and bisectors $-1= (XB; PC)= (YQ; DC)$ . So by the prism lemma $M \in XY$ , now $\angle MAC = \angle XAC = 90^\circ$ and we are finished.

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bin_sherlo

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bin_sherlo

#104 h Aug 11, 2024, 9:38 AM • 2 Y

Y by ehuseyinyigit, radian_51

$A$ is the intersection of $(QED)$ and $(PBE)$ hence $A$ is the miquel point of complete quadrilateral $BEQCDP$ . Invert the diagram from $A$ .
New Problem Statement After Inverting from $A$ : Let $P',B',Q',D'$ be a cyclic quadrilateral on a circle with this order. $P'Q'$ and $B'D'$ intersect at $C'$ . $D'A'$ intersects $(A'B'C')$ at $G'$ and $B'A'$ intersects $(A'C'D')$ at $F'$ . If $(A'F'P')$ and $(A'G'Q')$ intersect at $M',$ then $M'A'\perp A'C'$ .
Let's invert from $E'$ with radius $\sqrt{E'B'.E'P'}=\sqrt{E'Q'.E'D'}=\sqrt{E'C'.E'A'}$ .
New Problem Statement After Inverting the Second Diagram from $E'$ : $ABC$ is a triangle. We take a circle which intersects $AB,BC$ at $F,D$ respectively. $(ABP)$ and $BC$ intersect at $K$ and $(BDP)$ intersects $AB$ at $L$ . If $(BKF)$ crosses $(BCL)$ at $M$ , then $PM\perp BM$ .
Proof:
$\angle PLF=\angle PLB=\angle PDC= \angle ADC=\angle AFC=\angle LFP$ Thus, $PF=PL$ . Similarily,
$\angle PKC=180-\angle BKP=\angle PAB=\angle DAF=\angle DCF=\angle KCP$ Hence $PC=PK$ . Denote $S,T$ as the midpoints of $FL$ and $CK$ . Since $M$ is the miquel point of $FLCK,$ we get that $M$ is the center of spiral homothety taking $FSL$ to $KTC$ .
$\angle BSM=\angle FSM=\angle KTM=\angle BTM$ Which gives that $B,S,T,M$ are cyclic. $(BST)$ has diameter $BP$ hence $M$ lies on the circle whose diameter is $BP$ . Thus, $\angle PMB=90$ as desired. $\blacksquare$

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Ywgh1

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Ywgh1

#105 h Aug 13, 2024, 6:53 AM • 1 Y

Y by radian_51

Usamo 2018 p5
First of all we start off by showing that $P-E-Q$ are collinear and $AC$ bisect $\angle BAQ$ and $\angle DAP$ by angle chase.
Let $X$ and $Y$ as the intersection of the perpendicular to $AC$ at $A$ with $CF$ and $CG$ .
Then we get that.
$(CX; QF) = (CY; PG) = -1$ Which by prism lemma, we are done.

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Saucepan_man02

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Saucepan_man02

#106 h Nov 15, 2024, 7:24 AM • 1 Y

Y by radian_51

Here does DDIT :coolspeak:

:

Note that: $AC$ bisects $\angle BAQ$ and $\angle PAD$ as: $\angle PAC = \angle EBC = \angle DBC = \angle DAC$ $\angle BAE = \angle BAC = \angle BDC = \angle EDQ = \angle EAQ.$
Now, we proceed to apply DDIT. Consider the quadrilateral $GPFQ$ for DDIT.
Let line $\ell$ be perpendicular to $AC$ at $C$ . Now, project points $P, Q, D, F$ into line $\ell$ . Notice that: $P'C = G'C$ with $F'C=Q'C$ . Therefore, negative inversion at $C$ with $r = \sqrt{CP' \cdot CQ'}$ swaps: $(P', Q'), (G',F')$ (which is the halper line for DDIT). Let $M'$ denote the projection of $M$ (from $A$ ) onto $\ell$ . Therefore, we must have $(M', C)$ to swap each in the inversion which implies $M'$ to be point of infinity of line $\ell$ . Therefore, $AM \parallel \ell$ or $\angle MAC = 90^\circ$ .

This post has been edited 1 time. Last edited by Saucepan_man02, Nov 15, 2024, 7:26 AM
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L13832

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L13832

#107 h Nov 16, 2024, 3:27 AM • 1 Y

Y by radian_51

Solution

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peppapig_

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peppapig_

#108 h Nov 20, 2024, 3:47 AM • 2 Y

Y by dolphinday, radian_51

Solution w/ projective geometry (see attached).

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catherinexu-amo185-compressed.pdf (150kb)

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golue3120

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golue3120

#109 h Dec 7, 2024, 12:37 AM • 4 Y

Y by impromptuA, Sleepy_Head, qwerty123456asdfgzxcvb, radian_51

Here's a projective solution that seems at least somewhat different from previous solutions. All angles are directed modulo $180^\circ$ .

By DDIT with point $A$ and complete quadrilateral $CFGMPQ$ , we get that there exists an involution $\phi$ on the pencil of lines through $A$ sending $AC\leftrightarrow AM$ , $AB\leftrightarrow AD$ , and $AP\leftrightarrow AQ$ .

We have $\angle BAP-\angle DAQ=\angle BEP-\angle DEQ=0$ . Also, $\angle BAC=\angle BDC=\angle EDQ=\angle EAQ=\angle CAQ$ , so $AC$ bisects $\angle BAQ$ .

Thus there exists an involution $\psi$ on the pencil of lines through $A$ given by reflection across $AC$ , and it swaps $AB\leftrightarrow AQ$ , $AD\leftrightarrow AP$ , and $AC\leftrightarrow AC$ .

By MMP, we have that $\phi$ and $\psi$ commute because $\phi\circ\psi=\psi\circ\phi$ for $AB$ , $AD$ , $AP$ , and $AQ$ . Thus $\psi(AM)=\phi^{-1}(\psi(\phi(AM)))=AM$ . Thus $AM$ is the other angle bisector of $\angle BAQ$ , and so $\angle CAM=90^\circ$ .

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Mathandski

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Mathandski

#110 h Dec 19, 2024, 12:52 AM • 2 Y

Y by radian_51, ravengsd

Spiral sim where ;-;

Subjective Rating (MOHs) $\text{[math]}$

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cj13609517288

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cj13609517288

#111 h Mar 11, 2025, 6:30 PM

Y by

Note that $\angle ADQ=\angle ABP$ and
$\angle AQD=\angle AED=\angle APB.$ Therefore, there exists a spiral similarity centered at $A$ sending $DQ$ to $BP$ , so $P,E,Q$ are collinear.

Now let $X,Y$ be the intersections of $AM$ with $CF$ and $CG$ , respectively. Then the main claim is that $(FQ;XC)=-1$ . Indeed, this is true by the internal/external angle bisector configuration with $A$ , because $\angle CAB=\angle CAQ$ .

Similarly, $(GP;YC)=-1$ . So by projecting through $M$ , we get that $M$ also lies on $XY$ , as desired. $\blacksquare$

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