this hAOpefully shoudn't BE weird

The post below has been deleted. Click to close.

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

popop614

266 posts

popop614

#1 h Jul 17, 2024, 12:25 PM • 5 Y

Y by peace09, OronSH, MarkBcc168, kamatadu, Rounak_iitr

Let $ABCDE$ be a convex pentagon such that $\angle ABC = \angle AED = 90^\circ$ . Suppose that the midpoint of $CD$ is the circumcenter of triangle $ABE$ . Let $O$ be the circumcenter of triangle $ACD$ .

Prove that line $AO$ passes through the midpoint of segment $BE$ .

This post has been edited 1 time. Last edited by popop614, Jul 17, 2024, 12:31 PM
Reason: bu h

Z K Y

The post below has been deleted. Click to close.

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

Math-48

44 posts

Math-48

#2 h Jul 17, 2024, 12:27 PM • 2 Y

Y by Upwgs_2008, Muaaz.SY

Just bash it :yup:

Set $a=1~,~c=-1$ and so $|b|=1$

Let $M$ be the midpoint of $CD$ we see that it's collinear with the midpoint of both $AB$ and $AC$

So $\exists t \in \mathbb{R}$ such that $: m=t(b+1)$

Hence $d=2m-c=2tb+2t+1$

Now to evaluate $e$ we have:
$|ME|^2=|MA|^2\implies(e-m)(\overline{e}-\overline{m})=(m-1)(\overline{m}-1)\implies \overline{e}=\frac{(m-1)(m-b)+m(e-m)}{b(e-m)}\implies \overline{e}=\frac{(tb+t)e-(tb^2+2tb+t-b)}{b(e-(tb+t))}$ $\angle AED=90^\circ\implies\frac{\overline{e}-\overline{a}}{\overline{d}-\overline{e}}=\frac{e-a}{e-d}\implies\frac{\overline{e}-\overline{a}}{\overline{d}-\overline{a}}=\frac{e-a}{2e-a-d}\implies\overline{e}=\frac{(\overline{d}-\overline{a})(e-a)+\overline{a}(2e-a-d)}{2e-a-d}\implies\overline{e}=\frac{(tb+t+b)e-(tb^2+2tb+t+b)}{b(e-(tb+t+1))}$ By equating we get:
$(e-(tb+t+1))((tb+t)e-(tb^2+2tb+t-b))=(e-(tb+t))((tb+t+b)e-(tb^2+2tb+t+b))$ Which is a quadratic accept $1$ as a root, the other can be obtained by vieta's formula:
$e=\frac{(tb+t)(tb^2+2tb+t+b)-(tb+t+1)(tb^2+2tb+t-b)}{b}=\frac{tb^2+b-t}{b}$ define $N$ as the midpoint of $BE$ to get:
$n=\frac{tb^2+b^2+b-t}{2b}$ Now to find $o$ we first notice that it's pure imaginary and then:
$|OD|^2=|OA|^2\implies (o-d)(-o-\overline{d})=(o-1)(-o-1)\implies o= \frac{d\overline{d}-1}{\overline{d}-d}\implies o=\frac{2tb+2t+b+1}{1-b}$ And finally we finish with collinearity:
$\frac{n-a}{a-o}=\frac{(b-1)^2}{4b}\in\mathbb{R}~~~~~~~~~\blacksquare$

Z K Y

The post below has been deleted. Click to close.

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

MarkBcc168

1594 posts

MarkBcc168

#3 h Jul 17, 2024, 12:27 PM • 11 Y

Y by peace09, OronSH, khina, avisioner, GeoKing, Minkowsi47, Gato_combinatorio, Kingsbane2139, Rounak_iitr, MS_asdfgzxcvb, Sadigly

Just synthetic it

$[asy] size(6cm); defaultpen(fontsize(10pt)); import olympiad; import geometry; pair A = (1.3,1.0); pair B = (0,0); pair E = (4,0); pair O1 = circumcenter(A,B,E); pair X = 2*O1-A; pair C = extension(B,X,A,2*O1-E); pair D = A+X-C; pair M = (C+D)/2; pair N = (B+E)/2; pair O = circumcenter(A,C,D); fill(A--C--X--cycle, mediumgray); fill(A--B--E--cycle, mediumgray); draw(C--X--D, linewidth(0.7)); draw(A--M, linewidth(0.7), StickIntervalMarker(1, 2, p=linewidth(0.7), size=0.2cm)); draw(M--X, linewidth(0.7), StickIntervalMarker(1, 2, p=linewidth(0.7), size=0.2cm)); draw(C--M, linewidth(0.7), StickIntervalMarker(1, 1, p=linewidth(0.7), size=0.2cm)); draw(M--D, linewidth(0.7), StickIntervalMarker(1, 1, p=linewidth(0.7), size=0.2cm)); draw(A--O, dashed); draw(rightanglemark(A,B,C), black); draw(rightanglemark(A,E,D), black); draw(A--B--C--D--E--cycle, linewidth(1)); draw(A--C--X--cycle, black); draw(A--B--E--cycle, black); dot("$A$", A, dir(98)); dot("$B$", B, dir(172)); dot("$E$", E, dir(25)); dot("$X$", X, dir(-82)); dot("$C$", C, dir(-143)); dot("$D$", D, dir(-20)); dot("$M$", M, dir(-136)); dot("$N$", N, dir(62)); dot("$O$", O, dir(-56)); [/asy]$

Let $X = BC\cap DE$ , so by angle conditions, $X$ is the antipode of $A$ in $\odot(ABE)$ . In particular, $ACXD$ is a parallelogram.

Next, note that
$\left.\begin{array}{r} \angle CAX = \angle AXE = \angle ABE \\[2pt] \angle CXA = \angle BEA \end{array} \right\} \implies \triangle ABE\stackrel-\sim\triangle ACX.$ Thus, if $M$ and $N$ are midpoints of $CD$ and $BE$ , then
$\measuredangle ACM = -\measuredangle BAN = 90^\circ - \measuredangle(BX,AN) = 90^\circ - \measuredangle DAN,$ implying the conclusion.

This post has been edited 1 time. Last edited by MarkBcc168, Jul 17, 2024, 12:27 PM
Reason: I'm not the first post

Z K Y

The post below has been deleted. Click to close.

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

Om245

163 posts

Om245

#4 h Jul 17, 2024, 1:16 PM • 3 Y

Y by ehuseyinyigit, Rounak_iitr, poirasss

Lmao ratio lemma ......

Let $F = BC \cap ED$ , $X = AO \cap BE$ and $Y$ be midpoint of $BE$ and $CD$ . Observe that $\Box ABFE$ cyclic with diameter $AF$ .
Hence $Y$ lie on $AF$ . Which also give us $\Box ACFD$ parallelogram.
$\measuredangle EAO = \measuredangle EAD + \measuredangle DAO =90 - \measuredangle ADE + 90 - \measuredangle ACD = \measuredangle DCF$ And similarly $\measuredangle OAB = \measuredangle FDC$ .
Note from $\measuredangle CFD = \measuredangle ADE = \measuredangle BCA$ we have $\triangle ADE \sim \triangle ACB$ .

Now using ratio lemma in $\triangle CFD$ and $\triangle AEB$ be have
$\frac{sin \measuredangle DCF}{sin \measuredangle FDC}=\frac{FD}{FC}=\frac{AC}{AD}=\frac{AB}{AE}$ Hence $\frac{EX}{XB} = \frac{sin \measuredangle EAO \cdot AE}{sin \measuredangle OAB \cdot AB} = 1 \Rightarrow X$ is midpoint of $EB$ . $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

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

Marinchoo

407 posts

Marinchoo

#5 h Jul 17, 2024, 1:39 PM • 1 Y

Y by isomoBela

Let $A'$ be the reflection of $A$ over $M$ , the midpoint of $CD$ . Then $MA = MB = ME = MA'$ , so $\angle ABA' = \angle AEA' = 90^{\circ}$ . Therefore $C$ lies on $A'B$ and $D$ lies on $A'E$ . Additionally, $ACA'D$ is a parallelogram because $M$ is the midpoint of both $CD$ and $AA'$ .

We'll show $\angle NAC + \angle CDA = 90^{\circ}$ , which implies $N \in \overline{AO}$ where $N$ is the midpoint of $BE$ as $\angle CDA = 90^{\circ} - \angle OAC$ . To do so, use complex numbers with unit circle the circumcircle of $\triangle ABE$ . If $E'=-e$ , then $C = AE'\cap BA'$ as $AC\parallel A'D$ and $D = -C$ , so
$\begin{align*} c &= \frac{(a-e)(-ab)-(b-a)(-ae)}{-ae+ab} = \frac{2be-ae-ab}{b-e}\\ d &= \frac{-2be+ae+ab}{b-e}, \quad n = \frac{b+e}{2}. \end{align*}$ Finally, we compute:
$\begin{align*} \frac{n-a}{c-a}\cdot \frac{c-d}{a-d} &= \frac{\frac{b+e}{2}-a}{\frac{2be-ae-ab}{b-e}-a}\cdot \frac{2\left(\frac{2be-ae-ab}{b-e}\right)}{a+\frac{2be-ae-ab}{b-e}}\\ &=\frac{1}{4}(b+e-2a)(2/a-1/b-1/e)\cdot\frac{b-e}{4a(e-a)(b-a)} \end{align*}$ This can easily be checked to be purely imaginary, as desired.

Z K Y

The post below has been deleted. Click to close.

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

khina

993 posts

khina

#6 h Jul 17, 2024, 1:50 PM • 2 Y

Y by i3435, CyclicISLscelesTrapezoid

Areas!

Angle chase to find that $BC // AD$ , $AC // DE$ . Now let $A'$ be the antipode of $A$ on $(ABC)$ . We now have that
$\begin{align*} [AA'B] = [AA'C] = [ADC] \end{align*}$ which is symmetric, so $AA'$ bisects $BE$ as desired.

This post has been edited 1 time. Last edited by khina, Jul 17, 2024, 1:51 PM

Z K Y

The post below has been deleted. Click to close.

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

teoira

8 posts

teoira

#7 h Jul 17, 2024, 3:11 PM • 1 Y

Y by Assassino9931

Areas again https://dgrozev.wordpress.com/2024/07/17/an-easy-geometrical-problem-from-imo23-shortlist/

Z K Y

The post below has been deleted. Click to close.

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

EpicBird08

1740 posts

EpicBird08

#8 h Jul 17, 2024, 3:19 PM

Y by

Let $X$ be the intersection of $BC$ and $DE.$ Then if $M$ is the circumcenter of $\triangle ABE,$ then $M$ is the midpoint of both $CD$ and $AX,$ so $ACXD$ is a parallelogram. Moreover, if we draw the line $\ell$ through $A$ parallel to $CD$ , and let it intersect $DE$ and $BC$ at points $Y$ and $Z$ , respectively, this implies that $ACD$ is the medial triangle of $\triangle XYZ.$

Now, let $F$ and $G$ be the feet of the altitudes from $Y, Z$ to $ZX, XY,$ respectively, so that $YE = EG$ and $ZB = BF.$ Then if $N$ is the midpoint of $FG,$ then since $AENB$ is a parallelogram, $AN$ bisects $BE.$ It suffices to show that $A,N,O$ are collinear. Indeed, since $AF = AG,$ they all lie on the perpendicular bisector of $FG,$ so we are done.

This post has been edited 1 time. Last edited by EpicBird08, Jul 17, 2024, 3:23 PM

Z K Y

The post below has been deleted. Click to close.

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

Z4ADies

62 posts

Z4ADies

#9 h Jul 17, 2024, 6:26 PM

Y by

Let $BC \cap DE$ at $K$ . If $M$ is center of $ABE$ and $K$ is on that circle, then $M$ is center of $ABKE$ .Let $\angle EBM=\angle BEM=\alpha$ , $\angle MEK=\angle MKE=\angle ABE=\beta$ so, $\angle MBE=\angle MKB=90-\alpha-\beta$ .We know that $AM=MK$ and $CM=DM$ that means $ACKD$ is parallelogram.Thus, $AC,AD$ are parallel to $KD,KC$ respectively.Assume $AO \cap BC$ at $M'$ .Let $\angle OAM=\gamma$ .From parallelity $\angle KAC=\beta$ by cyclic condition $\implies$ $\angle BAC=\angle DAE=\alpha$ and $\angle OAD=90-\alpha-\beta-\gamma$ .From center $O$ condition $\implies$ $\angle OCA=\beta+\gamma$ , $\angle ODA=90-\alpha-\beta-\gamma$ and $\angle OCD=\angle ODC=\alpha$ . From paralellity $\angle DCK=90-\beta-\gamma$ and $\angle CDK=\alpha+\beta+\gamma$ .
Finally, median sine to $\triangle CKD=\triangle ABE$ equation is true.

Z K Y

The post below has been deleted. Click to close.

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

kamatadu

466 posts

kamatadu

#10 h Jul 17, 2024, 10:03 PM

Y by

Just MMP it :yup:

The crux of the problem is the following claim which we of course solve using MMP!

$\textbf{CLAIM:}$ $ABA'D$ is a cyclic quadrilateral where $A'$ is the $A$ -antipode. $E$ and $F$ are points on $BA'$ and $A'D$ respectively such that $AF \parallel A'D$ and $AE \parallel BA'$ . Then $EF$ , $AA$ and $BD$ are concurrent.

$\textbf{PROOF:}$

$[asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions are done using bubu-asy.py. This adds the dps, xmin, linewidth, fontsize and directions. https://github.com/Bubu-Droid/dotfiles/blob/master/bubu-scripts/bubu-asy.py */ pair O = (11.71916,6.19787); pair A = (6.12566,25.60504); pair B = (-5.57187,16.63555); pair D = (28.92364,16.77764); pair Ap = (17.31267,-13.20928); pair E = (23.54528,2.88728); pair F = (-0.10694,9.50847); pair T = (-25.27638,16.55438); import graph; size(10cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); real xmin = -5, xmax = 5, ymin = -5, ymax = 5; pen ffxfqq = rgb(1.,0.49803,0.); draw(circle(O, 20.19715), linewidth(0.6)); draw(A--B, linewidth(0.6)); draw(B--D, linewidth(0.6)); draw(D--A, linewidth(0.6)); draw(A--T, linewidth(0.6)); draw(T--B, linewidth(0.6)); draw(T--E, linewidth(0.6)); draw(B--Ap, linewidth(0.6) + ffxfqq); draw(Ap--D, linewidth(0.6) + blue); draw(F--A, linewidth(0.6) + blue); draw(A--E, linewidth(0.6) + ffxfqq); draw(A--Ap, linewidth(0.6) + linetype("4 4") + red); dot("$O$", O, SW); dot("$A$", A, NW); dot("$B$", B, NW); dot("$D$", D, NE); dot("$A'$", Ap, dir(280)); dot("$E$", E, dir(-15)); dot("$F$", F, SW); dot("$T$", T, NW); [/asy]$

We fix the circle, $A$ , $D$ and animate $B$ projectively on the circle. Clearly this fixes $A'$ and so the line $A'D$ is fixed. This further gives us that $A\infty_{A'D}$ is also fixed. Now we have that,
$B \mapsto A'B \mapsto A'B \cap A\infty_{A'D} \equiv F \text{ via } \text{circle} \mapsto \mathcal{P}(A') \mapsto A\infty_{A'D}$ is projective. So the degree of $F$ is $1$ .

Also, we have that,
$B \mapsto A'B \mapsto A'B \cap \ell_{\infty}\mapsto A\infty_{A'B} \mapsto A\infty_{A'B}\cap A'D \equiv E \text{ via } \text{ circle } \mapsto \mathcal{P}(A') \mapsto \ell_{\infty}\mapsto \mathcal{P}(A)\mapsto A'D$ is projective, where $\ell_{\infty}$ is the line at infinity. So the degree of $E$ is $1$ .

This gives us that the degree of line $EF$ is $=2$ . Also note that degree of line $AA$ is $0$ as the line is essentially fixed.

Now also note,
$B\mapsto DB \text{ via }\text{ circle }\mapsto \mathcal{P}(D)$ is projective. This gives us that the degree of line $DB$ is $=1$ .

So the concurrency condition of $AA \cap BD \cap EF$ has degree $0 + 1 + 2 = 3$ (can be rephrased as the collinearity of $\overline{AA\cap BD - E - F}$ ). Thus it suffices to check that it is true for $3 + 1 = 4$ cases.

$B = A$ .

In this case, note that $DB \equiv AD$ , $F\equiv A$ and $E\equiv A'$ which clearly shows that $A$ is the common concurrency point and done.
$B \equiv D$ -antipode.

In this case note that $F \equiv B$ , $E \equiv D$ , and so, $BD\equiv EF$ . This clearly finishes.
$B\equiv A'$ .

In this case, note that $E\equiv AA\cap BD$ which clearly finishes.
$B\equiv D$ .

In this case, note that $F\equiv \infty_{A'D}$ , $E\equiv \infty_{A'D}$ and so, we can remove this case using Zack's Lemma.

$[asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions are done using bubu-asy.py. This adds the dps, xmin, linewidth, fontsize and directions. https://github.com/Bubu-Droid/dotfiles/blob/master/bubu-scripts/bubu-asy.py */ pair A = (6.90289,21.50125); pair B = (-3.09670,12.35254); pair E = (23.50200,17.08615); pair D = (20.67305,6.45038); pair C = (2.76527,5.94536); pair T = (-46.98268,4.54241); pair F = (-4.31783,5.74561); pair G = (27.75616,6.65014); pair S = (5.80296,60.50422); pair R = (7.21464,10.44674); import graph; size(10cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); real xmin = -5, xmax = 5, ymin = -5, ymax = 5; pen ffxfqq = rgb(1.,0.49803,0.); draw(circle((11.71916,6.19787), 16.04337), linewidth(0.6)); draw(A--B, linewidth(0.6)); draw(B--E, linewidth(0.6)); draw(E--A, linewidth(0.6)); draw(A--T, linewidth(0.6)); draw(T--B, linewidth(0.6)); draw(T--D, linewidth(0.6)); draw(C--A, linewidth(0.6) + blue); draw(A--D, linewidth(0.6) + ffxfqq); draw(D--G, linewidth(0.6)); draw(F--S, linewidth(0.6)); draw(S--G, linewidth(0.6)); draw(B--G, linewidth(0.6)); draw(E--F, linewidth(0.6)); draw(B--C, linewidth(0.6) + ffxfqq); draw(D--E, linewidth(0.6) + blue); draw(S--R, linewidth(0.6)); dot("$A$", A, NW); dot("$B$", B, NW); dot("$E$", E, NE); dot("$D$", D, dir(270)); dot("$C$", C, dir(270)); dot("$T$", T, NW); dot("$F$", F, NW); dot("$G$", G, NE); dot("$S$", S, NW); dot("$R$", R, dir(270)); [/asy]$

Now back to the original problem note that we have to prove $AO$ passes through the midpoint of $BE$ . Let $M$ denote the midpoint of $BE$ . But note that this is equivalent to proving that the $A$ -altitude in $\triangle ACD$ and $AM$ are isogonal, i.e., the $A$ -altitude is the symmedian of $\triangle ABE$ .

Let $CD$ meet the circle again at $F$ and $G$ . Define $T=AA\cap BE\cap FG$ , $R = BG\cap FE$ and $S = FB\cap GE$ . Clearly $A$ lies on the polar of $T$ . Now by Brokard's Theorem, we also have that $SR$ is the polar of $T$ . This also gives us that $SR \perp FG$ since $FG$ is the diameter. This means that $AR\perp FG$ .

Now by Pascal on $BBGEEF$ , we finally get that $AR$ is the $A$ -altitude in $\triangle ACD$ which finishes.

This post has been edited 3 times. Last edited by kamatadu, Jul 18, 2024, 12:51 PM

Z K Y

The post below has been deleted. Click to close.

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

dkedu

180 posts

dkedu

#11 h Jul 17, 2024, 11:21 PM

Y by

We have $F,M,G,H,I$ as the midpoints of $\overline{AC}, \overline{CD}, \overline{AD}, \overline{AB}, \overline{AE}$ respectively.

Claim 1: $\overline{AC}$ and $\overline{MH}$ intersect at $F$ .

$F$ is the circumcenter of $\Delta ABC$ since $\Delta ABC$ is a right triangle, so we get that $\overline{FH} \perp \overline{AB}$ but since $M$ is the circumcenter of $\Delta ABE$ , we have that $\overline{MH} \perp \overline{AB}$ so we have the result.

Claim 2: $\overline{AC}\parallel \overline{DE}$ , and similarly $\overline{AD} \parallel \overline{BC}$ .

We have $\angle BCA = 90^\circ -\angle BAC = \angle HFA = \angle CFM = \angle CAB$ so they are parallel.

Let $OA$ intersect $BE$ at $N'$ . We have that
$\frac{N'B}{N'E} = \frac{AB\sin(\angle OAB)}{AC\sin(\angle OAE)} = \frac{AC}{AD}\cdot \frac{\sin(\angle OAB)}{\sin(\angle OAE)} = \frac{AC}{AD}\cdot \frac{\sin(\angle ACD)}{\sin(\angle ADC)} = 1$ Therefore, $N' = N$ , so we are done.

Z K Y

The post below has been deleted. Click to close.

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

Asynchrone

65 posts

Asynchrone

#12 h Jul 18, 2024, 12:00 AM • 2 Y

Y by Math_Kobekeye, Lorenzo_degli_atti

Solution using barycentric coordinates :
Let $BC$ and $ED$ intersect at point $F$ . Since $AF$ is a diameter, $M$ is the midpoint of $AF$ thus $ACFD$ is a parallelogram.
Further $\angle{CAE} = \angle{CAF} + \angle{FAE} = \angle{AFE} + \angle{FAE} = 180^\circ - \angle{AEF} = 90^\circ$ and $\angle{BAD} = 90^\circ$ in the same manner.
Let $A = (1,0,0), C = (0,1,0), D = (0,0,1)$ and let $CD = a, DA = c, AC = d$ , further we define the Conway notation as $S_a = \frac{c^2 + d^2 - a^2}{2}$ , $S_c$ and $S_d$ are defined in the same way.
Now since $A + F = C + D$ (parallelogram law, where addition is done component wise), we have that $F = (-1,1,1)$
We now compute point $B = (x,y,z)$ where $x + y + z = 1$ , since $B$ lies on $CF$ we have :
$\begin{vmatrix} 0 & 1 & 0 \\ -1 & 1 & 1 \\ x & y & z \notag \end{vmatrix} = x + z = 0.$
Hence, implying that $y = 1$ , and thus $B = (-z,1,z)$ .
Now using the fact that $\vec{AB} = (-z - 1, 1, z)$ is perpendicular to $\vec{AD} = (-1,0,1)$ , by EFFT, we have :
$0 = a^2 + c^2(-z - 1 - z) + d^2 \times - 1$ , and thus $z = \frac{-S_a}{c^2}$
We thus get that $B = ( \frac{S_a}{c^2},1, \frac{-S_a}{c^2})$ , and by symmetry that $E = ( \frac{S_a}{d^2},\frac{-S_a}{d^2}, 1)$
Hence the midpoint $N = (\frac{\frac{S_a}{c^2} + \frac{S_a}{d^2}}{2} = t, \frac{1 - \frac{S_a}{d^2}}{2}, \frac{1 - \frac{S_a}{c^2}}{2}) = (2t : \frac{d^2 - S_a}{d^2}:\frac{c^2 - S_a}{c^2}) = (2c^2d^2t = x : c^2S_c : d^2S_d)$ by the identities $c^2 - S_a = S_c$ and $d^2 - S_a = S_d$
But recall that $O = (a^2S_a : c^2S_c : d^2S_d)$ , thus since $N = (x : c^2S_c : d^2S_d)$ it must lie on the cevian $AO$ , as wanted.

Z K Y

The post below has been deleted. Click to close.

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

amuthup

779 posts

amuthup

#13 h Jul 18, 2024, 1:15 AM

Y by

Here's a weird synthetic solution.

Let $M,M_1,M_2$ be the midpoints of $\overline{CD},\overline{AB},$ and $\overline{AE}$ respectively. Since $\overline{CB}\perp\overline{AB}$ and $\overline{MM_1}\perp\overline{AB},$ we have that $\overline{DA}\perp\overline{AB}$ as well. Similarly, $\overline{CA}\perp\overline{AE}.$ Therefore, $\angle EAD=90^\circ-\angle DAC=\angle CAB,$ so $EAD$ and $BAC$ are similar right triangles.

Next, let $A'=\overline{ED}\cap\overline{BC}$ and let $N$ be the midpoint of $\overline{BE}.$ Since $A$ is the antipode of $A'$ with respect to $(A'BE),$ we know that $\overline{AN}$ passes through the orthocenter of triangle $A'BE,$ which we'll call $H.$ Let $X$ and $Y$ denote the feet of $H$ onto $\overline{AC}$ and $\overline{AD}$ respectively. Since $\overline{BH}\perp\overline{A'E},$ we in fact have $\overline{BH}\perp\overline{AC},$ so $X$ is the foot of $B$ onto $\overline{AC}.$ Similarly, $Y$ is the foot of $E$ onto $\overline{AD}.$ Now by similar triangles, $\tfrac{AX}{AY}=\tfrac{AC}{AD}.$ This implies (say, by homothety at $A$ ) that $H$ lies on $\overline{AO},$ which finishes because we already showed that $H$ lies on $\overline{AN}.$

Z K Y

The post below has been deleted. Click to close.

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

trk08

614 posts

trk08

#14 h Jul 18, 2024, 4:17 AM

Y by

Define $T=BC\cap DE$ , $M$ as the midpoint of $CD$ , and $N$ as the midpoint of $BE$ . Note that $AETB$ is cyclic with center $M$ , and $T$ being the antipode of $A$ . This implies $AM=MT$ and $CM=MD$ implying $ADTC$ is a parallelogram.

Claim:
$\angle TCD=\angle NAE$
Proof:
Note that it suffices to show $\triangle TCA\sim\triangle EAB$ . Clearly:
$\angle ATC=\angle ATB=\angle AEB$ $\angle TAC=\angle ATD=\angle ATE=\angle ABE,$ as desired $\square$

We can then finish by saying:
$\begin{align*} \angle EAO&=\angle OAD+\angle DAE\\ &=90-\angle ACD+\angle DAE\\ &=180-\angle CDT-\angle EDA\\ &=\angle ADC\\ &=\angle TCD\\ &=\angle EAN\\ \end{align*}$ implying $A,O,N$ are collinear $\blacksquare$

This post has been edited 1 time. Last edited by trk08, Jul 18, 2024, 4:18 AM

Z K Y

The post below has been deleted. Click to close.

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

GoodMorning

826 posts

GoodMorning

#15 h Jul 18, 2024, 4:52 AM

Y by

complete the cyclic quadrilateral and prove AGE \sim CMF following from ABE \sim ACF. angle chasing finishes

Attachments:

Z K Y

The post below has been deleted. Click to close.

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

Ege_Saribass

27 posts

Ege_Saribass

#33 h Sep 4, 2024, 10:29 AM • 1 Y

Y by Anancibedih

Let $AO \cap BE = \{N\}$ . One can easily conclude that $BC || AD$ and $AC || ED$ .
Hence, $\angle BAC = x$ , $\angle DAC = 90 - x$ , $\angle DAE = x$ .
Also let $\angle ADO = y$ . Hence, $\angle EAN = x+y$ , $\angle NAB = 90-y$ , $\angle ADC = x+y$ , $\angle ACD = 90-y$ .
By similarity and sine theorem:
$\frac{AE}{AB} = \frac{AD}{AC} = \frac{\sin(90-y)}{\sin(x)}$
Also by ratio lemma at $\triangle{AEB}$ :
$\frac{EN}{NB} = \frac{AE}{AB}.\frac{\sin(x+y)}{\sin(90-y)} = \frac{\sin(90-y)}{\sin(x+y)}.\frac{\sin(x+y)}{\sin(90-y)} = 1$
Hence, $EN = NB$ , we are done. (Similar solution when $\triangle ADC$ is non-acute.)
$\blacksquare$

Attachments:

Z K Y

The post below has been deleted. Click to close.

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

CyclicISLscelesTrapezoid

372 posts

CyclicISLscelesTrapezoid

#34 h Sep 15, 2024, 7:41 PM • 2 Y

Y by MathLuis, crazyeyemoody907

Use reference triangle $ACD$ . Let $M$ , $P$ , and $Q$ be the midpoints of segments $CD$ , $AC$ , and $AD$ , respectively. Notice that $MA=MB$ and $PA=PB$ , so $B$ is the reflection of $A$ over $\overline{MP}$ . Similarly, $E$ is the reflection of $A$ over $\overline{MQ}$ . Since $\overline{MP}$ and $\overline{MQ}$ are midlines, we have $\overline{CB} \parallel \overline{AD}$ and $\overline{DE} \parallel \overline{AC}$ . Thus, we have
$[ABO]=[ABQ]=[ACQ]=\frac{[ACD]}{2}=[ADP]=[AEP]=[AEO].$ Thus, $\overline{AO}$ bisects $\overline{BE}$ , as desired. $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

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

ezpotd

1250 posts

ezpotd

#35 h Sep 29, 2024, 2:55 AM

Y by

Let $(ABE)$ have center $M$ , $(AC)$ have center $X$ , $(AD)$ have center $Y$ . Since $M,X$ both lie on the perpendicular bisector of $AB$ , we know $MX \perp AB$ , since $MX$ is parallel to $AD$ (midline), we have $AD \perp AB$ , so $\angle CAB = 90 - \angle DAC$ . Likewise, $\angle EAD = 90 - \angle DAC$ , so $ABC$ and $AED$ are similar.

Let the foot from $A$ to $CD$ be $K$ . Observe $K$ lies on $(ABC)$ and $(ADE)$ . Note that $AO, AK$ are isogonal in $\angle DAC$ , and since $\angle EAD = \angle CAB$ , they are also isogonal in $\angle EAB$ . Thus it suffices to prove $AK$ is the $A$ -symmedian of $\triangle BAE$ . Since we know $AK$ is the radical axis of $(AC)$ and $(AD)$ , it suffices to find two points on the $A$ -symmedian that have equal power with respect to those two circles. One of these points is $A$ , obviously.

We then claim that the symmedian foot on $BE$ has equal power with respect to both circles. To do this, we use linearity of power of a point. Consider the function $f$ from points to reals that outputs the power of the point with respect to $(AC)$ minus the power of the point with respect to $(AD)$ . This function is known to be linear. We calculate this $f(B), f(E)$ . We know the power of $B$ with respect to $(AC)$ is just $0$ , then $AB \perp AY$ , so $AB$ is a tangent to $(AD)$ , so the power of $B$ with respect to $AD$ is $AB^2$ , so $f(B) = -AB^2$ . Likewise, $f(E) = AE^2$ . Thus, $f$ evaluated at the symmedian foot is just $\frac{f(B)AE^2 + f(E)AB^2}{AE^2 + AB^2} = 0$ , as desired. We are done.

Z K Y

The post below has been deleted. Click to close.

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

AshAuktober

933 posts

AshAuktober

#36 h Sep 29, 2024, 6:42 AM

Y by

Sketch: Insert $A'$ = A-antipode, $ACA'D$ parallelogram, do some angle chase and then sine rule.

Z K Y

The post below has been deleted. Click to close.

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

SatisfiedMagma

451 posts

SatisfiedMagma

#37 h Oct 19, 2024, 10:45 AM • 1 Y

Y by proxima1681

Why are G1's so hard... Took me three days to solve this one. Why lengths? Where is angle chasing? Anyways here is a solution with help from proxima1681 to help me find the length chase.

Solution: Let $X$ , $T$ and $M$ be the midpoints of $\overline{BE}$ , $\overline{MT}$ and $\overline{CD}$ respectively. Set $\triangle ACD$ to be the reference triangle. The main claim is as follows.
$[asy] import olympiad; import geometry; size(10cm); pair A = (-0.64,0.77); pair C = (0.93, -0.36); pair D = (-0.59, -0.81); pair O1 = (A+D)/2; pair O2 = (A+C)/2; pair E = intersectionpoints(line(D,(D+C-A)), circle((A+D)/2, length(O1--A)))[0]; pair B = intersectionpoints(line(C,(C+D-A)), circle(O2, length(O2--C)))[0]; pair M = (C+D)/2; pair O = circumcenter(A,C,D); pair X = extension(A,O,E,B); //(B+E)/2; sorry, diagram not accurate since coords of A,B,C, only up to 2 decimal points pair T = (A+E)/2; draw(A--B--C--D--E--A, blue); draw(M--E, red); draw(M--A, red); draw(M--B, red); draw(O--A, dashed+pink); draw(E--B); draw(A--D, blue); draw(A--C, blue); draw(M--T, deepgreen); dot("$A$", A, dir(A)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$B$", B, dir(B)); dot("$M$", M, dir(M)); dot("$X$", X, dir(X)); dot("$T$", T, dir(T)); dot("$O$", O, dir(O)); [/asy]$

Claim: $ED \parallel AC$ and $BC \parallel AD$ .

Proof: By symmetry, it is enough to show $ED \parallel AC$ . Start by noticing that $MT \parallel DE$ since $\measuredangle MTE = \measuredangle DET = 90^\circ$ . From here, instead of using midpoint theorem, we'll use complex numbers to finish. Note that $d - e \parallel (c+d)/2 - (e+a)/2$ . But, this means
$d - e \parallel \left(\frac{c-a}{2} + \frac{d-e}{2}\right) \parallel c-a$ which is as desired. $\square$

Now, from the above set of parallel lines, we can conclude that $\measuredangle(ED, AD) = \measuredangle(BC,AC)$ . Since $\measuredangle DEA = \measuredangle ABC = 90^\circ$ , we can easily show that $\triangle EDA \sim \triangle BCA$ .

Finally, let $\angle EAX = x$ and $\angle XAB = y$ . It is not hard to show that $x + y = \angle ADC + \angle ACD = D +C$ . From Ratio Lemma and $\triangle EDA \sim \triangle BCA$ , we get that
$\begin{align*} \frac{\sin x}{\sin y} = \frac{\overline{AB}}{\overline{AE}} &= \frac{\overline{AC}}{\overline{AD}} = \frac{\sin C}{\sin D} \\ \implies \sin x \sin D &= \sin C \sin y. \end{align*}$ Apply product to sum formula. Cancel $\cos(x-C)$ and $\cos(y-D)$ to get
$\cos(x+C) = \cos(y+D).$ Since $0 < x+C, y+D \le 2\pi$ , we either have $x + C = y + D$ , or $x + C + y +D = 2\pi$ . If the second condition holds, then we can conclude that $C+D = \pi/2$ . But then observe that in this case, $E \equiv D$ and $B \equiv C$ . The result is obviously true for the degenerate case.

Taking the next case, we can directly find out that $x = D$ and $y = C$ . Here, we can conclude easily. Note that
$\begin{align*} \measuredangle XAD &= \measuredangle XAE - \measuredangle DAE \\ &= \measuredangle ADC - (90^\circ - \measuredangle EDA) \\ &= \measuredangle ADC + \measuredangle CAD + 90^\circ \\ &= 90^\circ - \measuredangle DCA = \measuredangle OAD \end{align*}$ which shows $O-X-A$ are collinear as desired. $\blacksquare$

This post has been edited 1 time. Last edited by SatisfiedMagma, Dec 17, 2024, 6:51 AM
Reason: smol mistake

Z K Y

The post below has been deleted. Click to close.

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

HamstPan38825

8857 posts

HamstPan38825

#38 h Dec 7, 2024, 1:20 AM

Y by

This is a hard G1 but a satisfying one.

Let $F =\overline{BC} \cap \overline{DE}$ . Then $P$ is the center of $(ADFE)$ , i.e. it is the midpoint of $\overline{AF}$ , so $ACFD$ is a parallelogram. Let $M$ be the midpoint of $\overline{BE}$ and $G = \overline{BE} \cap \overline{CD}$ .

Claim: $AMPG$ is cyclic.

Proof: Clearly $\angle PMG = 90^\circ$ . Now let $H = \overline{AC} \cap (ADFE)$ . By Pascal on $AAHEBF$ , $\overline{AA} \cap \overline{BE}$ lies on $\overline{CP}$ , hence $\angle GAP = 90^\circ$ , which proves the result. $\blacksquare$

Then $\begin{align*} \measuredangle OAP &= \measuredangle OAD + \measuredangle DAP - 90^\circ - \measuredangle DCA + \measuredangle DAP \\ &= -\measuredangle\left(\overline{FD}, \overline{BE}\right) - \measuredangle\left(\overline{DC}, \overline{CA}\right) \\ &= -\measuredangle\left(\overline{DC}, \overline{BE}\right) = \measuredangle MGP = \measuredangle MAP. \end{align*}$ So $M$ lies on $\overline{OA}$ .

Z K Y

The post below has been deleted. Click to close.

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

shanelin-sigma

148 posts

shanelin-sigma

#39 h Dec 27, 2024, 8:23 AM

Y by

popop614 wrote:

Let $ABCDE$ be a convex pentagon such that $\angle ABC = \angle AED = 90^\circ$ . Suppose that the midpoint of $CD$ is the circumcenter of triangle $ABE$ . Let $O$ be the circumcenter of triangle $ACD$ .

Prove that line $AO$ passes through the midpoint of segment $BE$ .

Let $M,N$ be the midpoint of $\overline{CD},\overline{BE}$ ,respectively .
let $P$ be a point on the plane such that $\measuredangle PBA=\measuredangle ABE$ and $\measuredangle PEA=\measuredangle EAB$
Then $A$ is the incenter of $\triangle PBE$ , $M$ is the midpoint of arc $BE$ , and $\odot (ABE)$ is the chicken-feet circle.
then $AM,\odot(ABE),BC,DE$ meet at $P$ -excenter (Denote by $Q$ )
Since $\overline{AM}=\overline{MQ}$ and $\overline{CM}=\overline{MD}$ we know $ACQD$ is a parallelogram.( $\therefore AD\perp AB$ and $AC\perp AE$ )
$\therefore\measuredangle ACB=\measuredangle DQC=\measuredangle EDA$ , let this value be $\theta$ , so
$\begin{cases} \frac{AB}{AE}=\frac{AC\sin\theta}{AD\sin\theta}=\frac{AC}{AD}=\frac{AC}{CQ}\\ \measuredangle ACQ\overset{ACQD}{=}\measuredangle DQC=\measuredangle EQB\overset{\odot{ABCE}}{=}\measuredangle EAB \end{cases}$
$\implies\triangle EAB\overset{-}{\sim}\triangle QCA$ and since $M,N$ are the midpoints of $AQ,BE$ so $\triangle EAB\cup N\overset{-}{\sim}\triangle QCA\cup M$
$\implies \measuredangle BAN=-\measuredangle ACM=\measuredangle MCA=\measuredangle BCA+\measuredangle MCQ=(90^{\circ}-\measuredangle CAB)+\measuredangle MDA$
$\overset{90^{\circ}-\measuredangle ADM=\measuredangle CAO}{=}\measuredangle BAC+\measuredangle CAO=\measuredangle BAO$
$\implies A-N-O$

Attachments:

This post has been edited 1 time. Last edited by shanelin-sigma, Dec 27, 2024, 8:23 AM

Z K Y

The post below has been deleted. Click to close.

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

maths_enthusiast_0001

119 posts

maths_enthusiast_0001

#41 h Jan 13, 2025, 11:07 PM

Y by

Claim: Line $\color{blue}{AO}$ passes through the midpoint of segment $\color{blue}{BE}$ .
Proof: Define $N$ as the midpoint of $\overline{BE}$ and $X= BC \cap DE$ then clearly $A,B,X,E$ lie on a circle with diameter $\overline{AX}$ . Also $A,M,X$ are collinear where $M$ is the midpoint of $CD$ as well as the center of $(ABE)$ . Now toss the figure on the complex plane with $(ABE)$ as the unit circle. Let $a,b,c,d,e,o,x,m,n$ denote the complex numbers representing $A,B,C,D,E,O,X,M,N$ respectively.
Then, $\boxed{m=0}$ ; $\boxed{n=\frac{b+e}{2}}$ ; $\boxed{x=-a}$ and $\boxed{d=-c}$ . Now we have, $\boxed{c+(-ab)\overline{c}=-a+b}$ since $B-C-X$ and $\boxed{-c+(-ae)(-\overline{c})=-a+e}$ since $E-D-X$ .
On adding both the equations we get, $\boxed{\overline{c}=\frac{b+e-2a}{a(e-b)}}$ . On conjugating we get, $\boxed{c=\frac{a(b+e)-2be}{(b-e)}}$ .
Now on applying the circumcenter formula one can get, $\boxed{o=\frac{ac(1-c\overline{c})}{(c-a^{2}\overline{c})}}$ . Now,
$o-a=\frac{ac(1-c\overline{c})}{(c-a^{2}\overline{c})}-a=\frac{a\overline{c}(a^2-c^2)}{(c-a^{2}\overline{c})}$ Note that $c=a+\frac{2e(a-b)}{(b-e)}$ and $c-a^{2}\overline{c}=\frac{2(a-b)(e-a)}{(b-e)}$ . On plugging in and after some obvious calculations we get,
$(o-a)=\frac{2be(2a-(b+e))}{(b-e)^2} \implies \frac{(o-a)}{(2a-(b+e))}=\frac{2be}{(b-e)^2}$ Notice that $\alpha=\frac{2be}{(b-e)^2} \implies \overline{\alpha}=\frac{2/(be)}{(b-e)^2/(be)^2}=\frac{2be}{(b-e)^2}=\alpha$ implying $\alpha$ is real. Thus, $\frac{(o-a)}{(2a-(b+e))}$ is real.
Now, $A,N,O$ are collinear $\iff \frac{o-a}{\frac{b+e}{2}-a} \in \mathbb{R} \iff \frac{(o-a)}{(2a-(b+e))} \in\mathbb{R}$ which is indeed true. $\blacksquare$ ( $\mathcal{QED}$ )

Z K Y

The post below has been deleted. Click to close.

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

peace09

5416 posts

peace09

#42 h Jan 22, 2025, 7:43 PM • 3 Y

Y by dolphinday, centslordm, imagien_bad

https://cdn.artofproblemsolving.com/attachments/2/b/76731d272a32475715908a8cceb074811dded8.png

The circumcenter of $\triangle ABE$ is the midpoint of $A$ and $P:=BC\cap DE$ , by Thales; and so $ACPD$ is a parallelogram. In other words, $AC\parallel DE$ and $AD\parallel BC$ , giving the following problem statement:

Quote:

Reference $\triangle ACD$ with feet $C'$ and $D'$ and orthocentre $H$ . Construct rectangles $ABCC'$ and $AEDD'$ and parallelogram $ABA'E$ . Show that $A'$ and $H$ are isogonal in $\angle A$ i.e. $\angle BAE$ .

But $ACD\sim BAA'\cong EA'A$ by SAS, easy from there. $\blacksquare$

Lesson Learned. Keep it simple, stupid

This post has been edited 1 time. Last edited by peace09, Jan 25, 2025, 5:12 PM

Z K Y

The post below has been deleted. Click to close.

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

ErTeeEs06

32 posts

ErTeeEs06

#43 h Jan 23, 2025, 9:12 PM

Y by

Let $A'$ be the intersection of $BC$ and $DE$ . Then $A'$ is the antipode of $A$ on $(ABE)$ . Let $M$ be the midpoint of $CD$ , then $M$ also is the midpoint of $AA'$ , so $ACA'D$ is a parallelogram. Now $\angle ABE=\angle AA'D$ and $\angle BEA=\angle BA'A=\angle A'AD$ , so $\triangle ABE\sim \triangle DA'A$ . Let $N$ be the midpoint of $BE$ , then from the previous similarity we get that $\triangle ANE\sim \triangle DMA$ . Now we angle chase to finish.
$\begin{align*} \angle EAN& =\angle ADM \\ & =\angle DCA' \\ & =\angle A'CA-\angle ACD \\ & =\angle A'DA-\angle ACD \\ & =180^\circ-\angle ADE-\angle ACD \\ & =\angle EAD+\angle DAO \\ & =\angle EAO \end{align*}$ This implies that $A, N, O$ are collinear which finishes the proof.

Z K Y

The post below has been deleted. Click to close.

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

SomeonecoolLovesMaths

3146 posts

SomeonecoolLovesMaths

#44 h Jan 28, 2025, 6:59 PM

Y by

Helped by SilverBlaze_SY and Fibonacci_math.

Let $BC \cap ED = R$ .
Claim: $ABRE$ is cyclic.
Proof: As $\angle ABR + \angle AER = 180^{\circ}$ thus we are done. $\blacksquare$

Claim: $ACRD$ is a parallelogram.
Proof: As $R$ is the antipode of $A$ and $CM = MD$ , thus our claim is true. $\blacksquare$

Let $S$ be $AR \cap (ACD)$ .

Claim: $\triangle ABE \sim \triangle SDE$ .
Proof: $\angle SDC = \angle SAC$ (by angles in the same segment)
$= \angle SRD$ (by alternate angles)
$= \angle ABE$ . Hence we go the same for $\angle AEB$ and by $A-A$ criterion we are done!

Let $Q$ be the midpoint of $BE$ .
Claim: $\angle EAQ = \angle CSG$ .
Proof: From c.p.c.t by similar triangles.

Claim: $\angle EAO = \angle CSG$ .
Proof: $\angle EAO = \angle EAD + \angle DAO = \angle EAD + \angle ODA$
$\angle ACG = 90 - \angle DAO$
Say $\angle DAO = x$
Thus, $\angle CSA = 180^{\circ} - (\angle SCA + \angle CAS) = \angle BAE + x - 90^{\circ}$ .
$\angle DAB = \angle ABC = 90^{\circ}$ due to parallel lines.
$\angle CSA = \angle BAE - \angle DAB + x = \angle EAO$ .

As both of the angles lie on the same side and are equal, hence $A,Q,O$ are collinear.

This post has been edited 1 time. Last edited by SomeonecoolLovesMaths, Jan 28, 2025, 7:03 PM

Z K Y

The post below has been deleted. Click to close.

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

blueprimes

306 posts

blueprimes

#45 h Feb 15, 2025, 12:41 PM

Y by

Assume all angles directed. Denote $X = BC \cap ED$ , since $\angle BAE + \angle BXE = 90^\circ + 90^\circ = 180^\circ$ we have $ABXE$ is cyclic. Clearly $M$ is the center of the circle since the perpendicular bisectors of chords $AB$ and $AE$ of $(ABXE)$ intersect at $M$ , so $ACXD$ is a parallelogram.

$\textbf{Claim: } \triangle ACX \sim \triangle BAE$
We have $\angle CAX = \angle AXE = \angle ABE$ and $\angle CXA = \angle AEB$ so AA Similarity finishes.

Let $N$ be the midpoint of $\overline{BE}$ , the above claim implies $\angle EAN = \angle XCM = \angle XCD$ . Now $\angle CAO = 90^\circ - \angle ADC = 90^\circ - \angle XCD$ but
$\begin{align*} \angle CAN &= \angle BAE - \angle BAC - \angle EAN \\ &= (180^\circ - \angle CXD) - (90^\circ - \angle BCA) - \angle XCD \\ &= 90^\circ - \angle CXD + \angle CXD - \angle XCD \\ &= 90^\circ - \angle XCD \end{align*}$ thus $\angle CAO = \angle CAN$ , implying $A, N, O$ are collinear. We are done.

Z K Y

The post below has been deleted. Click to close.

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

Ilikeminecraft

302 posts

Ilikeminecraft

#46 h Mar 10, 2025, 7:51 PM • 1 Y

Y by Tsikaloudakis

my solution is pretty bad lmao
Claim: $AD\perp AB$
Proof: Let $K$ be the foot from $M$ to $AB.$ Clearly, $K$ is the midpoint of $AB.$
Let $L = AB\cap CD.$ Hence, there is a homothety mapping $BC\iff MK$ centered at $L,$ as $BC\parallel MK.$ Furthermore, $M, K$ are the midpoints of $AB, CD.$ Thus, there is a homothety centered at $L$ mapping $BC\iff AD.$ Thus, $AD\perp AB.$

Note that $\angle ACB = \angle CAD = \angle ADE.$ Let $T$ be the pole of $BE$ with respect to $(ABE).$
Claim: $A, T, F$ are collinear.
Proof: Note that $BFME$ is cyclic since $\angle BFE = \angle BFA +\angle AFE = 2\angle CAD$ while $\angle BME = \angle BMA + \angle AME = 360 - 2\angle BAE = 2\angle CAD.$ However, note that if $F’$ is the foot from $T$ to $CD,$ we have $BF’MET$ is cyclic. Thus, $F = F’.$

Thus, $AF$ is the $A$ -symmedian in $ABE.$ However, $AF, AO$ have to be isogonal with respect to $\angle BAE,$ as they are isogonal with respect to $\angle CAD.$ However, since $AF$ is symmedian, $AO$ is median, which finishes.

Z K Y

The post below has been deleted. Click to close.

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

Tsikaloudakis

1024 posts

Tsikaloudakis

#47 h Mar 11, 2025, 8:39 AM

Y by

αποθήκευση

Z K Y

The post below has been deleted. Click to close.

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

hgomamogh

35 posts

hgomamogh

#48 h Mar 18, 2025, 2:43 AM

Y by

This was a lot harder than I expected.

Let $F$ be the foot of the perpendicular from $C$ , $G$ the foot from $D$ , $M$ the midpoint of $CD$ , $H$ the midpoint of $FG$ , and $A'$ the reflection of $A$ about $M$ .

Because $AA'$ is a diameter of $(ABE)$ , we conclude that $\angle ABA' = 90^{\circ}$ . Hence, it follows that $A'$ , $B$ , and $C$ are collinear. Because $AC \parallel A'D$ , $\angle ACB = \angle CA'D = \angle CAD$ and therefore $\angle CAB = 90 - \angle CAD$ . Hence, $\angle DAB = 90^{\circ}$ , and therefore $AB \parallel FC$ . It easily follows that $ABCF$ is a rectangle. Similarly, $AEDG$ is a rectangle.

Additionally, we claim that $MH$ is parallel to $AO$ . This follows because $MF = MG$ , and hence both segments are perpendicular to $FG$ .

We now toss the problem onto the complex plane. Observe that $b = a + c - f$ and $e = a + d - g$ . Hence, the midpoint of $BE$ is equal to $\begin{align*} a + \frac{c + d}{2} - \frac{f + g}{2} = a + m - h. \end{align*}$
Furthermore, $MH$ is parallel to $AO$ , and hence the vector $MH$ is parallel to the vector $AO$ . It follows that the complex number $m - h$ can be written as $ta$ for some scalar $t$ . Therefore, the midpoint of $BE$ is $a(1 + t)$ , which clearly lies on line $AO$ , and we are done.

Z K Y