square geometry bisect $\angle ESB$

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square geometry bisect $\angle ESB$

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Source: BMO SL 2019, G1

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#1 h Nov 8, 2020, 12:52 AM

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Let $ABCD$ be a square of center $O$ and let $M$ be the symmetric of the point $B$ with respect to point $A$ . Let $E$ be the intersection of $CM$ and $BD$ , and let $S$ be the intersection of $MO$ and $AE$ . Show that $SO$ is the angle bisector of $\angle ESB$ .

This post has been edited 2 times. Last edited by GorgonMathDota, Nov 8, 2020, 1:10 AM

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#2 h Nov 8, 2020, 1:36 AM • 2 Y

Y by amar_04, Mathlover_1

Seems approachable by Coordinates
Here we go

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#3 h Nov 9, 2020, 5:42 PM

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Properties of this configuration :
$ASOB$ is cyclic, as well as $MDSA$
$AO$ is tangent to $(MAS)$ and $(DAS)$
Triangles $ASO$ and $ASD$ are similar, as well as $MAO$ and $MAC$

This post has been edited 1 time. Last edited by VicKmath7, Nov 9, 2020, 5:46 PM

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#4 h Nov 9, 2020, 7:37 PM • 1 Y

Y by mkomisarova

Assume that $MC \cap AD = X$ . It is easy to see that $AX$ is midline in the $\triangle MBC$ , therefore $X$ is midpoint of $AD$ and $OX \parallel MB$ .

Claim: Points $B,S,X$ are collinear.
Proof: Assume that $BS \cap MC = X'$ . By Ceva's theorem in $\triangle MBE$ , we have that $\frac{OE}{BO} =\frac{EX'}{X'M}$ . This forces to have that $OX' \parallel MB$ , which implies that $X = X'$

Let $Y$ be midpoint of $CD$ . It is easy to see that $G$ is centroid of $\triangle ACD$ , therefore points $A,S,E,Y$ are collinear. Observe that $\triangle BAX = \triangle DAY$ . Simple angle chasing reveals that $\angle BSE = 90^{\circ}$ . Since we have that $\angle BOA = \angle BSA =90^{\circ}$ , then quadrilateral $ABOS$ is cyclic and consequently we have that $\angle BAO =\angle BSO = \angle OSE =45^{\circ}$ . This proves that $SO$ is angle bisector of $\angle BSE$ as desired.

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#5 h Feb 25, 2022, 6:04 PM

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Let $BS$ meet $MC$ at $N$ .
Claim1 : $N$ is intersection of $AD$ and $MC$ .
Proof : By Ceva Theorem we have $\frac {MA}{AB} . \frac {BO}{OE} . \frac {EN}{NM} = 1$ so $\frac {BO}{OE} = \frac {NM}{EN}$ so $ON || MB$ .

Now we have $\angle EAD = \angle ECD = \angle NMA = \angle NBA$ so we have $\angle BSA = \angle 90 = \angle BOA$ so $ASOB$ is cyclic and we have $OA = OB$ so $\angle ASM = \angle BSO$ so $\angle ESO = \angle OSB$ .
we're Done.

This post has been edited 1 time. Last edited by Mahdi_Mashayekhi, Feb 25, 2022, 6:05 PM

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Iora

#6 h Aug 29, 2022, 7:43 PM

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Barycentric coordinates: $A(1,0,0),B(0,1,0),C(0,0,1),O(1:0:1),M(2,-1,0),$ where $a=c,a^2+c^2=b^2$
Calculate trivial determinants: $E(2:-1:2)$ , $S(4:-1:2)$ .
One method is calculating the ratio of sides $\frac{SE}{SB}=\frac{EO}{OB}$ . Or, using circle equation formula, $(AOB)=(ABS)\Rightarrow$ $AOSB$ is cyclic. By cyclic, $90=\angle BOA= \angle BSA \Rightarrow \angle BSE=90$ and $45=\angle BAC= \angle BSO \rightarrow \angle BSO=45$ .

Since $\angle BSE-\angle BSO=45=\angle BSO$ , we are done $\blacksquare$

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jayme

#7 h Aug 30, 2022, 5:45 AM

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Dear Mathlinkers,

https://artofproblemsolving.com/community/c6t48f6h2334497_square_geometry_bisect_angle_esb

Sincerely
Jean-Louis

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#8 h Jan 28, 2025, 12:58 PM

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λυση αργοτερα

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#9 h Jan 28, 2025, 4:31 PM

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see the figure

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#10 h Jan 28, 2025, 5:47 PM

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Let one side of the square, ie. $AB=6x$ . Quick calculation gives $MS=\dfrac{12x\sqrt{10}}{5}$ and $MO=3x\sqrt{10}$ . Hence,
$MA\cdot MB=72x^2=MS\cdot MO$ which implies $ASOB$ is cyclic, $AS\perp BS$ . And $\angle OSB=45^{\circ}$ giving $\angle ESO=\angle BSO$ as desired.

This post has been edited 1 time. Last edited by ehuseyinyigit, Jan 28, 2025, 5:47 PM

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#11 h Jan 28, 2025, 6:06 PM • 1 Y

Y by MihaiT

VicKmath7 wrote:

Properties of this configuration :
$ASOB$ is cyclic, as well as $MDSA$
$AO$ is tangent to $(MAS)$ and $(DAS)$
Triangles $ASO$ and $ASD$ are similar, as well as $MAO$ and $MAC$

It has been proved above that $ASOB$ cyclic.

Claim: MASD is cyclic.

PROOF. We know $MS=12x\sqrt{10}/5$ and $AP=2x$ . Thus $MP=2x\sqrt{10}$ and $PS=2x\sqrt{10}/5$ . Hence $MP\cdot PS=8x^2=AP\cdot PF$ holds, implying $MASD$ is cyclic.

Claim: $AO$ is tangent to $(MAS)$ and $(DAS)$ .

PROOF. Let $AO\cap SB=K$ . Then
$\angle AMS=45^{\circ}-\angle AOM=90^{\circ}-\angle AKS=\angle SAO$ proving our claim. $\angle AMO=\angle ADO$ shows the second tangency as well.

Claim: $ASD\sim OSA$ .

PROOF. We showed in tangency that $\angle ADS=\angle SAO$ . It sufficies to show $\angle DAS=\angle AOS$ . On the other hand,
$\angle DAS=45^{\circ}-\angle SAO=45^{\circ}+\angle AOS-45^{\circ}=\angle AOS$ giving the similarity between triangles.

Claim: $MAO\sim CAM$

PROOF. $MA^2=AO\cdot AC$ implies $(MOC)$ is tangent to $MA$ at $M$ giving the similarity.

This post has been edited 1 time. Last edited by ehuseyinyigit, Jan 28, 2025, 6:07 PM

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#12 h Yesterday at 8:03 AM • 1 Y

Y by MihaiT

Here's a different approach.

Claim: $MC$ is a symmedian in $\triangle BMD$ .
Proof: $(BMD)$ is a circle with center $A$ , therefore $CB$ and $CD$ are tangents.

Claim: $MASD$ is cyclic.
Proof: using the previous claim, $\angle{SAD} = \angle{EAD} = \angle{ECD} = \angle{MCD} = \pi/4 - \angle{CMD} = \pi/4 - \angle{BMD} = \angle{SMD}$ . Therefore, $\angle{OSD} = \pi/2$ .

Claim: $(B, E; O, D) = -1$ .
Proof: $(B, E; O, D) \stackrel{C}{=} (B, M, A, P_{\infty}) = -1$ .

The last two claims imply the desired result.

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#13 h Yesterday at 9:49 AM

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Coordinate bashing works.

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