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SCP

#1 h Aug 24, 2014, 3:24 PM • 2 Y

Y by Adventure10, Mango247

In the figure of http://www.artofproblemsolving.com/Forum/download/file.php?id=50643&mode=view
$\odot O_1$ and $\odot O_2$ intersect at two points $A$ , $B$ .
The extension of $O_1A$ meets $\odot O_2$ at $C$ , and the extension of $O_2A$ meets $\odot O_1$ at $D$ ,
and through $B$ draw $BE \parallel O_2A$ intersecting $\odot O_1$ again at $E$ .
If $DE \parallel O_1A$ , prove that $DC \perp CO_2$ .

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pi37

#2 h Aug 24, 2014, 3:59 PM • 2 Y

Y by Adventure10, Mango247

The existence of $E$ is equivalent to $\angle DAB=\angle DAC$ , so $AO_2$ bisects $\angle BAC$ . Thus $AC=AB$ , since this bisector passes through the circumcenter of $\triangle ABC$ . If we invert about $A$ , (marking points in the newly inverted diagram with $'$ s) we get that $D'A'$ is the perpendicular bisector of $B'C'$ , and $C'A'\perp B'D'$ , so $A'$ is the orthocenter of isosceles $B'D'C'$ . So additionally $D'$ is the orthocenter of $A'B'C'$ , which means if we invert about $D'$ , in the newest diagram $D"$ (we mark points in the twice inverted diagram with $"$ s) becomes the excenter of $B"A"C"$ . So since $B"A"C"$ is isosceles, the midpoint $X$ of minor arc $B"C"$ is the center of $(D"B"C"")$ , but then $XB"\perp B"A"$ and $XC"\perp C"A"$ . Thus $A"B"$ is tangent to $(D"B"C")$ , and $A"$ is be the intersection of the tangents from $B",C"$ to $(D"B"C")$ . Thus inverting back through $D"$ , $A'$ is the intersection of the circles through $D'$ tangent to $B'C'$ through $B',C'$ , and then inverting back into the first diagram through $A'$ , $DC$ is tangent to $(ABC)$ .

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Mikasa

#3 h Aug 25, 2014, 6:03 PM • 1 Y

Y by Adventure10

Note that $\angle AO_2O_1=\angle BO_2O_1=90^{\circ}-\angle BED$ . Again, $\angle ABC=\dfrac{1}{2}\angle AO_2C=90^{\circ}-\angle O_2AC=90^{\circ}-\angle O_1AD=90^{\circ}-\angle BED$ since $AO_1\parallel DE$ and $AD\parallel BE$ .
So $\angle ABC=\angle ACB\Rightarrow AC=AB$ . Thus $\angle AO_2C=\angle AO_2B$ . This means $\angle BAO_2=\angle CAO_2=\angle O_1AD=\angle O_1DA=\angle O_1DO_2\Rightarrow DO_1\parallel AB$ . But $AB\perp O_1O_2$ so $DO_1\perp O_1O_2$ . Now $\angle O_2CA=\angle O_2AC=\angle O_1AD=\angle O_1DA$ i.e $\angle O_1CO_2=\angle O_1DO_2$ which means $O_1DCO_2$ is cyclic. So, $\angle O_2CD=180^{\circ}-\angle O_2O_1D=180^{\circ}-90^{\circ}=90^{\circ}$ since $DO_1\perp O_1O_2$ . Thus $DC\perp CO_2$ as desired.

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#4 h May 11, 2015, 1:06 AM • 2 Y

Y by Adventure10, Mango247

the figure is not available, see here

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#5 h Jul 8, 2017, 7:46 PM • 2 Y

Y by Adventure10, Mango247

My solution for this nice problem:
We must to prove that $\angle DAC=2\angle AO_2C.$ We know $ABED$ and $EDCA$ are isoceles trapezoid and parallelogram, respectively.Then $AB=DE=AC.$ Then we can find easily $\angle O_2AB=\angle O_2CA=\angle O_2AC=\angle DAO_1=\angle ADO_1,$ then $O_1DCO_2$ is cyclic.Also we know $\angle DCA=\angle DEA=2\angle DO_1C=2\angle DO_2A=2\angle AO_2C.$ As desired.

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#6 h Jan 29, 2022, 8:05 AM

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First Note that ABED is isosceles trapezoid. Let DO1 meet $\odot O_1$ at S. ∠O2CA = ∠O2AC = ∠O1AD so we need to prove
∠O1AS = ∠DCA. ∠O1AS = ∠ASO1 = ∠AED so we need to prove DCAE is parallelogram. DE || AC and ∠CAO2 = ∠O1AD = ∠ABE = ∠BAO2 so CA = BA = DE so DCAE is parallelogram.
we're Done.

This post has been edited 1 time. Last edited by Mahdi_Mashayekhi, Jan 29, 2022, 8:06 AM

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#7 h May 4, 2024, 6:39 PM

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As shown in the figure, $\odot O_1$ and $\odot O_2$ intersect at two points $A$ , $B$ . The extension of $O_1A$ meets $\odot O_2$ at $C$ , and the extension of $O_2A$ meets $\odot O_1$ at $D$ , and through $B$ draw $BE \parallel O_2A$ intersecting $\odot O_1$ again at $E$ . If $DE \parallel O_1A$ , prove that $DC \perp CO_2$ .

https://cdn.artofproblemsolving.com/attachments/b/1/23ba65b40b9a35f17a91c7221eab724eddc83c.jpg

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