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Source: All Russian 2014 Grade 9 Day 2 P2

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mathuz

#1 h May 3, 2014, 2:53 PM • 2 Y

Y by Adventure10, Mango247

Let $ABCD$ be a trapezoid with $AB\parallel CD$ and $\Omega$ is a circle passing through $A,B,C,D$ . Let $\omega$ be the circle passing through $C,D$ and intersecting with $CA,CB$ at $A_1$ , $B_1$ respectively. $A_2$ and $B_2$ are the points symmetric to $A_1$ and $B_1$ respectively, with respect to the midpoints of $CA$ and $CB$ . Prove that the points $A,B,A_2,B_2$ are concyclic.

I. Bogdanov

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4148 posts

#2 h May 8, 2014, 10:32 PM • 3 Y

Y by nima1376, Adventure10, Mango247

mathuz wrote:

Let $ABCD$ is trapezoid with $AB\parallel CD,$ $\Omega$ is circle passes through $A,B,C,D.$ $\omega$ is circle passes through $C,D$ and intersects with $CA,CB$ at ${\color{red}A_1,B_1}$ respectively. $A_2$ and $B_2$ are the points symmetric to $A_1$ and $B_1$ respectively, with respect to the midpoints of $CA$ and $CB.$ Prove that the points $A,B,A_2,B_2$ are concyclic.

Typo corrected in red color. This is proved in the solution of the problem All Russian-2014, Grade 11, day 2, P2.

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mathuz

#3 h May 16, 2014, 8:08 PM • 2 Y

Y by Adventure10, Mango247

you are right!
Thank you Luis.

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#4 h May 18, 2014, 6:29 AM • 2 Y

Y by Adventure10, ehuseyinyigit

$D$ is a center of spiral similar which goes $BB_{1}$ to $AA_{1}$ $\Rightarrow \frac{AA_{1}}{BB_{1}}=\frac{AD}{BD}=\frac{BC}{CA}$
$AA_{1}.AC=BB_{1}.BC\Rightarrow CA_{2}.AC=CB_{2}.BC$
so $A_{2}B_{2}BA$ is cycle.
done

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#5 h May 19, 2014, 11:02 AM • 3 Y

Y by Mathematicalx, Adventure10, Mango247

$ABCD$ is cyclic in $\Omega$ . So, $\measuredangle BAC = \measuredangle DCA \Rightarrow BC = AD$
Similarly $BD = AC$ .
Now let $\left\{D,D'\right\} = AD \cap \omega$ . And by symmetry, $AD' = BB_1$
Now $A_1CDD'$ is cyclic(in $\omega$ ) and $A_1C \cap DD' = A$ . So(using power of point), $AA_1.AC = AD'.AD$ .
$\therefore CA_2/CB_2=AA_1/BB_1=AA_1/AD'=AD/AC=CB/CA \Rightarrow CA_2.CA=CB_2.CB$
$\therefore A_2, B_2, A, B$ are concyclic.

$[QED]$

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#6 h Sep 22, 2014, 6:50 PM • 1 Y

Y by Adventure10

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#7 h Dec 1, 2014, 1:36 AM • 4 Y

Y by A_Math_Lover, SSaad, Adventure10, Mango247

What a nice illustration of spiral similarity. Though I would have just said "isosceles trapezoid" in the problem statement.

$[asy] size(8cm); defaultpen(fontsize(9pt)); pair A = dir(110); pair B = dir(70); pair C = dir(-40); pair D = dir(220); pair B_1 = 0.3*C+0.7*B; pair O = circumcenter(B_1, C, D); pair A_1 = -C+2*foot(O, C, A); pair A_2 = A+C-A_1; pair B_2 = B+C-B_1; draw(CP(O, D), blue); draw(unitcircle, blue); draw(A--B--C--D--A--C); draw(circumcircle(A, B, A_2), red); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$B_1$", B_1, dir(B_1)); dot("$A_1$", A_1, dir(60)); dot("$A_2$", A_2, dir(-15)); dot("$B_2$", B_2, dir(B_2)); [/asy]$

We have $\triangle DAA_1 \sim \triangle DBB_1$ , but $DA = CB$ and $DB = AC$ . So $AA_1 \cdot AC = BB_1 \cdot BC$ , implying that $CA_2 \cdot CA = CB_2 \cdot CB$ .

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#8 h Jan 1, 2015, 7:48 AM • 1 Y

Y by Adventure10

Let $\omega$ intersect $BD$ in $A'$
Then it is easy to see $AA_1=BA'$
But since $A'B_1CD$ are concyclic,
$BB_1 \cdot BC = BA' \cdot BD$
$\implies BB_1 \cdot CB=AA_1 \cdot CA$
$\implies CB_2 \cdot CB= CA_2 \cdot CA$
QED

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#9 h Mar 22, 2015, 7:55 AM • 2 Y

Y by Adventure10, Mango247

easy!! but still posting!

let $\omega$ intersect $AD$ in $K$
than quite easily $\angle AKB_1=\angle ACD=180-\angle ABB_1$
and hence $ABB_1K$ is isosceles trapezium.
now by POP
we have $AD.AK=AA_1.AC=CA_2.AC$
on other note $AD.AK=BC.AK=BC.BB_1=BC.BB_2$
and hence $BC.BB_2=CA_2.CA$
and hence by POP we have $ABB_2A_2$ is cyclic quad.
thus we are done

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#10 h Oct 23, 2015, 12:46 PM • 2 Y

Y by Adventure10, Mango247

Another solution:

Let the circle $AA_2B$ intersect $AB$ again at $B'$ .

Now, $AB$ is the radical axis of $(ABCD) ; (AA_2B)$ and $CD$ is the radical axis of $(ABCD) ;(DCA_1B_1)$ .

Now, $AB \parallel CD$ and so $AB \parallel CD \parallel l$ where $l$ is the radical axis of $(AA_2B) ; (DCA_1B_1)$

Let $M$ and $N$ be the mid points of $CA,CB$ respectively. It is evident that $MN \parallel l$ and also,

$MA_1.MC=MA_2.MA$

so $M$ lies on $l$ . Therefore, $N$ lies on $l$ too and so by power of a point $B_2 \equiv B'$ thus, the result holds.

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#11 h Jul 10, 2016, 3:35 PM • 1 Y

Y by Adventure10

v_Enhance wrote:

What a nice illustration of spiral similarity. Though I would have just said "isosceles trapezoid" in the problem statement.

$[asy] size(8cm); defaultpen(fontsize(9pt)); pair A = dir(110); pair B = dir(70); pair C = dir(-40); pair D = dir(220); pair B_1 = 0.3*C+0.7*B; pair O = circumcenter(B_1, C, D); pair A_1 = -C+2*foot(O, C, A); pair A_2 = A+C-A_1; pair B_2 = B+C-B_1; draw(CP(O, D), blue); draw(unitcircle, blue); draw(A--B--C--D--A--C); draw(circumcircle(A, B, A_2), red); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$B_1$", B_1, dir(B_1)); dot("$A_1$", A_1, dir(60)); dot("$A_2$", A_2, dir(-15)); dot("$B_2$", B_2, dir(B_2)); [/asy]$

We have $\triangle DAA_1 \sim \triangle DBB_1$ , but $DA = CB$ and $DB = AC$ . So $AA_1 \cdot AC = BB_1 \cdot BC$ , implying that $CA_2 \cdot CA = CB_2 \cdot CB$ .

Can we just use symmetry to prove BB1×CB＝AA1×CA

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#12 h May 1, 2019, 4:58 AM • 2 Y

Y by Adventure10, Mango247

Nice problem for depicting the strength of power of a point

Here is my solution
Claim $AA_1\times AC=BB_1\times BC$
Proof It needs to be proven that $XA^2-R^2=XB^2-R^2$ where X is the center of the circle passing through the points $A_1,B_1,C$ and $D$ and R is its radius.Since a circle passes through this trapezoid, it is isosceles, so we get that the perpendicular bisectors of $AB$ and $CD$ coincide implying $XA=XB$ . Thus the powers of $A$ and $B$ with respect to this circle are equal!

The conclusion follows by just taking the isotomic conjugates.

This post has been edited 1 time. Last edited by mmathss, May 1, 2019, 4:59 AM

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#13 h Mar 6, 2023, 8:22 PM

Y by

Since $A$ and $B$ are symmetric with respect to $\omega$ , by power of a point we have
$CA_2 \cdot CA=AA_1 \cdot AC=\mathrm{Pow}_\omega (A)=\mathrm{Pow}_\omega (B)=BB_1\cdot BC=CB_2 \cdot CB,$ hence $ABA_2B_2$ is cyclic. $\blacksquare$

This post has been edited 1 time. Last edited by IAmTheHazard, Mar 6, 2023, 8:23 PM

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#14 h Jan 5, 2025, 8:25 PM

Y by

What we have:
$AA_1=A_2C$ , $BB_1=CB_2$ and an isosceles trapezoid.

Simple angle chase gives $\triangle DAB\sim \triangle DA_1B_1$ and $D$ is sipiral similarity center which goes $BB_1$ to $AA_1$ . Thus $AA_1\cdot DB=BB_1\cdot DA$ . Since $AA_1\dot AC=CA_2\cdot AC=BC\cdot CB_2$ . Hence $ABA_2B_2$ is cyclic.

This post has been edited 1 time. Last edited by ehuseyinyigit, Jan 7, 2025, 6:37 PM

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