About bicentric quadrilaterals

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About bicentric quadrilaterals

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perpendicular bisector

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Source: Brazil Math Olympiad 1995

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Johann Peter Dirichlet

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Johann Peter Dirichlet

#1 h Mar 16, 2006, 7:48 PM • 1 Y

Y by Adventure10

$ABCD$ is a quadrilateral with a circumcircle centre $O$ and an inscribed circle centre $I$ . The diagonals intersect at $S$ . Show that if two of $O,I,S$ coincide, then it must be a square.

This post has been edited 1 time. Last edited by Johann Peter Dirichlet, Mar 17, 2006, 4:43 AM

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Johann Peter Dirichlet

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Johann Peter Dirichlet

#2 h Mar 16, 2006, 7:50 PM • 1 Y

Y by Adventure10

A theorem guarrantees that in the general case these three points are in the same straight line.

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#3 h May 3, 2011, 3:29 PM • 2 Y

Y by Adventure10, Mango247

We know that $O$ is on every perpendicular bisector, and $I$ is on evry bisector. We easily get that $\triangle{IAB}$ and the other of the same type are isosceles.Using this we obtain that all of the angles of the quadrilateral are equal, so it is a rectangle, where the bisectors are diagonals, so it is a square.

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#4 h Jun 14, 2011, 5:24 PM • 2 Y

Y by Adventure10, Mango247

Let $O$ and $S$ coincide. Then since $ABCD$ is cyclic with center $O$ , we have $OA=OB=OC=OD$ , which means that the quadrilateral $ABCD$ is a parallelogram. Since it has an inscribed circle then $AB+CD=AD+BC$ , which means that $ABCD$ is a square. Let $S$ and $I$ coincide. Triangles $ABI$ and $CID$ are similar. Having the same altitude $r$ means they are congruent. Thus $ABCD$ is a square. Let $I$ and $O$ coincide. Again triangles $ABO$ and $CDO$ are similar. Having the same altitude $r$ means they are congruent. Hence done!

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#5 h Nov 2, 2018, 4:20 AM • 1 Y

Y by Adventure10

Case 1: Let $O$ and $S$ coincide at Point $X$ .

Proof: Note that $X$ is the circumcenter so $XA=XB=XC=XD$ and let $\angle XAB=\angle XBC=\angle XCD=\angle XDC=a$ and $\angle XBC=\angle XCB=\angle XAD=\angle XDA=b.$ Since $\angle A=a+b=\angle C,$ then cyclic quadrilateral $ABCD$ must have equal opposite angles, hence, $\angle A=\angle C=90^{\circ}$ . Similarly, we can get $\angle B=\angle D=90^{\circ}$ . Since the diagonals bisect each other, we have a square. $\Box$

Case 2: Let $O$ and $I$ coincide at Point $Y.$

Proof: Because $Y$ is the circumcenter, $\angle YAB=\angle YBA$ and $\angle YBC=\angle YCB.$ But, $Y$ is also the center of the inscribed circle, meaning that $Y$ lies on the angle bisectors of angle $A,B, C,D$ . Hence, $\angle YBA=\angle YBC,$ $\angle YAD=\angle YAB,$ and $\angle YCB=\angle YCD.$ Hence, $\angle DAY=\angle YAB=\angle YCB=\angle YCD$ and $ABCD$ is cyclic so $\angle A$ and $\angle C$ are both right angles. Similarly, $\angle B$ and $\angle D$ are right angles. Also, $\angle DAY=\angle YAB=\angle YCB=\angle YCD=45^{\circ}$ so $\angle AYB=\angle BYC=90^{\circ}$ so lines $AYC$ and $BYD$ are collinear and bisecting, which means that it is a square. $\Box$

Case 3: Let $I$ and $S$ coincide at Point $Z.$

Proof: In cyclic quadrilateral $ABCD$ , we get $\angle DAC=\angle DBC.$ But, $\angle DAC=\angle CAB$ and $\angle ABD=\angle CBD$ so $\angle A=\angle B.$ Hence, $\angle A=\angle B=90^{\circ}$ and similarly, we can prove that $\angle C=\angle D=90^{\circ}.$ Also, from equal base angles, we can prove that $AZ=ZB=ZC=ZD$ so $Z$ is also the circumcenter, proving that the diagonals bisect each other and it is a square. $\Box$

Hence, given two of the three points coincide , we can prove that it is a square. $\blacksquare$

This post has been edited 1 time. Last edited by Jzhang21, Nov 2, 2018, 4:21 AM

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