CentroAmerican 2015 #3

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CentroAmerican 2015 #3

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Source: 2015 CentroAmerican Math Olympiad #3

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#1 h Jun 27, 2015, 9:18 PM • 3 Y

Y by Math_CYCR, Adventure10, Mango247

Let $ABCD$ be a cyclic quadrilateral with $AB<CD$ , and let $P$ be the point of intersection of the lines $AD$ and $BC$ .The circumcircle of the triangle $PCD$ intersects the line $AB$ at the points $Q$ and $R$ . Let $S$ and $T$ be the points where the tangents from $P$ to the circumcircle of $ABCD$ touch that circle.

(a) Prove that $PQ=PR$ .

(b) Prove that $QRST$ is a cyclic quadrilateral.

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#2 h Jun 27, 2015, 9:50 PM • 2 Y

Y by CaptainFlint, Adventure10

a) Proof: Trivial, Angle Chasing

b) Proof: Easy to see: $PA*PD=PT^2=PS^2$

Since $\bigtriangleup$ $PQA$ ~ $\bigtriangleup$ $PDQ$ , $PQ^2=PA*PD$
Since $\bigtriangleup$ $PRA$ ~ $\bigtriangleup$ $PDR$ , $PR^2=PA*PD$

Now: $PA*PD=PQ^2=PR^2=PS^2=PT^2$

Then: $PQ=PR=PS=PT$

Now take the circumference with center $P$ and radius $PQ$ . Since here is easy to see that $Q$ , $R$ , $S$ and $T$ are concyclic.

$Q.E.D.$

This post has been edited 1 time. Last edited by Math_CYCR, Jul 20, 2015, 12:25 AM
Reason: I am Learning to use Latex

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#3 h Jun 27, 2015, 9:52 PM • 2 Y

Y by Adventure10, Mango247

Darn, sniped. Oh well; these solutions at least are non-isomorphic.

Solution

This post has been edited 3 times. Last edited by djmathman, Jun 27, 2015, 9:54 PM

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#4 h Sep 10, 2015, 8:44 PM • 2 Y

Y by Adventure10, Mango247

a.) Note that if $O$ is the circumcentre of $\triangle PCD$ then $PO$ is perpendicular to $AB$ and so $PQ=PR$ .
b.) Let $X$ be the intersection point of $AB,CD$
Then by Brokard's theorem on self polarity of the diagonal triangle in a complete quadrilateral we get that

$ST,AB,CD$ concur at $T$ .
By Power of a point,

$XQ.XR=XD.XC=XS.XT$ . so $T,S,Q,R$ are concyclic.(with circumcentre $P$ .)

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#5 h Jun 11, 2016, 9:11 PM • 5 Y

Y by isavl, samuraivader, AlastorMoody, Adventure10, Mango247

Let $\omega$ be the circumcircle of $PCD$
Lets consider inversion ( $\Psi$ ) with center $P$ and radius $PS$ .

By power of point we get that $PA*PD=PB*PC=PS^2$
So $\Psi(D)=A$ and $\Psi(C)=B$
Since $C$ and $D$ lie on $\omega \Rightarrow \Psi(\omega)=AB$

But since $Q$ and $R$ lie on $AB$ and on $\omega \Rightarrow \Psi(Q)=Q$ and $\Psi(R)=R$
So $Q$ and $R$ lie on the circle with center $P$ and radius $PS$

Then $PS=PT=PR=PQ$

$Q.E.D$

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#6 h Nov 29, 2018, 5:12 PM • 2 Y

Y by AlastorMoody, Adventure10

anantmudgal09 wrote:

b.) Let $X$ be the intersection point of $AB,CD$
Then by Brokard's theorem on self polarity of the diagonal triangle in a complete quadrilateral we get that

$ST,AB,CD$ concur at $T$ .

What if $AB||CD$ , and there's no intersection of these lines?

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#7 h Dec 24, 2022, 5:55 AM • 1 Y

Y by Mango247

Part a) Notice that $\angle PBR =\angle ABC = 180^{\circ} - \angle ADC = 180^{\circ} - \angle PDC = \angle PRC$ , so by case $AA$ , $\triangle PBR \sim \triangle PRC$ , which implies that $\angle PRB = \angle PCR$ , so $PR$ is tangent to $\odot(\triangle RBC)$ . Analogously $PQ$ is tangent to $\odot(\triangle QAD)$ .
It turns out that
$PQ^{2} = PA \cdot PD$ y $PR^{2} = PB \cdot PC$
In addition, by Pop at $P$ wrt $\odot(ABCD)$ ,
$PA \cdot PD = PB \cdot PC \Rightarrow PQ^{2} = PR^{2} \Rightarrow PQ = PR$ $\blacksquare$

Part b) Again by PoP at $P$ wrt $\odot(ABCD)$ ,
$PS^{2} = PT^{2} = PA \cdot PD = PQ^{2} = PR^{2}$
Therefore, $PQ = PR = PS = PT$ , then $P$ is the center of $\odot(QRST)$ and the conclusion follows $\blacksquare$

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