APRQ cyclic implies QRS collinear

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APRQ cyclic implies QRS collinear

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Source: ELMO 2009, Problem 2

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#1 h Dec 31, 2012, 12:35 PM • 2 Y

Y by son7, Adventure10

Let $ABC$ be a triangle such that $AB < AC$ . Let $P$ lie on a line through $A$ parallel to line $BC$ such that $C$ and $P$ are on the same side of line $AB$ . Let $M$ be the midpoint of segment $BC$ . Define $D$ on segment $BC$ such that $\angle BAD = \angle CAM$ , and define $T$ on the extension of ray $CB$ beyond $B$ so that $\angle BAT = \angle CAP$ . Given that lines $PC$ and $AD$ intersect at $Q$ , that lines $PD$ and $AB$ intersect at $R$ , and that $S$ is the midpoint of segment $DT$ , prove that if $A$ , $P$ , $Q$ , and $R$ lie on a circle, then $Q$ , $R$ , and $S$ are collinear.

David Rush

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#2 h Jan 2, 2013, 1:05 PM • 2 Y

Y by Adventure10, Mango247

Very interesting problem! The concyclicity of points $A,P,Q,R$ is not necessary and I do not use it below:

Let $R \equiv QS \cap PD$ . It is equivalent to prove that $A,B,R$ are collinear. First of all, it is obvious that $AD$ is a symmedian and that $AT$ is tangent to the circumcircle $(\omega)$ of $\triangle ABC$ . Let $J$ be the intersection point of the tangents to $(\omega)$ at $B,C$ and $F \equiv AJ \cap (\omega)$ . From La Hire's Theorem, $AF$ is the polar of $T$ so $(C,B;T,D) = -1$

Thus, $\frac{BD}{DC} = \frac{TB}{TC} = \frac{c^2}{b^2}$ . Observe that $\frac{SB}{SD} = \frac{\frac{TB+BD}{2} - BD}{\frac{TD}{2}} = \frac{TB-BD}{TD} = \frac{\frac{c^2}{b^2}TC - \frac{c^2}{b^2}DC}{TD} = \frac{c^2}{b^2}$ .

It follows that $\frac{SB}{SD} = \frac{BD}{DC} \implies \frac{SB}{BD} = \frac{SD}{DC} ~(1)$ . Applying Menelaus Theorem at $\triangle QCS$ we get

$\frac{CP}{PQ}\cdot \frac{SD}{DC} \cdot \frac{QR}{SR} = 1~(2)$ . Furthermore, $AP \parallel DC \implies \frac{CP}{PQ} = \frac{DA}{AQ}~(3)$ . Consequently,

$(1),(2),(3) \implies \frac{DA}{AQ} \cdot \frac{SB}{BD} \cdot \frac{QR}{SR} = 1$ . From reverse Menelaus Theorem at $\triangle SDQ$

it follows that $A,B,R$ are collinear and this completes the proof.

With Best Regards,
Grigoris

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#3 h Apr 16, 2013, 1:02 AM • 2 Y

Y by Adventure10, Mango247

As Grigoris mentioned, $A, P, Q, R$ need not be concyclic for $Q, R, S$ to be collinear.

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We use barycentric coordinates. Let $A = (1, 0, 0), B = (0, 1,0), C = (0, 0, 1)$ , and let $BC = a, CA = b, AB = c$ .

Since $\triangle ATB \sim \triangle CTA$ , we have $AT^2 = CT \cdot BT$ , which implies that $AT$ is tangent to the circumcircle of $\triangle ABC$ . It is well-known that the line tangent to the circumcircle at $A$ has equation $c^2y+b^2z = 0$ . Intersecting this line with $BC$ gives $T = (0:b^2:-c^2)$ . Now $M = \left(0, \tfrac{1}{2}, \tfrac{1}{2}\right)$ , and $D$ is the trace of the symmedian point $K = (a^2:b^2:c^2)$ wrt $A$ , so $D = (0:b^2:c^2)$ . Thus we also have $S = (0:b^4:-c^4)$ .

Denote by $\mathcal{P}_{\infty}$ the point of infinity of $BC$ . Then $\mathcal{P}_{\infty} = (0:-1:1)$ . Since $AP\parallel BC$ , $\mathcal{P}_{\infty}$ lies on $AP$ . Therefore $A\mathcal{P}_{\infty}$ has equation $y+z=0$ , and so we can let $P = (1, p, -p)$ for some $p$ .

Line $PD$ has equation $(-p(b^2+c^2))x+c^2y-b^2z = 0$ . $R = AB \cap PD$ , so intersecting $PD$ with $z = 0$ yields $R = (c^2:p(b^2+c^2):0)$ . In addition, line $PC$ has equation $px-y=0$ . Intersecting $PC$ with $AD: c^2y-b^2z = 0$ yields $Q = (b^2:pb^2:pc^2)$ .

Notice that
$\begin{align*} \det \left| \begin{array}{ccc}b^2 & pb^2 & pc^2 \\c^2 & p(b^2+c^2) & 0 \\0 & b^4 & -c^4 \end{array} \right| &= b^2(p(b^2+c^2)(-c^4))-pb^2(-c^6)+pc^2(c^2b^4) \\&= -pb^2c^4(b^2+c^2)+pb^2c^4(c^2+b^2) \\&=0\end{align*}$
implying that $Q, R, S$ are collinear.

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#4 h Dec 7, 2016, 5:30 PM • 1 Y

Y by Adventure10

We apply the Desargues Involution Theorem to quadrilateral $APRQ$ with respect to its circumcircle (or any conic through the four points) and line $BC$ . The the unique involution on line $BC$ which swaps $B,C$ and sends $D$ to itself also swaps the point at infinity along $BC$ and $QR \cap BC$ .

Since $AT$ is the tangent to $(ABC)$ at $A$ , we also have $(T,D;B,C)=-1$ .

Therefore, the involution we need is the inversion with respect to the circle $(TD),$ since it swaps $B,C$ . Since this involution sends the point at infinity to $S$ , it follows that $Q,R,S$ collinear.

This post has been edited 1 time. Last edited by zacchro, Dec 7, 2016, 5:36 PM

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#5 h Jan 6, 2022, 6:21 PM

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$A,P,Q,R$ I don't think the cyclical condition is necessary

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#6 h Jan 2, 2024, 6:15 PM

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Let $QR\cap BC=S'$

Claim: $B,C,Q,R$ are cyclic.
Proof: $\angle BRQ=180-\angle QPA=180-\angle QCB$

Claim: $S'D$ is tangent to $(DRQ)$ .
Proof: $\angle RDS'=\angle RPA=\angle RQA=\angle RQD$

$S'D^2=S'R.S'Q=S'B.S'C$ Also we know that $(T,D;B,C)=-1$ so $S'$ is the midpoint of $DT$ as desired. $\blacksquare$

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