AP ,BQ,CR are concurrent

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AP ,BQ,CR are concurrent

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Source: MEMO 2018 I3

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#1 h Sep 3, 2018, 2:05 AM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be an acute-angled triangle with $AB<AC,$ and let $D$ be the foot of its altitude from $A.$ Let $R$ and $Q$ be the centroids of triangles $ABD$ and $ACD$ , respectively. Let $P$ be a point on the line segment $BC$ such that $P \neq D$ and points $P$ $Q$ $R$ and $D$ are concyclic .Prove that the lines $AP$ $BQ$ and $CR$ are concurrent.

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#2 h Sep 3, 2018, 10:25 AM • 3 Y

Y by hhh07, Adventure10, Mango247

Let $M, A', B', C'$ be midpoints of $AD, BC, CA, AB$ respectively. Then a homothety with ratio $1.5$ takes $Q\to B'$ , $R\to C'$ . Since $D\in\odot(A'B'C')$ , this homothety takes $P$ to $A'$ or $DP : PA' = 2 : 1$ . However, if we let $T = BQ\cap CR\cap MA'$ , then Menelaus theorem on $\triangle MBA'$ and $\overline{CTR}$ gives $T$ is midpoint of $MA'$ . Hence using the converse of Menelaus theorem on $\triangle MDA'$ and $\overline{ATP}$ , we get the conclusion.

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#3 h Sep 12, 2018, 8:46 AM • 2 Y

Y by Adventure10, Mango247

Using Cartesian coordinates, we have $R=\frac13(A+B+D)$ and $Q=\frac13(A+C+D)$ . Since $R-Q=\frac13(B-C)$ , we have $RQ\parallel BC$ , so $DPQR$ is a cyclic trapezoid, whence the projection of the midpoint $\frac12(Q+R)=\frac16(2A+2D+B+C)$ onto $BC$ is exactly $\frac12(D+P)$ . Since projection is a linear function, or by just letting $DA,DB$ be the coordinate axes, we see that this projection is just
$\frac16(2D+2D+B+C)=\frac12(D+P),\quad \implies\quad P=\frac13(B+C+D).$ Therefore, the point-centroid lines $AP,BQ,CR$ concur at the centroid $\frac14(A+B+C+D)$ of quadrilateral $ABCD$ .

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#4 h Jan 10, 2022, 7:31 PM

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cheva sinusi

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#5 h Feb 2, 2022, 3:46 PM • 2 Y

Y by M.Roueintan, T.Mousavidin

Actually, there is no need for any computation in this problem:
Clearly $QR \parallel BC$ and a quick angle chase shows that $\angle PRQ=\angle CDQ=\gamma$ as well as $\angle RQP=\angle RPB=\beta$ . Hence triangles $PQR$ and $ABC$ are similar with parallel sides, hence the desired point of intersection is just the center of the homothety sending one to the other.

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#6 h Sep 30, 2023, 8:26 PM • 1 Y

Y by nargesrafi

Using Desargues's theorem we know that $AP$ , $BQ$ and $CR$ are concurrent if and only if $QR\cap BC$ , $QP\cap AB$ and $PR\cap AC$ are collinear. By angle chasing these points are all on the infinity line so they are collinear.

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#8 h Oct 3, 2023, 8:51 AM

Y by

Valencia_Ann wrote:

Using Desargues's theorem we know that $AP$ , $BQ$ and $CR$ are concurrent if and only if $QR\cap BC$ , $QP\cap AB$ and $PR\cap AC$ are collinear. By angle chasing these points are all on the infinity line so they are collinear.

Cheers $;)$

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#9 h Oct 3, 2023, 2:08 PM

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In general, it is true that for a triangle $ABC$ and point $D$ , if $P,Q,R$ are the centroids of triangles $BCD,CDA,ADB$ respectively, then $\triangle PQR$ is homothetic to $\triangle ABC$ . This follows since $Q-R=\tfrac{A+D+C}{3}-\tfrac{A+D+B}{3}=\tfrac{C-B}{3}$ , hence $\overline{QR} \parallel \overline{BC}$ , and similarly the other pairs of sides are parallel.

We apply this to the given problem. Set $\overline{BC}$ as the $x$ -axis. Since $Q=\tfrac{A+C+D}{3}$ and $R=\tfrac{A+B+D}{3}$ , $\overline{QR} \parallel \overline{BC}$ , hence the center of $(PQRD)$ has $x$ -coordinate equal to that of $\tfrac{A+D+\tfrac{B+C}{2}}{3}=\tfrac{2D+\tfrac{B+C}{2}}{3}$ . Thus $P$ has $x$ -coordinate equal to that of $\tfrac{D+B+C}{3}$ , and since its $y$ -coordinate is zero it equals $\tfrac{B+C+D}{3}$ , so $P$ is the centroid of degenerate triangle $BCD$ . $\blacksquare$

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