altitudes, concurrency wanted, incircle, BM + MC, CN + NA, AP + PB are min

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altitudes, concurrency wanted, incircle, BM + MC, CN + NA, AP + PB are min

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Source: 2018 Argentina Cono Sur Training List p40 https://artofproblemsolving.com/community/c6h2110311p15269631

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#1 h Aug 29, 2021, 10:58 AM

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The inscribed circumference of triangle $ABC$ touches sides $BC$ , $CA$ , $AB$ at points $D, E, F$ respectively. In the segments $EF$ , $FD$ , $DE$ we consider the points $M, N, P$ respectively such that $BM + MC$ , $CN + NA$ , $AP + PB$ are minimal.
$\bullet$ Prove that the lines $AM$ , $BN$ , $CP$ are concurrent.
$\bullet$ Prove that $DM$ , $EN$ , $FP$ are altitudes of the triangle $DEF$ .

This post has been edited 1 time. Last edited by parmenides51, Aug 11, 2024, 10:45 PM

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#2 h Aug 29, 2021, 11:12 AM

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It is well known that if you have a point $X$ on line $\ell$ and two points $A,B$ on the same side of $\ell$ then the minimum of $AX+BX$ is achieved when $\measuredangle(AX,\ell)=\measuredangle(\ell,BX)$ . Therefore $DM$ is the bisector of $BCM$ , so that $\frac{BD}{DC}=\frac{BM}{MC}$ . Let $X=EF\cap BC$ . Since lines $AD,BE,CF$ concur on Gergonnw point, it follows that $(BC,DX)=-1$ , i.e. $\frac{BD}{DC}=\frac{BX}{XC}$ . Therefore $(DX)$ is an Apollonian circle. Since $M$ lies on this circle with diameter $DX$ , it follows that $DM\perp MX$ , or in other words $DM$ is the height of $DEF$ , as wanted.
Now, since both $AD,BE,CF$ and the three altitudes of $DEF$ both concur (on the Gergonnw point and the orthocentre of $DEF$ respectively), it follows by Cevian nest that $AM,BN,CP$ concur.

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