Triangle is Isosceles

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Source: 2010 IrMO Paper 1 Problem 2

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#1 h Jan 30, 2018, 9:11 PM • 1 Y

Y by Adventure10

Let $ABC$ be a triangle and let $P$ denote the midpoint of the side $BC$ . Suppose that there exist two points $M$ and $N$ interior to the side $AB$ and $AC$ respectively, such that $|AD|=|DM|=2|DN|,$ where $D$ is the intersection point of the lines $MN$ and $AP$ . Show that $|AC|=|BC|$ .

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#2 h Jan 30, 2018, 10:11 PM • 2 Y

Y by Adventure10, Mango247

Let $X,Y$ be defined such that $H_{A,\frac{AD}{AP}}(M)=X$ and $H_{A,\frac{AD}{AP}}(N)=Y$ . Since $AD=DM=2DN$ we have that $AX=AP=2PY$ . Let $Z$ be a point on $XP$ such that $X-P-Y-Z$ and $YZ=PY$ . This implies $PZ=AP=XP$ so we have that $CZ\parallel AB$ ( $BP=PL$ and $XP=PZ$ so $XCZB$ is a parallelogram) and $AX \perp ZA$ . Take $R$ to be the midpoint of $XP$ . We have that $BR \parallel CY \parallel AC$ and $3XR=XY$ , which implies that $3BX=XA$ . From $BX=CZ\parallel AB$ we have that $CZ=AS$ where $S$ is the foot of the $C-$ altitude. Now we have that $3AS=XA=2AS+BS$ so $AS=BS$ which implies that $AC=BC$ .

This post has been edited 1 time. Last edited by Wolowizard, Jan 30, 2018, 10:11 PM

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#3 h Dec 23, 2020, 2:15 AM

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If such a set of points exist, we could take a homothety at $A$ which sends $M$ to $B$ , $D$ to $D'$ and $N$ to $N'$ . This would preserve the length ratios, and thus $B, D', N'$ would be another set of points that satisfy this condition.

Thus we may assume WLOG that $M = B$ . Then the median $AP$ cuts the segment $BN$ in the ratio $2:1$ , hence $BN$ is also a median, and since $AD = BD$ , we have $DN = DP$ , and $\triangle ADN \cong \triangle BDP$ , so $AC = BC$ as required.

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