Prove that S = 8S1 − 4S2 - [IMO LongList 1971]

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Prove that S = 8S1 − 4S2 - [IMO LongList 1971]

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#1 h Jan 1, 2011, 9:07 PM • 1 Y

Y by Adventure10

Let squares be constructed on the sides $BC,CA,AB$ of a triangle $ABC$ , all to the outside of the triangle, and let $A_1,B_1, C_1$ be their centers. Starting from the triangle $A_1B_1C_1$ one analogously obtains a triangle $A_2B_2C_2$ . If $S, S_1, S_2$ denote the areas of triangles $ABC,A_1B_1C_1,A_2B_2C_2$ , respectively, prove that $S = 8S_1 - 4S_2.$

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#2 h Jan 5, 2011, 2:52 AM • 2 Y

Y by Adventure10, Mango247

Let $ABPQ$ and $ARSC$ be the consecutive vertices of the squares constructed outwardly on $BA,AC.$ Thus, $\triangle QAC$ and $\triangle BAR$ are congruent by SAS criterion, due to $AQ=AB,$ $AC=AR$ and $\angle QAC=\angle BAR=90^{\circ}+\angle BAC$ $\Longrightarrow$ $\angle ABR=\angle AQC.$ If $D \equiv QC \cap BR,$ then it follows that $Q,A,D,B$ are concylic $\Longrightarrow$ $\angle QDR=90^{\circ},$ $(\star).$ Let $M,N,L$ be the midpoints of $BC,CA,AB.$ Since $MC_1$ and $MB_1$ become B- and C- midlines of $\triangle BCQ$ and $\triangle CBR,$ from $(\star)$ we deduce that $\triangle MC_1B_1$ is isosceles right at $M.$ In exactly the same way, we deduce that the midpoint of $RQ$ and $B_1,C_1$ form another isosceles right triangle $\Longrightarrow$ $A_2$ is the midpoint of $QR,$ i.e. $M$ is the reflection of $A_2$ about $B_1C_1.$ Likewise, $N,L$ are the reflections of $B_2,C_2$ about $C_1A_1,$ $A_1B_1$ $\Longrightarrow$ $\triangle A_2B_2C_2$ and the medial $\triangle MNL$ of $\triangle ABC$ become the outer and inner Vecten triangles of $\triangle A_1B_1C_1.$ Now, we know that the area of any triangle is the arithmetic mean between the areas of its outer and inner Vecten triangles. Therefore

$2|\triangle A_1B_1C_1|=| \triangle A_2B_2C_2|+ |\triangle MNL| \Longrightarrow 2S_1=S_2+\frac{_1}{^4}S.$

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