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Tangents forms triangle with two times less area
NO_SQUARES   1
N Apr 23, 2025 by Luis González
Source: Kvant 2025 no. 2 M2831
Let $DEF$ be triangle, inscribed in parabola. Tangents in points $D,E,F$ forms triangle $ABC$. Prove that $S_{DEF}=2S_{ABC}$. ($S_T$ is area of triangle $T$).
From F.S.Macaulay's book «Geometrical Conics», suggested by M. Panov
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Apr 23, 2025
Luis González
Apr 23, 2025
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Tangents forms triangle with two times less area
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Source: Kvant 2025 no. 2 M2831
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#1 h  Apr 23, 2025, 9:08 AM • 5 Y
Y by ehuseyinyigit, buratinogigle, kiyoras_2001, GeoKing, nguyenducmanh2705
Let $DEF$ be triangle, inscribed in parabola. Tangents in points $D,E,F$ forms triangle $ABC$. Prove that $S_{DEF}=2S_{ABC}$. ($S_T$ is area of triangle $T$).
From F.S.Macaulay's book «Geometrical Conics», suggested by M. Panov
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Luis González
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Luis González
#2 h  Apr 23, 2025, 8:37 PM • 1 Y
Y by NO_SQUARES
Lemma: Let $P$ be any point on the Steiner circumellipse of $\triangle {ABC}$ and let $\triangle DEF$ be the cevian triangle of $P$ WRT $\triangle{ABC}.$ Then $[DEF]=2 \cdot [ABC].$

Proof: Let $\triangle XYZ$ be the antimedial triangle of $\triangle {ABC}$ ($X,Y,Z$ against $A,B,C$). The Steiner circumellipse $\mathcal{S}$ of $\triangle {ABC}$ is obviously the Steiner inellipse of $\triangle XYZ$ tangent to its sides at $A,B,C$ $\Longrightarrow$ $X,E,F$ are collinear on the polar of $D$ WRT $\mathcal{S}$ and similarly $Y,F,D$ and $Z,D,E$ are respectively collinear. Thus by Pappus theorem for the hexagon $EAFZXY,$ it follows that $FZ \cap EY$ is at infinity, i.e. $FZ \parallel EY$ and similarly we get $DX \parallel FZ$ $\Longrightarrow$ $DX \parallel EY \parallel FZ$ $\Longrightarrow$ $[DEF]=[DYZ]=[CYZ]=2\cdot [ABC]. \ \blacksquare$

Back to the problem. Let $P \equiv AD \cap BE \cap CF$ be the perspector of the parabola. Since the center of any inconic is the isotomcomplement of its perspector (in the parabola case at infinity), then it follows then that the isotomic conjugate of $P$ WRT $\triangle {ABC}$ must be on the line at infinity $\Longrightarrow$ $P$ is on the Steiner circumellipse of $\triangle {ABC}$ and the result follows from the lemma.
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