Two perspective and inversely similar triangles

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Two perspective and inversely similar triangles

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#1 h Jun 6, 2015, 9:38 AM • 6 Y

Y by LeVietAn, Re1gnover, mineiraojose, enhanced, Adventure10, Mango247

Let $\triangle A_1B_1C_1, \triangle A_2B_2C_2$ be 2 perspective and inversely similar triangles .
Let $P \equiv A_1A_2 \cap B_1B_2 \cap C_1C_2$ and $T_1, T_2$ be the points s.t. $\triangle A_1B_1C_1 \cup T_1 \sim \triangle A_2B_2C_2 \cup T_2$ .

Prove that $T_1, T_2, P$ are collinear $\Longleftrightarrow A_1, B_1, C_1, T_1, P$ lie on a rectangular hyperbola

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#3 h Jan 12, 2016, 10:09 PM • 3 Y

Y by TelvCohl, enhanced, Adventure10

Let the perpendiculars from $B_1$ and $C_1$ to $C_2A_2$ and $A_2B_2,$ resp, intersect at $O_1.$ Then $\angle (O_1B_1,O_1C_1)=\angle (A_2C_2,A_2B_2)=\angle (A_1B_1,A_1C_1)$ $\Longrightarrow$ $O_1 \in \odot(A_1B_1C_1)$ $\Longrightarrow$ $\angle (O_1C_1,O_1A_1)=\angle (B_1C_1,B_1A_1)=\angle (B_2A_2,B_2C_2)$ $\Longrightarrow$ $A_1O_1 \perp B_2C_2$ $\Longrightarrow$ $\triangle A_1B_1C_1$ and $\triangle A_2B_2C_2$ are orthologic, so let $O_2$ be the orthology center of $\triangle A_2B_2C_2$ WRT $\triangle A_1B_1C_1.$ Since these triangles are also perspective with perspector $T,$ then by Sondat's theorem $T,O_1,O_2$ are collinear and the conics through $A_1,B_1,C_1,T,O_1$ and $A_2,B_2,C_2,T,O_2$ are rectangular hyperbolae $\mathcal{H}_1$ and $\mathcal{H}_2,$ resp (well-known for orthologic triangles).

Let $H_1 \in \mathcal{H}_1$ and $H_2 \in \mathcal{H}_2$ be the orthocenters of $\triangle A_1B_1C_1$ and $\triangle A_2B_2C_2,$ resp. Redefining $T_1$ as a point on $\mathcal{H}_1$ and $T_2$ the second intersection of $TT_1$ with $\mathcal{H}_2,$ then it's enough to prove that $\triangle A_1B_1C_1 \cup T_1 \sim \triangle A_2B_2C_2 \cup T_2.$

We know that $H_1,H_2$ and $T$ are collinear (see for instance the topic A Nice Result. post #7). Thus, using cross ratios on $\mathcal{H}_1$ and $\mathcal{H}_2,$ we obtain $C_1(A_1,B_1,O_1,H_1)=T(A_1,B_1,O_1,H_1) \equiv T(A_2,B_2,O_2,H_2)=C_2(A_2,B_2,O_2,H_2).$ But clearly $\triangle A_1B_1C_1 \cup \mathcal{H}_1,H_1,O_1 \sim \triangle A_2B_2C_2 \cup \mathcal{H}_2, H_2,O_2,$ thus we deduce that $\triangle A_1B_1C_1 \cup T_1 \sim \triangle A_2B_2C_2 \cup T_2,$ as desired.

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#4 h Jan 13, 2016, 3:12 AM • 2 Y

Y by Adventure10, Mango247

Actually the previous approach can be even simpler. Either the collinearity of $T,O_1,O_2$ or $T,H_1,H_2$ is sufficient to yield the result. So for example using the collinearity of $T,O_1,O_2$ found in the first paragraph, we have $C_1(A_1,B_1,O_1,T_1)=T(A_1,B_1,O_1,T_1) \equiv T(A_2,B_2,O_2,T_2)=C_2(A_2,B_2,O_2,T_2)$ and the conclusion follows.

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