Kiepert triangle problem

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#1 h Mar 28, 2015, 10:48 AM • 2 Y

Y by Adventure10, Mango247

Consider a triangle $ABC$ . Let $A_\alpha B_\alpha C_\alpha$ be the points satisfying
$\angle A_\alpha BC=\angle A_\alpha CB =\angle B_\alpha CA=\angle B_\alpha AC=\angle C_\alpha AB=\angle C_\alpha BA=\alpha$ ( $\alpha=0$ to $2\pi$ ).
Let $K_\alpha=AA_\alpha\cap BB_\alpha\cap CC_\alpha$ and $L_\alpha$ be the isogonal conjugate of $K_\alpha$ WRT $\triangle ABC$ . $\theta$ is fixed and $\beta$ is varying.
1. Prove that $L_{-\theta}$ , $K_\beta$ , $K_{\theta-\beta}$ are collinear.
2. Prove that $K_\beta K_{\theta+\beta}$ is tangent to a fixed conic $c_\theta$ .

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#4 h Mar 28, 2015, 12:59 PM • 4 Y

Y by VUThanhTung, Leooooo, Adventure10, Mango247

1.

My solution:

Lemma 1 :

$K_{\phi} L_{-\phi}$ pass through the Centroid $G$ of $\triangle ABC$

Proof of lemma 1:

From post #3 (remark) and post #4 in Property of Kariya point $\Longrightarrow K_{\phi}K_{-\phi}$ pass through the Symmedian point of $\triangle ABC$ ,
so from the lemma mentioned at post #3 in the topic Angle related to Fermat points and Isodynamic points we get $K_{\phi} L_{-\phi}$ pass through the isogonal conjugate of the Symmedian point of $\triangle ABC$ . i.e. $K_{\phi} L_{-\phi}$ pass through the Centroid $G$ of $\triangle ABC$
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Lemma 2 :

Let $A_{\phi}^*$ be the isogonal conjugate of $A_{\phi}$ WRT $\triangle ABC$ .

Then $A_{\phi}^* A_{90^{\circ}-\phi}$ pass through the orthocenter $H$ of $\triangle ABC$

Proof of lemma 2 :

Since $\angle A_{\phi}^*BA=\angle A_{\phi}^*CA=180^{\circ}-\phi, \angle A_{90^{\circ}-\phi} BC=\angle A_{90^{\circ}-\phi} CB=90^{\circ}-\phi$ ,
so we get $\angle A_{\phi}^*BA_{90^{\circ}-\phi}=90^{\circ}-\angle CBA=\angle HCB, \angle A_{\phi}^*CA_{90^{\circ}-\phi}=90^{\circ}-\angle ACB=\angle HBC$ ,
hence from the problem A useful collinearity we get $H, A_{90^{\circ}-\phi}, A_{\phi}^*$ are collinear .
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Lemma 3 :

$K_{\phi} L_{90^{\circ}-\phi}$ pass through $H$

Proof of lemma 3 :

Easy to see $A \in B_{90^{\circ}-\phi}C_{90^{\circ}-\phi}^*, A \in C_{90^{\circ}-\phi}B_{90^{\circ}-\phi}^*, B \in A_{90^{\circ}-\phi}C_{90^{\circ}-\phi}^*, C \in A_{90^{\circ}-\phi}B_{90^{\circ}-\phi}^*$ .

Since $CB_{\phi}$ is the isogonal conjugate of $CA_{\phi}$ WRT $\angle ACB$ ,
so we get $A_{\phi}^* \in CB_{\phi}$ (Similarly we can prove $A_{\phi}^* \in BC_{\phi}$ ) .

From lemma 2 we get $H \in A_{\phi}^*A_{90^{\circ}-\phi}$ .
Similarly we can prove $H$ lie on $B_{\phi}B_{90^{\circ}-\phi}^*$ and $C_{\phi}C_{90^{\circ}-\phi}^*$ ,
so from Desargue theorem ( for $\triangle BC_{\phi}C_{90^{\circ}-\phi}^*$ and $\triangle CB_{\phi}B_{90^{\circ}-\phi}^*$ ) we get $BC, B_{\phi}C_{\phi}, B_{90^{\circ}-\phi}^*C_{90^{\circ}-\phi}^*$ are concurrent ,
hence from Desargue theorem ( for $\triangle BB_{\phi}B_{90^{\circ}-\phi}^*$ and $\triangle CC_{\phi}C_{90^{\circ}-\phi}^*$ ) we get $K_{\phi}, L_{90^{\circ}-\phi}, H$ are collinear .
( notice that $K_{\phi} \equiv BB_{\phi} \cap CC_{\phi}, L_{90^{\circ}-\phi} \equiv BB_{90^{\circ}-\phi}^* \cap CC_{90^{\circ}-\phi}^*, H \equiv B_{\phi}B_{90^{\circ}-\phi}^* \cap C_{\phi}C_{90^{\circ}-\phi}^*$ )
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Back to the main problem :

Since $K_{\beta} \mapsto K_{\theta-\beta}$ form an involution on Kiepert hyperbola ,
so $K_{\beta} K_{\theta-\beta}$ pass through a fixed point $T$ (the pole of the involution) when $\beta$ varies .

When $\beta=0^{\circ}$ , from lemma 1 we get $K_{\beta} K_{\theta-\beta} \equiv G K_{\theta}$ pass through $L_{-\theta}$ . ... $(1)$
When $\beta=90^{\circ}$ , from lemma 3 we get $K_{\beta} K_{\theta-\beta} \equiv H K_{90^{\circ}+\theta}$ pass through $L_{-\theta}$ . ... $(2)$
From $(1), (2)$ we get $T \equiv L_{-\theta}$ . i.e. $K_{\beta} K_{\theta-\beta}$ always pass through $L_{-\theta}$ when $\beta$ varies

Q.E.D

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#5 h Mar 28, 2015, 1:04 PM • 4 Y

Y by VUThanhTung, Leooooo, Adventure10, Mango247

2.

Since $K_{\beta} \mapsto K_{\theta+\beta}$ is a homography on Kiepert hyperbola ,
so from the problem Homography on a Conic we get $K_{\beta} K_{\theta+\beta}$ is tangent to a fixed conic .

Q.E.D

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#6 h Mar 28, 2015, 2:26 PM • 1 Y

Y by Adventure10

wow, thank you for the solution

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#7 h Mar 28, 2015, 4:17 PM • 1 Y

Y by Adventure10

My solution for the part 1 is using Trilinear coordinates:
The trilinear coordinates of $K_\beta$ is $(\csc(A+\beta),\csc(B+\beta),\csc(C+\beta))$
The trilinear coordinates of $K_{\theta-\beta}$ is $(\csc(A+\theta-\beta),\csc(B+\theta-\beta),\csc(C+\theta-\beta))$
The trilinear coordinates of $K_{-\theta}$ is $(\csc(A-\theta),\csc(B-\theta),\csc(C-\theta))$ so the trilinear coordinates of $L_{-\theta}$ is $(1/(\csc(A-\theta),1/\csc(B-\theta),1/\csc(C-\theta))=(\sin(A-\theta),\sin(B-\theta),\sin(C-\theta))$ .

$L_{-\theta}$ , $K_\beta$ , $K_{\theta-\beta}$ are collinear if and only if $\det \begin{vmatrix}\csc(A+\beta)&\csc(B+\beta)&\csc(C+\beta)\\ \csc(A+\theta-\beta)&\csc(B+\theta-\beta)&\csc(C+\theta-\beta)\\\sin(A-\theta)&\sin(B-\theta)&\sin(C-\theta)\end{vmatrix}=0$ .
Although the computation of $\det ...$ has not been done yet but we all know that it must be equal to $0$ and we are done

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#8 h May 5, 2015, 11:34 AM • 2 Y

Y by Adventure10, Mango247

Here is an interesting collorary of 1. : $K_{\beta}L_{-2\beta}$ is tangent to the Kiepert hyperbola of $\triangle ABC$

This post has been edited 1 time. Last edited by VUThanhTung, May 5, 2015, 11:54 AM

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