Note about Morley Point and Kantor-Hervey Point

by TelvCohl, Mar 12, 2017, 2:29 PM

Notation

Given four lines $\ell_1,$ $\ell_2,$ $\ell_3,$ $\ell_4$ and let $P_{ij}$ be the intersection of $\ell_i$ and $\ell_j$ ( $1$ $\leq$ $i$ $<$ $j$ $\leq$ $4$ ). Let $O_w,$ $H_w$ be the circumcenter, orthocenter of the triangle formed by $\ell_x,$ $\ell_y,$ $\ell_z,$ respectively ( $\{ w, x, y, z \}$ $=$ $\{ 1, 2, 3, 4 \}$ ).

Preliminaries

Lemma : Given two inversely similar triangles $\triangle AB_1C_1,$ $\triangle AB_2C_2.$ Let $T$ be the intersection of the perpendicular bisector of $B_1B_2,$ $C_1C_2.$ Then $\measuredangle B_1TB_2$ $=$ $2\measuredangle B_1C_1A$ $=$ $2\measuredangle AC_2B_2.$

Proof : Let $O_1,$ $O_2$ be the circumcenter of $\triangle AB_1C_1,$ $\triangle AB_2C_2,$ respectively and let $\widetilde{T}$ be the point such that $AO_1$ $\parallel$ $\widetilde{T}O_2,$ $AO_2$ $\parallel$ $\widetilde{T}O_1.$ It's well-known that $\triangle B_1\widetilde{T}B_2$ and $\triangle AO_1B_1,$ $\triangle B_2O_2A$ are inversely similar, so $\widetilde{T}$ lies on the perpendicular bisector of $B_1B_2.$ Similarly, we can prove $\widetilde{T}$ lies on the perpendicular bisector of $C_1C_2,$ so we conclude that $\widetilde{T}$ $\equiv$ $T$ and $\measuredangle B_1TB_2$ $=$ $2\measuredangle B_1C_1A$ $=$ $2\measuredangle AC_2B_2.$ $\qquad \blacksquare$

Main result

Property 1 (Kantor-Hervey Point) : The perpendicular bisector of $O_1H_1,$ $O_2H_2,$ $O_3H_3,$ $O_4H_4$ are concurrent at $K.$

Proof : Let $K$ be the intersection of the perpendicular bisector of $O_1H_1,$ $O_4H_4.$ Since $\frac{P_{23}H_1}{P_{23}O_1}$ $=$ $\frac{P_{23}H_4}{P_{23}O_4}$ $=$ $2 \cos \angle P_{23},$ so notice $\{ P_{23}H_1,P_{23}O_1 \},$ $\{ P_{23}H_4, P_{23}O_4 \}$ are isogonal conjugate WRT $\angle P_{23}$ we get $\triangle P_{23}H_1H_4$ and $\triangle P_{23}O_1O_4$ are inversely similar. Let $\mathcal{S}$ be the Steiner line of the complete quadrilateral formed by $\ell_1,$ $\ell_2,$ $\ell_3,$ $\ell_4,$ then from Lemma we get $\measuredangle O_4KH_4$ $=$ $2 \measuredangle (\perp \ell_4, \mathcal{S} ).$ Similarly, we can prove $K$ lies on the the perpendicular bisector of $O_2H_2,$ $O_3H_3,$ so we conclude that the perpendicular bisector of $O_1H_1,$ $O_2H_2,$ $O_3H_3,$ $O_4H_4$ are concurrent at $K.$ $\qquad \blacksquare$

Property 2 (Morley Point) : The perpendicular from the midpoint of $O_1H_1,$ $O_2H_2,$ $O_3H_3,$ $O_4H_4$ to $\ell_1,$ $\ell_2,$ $\ell_3,$ $\ell_4,$ resp. are concurrent at $M.$

Proof : Let $M$ be the projection of $K$ on $\mathcal{S}$ and let $N_4$ be the midpoint of $O_4H_4.$ Clearly, $K,$ $M,$ $H_4,$ $N_4$ are concyclic, so $\measuredangle (MN_4,\mathcal{S})$ $=$ $\measuredangle N_4KH_4$ $\stackrel{\text{Lemma}}{=}$ $\measuredangle (\perp \ell_4, \mathcal{S})$ $\Longrightarrow$ $MN_4$ $\perp$ $\ell_4.$ Analogously, we can prove $M$ lies on the perpendicular from the midpoint of $O_1H_1,$ $O_2H_2,$ $O_3H_3$ to $\ell_1,$ $\ell_2,$ $\ell_3,$ respectively, so we conclude that the perpendicular from the midpoint of $O_tH_t$ to $\ell_t$ are concurrent at $M$ ( $1$ $\leq$ $t$ $\leq$ $4$ ). $\qquad \blacksquare$

Corollary : The line connecting the Kantor-Hervey point and the Morley point of a complete quadrilateral is perpendicular to its Steiner line.

This post has been edited 1 time. Last edited by TelvCohl, Aug 20, 2022, 12:25 AM
Reason: Thanks Vik956

Kantor-Hervey point

Morley Point

Comment

2 Comments

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Hello,

I try to understand your demonstration of property 1.
First of all, A is not defined. It should be P23.
Next I do not understand your "similarly" in the demonstration. K is assumed to be on perpendicular bissectors of O1H1 and O4H4. You computed the angle using the lemma. OK.
Yes we can do the same thing looking at O2H2 and O3H3 and will find a K', intersection of perpendicular bissectors of previous two segments. We can compute the angle O3K'H3.
But why K' should be same as K ?
Sorry if a missed some point.

Robert
______________________________________________________________________
Let $T_{x}$ ( $x\in\{1,2,3\}$ ) be the intersection of the perpendicular bisector of $O_{x}H_{x}$ with the perpendicular bisector of $O_{4}H_{4},$ then from my proof we get $\measuredangle O_{4}T_{x}H_{4} = 2\measuredangle (\perp \ell_{4}, \mathcal{S}).$ WLOG $T_{2}=T_{3}$ and apply the same argument for $O_{1}H_{1}$ …

~ Telv Cohl

This post has been edited 1 time. Last edited by TelvCohl, Aug 20, 2022, 12:57 AM

by Vik956, Aug 16, 2022, 3:49 PM

Report

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Got it ! Thank you.

by Vik956, Aug 24, 2022, 7:29 AM

Report

Telv Cohl's Geometry Blog

Note about Morley Point and Kantor-Hervey Point

by TelvCohl, Mar 12, 2017, 2:29 PM

2 Comments