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Given any triangle $\varDelta$, the perpendiculars from the vertices of its (primary) Morley triangle to their respective (nearest) side of $\varDelta$ intersect in a triangle $\varDelta'$, which is similar to $\varDelta$ but on a smaller scale—say with scale factor $s$. (If $\varDelta$ is isosceles, then $\varDelta'$ degenerates to a point, and $s=0$.)

What is the maximum value of $s$, and for what shape of $\varDelta$ is this value attained?

(This question was posted previously on Mathematics Stack Exchange, without response, but is probably more suitable for this site.)

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  • $\begingroup$ In my answer the core was maybe a computer algebra computation to get the expression for the scale factor $s$. If wanted, i can provide the (arguably simple) sage code, too. $\endgroup$
    – dan_fulea
    Apr 7, 2022 at 22:00

1 Answer 1

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We will perform computation using normed barycentric coordinates in the given triangle $\Delta ABC$ with side lengths $a,b,c$ and angles $\hat A=3\alpha$, $\hat B=3\beta$, $\hat C=3\gamma$. We denote by $R$ the circumradius of the circle $\odot(ABC)$, and the area $[ABC]$ by $S$.

Trilinear coordinates of the points $A'$, $B'$, $C'$ (vertices of the first Morley triangle of $\Delta ABC$) can be extracted from the rows of the matrix: $$ \begin{bmatrix} 1 & 2\cos \gamma & 2\cos\beta\\ 2\cos \gamma & 1 & 2\cos\alpha\\ 2\cos \beta & 2\cos\alpha & 1\\ \end{bmatrix} \ . $$ (See for instance https://mathworld.wolfram.com/FirstMorleyTriangle.html .)

The (not-normalized) barycentric coordinates can then be extracted from the rows of the matrix: $$ \tag{$*$} \begin{bmatrix} a & 2b\cos \gamma & 2c\cos\beta\\ 2a\cos \gamma & b & 2c\cos\alpha\\ 2a\cos \beta & 2c\cos\alpha & c\\ \end{bmatrix} \ . $$ Let now $P_A$, $P_B$, $P_C$ be three points with normalized barycentric coordinates $$ \begin{aligned} P_A &= (x_A, y_A, z_A)\ ,\\ P_B &= (x_B, y_B, z_B)\ ,\\ P_C &= (x_C, y_C, z_C)\ . \end{aligned} $$ Define $Q_A$ to be such that $Q_AP_B\perp AC$ and $Q_AP_C\perp AB$. Construct in a similar manner $Q_B,Q_C$. Computer algebra shows that $$ \begin{aligned} s^2 :=\ &\frac{[Q_AQ_BQ_C]}{[ABC]} =\frac{P_BP_C}{a^2} =\frac{P_CP_A}{b^2} =\frac{P_AP_B}{c^2} \qquad\text{ is given by the relation} \\ 16S^2\cdot s^2 =\ &\ \begin{pmatrix} + x_A(b^2-c^2) + a^2(y_A-z_A) \\ + y_B(c^2-a^2) + b^2(z_B-x_B) \\ + z_C(a^2-b^2) + c^2(x_C-y_C) \end{pmatrix} ^2 \ . \end{aligned} $$ (Keeping $A$ and exchanging $B\leftrightarrow C$ moves $s$ to $-s$.)

So we have to maximize the expression $$ \begin{aligned} s &= \frac 1{4S} \begin{pmatrix} + x_A(b^2-c^2) + a^2(y_A-z_A) \\ + y_B(c^2-a^2) + b^2(z_B-x_B) \\ + z_C(a^2-b^2) + c^2(x_C-y_C) \end{pmatrix} \\ &=\frac 1{4S}\sum x_A(b^2-c^2) + a^2(y_A-z_A) \\ &=\frac 1{4S}\sum (1-y_A-z_A)(b^2-c^2) + a^2(y_A-z_A) \\ &=\frac 1{4S}\sum y_A(a^2-b^2+c^2) -z_A(a^2+b^2-c^2) \\ &=\frac 1{4S}\sum y_A\cdot 2ac\sin B -z_A\cdot 2ab\sin C \\ &=\frac 1{4RS}\sum y_A\cdot abc -z_A\cdot abc \\ &=\sum(y_A - z_A)\ . \\[3mm] &\qquad\text{Now we plug in the values for $y_A$, $z_A$ from $(*)$:} \\[3mm] y_A &= \frac {2\cdot 2R\sin B\cos\gamma} {2R(\sin A +2\cdot \sin B\cos \gamma +2\cdot \sin C\cos \beta)}\ , \\ z_A &= \frac {2\cdot 2R\sin C\cos\beta} {2R(\sin A +2\cdot \sin B\cos \gamma +2\cdot \sin C\cos \beta)}\ , \\[3mm] &\qquad\text{getting:} \\[3mm] s &=\sum \frac {2\cdot \sin B\cos \gamma - 2\sin C\cos\beta} {\sin A +2\cdot \sin B\cos \gamma +2\cdot \sin C\cos \beta} % \\ % &=-3 + \sum % \frac % {\sin A + 2\cdot \sin B\cos \gamma} % {\sin A + 2\cdot \sin B\cos \gamma + 2\cdot \sin C\cos \beta} \ . \end{aligned} $$ This is not an easy task now. (Either for the last expression of $s$, or for the formula in between with a cyclic sum $s=\sum(y_A-z_A)$, if there is some better geometric interpretation.) So we are starting to solve a new problem in the problem.


Since time is an issue for me, instead of starting to compute using Lagrange multipliers, in order to proceed in some few lines here is a picture of the function to be maximized:

mathoverflow 417175

The plot uses $x,y\in[0,1]$ to parametrize the angles $\alpha,\beta,\gamma$ as follows:

$\displaystyle \alpha = \frac\pi3\cdot x$, $\displaystyle \beta = \frac\pi3\cdot y(1-x)$, and $\displaystyle \gamma = \frac\pi3\cdot (1-y)(1-x)$.

It turns out that the maximum is in fact a supremum, taken for the case when

$\displaystyle \alpha\nearrow\frac \pi 3$, and then $\displaystyle \beta,\gamma\searrow 0$, so that $\displaystyle \alpha+\beta+\gamma=\frac \pi 3$.

In terms of the angles of $\Delta ABC$ we have $\hat A\nearrow\pi$, $\hat B,\hat C\searrow 0$. (But $\hat B$, $\hat C$ may need to be still correlated.)

To compute this supremum, i need a notation that i can better type. So i will switch from $\alpha,\beta,\gamma$ to $x,y,z$. Then $\hat B=3y$, $\hat C=3z$, and corresponding sine values will be approximated by hand waving with $3y+O(y^3)$ and $3z+O(z^3)$, so that for $\hat A=\pi-(3x+3y)$ we have also a sine value in the shape $3y+3z+O(\dots)$. Computations will omit below terms in total monomial degree bigger / equal two. So $$ \begin{aligned} s &= \frac {2\sin B\cos\frac C3 - 2\sin C\cos \frac B3} {\sin A + 2\sin B\cos\frac C3 + 2\sin C\cos \frac B3} \\ &\qquad +\frac {2\sin C\cos\frac A3 - 2\sin A\cos \frac C3} {\sin B + 2\sin C\cos\frac A3 + 2\sin A\cos \frac C3} \\ &\qquad\qquad +\frac {2\sin A\cos\frac B3 - 2\sin B\cos \frac A3} {\sin C + 2\sin A\cos\frac B3 + 2\sin B\cos \frac A3} \\ &\sim \frac{2(3y-3z)}{3y+3z + 2(3y+3z)}\\ &\qquad +\frac {2\left(3z\cdot\frac 12 -(3y+3z)\right)}{3y + 2\left(3z\cdot\frac 12 +(3y+3z)\right)}\\ &\qquad\qquad +\frac {2\left((3y+3z) - 3y\cdot\frac 12\right)}{3z + 2\left((3x+3y) + 3y\cdot\frac 12\right)} \\ &= \frac {2y-2z}{3y+3z} - \frac {2y+z}{3y+3z} + \frac {y+2z}{3y+3z} \\ &=\frac{y-z}{3y+3z}\ . \end{aligned} $$ So we expect $\displaystyle\color{blue}{\frac 13}$ as a supremum value.

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  • $\begingroup$ Thank you very much. As a check, the contours at the level $z=0$ (i.e. $s=0$) on the illustrated surface should include—perhaps comprise—the three curves $y=\frac12$, $y=x/(1-x)$, and $y=(1-2x)/(1-x)$. $\endgroup$ Apr 7, 2022 at 23:20

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