geometry problem where the solution involves the use of some properties of complex numbers in geometry

Question

The statement of the problem : The triangle $\mathcal {ABC} $ is inscribed in circle $\mathcal C $ with center $\mathcal O $ and radius 1. For any point $\mathcal M $ on circle $\mathcal C $ {$\mathcal A $,$\mathcal B $,$\mathcal C $} we note s($\mathcal M $)= $OH_1^2+OH_2^2+OH_3^2$, where $H_1, H_2, H_3$ are the orthocenters of the triangles $\mathcal {MAB} $, $\mathcal {MBC} $ respectively $\mathcal {MCA} $.

a) Show that, if $\mathcal {ABC} $ is an equilateral triangle, then s($\mathcal M$) = 6, for any point $\mathcal M $ on circle $\mathcal C $ \ {$\mathcal A $,$\mathcal B $,$\mathcal C $}

b) Determine the smallest natural number k with the property that if there are distinct points $M_1,M_2,...,M_k$ on circle $\mathcal C $ \ {$\mathcal A $,$\mathcal B $,$\mathcal C $} such that $s(M_1)=s(M_2)=... s(M_k)=6$ implies that ABC is an equilateral triangle.

My approach: For the point a) I used the point affixes and Sylvester's relation for the orthocenter. I proved that $s(M)=9 + 2m(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) + \frac{2}{m}(a+b+c) + \frac{a+c}{b} + \frac{a+b}{c} + \frac{c+b}{a} $ . Using the fact that the triangle is equilateral, we extract the fact that $a+b+c=0$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 0$ which helps us to reach the conclusion that s($\mathcal M$) = 6. I encountered the problem at point b), where I have no idea what strategy I should take, but I probably still have to use complex numbers because they were quite helpful at a) .

Any and all proofs will be helpful. Thanks a lot!

Answer 1

Let me first replace this issue in a classical context.

Let $t:=a+b+c \tag{1}$

Alignment relationships $2O+H_k=3G_k$ give :

$$\begin{cases}h_1&=&a+b+m\\ h_2&=&b+c+m\\ h_3&=&c+a+m\end{cases}$$

$$s(m)=|h_1|^2+|h_2|^2+|h_3|^2$$

$$s(m)=|m-(-a-b)|^2+|m-(-b-c)|^2+|m-(-b-c)|^2\tag{2}$$

The locus of points $m$ such that $s(m)=6$ enters into a classical configuration which is as follows :

Let $A_1,A_2,\cdots A_n$ be a set of points in the plane with centroid $G$. The set of points $M$ such that $\varphi(M):=\sum_{p=1}^n (MA_p)^2 = k$ (k being a constant) is (according to the sign of $k-\varphi(G)$) :

• either the circle with center $G$ and radius $r=\sqrt{\frac{1}{n}(k-\varphi(G))}$

• or the void set.

(see my answer here)

We are now able to answer to the two questions.

Question a)

In our case, referring to (2), the points are $-a-b,-a-c,-b-c$ ; their centroid is $$g=\frac13(-2a-2b-2c)=-\frac23 t$$ and the constant is $k=6$.

$$s(g)=|-\tfrac23 t+a+b|^2+|-\tfrac23 t+b+c|^2+|-\tfrac23 t + c+a|^2=\tfrac19[|a+b-2c|^2+|b+c-2a|^2+|c+a-2b|^2]\tag{4}$$

If $a,b,c$ is an equilateral triangle, we have $a+b+c=0$, otherwise said $g=0$ is the center of the circle. Besides, (4) becomes :

$$s(g)=\tfrac19[|3c|^2+|3a|^2+|3b|^2]=3$$

As $k-s(g)>0$, we can conclude that the locus is the circle with center $g=0$ and radius

$$r=\sqrt{\tfrac13(6-s(g))}=1\tag{3}$$

i.e., the unit circle.

Question b) :

When $a,b,c$ isn't an equilateral triangle, let us exhibit a case where there exist $2$ distinct points $M_1,M_2$ on the unit circle such that $s(M)=6$.

Let us take $a=1,b=i,c=-i$ which is a right isosceles triangle. Using the results above, the locus of points $M$ is a circle with center $-2t=-2$ and radius (see formulas (3) and (4)) : $r=\sqrt{\tfrac13(6-\tfrac{25}{9})}$ slightly larger than $1$, therefore having 2 distinct intersection points with the unit circle.

But one cannot have more than 2 such points. It is impossible to have $a,b,c$ not equilateral triangle and $3$ distinct points $M=M_1,M_2,M_3$ belonging to the unit circle and such that the affix of one of the points $M_k$ belongs to the unit circle. This is plainly impossible because the number of intersection points of 2 circles with different centers is at most $2$.

Therefore, the maximum number of points is $2$.

Answer 2

a)

You have already written a proof using $$s(M)=9 + 2m\bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\bigg) + \frac{2}{m}(a+b+c) + \frac{a+c}{b} + \frac{a+b}{c} + \frac{c+b}{a}\tag1$$

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b)

You can use $(1)$ again.

We see that $s(M)=6$ is equivalent to $$\begin{align}& 6=9 + 2m\bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\bigg) + \frac{2}{m}(a+b+c) + \frac{a+c+\color{red}{b-b}}{b} + \frac{a+b+\color{red}{c-c}}{c} + \frac{c+b+\color{red}{a-a}}{a} \\\\&\iff 6=9-3 + 2m\bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\bigg) + \frac{2}{m}(a+b+c) + \frac{a+c+b}{b} + \frac{a+b+c}{c} + \frac{c+b+a}{a} \\\\&\iff 0= 2m\bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\bigg) + \frac{2}{m}(a+b+c) + (a+b+c)\bigg(\frac 1a+\frac 1b+\frac 1c\bigg) \\\\&\iff 0= 2m\bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\bigg) + \frac{2}{m}(a+b+c) +|a+b+c|^2\tag2\end{align}$$

Here, let $m=x+yi,a+b+c=s+ti$ where $x,y,s,t\in\mathbb R$ and $x^2+y^2=1$.

Then, $$\begin{align}(2)&\iff 0=2(x+yi)(s-ti)+2(x-yi)(s+ti)+s^2+t^2 \\\\&\iff 0=4sx+4ty+s^2+t^2\tag3\end{align}$$ If $\triangle{ABC}$ is not an equilateral triangle, then its centroid is not the same as its circumcenter, so we have $(s,t)\not=(0,0)$, and so the line $(3)$ has at most two intersection points with the circle $x^2+y^2=1$.

Therefore, the smallest natural number $k$ with the property is $\color{red}3$.

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Added :

"so we have $(s,t)≠(0,0)$, and so the line $(3)$ has at most two intersection points with the circle $x^2+y^2=1$." can you please explain this part more clearly and why it implies $k = 3$

If $\triangle{ABC}$ is not an equilateral triangle, then its centroid is not the same as its circumcenter, so we have $\frac{a+b+c}{3}\not=0$, i.e. $s+ti\not=0$, i.e. $(s,t)\not=(0,0)$. So, $(3)$ does represent a line.

A point $(x,y)$ has to be both on the circle $x^2+y^2=1$ and on the line $4sx+4ty+s^2+t^2=0$. A line and a circle have at most two intersection points. So, the number of such $(x,y)$ is at most $2$. This means that the number of such $M$ is at most $2$.

In conclusion, we can say that if $\triangle{ABC}$ is not an equilateral triangle, then $k\le 2$.

Considering the contraposition, we can say that if $k\ge 3$, then $\triangle{ABC}$ is an equilateral triangle.

Therefore, the smallest natural number $k$ with the property is $3$.