Given points $u,o,g$, find points $v,w$ such that $o$ is circumcenter and $g$ is centroid

Question

Let $u=(7,4), o=(1,1), g=(1,2)$. Find $v,w \in \mathbb{R}^2$ such that $o$ is the circumcenter and $g$ is the centroid of $[u,v,w]$.

I understand that $g$ must be the intersection of the three medians, which can be written out as follows:

1. $u-\frac{1}{2}(v-w)$ or $t(u-\frac{1}{2}(v-w))+o$ ($t\in\mathbb{R}$)
2. $v-\frac{1}{2}(w-u)$ or $s(v-\frac{1}{2}(w-u))+o$ ($s\in\mathbb{R}$)
3. $w-\frac{1}{2}(v-u)$ or $r(w-\frac{1}{2}(v-u))+o$ ($r\in\mathbb{R}$)

However, I'm not sure what to do in order to find the points $v$ and $w$. Namely, what should the lines be equated to in order to come up with some equations for $v$ and $w$?

Answer 1

The approach using medians would be very messy and involve non-linear equations. Below is a far simpler approach.

The distance between $o$ and $u$ is $\sqrt{(7-1)^2+(4-1)^2}=\sqrt{45}$. Since $o$ is the circumcentre, $v$ and $w$ also lie on the circle $(x-1)^2+(y-1)^2=45$ or $y=\pm\sqrt{45-(x-1)^2}+1$. Now let the $x$-coordinates of $v$ and $w$ be $a$ and $b$ respectively.

The centroid of a triangle is the geometric mean of its vertices, so we have (up to the signs of the square roots) $$7+a+b=3\cdot1=3\tag1$$ $$4\pm\sqrt{45-(a-1)^2}+1\pm\sqrt{45-(b-1)^2}+1=3\cdot2=6\tag2$$ Here we find ourselves in luck, as $(2)$ becomes $$\pm\sqrt{45-(a-1)^2}\pm\sqrt{45-(b-1)^2}=0$$ $$45-(a-1)^2=45-(b-1)^2$$ $$a-1=\pm(b-1)$$ $$a=b\lor a+b=2$$ Obviously $a+b=2$ is incorrect, as substituting this into $(1)$ yields the contradiction $9=3$, so $a=b=-2$. The square roots must then have opposite signs for a non-degenarate triangle: $$+\sqrt{45-(-2-1)^2}+1=7\qquad-\sqrt{45-(-2-1)^2}+1=-5$$ Therefore $v=(-2,7)$ and $w=(-2,-5)$ (or the other way around).

Answer 2

If $h$ is the affix of the orthocenter, since $g$ divides $oh$ internally in the ratio $1:2$, we find the coordinates of $h$ as $(1,4)$. So $uh$ lies along the line $y=4$, and hence $vw$ will be parallel to x-axis. Hence the x-coordinate of the midpoint $m$ of $vw$ will be the x-coordinate of $v,w$.

Now we know that $om \parallel uh, \text{and} \ |om|=2|uh|$, and hence we get $m = (-2,1)$.

$v,w$ lie on the circumcircle $(x-1)^2+(y-1)^2 = 45$ and substituting $x=-2$ we get $y= \pm 6$ giving us the coordinates of $v,w$ as $(-2,7)$ and $(-2,-5)$

Answer 3

HINT:

Since your domain is $\mathbb{R}^2$, you have 4 unknowns ($v_x$, $v_y$, $w_x$ and $w_y$)

Given the coordinates of the circumcentre and one vertex, you already know the circumcircle radius. So you have two equations linking the (known) circumcentre to the other two vertices, through the (again, known) circumcircle radius.

Since the three medians intersect, you only need to consider any 2 of 3. So now, you get 4 equations from the intersections (one each for $x$ and $y$) of the medians but there are also 2 additional unknowns in terms of the distance fractions from each vertex to its opposite side bisector.

You now have a system of 6 e for the 6 unknowns.

Answer 4

The approach that I used is as follows. Set up the following five equations and then solve them (preferably on a computer algebra system like Wolfram Alpha).

$$v_1+w_1=-4$$ $$v_2+w_2=2$$ $$(1-w_1)^2+(1-w_2)^2=45$$ $$(1-v_1)^2+(1-v_2)^2=45$$ $$(v_1-1)(w_1-7)+(v_2-4)(w_2-4)=0$$

This approach works, but it is a bit too tedious.

Answer 5

Hint: in complex numbers $u=7+4i, o=1+i, g=1+2i\,$, hence the circumradius is $R=|u-o|=3 \sqrt{5}\,$. Let $g'=g-o=i,u'=u-o=6+3i, v'=v-o, w'=w-o\,$ then the system to solve for $v',w'$ can be written as:

$$ \begin{cases} |v'|=|w'| = R \\ u'+v'+w' = 3g' \end{cases} $$

Writing the latter equation as $w' = 3g' - u'-v' = z - v'$ with $z=3g'-u'=-6\,$, then taking the magnitudes on both sides and using that $z \in \mathbb{R}\,$, so $z=\bar z\,$, gives:

$$ R^2=|w'|^2=|v'-z|^2=(v'-z)(\bar v' - z) = |v'|^2-z(v'+\bar v')+|z|^2=R^2-z(v'+\bar v')+z^2 $$

After canceling out $R^2$ and dividing by $z \ne 0$ it follows that $v'+\bar v'=z=-6\,$, so $\operatorname{Re}(v')=-3\,$, then $\operatorname{Im}(v')=\pm\sqrt{R^2-\operatorname{Re}(v')^2}=\pm\sqrt{45-9}=\pm 6\,$. The two cases give:

• $v'= -3 +6i \implies w'=z-v'=-6-(-3 +6i)=-3-6i$

• $v'= -3 -6i \implies w'=z-v'=-6-(-3 -6i)=-3+6i$

Reverting to the original variables, $v=v'+o, w=w'+o\,$ gives $\{v,w\}=\{-2-5i, -2+7i\}\,$.