how to find three vertices of a triangle.

Question

Where s=circumcenter, H= orthocenter, and A'= midpoint of one side of triangle.

How can can I determine the location of the three vertices of the triangle?

Answer 1

Let us impose coordinates to facilitate the discussion. We will place $H$ at the origin which then gives $A' = (-1,\ 6)$ and $S=(-6,6)$.

From the geometry of the Euler line, we know that the centroid of the triangle $G$ lies on the line $SH$ in the ratio $|SG|:|GH| = 1:2$. Therefore the coordinate of $G$ is given by $G = (4,4)$.

We also know that $A'$ is the midpoint of a side, call it $a$. Since $S$ is the circumcenter of the triangle, it follows that the line $\overline{SA'}$ must be the perpendicular bisector of side $a$. This means that $a$ is perpendicular to $\overline{SA'}$ and in particular $a$ is parallel to the $y$-axis. Then the line perpendicular to $a$ and through $H$ must be an altitude of the triangle, i.e. the $x$-axis is an altitude of the triangle. Therefore the vertex $A$ opposite side $a$ must lie on the $x$-axis.

But $A$ must also lie on the line $\overline{A'G}$ for the centroid lies on lines joining vertices and midpoints of opposite sides. It follows that $A$ is the intersection of $\overline{A'G}$ with the $x$-axis and in particular we have $A = (-14,\ 0)$ in our case.

Finally, the circumcircle $\Gamma$ is the circle with center $S$ passing through $A$. The intersection of $\Gamma$ with the side $a$ then determines the remaining two vertices of the triangle. This determines the triangle uniquely given $A',\ S$ and $H$ and we can verify that each of the points are indeed the corresponding special points of the triangle.

Answer 2

Another way of solving the problem is to use vector algebra.

Let the coordinate system have origin at the circumcenter $\mathbf{S}=(x_s,y_s)=(0,0)$. The orthocenter has position vector $\mathbf{H}=(x_h,y_h)=(6,-6)$. Let $\mathbf{X}_1=(x_1,y_1), \mathbf{X}_2=(x_2,y_2),\mathbf{X}_3=(x_3,y_3)$ be the three vertices and let $\mathbf{A}=(x_a,y_a)=(7,0)$ be the the midpoint of line $23$.

Then $\mathbf{A} = 0.5 (\mathbf{X}_2+\mathbf{X}_3)$ and we can express $x_2$ and $y_2$ as $$ x_2 = 14 - x_3 ~,~~ y_2 = -y3 \,. $$ Also, using the definition of the orthocenter and the vector scalar product, $$ (\mathbf{X}_3-\mathbf{H})\cdot(\mathbf{X}_1-\mathbf{X}_2) = 0 ~,~~ (\mathbf{X}_2-\mathbf{H})\cdot(\mathbf{X}_1-\mathbf{X}_3) = 0 \,. $$ We can solve these two equations to express $(x_1,y_1)$ in terms of $(x_3,y_3)$ after eliminating $(x_2,y_2)$ to get $$ \begin{align} x_1 &= \frac{y_3^3 + [x_3(x_3-14)+6]\,y_3 + 26(x_3-7)}{6x_3-y_3-42} \\ y_1 &= \frac{(7-x_3)y_3^2 + 6y_3- x_3^3 + 21 x_3^2 - 146x_3 + 336}{6x_3-y_3-42} \end{align} $$ To solve for $x_3$ we can use the relation $$ (\mathbf{S}-\mathbf{A})\cdot(\mathbf{X}_3-\mathbf{X}_2) = 0 \,. $$ The solution is $x_3 = 7$. To find $y_3$ we can use the definition of the circumcenter to get $$ (\mathbf{X}_1-\mathbf{S})\cdot(\mathbf{X}_1-\mathbf{S}) = (\mathbf{X}_3-\mathbf{S})\cdot(\mathbf{X}_3-\mathbf{S}) $$ The result is the equation $$ y_3^4 - 87 y_3^2 + 1836 = 0 $$ with solutions $(-7.14,-6,6,7.14)$. If we assume that $y_3 >0$ we have two possible values only one of which, $y_3 = 7.14$, is a valid solution. Using this solution we get $(x_1,y_1)=(-8,-6)$, $(x_2,y_2)=(7,-7.14)$, $(x_3,y_3)=(7,7.14)$.