algebra-precalculuseuclidean-geometrylinear algebravectors
Pentagon $ABCDE$ is inscribed in a circle. For any edge of $ABCDE$, we can draw the line perpendicular to that edge that contains the centroid of the remaining three vertices. Show that these 5 lines are concurrent.
We just finished a unit on vectors in my Pre-Calculus class but I have no idea on how to do this final problem. Any help would be greatly appreciated!
claim: all ten lines go through the point $\frac{1}{3}(a+b+c+d+e).$
proof: i will use complex numbers instead of vectors. chooses a coordinate system so that the circle has unit radius and the point $D$ is at $1.$ let the complex numbers $a, b, c=e^{2i\gamma}, d = 1, e = e^{2i\epsilon}.$ the line through the center of the triangle $ABC$ and orthogonal to line $DE$ has the parametric form $$\mbox{ line1:} \frac{1}{3}(a+b+c) +\frac{s}{3}e^{i\epsilon} , s \mbox{ real}$$ in the same way the line through the center of the triangle $ABE$ orthogonal to $DC$ has the parametric form $$\mbox{ line2:} \frac{1}{3}(a+b+e) + \frac{t}{3}e^{i\gamma}, t \mbox{ real}$$
solving the two equation and their complex conjugates, we find that $$ s =e^{i\epsilon} + e^{-i\epsilon} \mbox{ and } t = e^{i\gamma} + e^{-i\gamma}$$ and the common points of intersection as claimed.
Based on the described construction and using the OP's notations, we have the following figure:
It is enough to demonstrate that $DG=r$, the radius of the circle.
By the construction, the triangle $ABD$ is equilateral and then $AB=BD=AF$. $BDG$ is an isosceles triangle because the triangles $ABG$ and $ADG$ are congruent. $\angle AGD = \angle BGA$ and $AB=AD=AF$.
Now, connecting $A$ and $F$ with the center $O$ we get the green triangle $AFO$. To show that $DG=FO=r$, it is enough to prove that the triangles $BDG$ and $AFO$ are congruent.
We have already seen that the bases of these triangles equal. What remains is to show that $\angle AOF = \angle BGD$.
This is easy. Just look at the figure and you will see that the $\angle BDG =2\times\angle AGD = \angle AOF$. (See the Wiki article called Inscribed angle.)
Finally, this is how to use the construction at stake to find the center of the circle:
Note
This construction is not that bad compared to the most frequently used one in the case of which you construct two bisecting perpendiculars. In the case of this present construction it is enough to draw the line through $F$ and $D$ which already exist after constructing the first bisecting perpendicular.
Best Answer
WOLOG, assume the pentagon is inscribed in a circle of radius $R$ centered at origin $O$.
Let $p_1, p_2, p_3, p_4, p_5$ be the vectors correspond to vertices $A,B,C,D,E$ respectively. We have
$$|p_1|^2 = |p_2|^2 = |p_3|^2 = |p_4|^2 = |p_5|^2 = R^2$$
Let $G$ be the vertex centroid of the pentagon and $P$ be a point at the five-third mark on the ray $OG$. If $p$ is the corresponding vector, we have $$p = \frac53 \left(\frac{p_1 + p_2 + p_3 + p_4 + p_5}{5}\right) = \frac{p_1 + p_2 + p_3 + p_4 + p_5}{3}$$ Let $[5] = \{ 1, 2, 3, 4, 5 \}$. For any distinct $i, j \in [5]$, let $\ell, m, n$ be the rest of indices from $[5]$.
i.e. $[5]$ is a disjoint union of $\{ i, j \}$ and $\{ \ell, m, n \}$. We have
$$(p_i - p_j) \cdot \left( p - \frac{p_\ell + p_m + p_n}{3}\right) = \frac{(p_i - p_j) \cdot (p_i + p_j)}{3} = \frac{|p_i|^2 - |p_j|^2}{3} = 0$$
This is the equation for point $p$ lying on a line passing through $\frac{p_\ell + p_m + p_n}{3}$ (the centroid of triangle with vertices $p_\ell$, $p_m$, $p_n$ ) perpendicular to line $p_i p_j$.
Substitute $p_i,p_j$ by $p_1,p_2$, we find $P$ lie on a line perpendicular to $AB$ which passes through the centroid of triangle $CDE$. Same thing happens to other sides of the pentagon. As a result, the $5$ lines mentioned in question are concurrent and meet at this specific point $P$.