We need to locate the circumcenter, incenter, centroid, and orthocenter of each triangle. Let's deal with one thing at a time.
Circumcenter
We can recall that a circumcenter is a point of intersection of perpendicular bisectors of a triangle. A perpendicular bisector intersects a segment at its midpoint and is perpendicular to the segment. Let's draw perpendicular bisectors to each side of each triangle. Points of their intersection are the circumcenters of the triangle. We can name them C.
Incenter
An incenter of a triangle is a point of concurrency of the triangle angle bisectors. Thus, let's draw angle bisectors in each triangle and find their point of concurrency. We can name these incenters I.
Centroid
A centroid of a triangle is a point of intersection of the triangle medians. We can review that a median of a triangle is a segment that connects a vertex of the triangle with the midpoint of the opposite to the vertex side. Let's draw medians in each triangle. Points of their intersection are the centroids of the triangles, which we can name B.
Orthocenter
An orthocenter of a triangle is a point of intersection of the triangle altitudes. Let's recall that an altitude is a segment from a vertex to the line containing the opposite side and perpendicular to the line containing that side. We are going to draw altitudes to each side of each triangle. Points of their intersection are the orthocenters, which we can name O.
b
Analyzing the diagrams from Part A, we can see that four points of concurrency are situated at the same point for each triangle. This allows us to make a conjecture that a circumcenter, incenter, centroid, and orthocenter are the same point in equilateral triangles.
c
Let's draw an equilateral triangle â–³ ABC on a coordinate plane. We will place A at the origin, so that AC lies on x-axis. Let the length of one unit be a. Point C lies four units to the right from A, so its coordinates are (4a,0).
Now we can draw an altitude BK to the side AC, and consider the right triangle â–³ BKC. In an equilateral triangle, an altitude of a side is also a median of that side. Thus, K is a midpoint of AC, so it has the coordinates (2a,0).
Calculating the difference of the x-coordinates of K and C, we can find the length of KC. It is 4a-2a= 2a. The measure of AC is 4a. Since â–³ ABC is an equilateral triangle, segment BC also measures 4a.
We want to find the measure of the altitude BK. â–³ BKC is a right triangle, so we can use the Pythagorean Theorem.
BK^2+KC^2=BC^2
Since we know the measures of KC and BC, we can substitute KC with 2a and BC with 4a. This way we get an equation where the only unknown is BK. Let's solve the equation and find its value.
The measure of BK is 2asqrt(3), which means that point B is located 2asqrt(3) units above the x-axis. Thus, its coordinates are (2a,2asqrt(3)).
To locate the point of concurrency, we can use the Centroid Theorem. Let's recall what it states.
The medians of a triangle inersect
at a point called the centroid that is
two thirds of the distance from each vertex
to the midpoint of the opposite side.
According to this theorem, a centroid of â–³ ABC is situated two thirds of the distance BK. Thus, multiplying 2asqrt(3) by 23, we get the distance from B to the centroid D.
BD=2/3* 2asqrt(3)=4asqrt(3)/3
Let's add this information on our diagram.
As we can see, D is situated 4asqrt(3)3 lower than point B. Hence, if we subtract 4asqrt(3)3 from point's B y-coordinate 2asqrt(3), we will find the y-coordinate of D.
The coordinates of the centroid D of â–³ ABC are (2a, 2asqrt(3)3). From Part B, we know that a circumcenter, incenter, centroid, and orthocenter are located at the same point in an equilateral triangle. Therefore, these are the coordinates for each of these points of concurrency.
