Problem #11

Consider any triangle ABC. Find a construction for a point P such that the sum of the
distances from P to each of the three vertices is a minimum.

LetÕs explore the triangle ABC below with the indicated
medians:

__INITIAL HYPOTHESES__

1) The point P would be inside the triangle, because any
point outside the triangle would be equidistant to the nearest vertex(es)
inside and would also have a greater distance from the farthest
vertex(es). The extra distance
will yield a greater sum than what is required.

2) The point P is a center of a triangle and is equidistant
from each vertex.

LetÕs consider four specific centers for a triangle:

__Centroid__ (G)- the common intersection of the three
medians, which are formed from segments, constructed using a vertex and the
midpoint of the opposite side.

__Orthocenter (H)__- The common intersection of the three
lines containing the altitudes.

__Circumcenter__ (C)- is the point in the plane that is
equidistant from the three vertices of the triangle.

__Incenter__ (I) is the point on the interior of the
triangle that is equidistant from the three sides.

The points C, G, I, and H are all centers of the acute
triangle ABC. Because all of the
angels at each vertex are acute, all of the centers are located within the
interior of the triangle.

Now letÕs reconstruct the perpendicular bisectors which will
have a common point C which is the Circumcenter (C).

The red dashed lines indicate the perpendicular bisectors to the sides of the triangle. The segments ACÕ, CCÕ, and BCÕ which are represented by the green dashed lines, are all congruent. Since C is equidistant to all of the vertices, then CÕ is my point ÔPÕ where the sum of the segments from that point to the vertices are a minimum.

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