Download 5/1/16 - Art of Problem Solving

5/1/16
Jesse Cohen
The three medians of a triangle all intersect at a single point which is called the
centroid. The centroid divides each median such that the distance from the centroid to the
vertex is twice the distance along the median from the centroid to the opposite edge. By
applying this to the diagram we get AG = 2 and GE = ½.
Looking at the angles, since BDG is equilateral, each angle in the triangle is 60
degrees. GDC = 120 degrees because it and BDG are supplementary to each other. Since D
is the midpoint of BC then DC must equal BD which equals 1, making GDC isosceles.
Therefore angles G and C of GDC must be equal and since the third angle of the
triangle is 120, G and C must each measure 30 degrees. Now if we drop a perpendicular
from Vertex D of triangle DGC intersecting GC at P, it will divide DGC into two
congruent 30-60-90 right triangles (GDP and CDP). Since we know the ratios of the
sides we can easily figure out that GP=CP= sqrt(3)/2 and that DP = ½.
Since we know that BGD = 60 and DGC = 30, we can figure out that CGE
which is the supplement of the two angles measures 90 degrees. Since EGC is a right
triangle and we know EG (1/2) and GC (sqrt(3)), EC must equal sqrt(13)/2. Since E is the
midpoint of AC EC=AE and therefore AC, the side of the triangle is equal to sqrt(13).
We now know the length of GC ( sqrt (3) ). Since G is the centroid GF is half GC or,
sqrt(3)/2. Since FGB is a vertical angle to EGC, it measures 90 degrees and we now
A have another right triangle, triangle FGB. We know side GB (1) and we just
found the value of FG. Therefore side FB = sqrt(7)/2. Since F is the midpoint
of AB, FBAF=sqrt(7)/2. Therefore, side FB measures sqrt(7).
The lengths of the sides of the triangle are: AB = sqrt (7), BC = 2, and
CA = sqrt (13)
2
13
7
E
F
3
1/2
30°
60°
2
G 90°
90°
60°
30°
3
1
3
2
1
P
1/2
60°
60°
B
1
60°
60°
D
1
30°
C