## Friday, September 23, 2016

### Geometry Problem 1263: Triangle, Medians, Midpoint, 90 Degrees, Perpendicular

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

Click on the figure below to view more details of problem 1263.

1. Problem 1263
Let K is centroid of the triangle ABC with BF=3.KF. Then
3.KF=3.AC/2 or KF=AC/2 (AH=HC).Therefore AD is perpendicular CE.
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

2. Let G be the centroid of the triangle ABC
and AC = 2 units => BF = 3
Since G divides BG:GF = 2:1 => GF = 1
So triangle AFG and ACG are isosceles triangles.
Consequitively we can form a circle with Center F and radius = AF = FC = FG = 1 unit.
Thus AC is the diameter of the circle and G is a point on the circle
So Angle AGC is always 90

3. Interestingly Area(ABC) = 3AD.CE because the 6 smaller triangles around the centroid are equal in area

1. While the smaller triangles are the same area, they are not equal in area to 1/2 AD.CE , hence the statement above is incorrect :(

4. It will be more challenging if asked what should be ratio AC/BF in-order 2 medians be perpendicular ?

1. In a way. For the medians to be perpendicular GF = AF = FC. It follows that AC/BF must be 2/3.

5. Each of the triangles about the centroid has area AD.CE/9 and the area of Tr. ABC = (2/3)AD.CE

Regret the error

6. Let G be the centroid of the triangle, Let AC=2a, since BF=1.5AC, then BF=3a, by median concurrency theorem, BG=2a, and GF=AF=CF=a, and GF, AF, and CF are radii of a semi-circle with center F and diameter AC, triangle AGC is inscribed in this semi-circle, therefore AGC is always a right angle.