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.

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## Friday, September 23, 2016

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

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Problem 1263

ReplyDeleteLet 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

Let G be the centroid of the triangle ABC

ReplyDeleteand 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

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

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

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

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

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

ReplyDeleteRegret the error

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.

ReplyDelete