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