Geometry (with some Algebra, too)
Geometry
Level
3
Given a triangle ABC. Draw 3 points D, E, F on AB, AC, BC (in order). 3 segments AF, BE, CD intersect at point K.
Which condition must happen so that the expression below happens?
ABC is a right triangle
All are incorrect (except this one)
ABC is an isosceles triangle
K is the centroid of triangle ABC.
ABC is an equilateral triangle
K is the circumcenter of triangle ABC.
K is the orthocenter of triangle ABC.
K is the incenter of triangle ABC.
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1 solution
Symbols meaning:
S < n a m e − o f − a − t r i a n g l e > refers to the area of that triangle.
S 1 refers to the area of the triangle AKC. S 2 refers to the area of the triangle BKC. S 3 refers to the area of the triangle AKB (as shown in the picture above).
Solution:
We know that if 2 triangles share the same altitude, the ratio between their corresponding bases is equal to the ratio between their areas.
Look at the picture, we can see that the 2 triangles AKB and KBF share the same altitude. Therefore, K F A K = S K B F S 3 .
2 triangles AKC and CKF also share the same altitude. Therefore, K F A K = S K C F S 1 .
Hence, K F A K = S K C F S 1 = S K B F S 3 = S K B F + S K C F S 3 + S 1 = S 2 S 3 + S 1 .
If you keep on using this method, we can find K E B K = S 1 S 3 + S 2 and K D C K = S 3 S 2 + S 1 .
Therefore, K F A K + K E B K + K D C K = S 2 S 3 + S 1 + S 1 S 3 + S 2 + S 3 S 2 + S 1 = ( S 2 S 3 + S 3 S 2 ) + ( S 2 S 1 + S 1 S 2 ) + ( S 3 S 1 + S 1 S 3 )
Now we use the inequality b a + a b ≥ 2 (an example of Direct Application of AM-GM to an Inequality )
This results in K F A K + K E B K + K D C K ≥ 2 + 2 + 2 = 6
The equality K F A K + K E B K + K D C K = 6 happens if and only if S 1 = S 2 = S 3 .
We note that the 3 segments AK, BK and CK divide the triangle ABC into 3 smaller triangles with the same area. Therefore, K must be the c e n t r o i d of triangle ABC (because 3 segments made by connecting the vertices with the centroid divide the triangle into 3 smaller triangles with the same area)