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# Centroid of Triangle | SMO, 2009 | Problem 1

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2009 based on Centroid of Triangle.

## Problem - Centroid of Triangle (SMO Entrance)

Let M and N be points on sides AB and AC of triangle ABC respectively. If $\frac {BM}{MA} + \frac {CN}{NA} = 1$ . Can we show that MN passes through the centroid of ABC?

### Key Concepts

2D - Geometry

Triangle

Menelaus's Theorem

## Check the Answer

Answer: Yes , we can.

Challenges and Thrills - Pre College Mathematics

## Try with Hints

If we got stuck in this problem then we can start this problem by applying Menelaus's Theorem.

It states : if a line intersects $\triangle ABC$ or extended sides at points D, E, and F, the following statement holds: $\frac {AD}{BD} \times \frac {BE}{EC} \times \frac {CF}{AF} = 1$

Again let D is the mid point of AC. As $\frac {BM}{MA} + \frac {CN}{NA} = 1$ then $\frac {CN}{NA}<1$ where N lies in the line segment CD.From the picture above we can see g is the intersection point between two lines BD and MN. So if we apply Menelaus's Theorem we get :

$\frac {DG}{GB} . \frac {BM}{MA} .\frac {AN}{ND} = 1$

Now try the rest of the problem.....................................

After the 1st hint again, $\frac {BG}{GD} = \frac{BM}{MA} . \frac {AN}{ND} = ( 1 - \frac {CN}{NA}). \frac {AN}{ND}$

= $\frac {NA - CN}{ND} = \frac {(2CD - CN) - CN}{ND}$

=$\frac {2 ND}{ND}$ = 2

Thus G is the centroid .(Proved)

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