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Singapore Math Olympiad

Centroid of Triangle | SMO, 2009 | Problem 1

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2009 based on Centroid of Triangle. You may use sequential hints to solve the problem.

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.

Singapore Mathematics Olympiad

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.

Centroid of Triangle- Problem

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

Centroid of Triangle

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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