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National BDMO 2016 : Junior 9

Posted: Wed Jan 11, 2017 11:47 pm
Area of $\bigtriangleup ABC$ is $2016$. $D,E,F$ are three points on the sides $BC,AB,AC$ respectively. Show that, the area of at least one triangle among $\bigtriangleup AEF$, $\bigtriangleup BDE$, $\bigtriangleup CDF$ is not larger than $504$ square units.

Re: National BDMO 2016 : Junior 9

Posted: Thu Jan 12, 2017 6:51 pm
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Corrected statement: Area of $\triangle ABC$ is 2016 square units. $D,E,F$ are three points on the sides $BC,AB,AC$ respectively. Show that, the area of at least on triangle among $\triangle AEF$, $\triangle BDE$, $\triangle CDF$ is not larger than $504$ square units.

We have to show that at least one of the colored triangles is less than or equal to one fourth of the original triangle. Now, let $R=\dfrac{\text{Area of red colored triangle}}{\text{Area of full triangle}}$. Define $B$ and $G$ similarly. Now, define $p,q,r,u,v,w$ as in the diagram. Check the following equalities:

$$R=\dfrac{pw}{(r+w)(p+u)}$$

$$G=\dfrac{qu}{(p+u)(q+v)}$$

$$B=\dfrac{rv}{(q+v)(r+w)}$$

Multiplying them, we get $RGB=\dfrac{pqruvw}{(p+u)^2(q+v)^2(r+w)^2}$. But we know that $(a+b)^2=4ab+(a-b)^2\ge 4ab$. So, it is clear that, $RGB\le\dfrac{1}{4^3}$. But if each of $R,G$ and $B$ are greater than $\dfrac{1}{4}$ then this won't be true. So, at least one of them is less than or equal to $\dfrac{1}{4}$ and we are done.