Triangles (2).

Geometry Level 5

Let $P$ be a variable point inside $\triangle ABC$ and let $D,E,$ and $F$ be the feet of perpendiculars from $P$ to sides $BC,CA,$ and $AB,$ respectively. If $\dfrac{BC}{PD} + \dfrac{CA}{PE} + \dfrac{AB}{PF} \ge \dfrac{a s^2}{A},$ where $s$ and $A$ denote the semi-perimeter and the area of $\triangle ABC,$ respectively, then find the maximum value of $a.$

The answer is 2.

2 solutions

A Former Brilliant Member
Mar 31, 2018

Let : $\triangle _{3} , \triangle _{2} , \triangle _ {1}$ be the areas of triangles $\text{BPC, CPA \& BPA}$ respectively .

So,

$2 \triangle _{3}=\text{BC} \times \text{PD}$

$2 \triangle _{2}=\text{AC} \times \text{PE}$

$2 \triangle _{1}=\text{AB} \times \text{PF}$

Using above relations , we get :

$\dfrac{\text{BC}}{\text{PD}} + \dfrac{\text{CA}}{\text{PE}} + \dfrac{\text{AB}}{\text{PF}} = \dfrac{\text{AB}^2}{2\triangle _{1}} +\dfrac{\text{BC}^2}{2\triangle _{3}} + \dfrac{\text{CA}^2}{2\triangle _{2}}$

Using Titu's Lemma , we get :

$\dfrac{\text{AB}^2}{2\triangle _{1}} +\dfrac{\text{BC}^2}{2\triangle _{3}} + \dfrac{\text{CA}^2}{2\triangle _{2}} \ge \dfrac{(\text{AB}+\text{BC}+\text{CA})^2}{2(\triangle _{1} + \triangle _{2} + \triangle _{3})}$

$\implies \dfrac{\text{AB}^2}{2\triangle _{1}} +\dfrac{\text{BC}^2}{2\triangle _{3}} + \dfrac{\text{CA}^2}{2\triangle _{2}} \ge \dfrac{2s^2}{A}$

Hence , $\boxed{a=2}$

Sahil Silare
Mar 30, 2018

Assume the triangle to be equilateral with side $x$ , then just plug in the values.

Triangles (2).

The answer is 2.

2 solutions

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