# Some Standout methods of approach

## The Area Way

Incases , where a relation with areas is possible, It is quite efficient to use it to solve the problem.

Example –

If $h_a , h_b , h_c$ are the altitudes of a triangle from A,B,C to the opposite sides and r is the inradius. Prove that –

1. $\frac{1}{r} = \frac{1}{h_a} + \frac{1}{h_b} + \frac{1}{h_c}$

2. $h_a + h_b + h_c \ge 9r$

### Proof

Let I be the incentre.

Note that ,

$\frac{(IBC)}{(ABC)} = \frac{ra}{h_a}$

and that ,

$\sum (IBC) = (ABC)$

Hence proved.

2. By  Cauchy ,

$(h_a + h_b + h_c)(\frac{1}{h_a} + \frac{1}{h_b} + \frac{1}{h_c}) \ge 9$

And using the identity, we are done.

3. Let ABC be a triangle with H as its orthocentre , and AD,BE,CF its altitudes.

Prove that –

$\frac{AD}{HD} + \frac{BE}{HE} + \frac{CF}{HF} \ge 9$

### Proof

Both AD and HD are altitudes to the same base BC , Hence the areas of triangles ABC and HBC are in the ratio of their altitudes (*)

Let, $(ABC) = S , (HBC) = S_1 , (HCA) = S_2 , (HAB) = S_3$

Thus , the inequality is equivalent to –

$\frac{1}{S_1}+\frac{1}{S_2}+\frac{1}{S_3} \ge \frac{9}{S}$

Note that – $S = S_1 + S_2 + S_3$

Hence it is true by Cauchy-Schwarz Inequality.

4. With the Same notations prove that –

$\frac{HD}{HA}+\frac{HE}{HB}+\frac{HF}{HC} \ge \frac{3}{2}$

### Proof

Note that –

$latex \frac{HD}{HA} = \frac{HD}{AD – HD} = \frac{S_1}{S-S_1} = \frac{S_1}{S_2+S_3}$

Thus ,

$LHS = \sum \frac{S_1}{S_2+S_3} \ge \frac{3}{2}$

The last inequality is the famous Nesbitt’s Inequality. Hence Proved.

Also note that – equality occurs when $S_1 = S_2 = S_3 = \frac{S}{3}$

Or, when the triangle is equilateral.

## Exercises

1. ABC triangle, L,M,N are points on BC,CA,AB and P,Q,R are intersections of AL,BM,CN with the circum-circle of the triangle. Prove that –

$\frac{AL}{LP} + \frac{BM}{MQ} + \frac{CN}{NR} \ge 9$

(Source : Korea 1995)

2.  Let AD,BE,CF be altitudes from A,B,C of triangle ABC and PQ , PR, PS be projections of P on BC,CA,AB . Prove that –

$\frac{AD}{PQ} + \frac{BE}{PR} + \frac{CF}{PS} \ge 9$

3. The altitudes AD , BE , CF intersect the circum-circles at D’ , E’ , F’ . The prove that –

(a) $\frac{AD}{DD'} + \frac{BE}{EE'} + \frac{CF}{FF'} \ge 9$

(b) $\frac{AD}{AD'} + \frac{BE}{BE'} + \frac{CF}{CF'} \ge \frac{9}{4}$

## The Equilateral Usage

In almost every geometrical inequality, we would note that , the equality condition refers to the equilateral condition – triangle or any polygon.

There has to be some way of utilising this fact right ? , Well there IS .

We can illustrate by a different and beautiful proof of our Weitzenbock’s Inequality –

For a triangle ABC , Prove that –

$a^2 + b^2 + c^2 \ge 4\sqrt{3}\Delta$

where, $\Delta$ is the Area of the triangle.

### Proof

Consider the equilateral triangle of length $a$ , It has altitude – $\frac{\sqrt{3}a}{2}$ and area – $\frac{\sqrt{3}a^2}{4}$

Consider the Given triangle – ABC , with altitude – AD – h.

Let AD divide BC into – $\frac{a}{2} - x$ and $\frac{a}{2} + x$

For our triangle ABC , let , $h = \frac{\sqrt{3}a}{2} + y$ .

Therefore,

$A = a^2 + b^2 + c^2 - 4\sqrt{3}\Delta$

$A = a^2 + h^2 + (\frac{a}{2} - x)^2 + h^2 + (\frac{a}{2} + x)^2 - 4\sqrt{3}\cdot \frac{ah}{2}$

$A = \frac{3a^2}{2} + 2(\frac{\sqrt{3}a}{2}+y)^2 + 2x^2 - 3a^2 - 2\sqrt{3}ay$

$A = 2(x^2+y^2) \ge 0$

with equality occuring only when – $x=y=0 \implies a=b=c$