Name: Sum of Infinite Terms of G.P.
Uploaded: 2019-07-19
Duration: 3 min 36 s
Description: Do you know that we can find the sum of an infinite number of terms of G.P.? Of course, there will be some conditions. Check this video to learn more.

Question

Let

P

is any point inside the triangle

A B C

of side lengths

6, 5, 5

units and

p_{1}, p_{2}, p_{3}

be the lengths of perpendiculars drawn from

P

to the sides of triangle. Find the maximum value of

p_{1} \cdot p_{2} \cdot p_{3} .

Accepted Answer

Let, $∆ A B C,$ be the given triangle with sides $A B = 5, B C = 6, C A = 5$ units, and $p_{1} \cdot p_{2} \cdot p_{3} .$ is a point in the interior of $∆ A B C,$ such that the perpendiculars from $P$ on the sides $B C, C A & A B$ are respectively of lengths $p_{1}, p_{2} & p_{3} .$

Join the point $P$ to $A, B$ and $C,$ so that the given triangle $A B C$ is divided into three triangles $P A B, P B C & P C A .$

Now, we know that, the area of a triangle is $= \frac{1}{2} (b \times h),$ where $b$ is the length of a side and $h$ is the length of perpendicular from the opposite vertex.

Hence, we have $a r (∆ P A B) = \frac{1}{2} (5 p_{3}), a r (∆ P B C) = \frac{1}{2} (6 p_{1})$ and $a r (∆ P C A) = \frac{1}{2} (5 p_{3})$ square units.

Also, the area of a triangle whose sides are of length $a, b & c,$ using Heron's formula is $= \sqrt{s (s - a) (s - b) (s - c)},$ where $s$ is the semi-perimeter of the triangle.

Hence, for the area of $∆ A B C,$ we have $a = 6, b = c = 5$ units, thus $s = \frac{a + b + c}{2} = \frac{6 + 5 + 5}{2} = 8$ units.

And, $a r (∆ A B C) = \sqrt{8 (8 - 6) (8 - 5) (8 - 5)}$

$\Rightarrow a r (∆ A B C) = \sqrt{8 (2) (3) (3)} = 12$ square units.

Now, $a r (∆ A B C) = a r (∆ P A B) + a r (∆ P B C) + a r (∆ P C A)$

$\Rightarrow 12 = \frac{5 p_{3}}{2} + 3 p_{1} + \frac{5 p_{2}}{2}$

$\Rightarrow 6 p_{1} + 5 p_{2} + 5 p_{3} = 24$

$\Rightarrow \frac{6 p_{1}}{5} + p_{2} + p_{3} = \frac{24}{5} . . . (i)$

We know that for positive numbers $A . M . \geq G . M .,$ where $A . M .$ is the arithmetic mean and $G . M .$ is the geometric mean of the numbers.

Also, for $n$ positive numbers, we have $A . M . = \frac{(s u m o f n u m b e r s)}{n}$ and $G . M . = {(p r o d u c t o f n u m b e r s)}^{\frac{1}{n}} .$

Thus, we have $\frac{\frac{6 p_{1}}{5} + p_{2} + p_{3}}{3} \geq {(\frac{6 p_{1}}{5} \times p_{2} \times p_{3})}^{\frac{1}{3}}$

$\Rightarrow \frac{\frac{24}{5}}{3} \geq {(\frac{6}{5} \times p_{1} \cdot p_{2} \cdot p_{3})}^{\frac{1}{3}}$

$\Rightarrow \frac{8}{5} \geq {(\frac{6}{5} \times p_{1} \cdot p_{2} \cdot p_{3})}^{\frac{1}{3}}$

$\Rightarrow \frac{6}{5} \times p_{1} \cdot p_{2} \cdot p_{3} \leq {(\frac{8}{5})}^{3}$

$\Rightarrow p_{1} \cdot p_{2} \cdot p_{3} \leq \frac{512}{125} \times \frac{5}{6}$

$\Rightarrow p_{1} \cdot p_{2} \cdot p_{3} \leq \frac{256}{75} .$

Hence, the maximum value of $p_{1} \cdot p_{2} \cdot p_{3}$ is $\frac{256}{75} .$

Let $P$ is any point inside the triangle $A B C$ of side lengths $6, 5, 5$ units and $p_{1}, p_{2}, p_{3}$ be the lengths of perpendiculars drawn from $P$ to the sides of triangle. Find the maximum value of $p_{1} \cdot p_{2} \cdot p_{3} .$

Important Questions on Sequences and Series

Learn and Prepare for any exam you want !

Let P is any point inside the triangle ABC of side lengths 6,5,5 units and p1,p2,p3 be the lengths of perpendiculars drawn from P to the sides of triangle. Find the maximum value of p1·p2·p3.

Important Questions on Sequences and Series

Learn and Prepare for any exam you want !

Let $P$ is any point inside the triangle $A B C$ of side lengths $6, 5, 5$ units and $p_{1}, p_{2}, p_{3}$ be the lengths of perpendiculars drawn from $P$ to the sides of triangle. Find the maximum value of $p_{1} \cdot p_{2} \cdot p_{3} .$