Let a - b and c be the side lengths of a triangle ABC and assume that a - b and a - c - If x - b - c - a -2- then the minimum value of ax - rR- where r and R denote the inradius and circumradius- respectively of triangle ABC- is

Let y=axrR=a(b+c a)2·s·abc4 =a(b+c a)(b+c+a)abc =(b+c)^2 a^2bc ⇒ y=(bc+cb+2) a· abc As a≤ b and a≤ c ⇒ab≤1 and ac≤1 ∴a^2bc≤1 ∴ y≥(bc+cb+2) 1 Using A.M. ≥ G.M., ...

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Standard XI

Mathematics

Single Point Continuity

Let a , b and...

Question

Let $a, b$ and $c$ be the side lengths of a triangle $A B C$ and assume that $a \leq b$ and $a \leq c .$ If $x = \frac{b + c - a}{2},$ then the minimum value of $\frac{a x}{r R},$ where $r$ and $R$ denote the inradius and circumradius, respectively of triangle $A B C,$ is

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Solution

Let $y = \frac{a x}{r R} = \frac{a (b + c - a)}{2 \cdot \frac{△}{s} \cdot \frac{a b c}{4 △}}$
$= \frac{a (b + c - a) (b + c + a)}{a b c}$
$= \frac{(b + c)^{2} - a^{2}}{b c}$
$\Rightarrow y = (\frac{b}{c} + \frac{c}{b} + 2) - \frac{a \cdot a}{b c}$
As $a \leq b$ and $a \leq c$
$\Rightarrow \frac{a}{b} \leq 1$ and $\frac{a}{c} \leq 1$
$∴ \frac{a^{2}}{b c} \leq 1$
$∴ y \geq (\frac{b}{c} + \frac{c}{b} + 2) - 1$
Using A.M. $\geq$ G.M.,
$\frac{b}{c} + \frac{c}{b} \geq 2$
$\Rightarrow y \geq 2 + 2 - 1$ or $y \geq 3$
$\Rightarrow y_{min} = 3$ (equality holds when $a = b = c$ )

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Let a,b and c be the side lengths of a triangle ABC and assume that a≤b and a≤c. If x=b+c−a2, then the minimum value of axrR, where r and R denote the inradius and circumradius, respectively of triangle ABC, is

Let y=axrR=a(b+c−a)2⋅△s⋅abc4△ =a(b+c−a)(b+c+a)abc =(b+c)2−a2bc ⇒y=(bc+cb+2)−a⋅abc As a≤b and a≤c ⇒ab≤1 and ac≤1 ∴a2bc≤1 ∴y≥(bc+cb+2)−1 Using A.M. ≥ G.M., bc+cb≥2 ⇒y≥2+2−1 or y≥3 ⇒ymin=3 (equality holds when a=b=c)

Let $a, b$ and $c$ be the side lengths of a triangle $A B C$ and assume that $a \leq b$ and $a \leq c .$ If $x = \frac{b + c - a}{2},$ then the minimum value of $\frac{a x}{r R},$ where $r$ and $R$ denote the inradius and circumradius, respectively of triangle $A B C,$ is