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MATHMAA

# 16th KVJ Solutions10

Here you find the solutions of  16th KVJ Solutions10 in detail explanation .

10Q) Let PQRS be a rectangle such that PQ= a and QR =b. Suppose $$r_{1}$$ is the radius of the circle passing through P and Q and touching RS and $$r_{2}$$ is the radius of the circle passing through Q and R and touching PS. Show that :

$$5(a+b) \leq 8(r_{1}+r_{2} )$$.

Solution:

Please see the figure below for this Question.

PQRS be a rectangle .

PQ= a , QR =b .

Let B be the center , $$r_{1}$$ be the radius , draw a circle passing through P , Q and touches RS.

Let E be the center , $$r_{2}$$ be the radius , draw a circle passing through Q, R and touches PS.

BC $$\perp$$ RS and C is the midpoint of RS, extend CB to meet PQ at D.

PD=DQ=$$\frac{a}{2}$$

CD =b, BD+BC =b

BD =b - $$r_{1}$$ ---------(1)

From triangle BDQ ,

$$BD^{2}= BQ^{2} - DQ^{2}$$

BD = $$\sqrt{(r_{1})^{2} - \frac{a^{2}}{4}}$$ -----(2)

From (1) and (2)

b - $$r_{1}$$ = $$\sqrt{(r_{1})^{2} - \frac{a^{2}}{4}}$$

Squaring on both sides, we get

$$b^{2} -2br_{1} +(r_{1})^{2} = (r_{1})^{2} - \frac{a^{2}}{4}$$

$$2br_{1} = b^{2} + \frac{a^{2}}{4}$$

$$r_{1} = \frac{b}{2} + \frac{a^{2}}{8b}$$ -------(3)

Similarly we get,

$$r_{2} = \frac{a}{2} + \frac{b^{2}}{8a}$$ -------(4)

From (3) and (4) , we add

$$r_{1} + r_{2} = \frac{a+b}{2} + \frac {a^{3}+b^{3}}{8ab}$$

$$r_{1} + r_{2} = \frac{a+b}{2} + \frac{(a+b)(a^{2}-ab+b^{2}}{8ab}$$

$$r_{1} +r_{2} = (a+b)(\frac{1}{2} + \frac{a^{2}-ab+b^{2}}{8ab})$$

$$r_{1} + r_{2} =(a+b)(\frac{1}{2} + \frac{(a-b)^{2}+ab}{8ab})$$

$$r_{1} +r_{2} = (a+b)(\frac{1}{2} + \frac{(a-b)^{2}}{8ab} +\frac{1}{8})$$

$$r_{1} + r_{2} = (a+b)(\frac{5}{8} + \frac{(a-b)^{2}}{8ab})$$

$$r_{1} +r_{2} \geq \frac{5(a+b)}{8}$$

$$8(r_{1}+r_{2}) \geq 5(a+b)$$

Therefore $$5(a+b) \leq 8(r_{1}+r_{2})$$.

Hence proved.

This  is the  explanation for  16th KVJ Solutions10 . Please feel free to share you opinion on this .

Figure 10