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Question-15434




Question Number 15434 by mrW1 last updated on 10/Jun/17
Commented by mrW1 last updated on 10/Jun/17
Two circles with radius 6 and 10 and  center at A and B intersect at point C  and D. The distance between A and B  is 12. What is the max. length of  segment EF through C?
$$\mathrm{Two}\:\mathrm{circles}\:\mathrm{with}\:\mathrm{radius}\:\mathrm{6}\:\mathrm{and}\:\mathrm{10}\:\mathrm{and} \\ $$$$\mathrm{center}\:\mathrm{at}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{intersect}\:\mathrm{at}\:\mathrm{point}\:\mathrm{C} \\ $$$$\mathrm{and}\:\mathrm{D}.\:\mathrm{The}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B} \\ $$$$\mathrm{is}\:\mathrm{12}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{max}.\:\mathrm{length}\:\mathrm{of} \\ $$$$\mathrm{segment}\:\mathrm{EF}\:\mathrm{through}\:\mathrm{C}? \\ $$
Commented by ajfour last updated on 10/Jun/17
(EF)_(max) =2×(AB)
$$\left({EF}\right)_{{max}} =\mathrm{2}×\left({AB}\right)\:\: \\ $$
Commented by mrW1 last updated on 10/Jun/17
please explain your proof.
$$\mathrm{please}\:\mathrm{explain}\:\mathrm{your}\:\mathrm{proof}. \\ $$
Commented by ajfour last updated on 10/Jun/17
Please view Q. 15480 , Sir .
$${Please}\:{view}\:{Q}.\:\mathrm{15480}\:,\:{Sir}\:. \\ $$
Commented by Tinkutara last updated on 11/Jun/17
Sir, will EF be maximum when it is  parallel to AB? Because then I can  prove EF = 2AB.
$$\mathrm{Sir},\:\mathrm{will}\:{EF}\:\mathrm{be}\:\mathrm{maximum}\:\mathrm{when}\:\mathrm{it}\:\mathrm{is} \\ $$$$\mathrm{parallel}\:\mathrm{to}\:{AB}?\:\mathrm{Because}\:\mathrm{then}\:\mathrm{I}\:\mathrm{can} \\ $$$$\mathrm{prove}\:{EF}\:=\:\mathrm{2}{AB}. \\ $$
Commented by mrW1 last updated on 11/Jun/17
Yes, EF is max. when EF is parallel  to AB.  We can prove this without much  calculation:  We look at the triangle ΔEDF. We  know no matter where the point E  and F lie, the angle ∠CED and ∠CFD  remain constant. That means the  angle ∠EDF is constant when the  point E and F change. With constant  angle ∠EDF we know EF is maximum  when DE and DF are maximum.  The maximum of DE and DF are  the diameters of the circles. So  the maximum of EF occurs when  DE and DF pass through the centers  of the circles, in this case EF∣∣AB and  EF=2AB.
$$\mathrm{Yes},\:\mathrm{EF}\:\mathrm{is}\:\mathrm{max}.\:\mathrm{when}\:\mathrm{EF}\:\mathrm{is}\:\mathrm{parallel} \\ $$$$\mathrm{to}\:\mathrm{AB}. \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{prove}\:\mathrm{this}\:\mathrm{without}\:\mathrm{much} \\ $$$$\mathrm{calculation}: \\ $$$$\mathrm{We}\:\mathrm{look}\:\mathrm{at}\:\mathrm{the}\:\mathrm{triangle}\:\Delta\mathrm{EDF}.\:\mathrm{We} \\ $$$$\mathrm{know}\:\mathrm{no}\:\mathrm{matter}\:\mathrm{where}\:\mathrm{the}\:\mathrm{point}\:\mathrm{E} \\ $$$$\mathrm{and}\:\mathrm{F}\:\mathrm{lie},\:\mathrm{the}\:\mathrm{angle}\:\angle\mathrm{CED}\:\mathrm{and}\:\angle\mathrm{CFD} \\ $$$$\mathrm{remain}\:\mathrm{constant}.\:\mathrm{That}\:\mathrm{means}\:\mathrm{the} \\ $$$$\mathrm{angle}\:\angle\mathrm{EDF}\:\mathrm{is}\:\mathrm{constant}\:\mathrm{when}\:\mathrm{the} \\ $$$$\mathrm{point}\:\mathrm{E}\:\mathrm{and}\:\mathrm{F}\:\mathrm{change}.\:\mathrm{With}\:\mathrm{constant} \\ $$$$\mathrm{angle}\:\angle\mathrm{EDF}\:\mathrm{we}\:\mathrm{know}\:\mathrm{EF}\:\mathrm{is}\:\mathrm{maximum} \\ $$$$\mathrm{when}\:\mathrm{DE}\:\mathrm{and}\:\mathrm{DF}\:\mathrm{are}\:\mathrm{maximum}. \\ $$$$\mathrm{The}\:\mathrm{maximum}\:\mathrm{of}\:\mathrm{DE}\:\mathrm{and}\:\mathrm{DF}\:\mathrm{are} \\ $$$$\mathrm{the}\:\mathrm{diameters}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circles}.\:\mathrm{So} \\ $$$$\mathrm{the}\:\mathrm{maximum}\:\mathrm{of}\:\mathrm{EF}\:\mathrm{occurs}\:\mathrm{when} \\ $$$$\mathrm{DE}\:\mathrm{and}\:\mathrm{DF}\:\mathrm{pass}\:\mathrm{through}\:\mathrm{the}\:\mathrm{centers} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{circles},\:\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{EF}\mid\mid\mathrm{AB}\:\mathrm{and} \\ $$$$\mathrm{EF}=\mathrm{2AB}. \\ $$
Commented by mrW1 last updated on 11/Jun/17
Commented by ajfour last updated on 11/Jun/17
i found a simpler way , Sir.  simpler to my earlier lengthy  algebraic method. Kindly view it  (herein as another answer).
$${i}\:{found}\:{a}\:{simpler}\:{way}\:,\:{Sir}. \\ $$$${simpler}\:{to}\:{my}\:{earlier}\:{lengthy} \\ $$$${algebraic}\:{method}.\:{Kindly}\:{view}\:{it} \\ $$$$\left({herein}\:{as}\:{another}\:{answer}\right). \\ $$
Answered by ajfour last updated on 11/Jun/17
Commented by ajfour last updated on 11/Jun/17
length APB = AP+PB                            = 2(MP+PN)                            = 2(MN)                            = 2(d cos θ)                            = ((2d)/( (√(1+tan^2 θ))))     maximum length of APB                            = 2d    (for θ=0)      that is when it is // to C_1 C_2  .
$${length}\:{APB}\:=\:{AP}+{PB} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\left({MP}+{PN}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\left({MN}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\left({d}\:\mathrm{cos}\:\theta\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{2}{d}}{\:\sqrt{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \theta}}\: \\ $$$$\:\:{maximum}\:{length}\:{of}\:{APB} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}{d}\:\:\:\:\left({for}\:\theta=\mathrm{0}\right)\: \\ $$$$\:\:\:{that}\:{is}\:{when}\:{it}\:{is}\://\:{to}\:{C}_{\mathrm{1}} {C}_{\mathrm{2}} \:. \\ $$
Commented by mrW1 last updated on 11/Jun/17
you have shown in Q15480 that  the maximum is when m=0, m  is the inclination of line EF.
$$\mathrm{you}\:\mathrm{have}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{Q15480}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{maximum}\:\mathrm{is}\:\mathrm{when}\:\mathrm{m}=\mathrm{0},\:\mathrm{m} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{inclination}\:\mathrm{of}\:\mathrm{line}\:\mathrm{EF}. \\ $$
Commented by ajfour last updated on 11/Jun/17
yes,but the same expression for  length of EF =((2d)/( (√(1+m^2 )))) is arrived  at by this method with less effort.
$${yes},{but}\:{the}\:{same}\:{expression}\:{for} \\ $$$${length}\:{of}\:{EF}\:=\frac{\mathrm{2}{d}}{\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }}\:{is}\:{arrived} \\ $$$${at}\:{by}\:{this}\:{method}\:{with}\:{less}\:{effort}. \\ $$
Commented by mrW1 last updated on 11/Jun/17
this is better and clear!
$$\mathrm{this}\:\mathrm{is}\:\mathrm{better}\:\mathrm{and}\:\mathrm{clear}! \\ $$

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