Menu Close

Question-15434

Tinku Tara 0 Comments
Pin




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!
$$\mathrm{this}\:\mathrm{is}\:\mathrm{better}\:\mathrm{and}\:\mathrm{clear}! \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *