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Question Number 121615 by ajfour last updated on 10/Nov/20

Commented by ajfour last updated on 10/Nov/20

The blue triangle is right angled and isosceles with AF=BF=3. Radius of circle r=2. Find maximum side length of equilateral △DEF.

$${The}\:{blue}\:{triangle}\:{is}\:{right}\:{angled}\:{and} \\ $$$${isosceles}\:{with}\:\:{AF}={BF}=\mathrm{3}. \\ $$$${Radius}\:{of}\:{circle}\:{r}=\mathrm{2}. \\ $$$${Find}\:{maximum}\:{side}\:{length}\:{of} \\ $$$${equilateral}\:\bigtriangleup{DEF}. \\ $$

Commented by MJS_new last updated on 10/Nov/20

(1) start with the blue triangle (2) D on line AB (3) construct E ⇒ we have the minimal side s when D=((A+B)/2) ⇒ the maximum depends on the possibility of the circle which exists for ∣AE∣≤4 if D is between A and B and closer to A the maximum is at D=A if we allow D anywhere on the line through A and B it′s harder to calculate...

$$\left(\mathrm{1}\right)\:\mathrm{start}\:\mathrm{with}\:\mathrm{the}\:\mathrm{blue}\:\mathrm{triangle} \\ $$$$\left(\mathrm{2}\right)\:{D}\:\mathrm{on}\:\mathrm{line}\:{AB} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{construct}\:{E} \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{have}\:\mathrm{the}\:\mathrm{minimal}\:\mathrm{side}\:{s}\:\mathrm{when}\:{D}=\frac{{A}+{B}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{depends}\:\mathrm{on}\:\mathrm{the}\:\mathrm{possibility} \\ $$$$\:\:\:\:\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{which}\:\mathrm{exists}\:\mathrm{for}\:\mid{AE}\mid\leqslant\mathrm{4} \\ $$$$\:\:\:\:\:\mathrm{if}\:{D}\:\mathrm{is}\:\mathrm{between}\:{A}\:\mathrm{and}\:{B}\:\mathrm{and}\:\mathrm{closer}\:\mathrm{to}\:{A} \\ $$$$\:\:\:\:\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{is}\:\mathrm{at}\:{D}={A} \\ $$$$\:\:\:\:\:\mathrm{if}\:\mathrm{we}\:\mathrm{allow}\:{D}\:\mathrm{anywhere}\:\mathrm{on}\:\mathrm{the}\:\mathrm{line}\:\mathrm{through} \\ $$$$\:\:\:\:\:{A}\:\mathrm{and}\:{B}\:\mathrm{it}'\mathrm{s}\:\mathrm{harder}\:\mathrm{to}\:\mathrm{calculate}... \\ $$

Commented by ajfour last updated on 10/Nov/20

wont be as tough sir, please arrive at an answer, i shall try the normal easy way, howsoever lengthy..

$${wont}\:{be}\:{as}\:{tough}\:{sir},\:{please}\:{arrive} \\ $$$${at}\:{an}\:{answer},\:{i}\:{shall}\:{try}\:{the}\:{normal} \\ $$$${easy}\:{way},\:{howsoever}\:{lengthy}.. \\ $$

Commented by MJS_new last updated on 10/Nov/20

ok but later

$$\mathrm{ok}\:\mathrm{but}\:\mathrm{later} \\ $$

Commented by ajfour last updated on 10/Nov/20

i think for s_(max) D is at A, s_(max) =3.

$${i}\:{think}\:{for}\:{s}_{{max}} \:\:{D}\:{is}\:{at}\:{A},\:\:{s}_{{max}} =\mathrm{3}. \\ $$

Commented by MJS_new last updated on 10/Nov/20

yes. if D is allowed only between A and B. else s_(max) =4

$$\mathrm{yes}.\:\mathrm{if}\:{D}\:\mathrm{is}\:\mathrm{allowed}\:\mathrm{only}\:\mathrm{between}\:{A}\:\mathrm{and}\:{B}. \\ $$$$\mathrm{else}\:{s}_{{max}} =\mathrm{4} \\ $$

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