Question-182006

Question Number 182006 by mr W last updated on 03/Dec/22

Commented by mr W last updated on 03/Dec/22

find the largest area of an inscribed right−angle triangle in a given acute triangle with sides a,b,c. (a≥b≥c)

$${find}\:{the}\:{largest}\:{area}\:{of}\:{an}\:{inscribed} \\ $$$${right}−{angle}\:{triangle}\:{in}\:{a}\:{given}\:{acute} \\ $$$${triangle}\:{with}\:{sides}\:{a},{b},{c}.\:\left({a}\geqslant{b}\geqslant{c}\right) \\ $$

Commented by Acem last updated on 03/Dec/22

$${Interesting}\:{one}! \\ $$

Answered by mr W last updated on 03/Dec/22

Commented by mr W last updated on 03/Dec/22

A_(max) =((a^2 sin γ cos γ)/2)=((a^2 sin 2γ)/4)

$${A}_{{max}} =\frac{{a}^{\mathrm{2}} \:\mathrm{sin}\:\gamma\:\mathrm{cos}\:\gamma}{\mathrm{2}}=\frac{{a}^{\mathrm{2}} \:\mathrm{sin}\:\mathrm{2}\gamma}{\mathrm{4}} \\ $$

Commented by Acem last updated on 03/Dec/22

The 2nd one_(below) = ((a^2 sin β cos β)/2)= ((a^2 sin 2β)/4) β> γ then the 2nd triangle is larger than the 1st

$${The}\:\mathrm{2}{nd}\:{one}_{{below}} =\:\frac{{a}^{\mathrm{2}} \:\mathrm{sin}\:\beta\:\mathrm{cos}\:\beta}{\mathrm{2}}=\:\frac{{a}^{\mathrm{2}} \:\mathrm{sin}\:\mathrm{2}\beta}{\mathrm{4}} \\ $$$$\:\beta>\:\gamma\:{then}\:{the}\:\mathrm{2}{nd}\:{triangle}\:{is}\:{larger}\:{than}\:{the}\:\mathrm{1}{st} \\ $$

Commented by mr W last updated on 03/Dec/22

don′t you see that β>γ doesn′t automatically mean that the second triangle is larger than the first one? look at the diagram below.

$${don}'{t}\:{you}\:{see}\:{that}\:\beta>\gamma\:{doesn}'{t}\: \\ $$$${automatically}\:{mean}\:{that}\:{the}\:{second}\: \\ $$$${triangle}\:{is}\:{larger}\:{than}\:{the}\:{first}\:{one}?\: \\ $$$${look}\:{at}\:{the}\:{diagram}\:{below}. \\ $$

Commented by mr W last updated on 03/Dec/22

Commented by mr W last updated on 03/Dec/22

Commented by mr W last updated on 03/Dec/22

this picture shows that the first triangle (=((a^2 sin 2γ)/4)) is always larger than the second one (=((a^2 sin 2β)/4)).

$${this}\:{picture}\:{shows}\:{that}\:{the}\:{first} \\ $$$${triangle}\:\left(=\frac{{a}^{\mathrm{2}} \:\mathrm{sin}\:\mathrm{2}\gamma}{\mathrm{4}}\right)\:{is}\:{always}\:{larger} \\ $$$${than}\:{the}\:{second}\:{one}\:\left(=\frac{{a}^{\mathrm{2}} \mathrm{sin}\:\mathrm{2}\beta}{\mathrm{4}}\right). \\ $$

Commented by Acem last updated on 03/Dec/22

Commented by Acem last updated on 03/Dec/22

We may say here that △MAC > BAN It′s ok But, Area_(MAC) = (1/2) b ∣AM∣= (1/2) b^( 2) tan γ So how can we judge? cause all your posts made me think that the maximum area is when the right angle be on the 2nd side “middle length” & the hypotenuse is whole of the side “a” with a formula ((a^2 sin 2γ)/4) ∣^( ?) (1/2) b^( 2) tan γ Note: I thought through diagram 1 that you say that there′s 2 triangles as maximum

$${We}\:{may}\:{say}\:{here}\:{that}\:\bigtriangleup{MAC}\:>\:{BAN} \\ $$$$\:{It}'{s}\:{ok}\:{But},\:{Area}_{{MAC}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\:{b}\:\mid{AM}\mid=\:\frac{\mathrm{1}}{\mathrm{2}}\:{b}^{\:\mathrm{2}} \:\mathrm{tan}\:\gamma \\ $$$$\:{So}\:{how}\:{can}\:{we}\:{judge}?\:{cause}\:{all}\:{your}\:{posts}\:{made} \\ $$$$\:{me}\:{think}\:{that}\:{the}\:{maximum}\:{area}\:{is}\:{when}\:{the} \\ $$$$\:{right}\:{angle}\:{be}\:{on}\:{the}\:\mathrm{2}{nd}\:{side}\:“{middle}\:{length}''\:\& \\ $$$$\:{the}\:{hypotenuse}\:{is}\:{whole}\:{of}\:{the}\:{side}\:“{a}''\:{with} \\ $$$$\:{a}\:{formula}\:\frac{{a}^{\mathrm{2}} \:\mathrm{sin}\:\mathrm{2}\gamma}{\mathrm{4}}\:\:\overset{\:?} {\mid}\:\frac{\mathrm{1}}{\mathrm{2}}\:{b}^{\:\mathrm{2}} \:\mathrm{tan}\:\gamma \\ $$$$ \\ $$$${Note}:\:{I}\:{thought}\:{through}\:{diagram}\:\mathrm{1}\:{that}\:{you}\:{say} \\ $$$$\:{that}\:{there}'{s}\:\mathrm{2}\:{triangles}\:{as}\:{maximum} \\ $$

Commented by Acem last updated on 04/Dec/22

I back to your question, you mentioned that it is an acute triangle, well can now say? : For a≥ b ≥ c then: maximum { ((in an cute tri. ((a^2 sin 2γ)/4))),((in obtuse tri. ((b^2 tan γ)/2) _(You agree?) )) :}

$${I}\:{back}\:{to}\:{your}\:{question},\:{you}\:{mentioned}\:{that}\:{it}\:{is} \\ $$$$\:{an}\:{acute}\:{triangle},\:{well}\:{can}\:{now}\:{say}?\:: \\ $$$$\:{For}\:\:{a}\geqslant\:{b}\:\geqslant\:{c}\:\:\:{then}: \\ $$$$\:{maximum}\:\begin{cases}{{in}\:{an}\:{cute}\:{tri}.\:\:\:\:\:\:\frac{{a}^{\mathrm{2}} \:\mathrm{sin}\:\mathrm{2}\gamma}{\mathrm{4}}}\\{{in}\:{obtuse}\:{tri}.\:\:\:\:\:\:\:\frac{{b}^{\mathrm{2}} \:\mathrm{tan}\:\gamma}{\mathrm{2}}\:\:_{{You}\:{agree}?} }\end{cases} \\ $$$$ \\ $$

Commented by mr W last updated on 04/Dec/22

$${agree}. \\ $$

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