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Question-186464




Question Number 186464 by mr W last updated on 04/Feb/23
Commented by mr W last updated on 04/Feb/23
general case for Q186302:  find the maximum area of the  triangle.
$${general}\:{case}\:{for}\:{Q}\mathrm{186302}: \\ $$$${find}\:{the}\:{maximum}\:{area}\:{of}\:{the} \\ $$$${triangle}. \\ $$
Commented by Frix last updated on 05/Feb/23
Still that left vertex must be on the horizontal  diameter. (The area seems to be rh?)  Another question: What if the right angle  has got a different fixed value?
$$\mathrm{Still}\:\mathrm{that}\:\mathrm{left}\:\mathrm{vertex}\:\mathrm{must}\:\mathrm{be}\:\mathrm{on}\:\mathrm{the}\:\mathrm{horizontal} \\ $$$$\mathrm{diameter}.\:\left(\mathrm{The}\:\mathrm{area}\:\mathrm{seems}\:\mathrm{to}\:\mathrm{be}\:{rh}?\right) \\ $$$$\mathrm{Another}\:\mathrm{question}:\:\mathrm{What}\:\mathrm{if}\:\mathrm{the}\:\mathrm{right}\:\mathrm{angle} \\ $$$$\mathrm{has}\:\mathrm{got}\:\mathrm{a}\:\mathrm{different}\:\mathrm{fixed}\:\mathrm{value}? \\ $$
Commented by Frix last updated on 05/Feb/23
No...  Maybe we can try to see it the other way  round: Find the minimum circle for a given  rectangular triangle with center on the  intersection of a line through the vertex  at the right angle and a line in a right  angle to the first one which goes through  that other vertex. This seems to be a  special group of circles...
$$\mathrm{No}… \\ $$$$\mathrm{Maybe}\:\mathrm{we}\:\mathrm{can}\:\mathrm{try}\:\mathrm{to}\:\mathrm{see}\:\mathrm{it}\:\mathrm{the}\:\mathrm{other}\:\mathrm{way} \\ $$$$\mathrm{round}:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{circle}\:\mathrm{for}\:\mathrm{a}\:\mathrm{given} \\ $$$$\mathrm{rectangular}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{center}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{intersection}\:\mathrm{of}\:\mathrm{a}\:\mathrm{line}\:\mathrm{through}\:\mathrm{the}\:\mathrm{vertex} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{right}\:\mathrm{angle}\:\mathrm{and}\:\mathrm{a}\:\mathrm{line}\:\mathrm{in}\:\mathrm{a}\:\mathrm{right} \\ $$$$\mathrm{angle}\:\mathrm{to}\:\mathrm{the}\:\mathrm{first}\:\mathrm{one}\:\mathrm{which}\:\mathrm{goes}\:\mathrm{through} \\ $$$$\mathrm{that}\:\mathrm{other}\:\mathrm{vertex}.\:\mathrm{This}\:\mathrm{seems}\:\mathrm{to}\:\mathrm{be}\:\mathrm{a} \\ $$$$\mathrm{special}\:\mathrm{group}\:\mathrm{of}\:\mathrm{circles}… \\ $$
Commented by mr W last updated on 05/Feb/23
this is true. but it′s not obvious to me.  or can you simply explain that the   left vertex must lie on the horizontal  diameter such that the area is   maximum?
$${this}\:{is}\:{true}.\:{but}\:{it}'{s}\:{not}\:{obvious}\:{to}\:{me}. \\ $$$${or}\:{can}\:{you}\:{simply}\:{explain}\:{that}\:{the}\: \\ $$$${left}\:{vertex}\:{must}\:{lie}\:{on}\:{the}\:{horizontal} \\ $$$${diameter}\:{such}\:{that}\:{the}\:{area}\:{is}\: \\ $$$${maximum}? \\ $$
Commented by mr W last updated on 05/Feb/23
it is proved as shown below that for  the maximum triangle the right  angle vertex must lie on the horizontal  diameter. the question is only that  this result is not obvious. also for  the question for minimum circle  as you segguested above it is not  obvious where its center should lie.  so it doesn′t help, because we don′t  need a calculation proof (which we  have already), but an obvious   explanation.
$${it}\:{is}\:{proved}\:{as}\:{shown}\:{below}\:{that}\:{for} \\ $$$${the}\:{maximum}\:{triangle}\:{the}\:{right} \\ $$$${angle}\:{vertex}\:{must}\:{lie}\:{on}\:{the}\:{horizontal} \\ $$$${diameter}.\:{the}\:{question}\:{is}\:{only}\:{that} \\ $$$${this}\:{result}\:{is}\:{not}\:{obvious}.\:{also}\:{for} \\ $$$${the}\:{question}\:{for}\:{minimum}\:{circle} \\ $$$${as}\:{you}\:{segguested}\:{above}\:{it}\:{is}\:{not} \\ $$$${obvious}\:{where}\:{its}\:{center}\:{should}\:{lie}. \\ $$$${so}\:{it}\:{doesn}'{t}\:{help},\:{because}\:{we}\:{don}'{t} \\ $$$${need}\:{a}\:{calculation}\:{proof}\:\left({which}\:{we}\right. \\ $$$$\left.{have}\:{already}\right),\:{but}\:{an}\:{obvious}\: \\ $$$${explanation}. \\ $$
Commented by Frix last updated on 05/Feb/23
Start with any triangle ABC.  The centers of circles through A and C lie  on the symmetry axe of AC.  Let D the center of the circle we′re looking  for. The additional condition is CD⊥BD.  There are only 2 possible centers D.  Draw it, you′ll see it.
$$\mathrm{Start}\:\mathrm{with}\:\mathrm{any}\:\mathrm{triangle}\:{ABC}. \\ $$$$\mathrm{The}\:\mathrm{centers}\:\mathrm{of}\:\mathrm{circles}\:\mathrm{through}\:{A}\:\mathrm{and}\:{C}\:\mathrm{lie} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{symmetry}\:\mathrm{axe}\:\mathrm{of}\:{AC}. \\ $$$$\mathrm{Let}\:{D}\:\mathrm{the}\:\mathrm{center}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{we}'\mathrm{re}\:\mathrm{looking} \\ $$$$\mathrm{for}.\:\mathrm{The}\:\mathrm{additional}\:\mathrm{condition}\:\mathrm{is}\:{CD}\bot{BD}. \\ $$$$\mathrm{There}\:\mathrm{are}\:\mathrm{only}\:\mathrm{2}\:\mathrm{possible}\:\mathrm{centers}\:{D}. \\ $$$$\mathrm{Draw}\:\mathrm{it},\:\mathrm{you}'\mathrm{ll}\:\mathrm{see}\:\mathrm{it}. \\ $$
Commented by mr W last updated on 05/Feb/23
i got the simple explanation why  for the maximum triangle the vertex  must lie on the horizontal diameter.
$${i}\:{got}\:{the}\:{simple}\:{explanation}\:{why} \\ $$$${for}\:{the}\:{maximum}\:{triangle}\:{the}\:{vertex} \\ $$$${must}\:{lie}\:{on}\:{the}\:{horizontal}\:{diameter}. \\ $$
Commented by mr W last updated on 05/Feb/23
Commented by mr W last updated on 05/Feb/23
it′s easy to see   Δ=[BCA]=[BCA′]=((hd)/2)  d_(max) =diameter=2r  ⇒Δ_(max) =((h×2r)/2)=hr  when vertex C lies on the horizontal  diameter.
$${it}'{s}\:{easy}\:{to}\:{see}\: \\ $$$$\Delta=\left[{BCA}\right]=\left[{BCA}'\right]=\frac{{hd}}{\mathrm{2}} \\ $$$${d}_{{max}} ={diameter}=\mathrm{2}{r} \\ $$$$\Rightarrow\Delta_{{max}} =\frac{{h}×\mathrm{2}{r}}{\mathrm{2}}={hr} \\ $$$${when}\:{vertex}\:{C}\:{lies}\:{on}\:{the}\:{horizontal} \\ $$$${diameter}. \\ $$
Commented by ajfour last updated on 05/Feb/23
Δ=[BCA]=[BCA′]   understood  but =((hd)/2)    (how?)
$$\Delta=\left[{BCA}\right]=\left[{BCA}'\right]\:\:\:{understood} \\ $$$${but}\:=\frac{{hd}}{\mathrm{2}}\:\:\:\:\left({how}?\right) \\ $$$$ \\ $$
Commented by mr W last updated on 05/Feb/23
Commented by mr W last updated on 05/Feb/23
[BCA′]=[BCO]+[BOA′]                 =((BO×d_1 )/2)+((BO×d_1 )/2)                 =((BO×2d_1 )/2)                 =((h×d)/2)
$$\left[{BCA}'\right]=\left[{BCO}\right]+\left[{BOA}'\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{BO}×{d}_{\mathrm{1}} }{\mathrm{2}}+\frac{{BO}×{d}_{\mathrm{1}} }{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{BO}×\mathrm{2}{d}_{\mathrm{1}} }{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{h}×{d}}{\mathrm{2}} \\ $$
Commented by ajfour last updated on 05/Feb/23
i guess you mean  =(((h+s)d_1 )/2)+(((h−s)d_1 )/2)=((h(d_1 +d_2 ))/2)=((hd)/2)  Thanks sir, i understand the  explanation pretty well now,  awesome reasoning!
$${i}\:{guess}\:{you}\:{mean} \\ $$$$=\frac{\left({h}+{s}\right){d}_{\mathrm{1}} }{\mathrm{2}}+\frac{\left({h}−{s}\right){d}_{\mathrm{1}} }{\mathrm{2}}=\frac{{h}\left({d}_{\mathrm{1}} +{d}_{\mathrm{2}} \right)}{\mathrm{2}}=\frac{{hd}}{\mathrm{2}} \\ $$$${Thanks}\:{sir},\:{i}\:{understand}\:{the} \\ $$$${explanation}\:{pretty}\:{well}\:{now}, \\ $$$${awesome}\:{reasoning}! \\ $$
Commented by mr W last updated on 06/Feb/23
i meant ((hd_1 )/2)+((hd_1 )/2)=((hd)/2) indeed.  i had a typo with CO×d_1 . it is BO×d_1 .
$${i}\:{meant}\:\frac{{hd}_{\mathrm{1}} }{\mathrm{2}}+\frac{{hd}_{\mathrm{1}} }{\mathrm{2}}=\frac{{hd}}{\mathrm{2}}\:{indeed}. \\ $$$${i}\:{had}\:{a}\:{typo}\:{with}\:{CO}×{d}_{\mathrm{1}} .\:{it}\:{is}\:{BO}×{d}_{\mathrm{1}} . \\ $$
Commented by mr W last updated on 05/Feb/23
Answered by mr W last updated on 04/Feb/23
Commented by mr W last updated on 05/Feb/23
a_1 a=c_1 c=(h−r)(h+r)=h^2 −r^2   a_1 =((h^2 −r^2 )/a)  b^2 =c^2 −a^2   b^2 =(2r)^2 −(a−a_1 )^2   c^2 −a^2 =4r^2 −(a−a_1 )^2   c^2 =4r^2 +(2a−a_1 )a_1   c^2 =4r^2 +(2a−((h^2 −r^2 )/a))×((h^2 −r^2 )/a)  ⇒c^2 =2(h^2 +r^2 )−(((h^2 −r^2 )^2 )/a^2 )  ⇒b^2 =c^2 −a^2 =2(h^2 +r^2 )−(((h^2 −r^2 )^2 )/a^2 )−a^2   area Δ=((ab)/2)  let Φ=4Δ^2 =a^2 b^2   Φ=a^2 [2(h^2 +r^2 )−(((h^2 −r^2 )^2 )/a^2 )−a^2 ]  Φ=2(h^2 +r^2 )a^2 −a^4 −(h^2 −r^2 )^2   Φ=2(h^2 +r^2 )a^2 −a^4 −(h^2 +r^2 )^2 +4h^2 r^2   ⇒Φ=4h^2 r^2 −(a^2 −h^2 −r^2 )^2   we see the maximum of Φ is 4h^2 r^2   when a^2 =h^2 +r^2 .  that means the maximum area is  Δ_(max) =hr, when the right angle vertex  lies on the horizontal diameter.
$${a}_{\mathrm{1}} {a}={c}_{\mathrm{1}} {c}=\left({h}−{r}\right)\left({h}+{r}\right)={h}^{\mathrm{2}} −{r}^{\mathrm{2}} \\ $$$${a}_{\mathrm{1}} =\frac{{h}^{\mathrm{2}} −{r}^{\mathrm{2}} }{{a}} \\ $$$${b}^{\mathrm{2}} ={c}^{\mathrm{2}} −{a}^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} =\left(\mathrm{2}{r}\right)^{\mathrm{2}} −\left({a}−{a}_{\mathrm{1}} \right)^{\mathrm{2}} \\ $$$${c}^{\mathrm{2}} −{a}^{\mathrm{2}} =\mathrm{4}{r}^{\mathrm{2}} −\left({a}−{a}_{\mathrm{1}} \right)^{\mathrm{2}} \\ $$$${c}^{\mathrm{2}} =\mathrm{4}{r}^{\mathrm{2}} +\left(\mathrm{2}{a}−{a}_{\mathrm{1}} \right){a}_{\mathrm{1}} \\ $$$${c}^{\mathrm{2}} =\mathrm{4}{r}^{\mathrm{2}} +\left(\mathrm{2}{a}−\frac{{h}^{\mathrm{2}} −{r}^{\mathrm{2}} }{{a}}\right)×\frac{{h}^{\mathrm{2}} −{r}^{\mathrm{2}} }{{a}} \\ $$$$\Rightarrow{c}^{\mathrm{2}} =\mathrm{2}\left({h}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)−\frac{\left({h}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)^{\mathrm{2}} }{{a}^{\mathrm{2}} } \\ $$$$\Rightarrow{b}^{\mathrm{2}} ={c}^{\mathrm{2}} −{a}^{\mathrm{2}} =\mathrm{2}\left({h}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)−\frac{\left({h}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }−{a}^{\mathrm{2}} \\ $$$${area}\:\Delta=\frac{{ab}}{\mathrm{2}} \\ $$$${let}\:\Phi=\mathrm{4}\Delta^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$\Phi={a}^{\mathrm{2}} \left[\mathrm{2}\left({h}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)−\frac{\left({h}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }−{a}^{\mathrm{2}} \right] \\ $$$$\Phi=\mathrm{2}\left({h}^{\mathrm{2}} +{r}^{\mathrm{2}} \right){a}^{\mathrm{2}} −{a}^{\mathrm{4}} −\left({h}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\Phi=\mathrm{2}\left({h}^{\mathrm{2}} +{r}^{\mathrm{2}} \right){a}^{\mathrm{2}} −{a}^{\mathrm{4}} −\left({h}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}{h}^{\mathrm{2}} {r}^{\mathrm{2}} \\ $$$$\Rightarrow\Phi=\mathrm{4}{h}^{\mathrm{2}} {r}^{\mathrm{2}} −\left({a}^{\mathrm{2}} −{h}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${we}\:{see}\:{the}\:{maximum}\:{of}\:\Phi\:{is}\:\mathrm{4}{h}^{\mathrm{2}} {r}^{\mathrm{2}} \\ $$$${when}\:{a}^{\mathrm{2}} ={h}^{\mathrm{2}} +{r}^{\mathrm{2}} . \\ $$$${that}\:{means}\:{the}\:{maximum}\:{area}\:{is} \\ $$$$\Delta_{{max}} ={hr},\:{when}\:{the}\:{right}\:{angle}\:{vertex} \\ $$$${lies}\:{on}\:{the}\:{horizontal}\:{diameter}. \\ $$
Commented by mr W last updated on 05/Feb/23

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