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

$${this}\:{is}\:{true}.\:{but}\:{it}'{s}\:{not}\:{obvious}\:{to}\:{me}. \\ $$$${or}\:{can}\:{you}\:{simply}\:{explain}\:{that}\:{the}\: \\ $$$${left}\:{vertex}\:{must}\:{lie}\:{on}\:{the}\:{horizontal} \\ $$$${diameter}\:{such}\:{that}\:{the}\:{area}\:{is}\: \\ $$$${maximum}? \\ $$

Commented by mr W last updated on 05/Feb/23

$${it}\:{is}\:{proved}\:{as}\:{shown}\:{below}\:{that}\:{for} \\ $$$${the}\:{maximum}\:{triangle}\:{the}\:{right} \\ $$$${angle}\:{vertex}\:{must}\:{lie}\:{on}\:{the}\:{horizontal} \\ $$$${diameter}.\:{the}\:{question}\:{is}\:{only}\:{that} \\ $$$${this}\:{result}\:{is}\:{not}\:{obvious}.\:{also}\:{for} \\ $$$${the}\:{question}\:{for}\:{minimum}\:{circle} \\ $$$${as}\:{you}\:{segguested}\:{above}\:{it}\:{is}\:{not} \\ $$$${obvious}\:{where}\:{its}\:{center}\:{should}\:{lie}. \\ $$$${so}\:{it}\:{doesn}'{t}\:{help},\:{because}\:{we}\:{don}'{t} \\ $$$${need}\:{a}\:{calculation}\:{proof}\:\left({which}\:{we}\right. \\ $$$$\left.{have}\:{already}\right),\:{but}\:{an}\:{obvious}\: \\ $$$${explanation}. \\ $$

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

$${i}\:{got}\:{the}\:{simple}\:{explanation}\:{why} \\ $$$${for}\:{the}\:{maximum}\:{triangle}\:{the}\:{vertex} \\ $$$${must}\:{lie}\:{on}\:{the}\:{horizontal}\:{diameter}. \\ $$

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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