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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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