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Question-162071




Question Number 162071 by mr W last updated on 26/Dec/21
Commented by mr W last updated on 27/Dec/21
if the distances from a point to the  vertices of a triangle are p, q, r  respectively, find the maximum area  of the triangle and its side lengthes.
$${if}\:{the}\:{distances}\:{from}\:{a}\:{point}\:{to}\:{the} \\ $$$${vertices}\:{of}\:{a}\:{triangle}\:{are}\:{p},\:{q},\:{r} \\ $$$${respectively},\:{find}\:{the}\:{maximum}\:{area} \\ $$$${of}\:{the}\:{triangle}\:{and}\:{its}\:{side}\:{lengthes}. \\ $$
Commented by mr W last updated on 26/Dec/21
Commented by mr W last updated on 27/Dec/21
it is to find the triangle whose vertices  lie on the three circles respectively  and which has the maximum area.
$${it}\:{is}\:{to}\:{find}\:{the}\:{triangle}\:{whose}\:{vertices} \\ $$$${lie}\:{on}\:{the}\:{three}\:{circles}\:{respectively} \\ $$$${and}\:{which}\:{has}\:{the}\:{maximum}\:{area}. \\ $$
Commented by Tawa11 last updated on 27/Dec/21
Wow, great sir.
$$\mathrm{Wow},\:\mathrm{great}\:\mathrm{sir}. \\ $$
Commented by mr W last updated on 29/Dec/21
thanks for this vector way sir!  i came to the same conclusion as  stated below.
$${thanks}\:{for}\:{this}\:{vector}\:{way}\:{sir}! \\ $$$${i}\:{came}\:{to}\:{the}\:{same}\:{conclusion}\:{as} \\ $$$${stated}\:{below}. \\ $$
Commented by aleks041103 last updated on 29/Dec/21
Let p,q,r represent the vectors.  S^→ =Se_z ^(→) =(1/2)(p^→ ×q^→ +q^→ ×r^→ +r^→ ×p^→ )  Since S is extremal, then we may  variate the vectors as  p^→ →p^→ +δp^→   q^→ →q^→ +δq^→   r^→ →r^→ +δr^→   And search for p^→ ,q^→ ,r^→  which make δS=0 and  so δS^→ =0.  Also the variations must be such that  they preserve ∣p∣,∣q∣ and ∣r∣.  δ(p^→ .p^→ )=2p^→ .δp^→ =0⇒p^→ ⊥δp^→ .  Without restriction we may choose  δq^→ =δr^→ =0 since the variations are arbitrary  ⇒δSe_z ^(→) =(1/2)(δp^→ ×q^→ +r^→ ×δp^→ )=0  ⇒δp^→ ×(q^→ −r^→ )=0  ⇒p^→ ⊥δp^→ ∥q^→ −r^→ ⇒p^→ ⊥(q^→ −r^→ )  But (q^→ −r^→ ) is the side of the triangle  opposite the vertex defined by p^→ .  ⇒p^→  must be colinear to the height  This must be true for q^→  and r^→  too.  ⇒ The starting point is the orthocenter.
$${Let}\:{p},{q},{r}\:{represent}\:{the}\:{vectors}. \\ $$$$\overset{\rightarrow} {{S}}={S}\overset{\rightarrow} {{e}_{{z}} }=\frac{\mathrm{1}}{\mathrm{2}}\left(\overset{\rightarrow} {{p}}×\overset{\rightarrow} {{q}}+\overset{\rightarrow} {{q}}×\overset{\rightarrow} {{r}}+\overset{\rightarrow} {{r}}×\overset{\rightarrow} {{p}}\right) \\ $$$${Since}\:{S}\:{is}\:{extremal},\:{then}\:{we}\:{may} \\ $$$${variate}\:{the}\:{vectors}\:{as} \\ $$$$\overset{\rightarrow} {{p}}\rightarrow\overset{\rightarrow} {{p}}+\delta\overset{\rightarrow} {{p}} \\ $$$$\overset{\rightarrow} {{q}}\rightarrow\overset{\rightarrow} {{q}}+\delta\overset{\rightarrow} {{q}} \\ $$$$\overset{\rightarrow} {{r}}\rightarrow\overset{\rightarrow} {{r}}+\delta\overset{\rightarrow} {{r}} \\ $$$${And}\:{search}\:{for}\:\overset{\rightarrow} {{p}},\overset{\rightarrow} {{q}},\overset{\rightarrow} {{r}}\:{which}\:{make}\:\delta{S}=\mathrm{0}\:{and} \\ $$$${so}\:\delta\overset{\rightarrow} {{S}}=\mathrm{0}. \\ $$$${Also}\:{the}\:{variations}\:{must}\:{be}\:{such}\:{that} \\ $$$${they}\:{preserve}\:\mid{p}\mid,\mid{q}\mid\:{and}\:\mid{r}\mid. \\ $$$$\delta\left(\overset{\rightarrow} {{p}}.\overset{\rightarrow} {{p}}\right)=\mathrm{2}\overset{\rightarrow} {{p}}.\delta\overset{\rightarrow} {{p}}=\mathrm{0}\Rightarrow\overset{\rightarrow} {{p}}\bot\delta\overset{\rightarrow} {{p}}. \\ $$$${Without}\:{restriction}\:{we}\:{may}\:{choose} \\ $$$$\delta\overset{\rightarrow} {{q}}=\delta\overset{\rightarrow} {{r}}=\mathrm{0}\:{since}\:{the}\:{variations}\:{are}\:{arbitrary} \\ $$$$\Rightarrow\delta{S}\overset{\rightarrow} {{e}_{{z}} }=\frac{\mathrm{1}}{\mathrm{2}}\left(\delta\overset{\rightarrow} {{p}}×\overset{\rightarrow} {{q}}+\overset{\rightarrow} {{r}}×\delta\overset{\rightarrow} {{p}}\right)=\mathrm{0} \\ $$$$\Rightarrow\delta\overset{\rightarrow} {{p}}×\left(\overset{\rightarrow} {{q}}−\overset{\rightarrow} {{r}}\right)=\mathrm{0} \\ $$$$\Rightarrow\overset{\rightarrow} {{p}}\bot\delta\overset{\rightarrow} {{p}}\parallel\overset{\rightarrow} {{q}}−\overset{\rightarrow} {{r}}\Rightarrow\overset{\rightarrow} {{p}}\bot\left(\overset{\rightarrow} {{q}}−\overset{\rightarrow} {{r}}\right) \\ $$$${But}\:\left(\overset{\rightarrow} {{q}}−\overset{\rightarrow} {{r}}\right)\:{is}\:{the}\:{side}\:{of}\:{the}\:{triangle} \\ $$$${opposite}\:{the}\:{vertex}\:{defined}\:{by}\:\overset{\rightarrow} {{p}}. \\ $$$$\Rightarrow\overset{\rightarrow} {{p}}\:{must}\:{be}\:{colinear}\:{to}\:{the}\:{height} \\ $$$${This}\:{must}\:{be}\:{true}\:{for}\:\overset{\rightarrow} {{q}}\:{and}\:\overset{\rightarrow} {{r}}\:{too}. \\ $$$$\Rightarrow\:{The}\:{starting}\:{point}\:{is}\:{the}\:{orthocenter}. \\ $$
Commented by papa last updated on 04/May/22
hi plz tell me how to make p vector
$${hi}\:{plz}\:{tell}\:{me}\:{how}\:{to}\:{make}\:{p}\:{vector} \\ $$
Commented by mr W last updated on 04/May/22
Commented by mr W last updated on 04/May/22
Answered by mr W last updated on 27/Dec/21
Commented by mr W last updated on 27/Dec/21
thanks for reviewing sir! you seem  to have been absent for a long time,  welcome back!
$${thanks}\:{for}\:{reviewing}\:{sir}!\:{you}\:{seem} \\ $$$${to}\:{have}\:{been}\:{absent}\:{for}\:{a}\:{long}\:{time}, \\ $$$${welcome}\:{back}! \\ $$
Commented by mr W last updated on 27/Dec/21
the area of ABC is  Δ=(1/2)(pq sin α+qr sin β+rp sin γ)  α+β+γ=2π  Δ=(1/2)[pq sin α+qr sin β−rp sin (α+β)]  such that Δ is maximum,  (∂Δ/∂α)=(1/2)[pq cos α−rp cos (α+β)]=0  ⇒pq cos α=rp cos (α+β)=rp cos γ  (∂Δ/∂β)=(1/2)[qr cos β−rp cos (α+β)]=0  ⇒qr cos β=rp cos (α+β)=rp cos γ    so we have  pq cos α=qr cos β=rp cos γ=k say  cos α=(k/(pq)) ⇒sin α=(√(1−(k^2 /(p^2 q^2 ))))  cos β=(k/(qr)) ⇒sin β=(√(1−(k^2 /(q^2 r^2 ))))  cos γ=(k/(rp)) ⇒sin γ=(√(1−(k^2 /(r^2 p^2 ))))  k−rp [(k^2 /(pq^2 r))−(√((1−(k^2 /(p^2 q^2 )))(1−(k^2 /(q^2 r^2 )))))]=0  (√((p^2 q^2 −k^2 )(q^2 r^2 −k^2 )))=k^2 −q^2 k  p^2 q^4 r^2 −(p^2 q^2 +q^2 r^2 )k^2 +k^4 =k^4 +q^4 k^2 −2q^2 k^3   p^2 q^2 r^2 −(p^2 +q^2 +r^2 )k^2 +2k^3 =0   determinant ((((1/k^3 )−(((p^2 +q^2 +r^2 )/(p^2 q^2 r^2 )))(1/k)+(2/(p^2 q^2 r^2 ))=0)))  p^2 q^2 r^2 −(1/(27))(p^2 +q^2 +r^2 )^3 ≤0 ⇒three roots  (1/k)=(2/(pqr))(√((p^2 +q^2 +r^2 )/3)) sin {((2nπ)/3)+(1/3) sin^(−1)  pqr((3/(p^2 +q^2 +r^2 )))^(3/2) }  (n=0,1,2)  but suitable for maximum Δ is only   the root with n=2:   determinant ((((1/k)=−(2/(pqr))(√((p^2 +q^2 +r^2 )/3)) sin {(π/3)+(1/3) sin^(−1)  pqr((3/(p^2 +q^2 +r^2 )))^(3/2) })))    Δ_(max) =(1/2)((√(p^2 q^2 −k^2 ))+(√(q^2 r^2 −k^2 ))+(√(r^2 p^2 −k^2 )))  a^2 =q^2 +r^2 −2qr cos β=q^2 +r^2 −2k  a=(√(q^2 +r^2 −2k))  b=(√(r^2 +p^2 −2k))  c=(√(p^2 +q^2 −2k))    example: p=10, q=8, r=4  Δ_(max) ≈67.303409  a≈11.084677  b≈12.604367  c≈14.382978  coresponding equilateral triangle:  s≈11.943609, Δ≈61.769169
$${the}\:{area}\:{of}\:{ABC}\:{is} \\ $$$$\Delta=\frac{\mathrm{1}}{\mathrm{2}}\left({pq}\:\mathrm{sin}\:\alpha+{qr}\:\mathrm{sin}\:\beta+{rp}\:\mathrm{sin}\:\gamma\right) \\ $$$$\alpha+\beta+\gamma=\mathrm{2}\pi \\ $$$$\Delta=\frac{\mathrm{1}}{\mathrm{2}}\left[{pq}\:\mathrm{sin}\:\alpha+{qr}\:\mathrm{sin}\:\beta−{rp}\:\mathrm{sin}\:\left(\alpha+\beta\right)\right] \\ $$$${such}\:{that}\:\Delta\:{is}\:{maximum}, \\ $$$$\frac{\partial\Delta}{\partial\alpha}=\frac{\mathrm{1}}{\mathrm{2}}\left[{pq}\:\mathrm{cos}\:\alpha−{rp}\:\mathrm{cos}\:\left(\alpha+\beta\right)\right]=\mathrm{0} \\ $$$$\Rightarrow{pq}\:\mathrm{cos}\:\alpha={rp}\:\mathrm{cos}\:\left(\alpha+\beta\right)={rp}\:\mathrm{cos}\:\gamma \\ $$$$\frac{\partial\Delta}{\partial\beta}=\frac{\mathrm{1}}{\mathrm{2}}\left[{qr}\:\mathrm{cos}\:\beta−{rp}\:\mathrm{cos}\:\left(\alpha+\beta\right)\right]=\mathrm{0} \\ $$$$\Rightarrow{qr}\:\mathrm{cos}\:\beta={rp}\:\mathrm{cos}\:\left(\alpha+\beta\right)={rp}\:\mathrm{cos}\:\gamma \\ $$$$ \\ $$$${so}\:{we}\:{have} \\ $$$${pq}\:\mathrm{cos}\:\alpha={qr}\:\mathrm{cos}\:\beta={rp}\:\mathrm{cos}\:\gamma={k}\:{say} \\ $$$$\mathrm{cos}\:\alpha=\frac{{k}}{{pq}}\:\Rightarrow\mathrm{sin}\:\alpha=\sqrt{\mathrm{1}−\frac{{k}^{\mathrm{2}} }{{p}^{\mathrm{2}} {q}^{\mathrm{2}} }} \\ $$$$\mathrm{cos}\:\beta=\frac{{k}}{{qr}}\:\Rightarrow\mathrm{sin}\:\beta=\sqrt{\mathrm{1}−\frac{{k}^{\mathrm{2}} }{{q}^{\mathrm{2}} {r}^{\mathrm{2}} }} \\ $$$$\mathrm{cos}\:\gamma=\frac{{k}}{{rp}}\:\Rightarrow\mathrm{sin}\:\gamma=\sqrt{\mathrm{1}−\frac{{k}^{\mathrm{2}} }{{r}^{\mathrm{2}} {p}^{\mathrm{2}} }} \\ $$$${k}−{rp}\:\left[\frac{{k}^{\mathrm{2}} }{{pq}^{\mathrm{2}} {r}}−\sqrt{\left(\mathrm{1}−\frac{{k}^{\mathrm{2}} }{{p}^{\mathrm{2}} {q}^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{{k}^{\mathrm{2}} }{{q}^{\mathrm{2}} {r}^{\mathrm{2}} }\right)}\right]=\mathrm{0} \\ $$$$\sqrt{\left({p}^{\mathrm{2}} {q}^{\mathrm{2}} −{k}^{\mathrm{2}} \right)\left({q}^{\mathrm{2}} {r}^{\mathrm{2}} −{k}^{\mathrm{2}} \right)}={k}^{\mathrm{2}} −{q}^{\mathrm{2}} {k} \\ $$$${p}^{\mathrm{2}} {q}^{\mathrm{4}} {r}^{\mathrm{2}} −\left({p}^{\mathrm{2}} {q}^{\mathrm{2}} +{q}^{\mathrm{2}} {r}^{\mathrm{2}} \right){k}^{\mathrm{2}} +{k}^{\mathrm{4}} ={k}^{\mathrm{4}} +{q}^{\mathrm{4}} {k}^{\mathrm{2}} −\mathrm{2}{q}^{\mathrm{2}} {k}^{\mathrm{3}} \\ $$$${p}^{\mathrm{2}} {q}^{\mathrm{2}} {r}^{\mathrm{2}} −\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} \right){k}^{\mathrm{2}} +\mathrm{2}{k}^{\mathrm{3}} =\mathrm{0} \\ $$$$\begin{array}{|c|}{\frac{\mathrm{1}}{{k}^{\mathrm{3}} }−\left(\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} }{{p}^{\mathrm{2}} {q}^{\mathrm{2}} {r}^{\mathrm{2}} }\right)\frac{\mathrm{1}}{{k}}+\frac{\mathrm{2}}{{p}^{\mathrm{2}} {q}^{\mathrm{2}} {r}^{\mathrm{2}} }=\mathrm{0}}\\\hline\end{array} \\ $$$${p}^{\mathrm{2}} {q}^{\mathrm{2}} {r}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{27}}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)^{\mathrm{3}} \leqslant\mathrm{0}\:\Rightarrow{three}\:{roots} \\ $$$$\frac{\mathrm{1}}{{k}}=\frac{\mathrm{2}}{{pqr}}\sqrt{\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} }{\mathrm{3}}}\:\mathrm{sin}\:\left\{\frac{\mathrm{2}{n}\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sin}^{−\mathrm{1}} \:{pqr}\left(\frac{\mathrm{3}}{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} }\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right\}\:\:\left({n}=\mathrm{0},\mathrm{1},\mathrm{2}\right) \\ $$$${but}\:{suitable}\:{for}\:{maximum}\:\Delta\:{is}\:{only}\: \\ $$$${the}\:{root}\:{with}\:{n}=\mathrm{2}: \\ $$$$\begin{array}{|c|}{\frac{\mathrm{1}}{{k}}=−\frac{\mathrm{2}}{{pqr}}\sqrt{\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} }{\mathrm{3}}}\:\mathrm{sin}\:\left\{\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sin}^{−\mathrm{1}} \:{pqr}\left(\frac{\mathrm{3}}{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} }\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right\}}\\\hline\end{array} \\ $$$$ \\ $$$$\Delta_{{max}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{{p}^{\mathrm{2}} {q}^{\mathrm{2}} −{k}^{\mathrm{2}} }+\sqrt{{q}^{\mathrm{2}} {r}^{\mathrm{2}} −{k}^{\mathrm{2}} }+\sqrt{{r}^{\mathrm{2}} {p}^{\mathrm{2}} −{k}^{\mathrm{2}} }\right) \\ $$$${a}^{\mathrm{2}} ={q}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{qr}\:\mathrm{cos}\:\beta={q}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{k} \\ $$$${a}=\sqrt{{q}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{k}} \\ $$$${b}=\sqrt{{r}^{\mathrm{2}} +{p}^{\mathrm{2}} −\mathrm{2}{k}} \\ $$$${c}=\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} −\mathrm{2}{k}} \\ $$$$ \\ $$$$\underline{{example}:\:{p}=\mathrm{10},\:{q}=\mathrm{8},\:{r}=\mathrm{4}} \\ $$$$\Delta_{{max}} \approx\mathrm{67}.\mathrm{303409} \\ $$$${a}\approx\mathrm{11}.\mathrm{084677} \\ $$$${b}\approx\mathrm{12}.\mathrm{604367} \\ $$$${c}\approx\mathrm{14}.\mathrm{382978} \\ $$$${coresponding}\:{equilateral}\:{triangle}: \\ $$$${s}\approx\mathrm{11}.\mathrm{943609},\:\Delta\approx\mathrm{61}.\mathrm{769169} \\ $$
Commented by mr W last updated on 27/Dec/21
Commented by mr W last updated on 27/Dec/21
Commented by mr W last updated on 27/Dec/21
Commented by mr W last updated on 27/Dec/21
Commented by mr W last updated on 27/Dec/21
we see generally the coresponding  equilateral triangle (blue marked in  diagrams) (see also Q123040) is not  the triangle with the maximum   area.
$${we}\:{see}\:{generally}\:{the}\:{coresponding} \\ $$$${equilateral}\:{triangle}\:\left({blue}\:{marked}\:{in}\right. \\ $$$$\left.{diagrams}\right)\:\left({see}\:{also}\:{Q}\mathrm{123040}\right)\:{is}\:{not} \\ $$$${the}\:{triangle}\:{with}\:{the}\:{maximum}\: \\ $$$${area}. \\ $$
Commented by mr W last updated on 27/Dec/21
from pq cos α=qr cos β=rp cos γ  we get q cos α=r cos γ, that means  AO⊥BC. similarly BO⊥CA, CO⊥AB.  that means O is the orthocenter of  triangle ABC when the triangle  has the maximum area.  example: p=7, q=6, r=5  a≈9.7924505  b≈10.4351372  c≈10.9495245  this is the same result as in Q161015.
$${from}\:{pq}\:\mathrm{cos}\:\alpha={qr}\:\mathrm{cos}\:\beta={rp}\:\mathrm{cos}\:\gamma \\ $$$${we}\:{get}\:{q}\:\mathrm{cos}\:\alpha={r}\:\mathrm{cos}\:\gamma,\:{that}\:{means} \\ $$$${AO}\bot{BC}.\:{similarly}\:{BO}\bot{CA},\:{CO}\bot{AB}. \\ $$$${that}\:{means}\:{O}\:{is}\:{the}\:{orthocenter}\:{of} \\ $$$${triangle}\:{ABC}\:{when}\:{the}\:{triangle} \\ $$$${has}\:{the}\:{maximum}\:{area}. \\ $$$${example}:\:{p}=\mathrm{7},\:{q}=\mathrm{6},\:{r}=\mathrm{5} \\ $$$${a}\approx\mathrm{9}.\mathrm{7924505} \\ $$$${b}\approx\mathrm{10}.\mathrm{4351372} \\ $$$${c}\approx\mathrm{10}.\mathrm{9495245} \\ $$$${this}\:{is}\:{the}\:{same}\:{result}\:{as}\:{in}\:{Q}\mathrm{161015}. \\ $$
Commented by Tawa11 last updated on 27/Dec/21
Great sir.
$$\mathrm{Great}\:\mathrm{sir}. \\ $$
Commented by behi834171 last updated on 27/Dec/21
greaaaaaaaaaat dear master.  thanks a lot.
$${greaaaaaaaaaat}\:{dear}\:{master}. \\ $$$${thanks}\:{a}\:{lot}. \\ $$
Commented by behi834171 last updated on 27/Dec/21
  Thank you so much,dear master,for your favor.  I'm here,from now on.
$$ \\ $$Thank you so much,dear master,for your favor.
I'm here,from now on.
Commented by behi834171 last updated on 27/Dec/21
γ=90^•  is an  interesting case.I think.
$$\gamma=\mathrm{90}^{\bullet} \:{is}\:{an}\:\:{interesting}\:{case}.{I}\:{think}. \\ $$
Commented by mr W last updated on 27/Dec/21
when one of p,q,r is zero, then the  coresponding vertex is also the  orthocenter. in this case the triangle  is right angled.
$${when}\:{one}\:{of}\:{p},{q},{r}\:{is}\:{zero},\:{then}\:{the} \\ $$$${coresponding}\:{vertex}\:{is}\:{also}\:{the} \\ $$$${orthocenter}.\:{in}\:{this}\:{case}\:{the}\:{triangle} \\ $$$${is}\:{right}\:{angled}. \\ $$
Commented by mr W last updated on 27/Dec/21
Commented by behi834171 last updated on 27/Dec/21
Commented by behi834171 last updated on 27/Dec/21
γ=90^• ,p,q,r≠0
$$\gamma=\mathrm{90}^{\bullet} ,{p},{q},{r}\neq\mathrm{0} \\ $$
Commented by mr W last updated on 27/Dec/21
the task here is to find the triangle  with maximum area when p,q,r are  given. it is not to find the triangle  with given α,β or γ.   so far as p,q,r≠0, the triangle with   maxixum area can never be a triangle  with γ=90° !  the triangle you gave with γ=90° is  not the triangle which has the  maximum area.
$${the}\:{task}\:{here}\:{is}\:{to}\:{find}\:{the}\:{triangle} \\ $$$${with}\:{maximum}\:{area}\:{when}\:{p},{q},{r}\:{are} \\ $$$${given}.\:{it}\:{is}\:{not}\:{to}\:{find}\:{the}\:{triangle} \\ $$$${with}\:{given}\:\alpha,\beta\:{or}\:\gamma.\: \\ $$$${so}\:{far}\:{as}\:{p},{q},{r}\neq\mathrm{0},\:{the}\:{triangle}\:{with}\: \\ $$$${maxixum}\:{area}\:{can}\:{never}\:{be}\:{a}\:{triangle} \\ $$$${with}\:\gamma=\mathrm{90}°\:! \\ $$$${the}\:{triangle}\:{you}\:{gave}\:{with}\:\gamma=\mathrm{90}°\:{is} \\ $$$${not}\:{the}\:{triangle}\:{which}\:{has}\:{the} \\ $$$${maximum}\:{area}. \\ $$
Commented by mr W last updated on 27/Dec/21
Commented by behi834171 last updated on 27/Dec/21
thank you very much for your attention.  can we find the sides of such triangle  whit:γ=90^•  ?  i have the such unsolved problem in forum  but i don′t remember its q.ID...
$${thank}\:{you}\:{very}\:{much}\:{for}\:{your}\:{attention}. \\ $$$${can}\:{we}\:{find}\:{the}\:{sides}\:{of}\:{such}\:{triangle} \\ $$$${whit}:\gamma=\mathrm{90}^{\bullet} \:? \\ $$$${i}\:{have}\:{the}\:{such}\:{unsolved}\:{problem}\:{in}\:{forum} \\ $$$${but}\:{i}\:{don}'{t}\:{remember}\:{its}\:{q}.{ID}… \\ $$
Commented by mr W last updated on 27/Dec/21
you must specify exactly what is your  task. in the current question the  point O is the orthocenter of the  triangle. we can not find a triangle  whose orthocenter lies inside the  triangle and makes an angle γ=90°.  this is only the case when the   orthocenter lies at one vertex of a  right angled triangle, but in this  case one of p,q,r is zero.
$${you}\:{must}\:{specify}\:{exactly}\:{what}\:{is}\:{your} \\ $$$${task}.\:{in}\:{the}\:{current}\:{question}\:{the} \\ $$$${point}\:{O}\:{is}\:{the}\:{orthocenter}\:{of}\:{the} \\ $$$${triangle}.\:{we}\:{can}\:{not}\:{find}\:{a}\:{triangle} \\ $$$${whose}\:{orthocenter}\:{lies}\:{inside}\:{the} \\ $$$${triangle}\:{and}\:{makes}\:{an}\:{angle}\:\gamma=\mathrm{90}°. \\ $$$${this}\:{is}\:{only}\:{the}\:{case}\:{when}\:{the}\: \\ $$$${orthocenter}\:{lies}\:{at}\:{one}\:{vertex}\:{of}\:{a} \\ $$$${right}\:{angled}\:{triangle},\:{but}\:{in}\:{this} \\ $$$${case}\:{one}\:{of}\:{p},{q},{r}\:{is}\:{zero}. \\ $$
Commented by behi834171 last updated on 27/Dec/21
the task is:to find sides of a triangle  with given:p,q,r and: γ=90^(• ) .  is it possible?
$${the}\:{task}\:{is}:{to}\:{find}\:{sides}\:{of}\:{a}\:{triangle} \\ $$$${with}\:{given}:{p},{q},{r}\:{and}:\:\gamma=\mathrm{90}^{\bullet\:} . \\ $$$${is}\:{it}\:{possible}? \\ $$
Commented by mr W last updated on 27/Dec/21
then infinite triangles are possible.  only AC=(√(r^2 +p^2 )) is fixed.
$${then}\:{infinite}\:{triangles}\:{are}\:{possible}. \\ $$$${only}\:{AC}=\sqrt{{r}^{\mathrm{2}} +{p}^{\mathrm{2}} }\:{is}\:{fixed}. \\ $$
Commented by mr W last updated on 27/Dec/21

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