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Question Number 52911 by ajfour last updated on 15/Jan/19

Commented by ajfour last updated on 15/Jan/19

let for example: q=5, p=7, r=4 .

$${let}\:{for}\:{example}:\:\:{q}=\mathrm{5},\:{p}=\mathrm{7},\:{r}=\mathrm{4}\:. \\ $$

Commented by MJS last updated on 16/Jan/19

not sure if it′s getting easier, but there are also pairs of triangles with equal areas △bcp ≡ △cqr =((bc)/2) △abq ≡ △apr =((ab)/2)

$$\mathrm{not}\:\mathrm{sure}\:\mathrm{if}\:\mathrm{it}'\mathrm{s}\:\mathrm{getting}\:\mathrm{easier},\:\mathrm{but}\:\mathrm{there}\:\mathrm{are} \\ $$$$\mathrm{also}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{triangles}\:\mathrm{with}\:\mathrm{equal}\:\mathrm{areas} \\ $$$$\bigtriangleup{bcp}\:\equiv\:\bigtriangleup{cqr}\:=\frac{{bc}}{\mathrm{2}} \\ $$$$\bigtriangleup{abq}\:\equiv\:\bigtriangleup{apr}\:=\frac{{ab}}{\mathrm{2}} \\ $$

Answered by mr W last updated on 15/Jan/19

a^2 +b^2 =q^2 b^2 +c^2 =p^2 b^2 +(c−a)^2 =r^2 c^2 −(c−a)^2 =p^2 −r^2 2ac−a^2 =p^2 −r^2 ⇒c=((a^2 +p^2 −r^2 )/(2a)) q^2 −a^2 +(((a^2 +p^2 −r^2 )/(2a)))^2 =p^2 a^2 −(((a^2 +p^2 −r^2 )/(2a)))^2 =q^2 −p^2 3a^4 +2(p^2 +r^2 −2q^2 )a^2 −(p^2 −r^2 )^2 =0 a^2 =(((√((p^2 +r^2 −2q^2 )^2 +3(p^2 −r^2 )^2 ))−(p^2 +r^2 −2q^2 ))/3) ⇒a=(√(((√((p^2 +r^2 −2q^2 )^2 +3(p^2 −r^2 )^2 ))−(p^2 +r^2 −2q^2 ))/3)) ⇒c=... ⇒b=(√(q^2 −a^2 ))=... example: p=7, q=5, r=4 ⇒a=(√(((√((49+16−50)^2 +3(49−16)^2 ))−(49+16−50))/3)) ⇒a=(√((6(√(97))−15)/3))≈3.834 ...

$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={q}^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} +{c}^{\mathrm{2}} ={p}^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} +\left({c}−{a}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$ \\ $$$${c}^{\mathrm{2}} −\left({c}−{a}\right)^{\mathrm{2}} ={p}^{\mathrm{2}} −{r}^{\mathrm{2}} \\ $$$$\mathrm{2}{ac}−{a}^{\mathrm{2}} ={p}^{\mathrm{2}} −{r}^{\mathrm{2}} \\ $$$$\Rightarrow{c}=\frac{{a}^{\mathrm{2}} +{p}^{\mathrm{2}} −{r}^{\mathrm{2}} }{\mathrm{2}{a}} \\ $$$${q}^{\mathrm{2}} −{a}^{\mathrm{2}} +\left(\frac{{a}^{\mathrm{2}} +{p}^{\mathrm{2}} −{r}^{\mathrm{2}} }{\mathrm{2}{a}}\right)^{\mathrm{2}} ={p}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} −\left(\frac{{a}^{\mathrm{2}} +{p}^{\mathrm{2}} −{r}^{\mathrm{2}} }{\mathrm{2}{a}}\right)^{\mathrm{2}} ={q}^{\mathrm{2}} −{p}^{\mathrm{2}} \\ $$$$\mathrm{3}{a}^{\mathrm{4}} +\mathrm{2}\left({p}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{q}^{\mathrm{2}} \right){a}^{\mathrm{2}} −\left({p}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{0} \\ $$$${a}^{\mathrm{2}} =\frac{\sqrt{\left({p}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{q}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{3}\left({p}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)^{\mathrm{2}} }−\left({p}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{q}^{\mathrm{2}} \right)}{\mathrm{3}} \\ $$$$\Rightarrow{a}=\sqrt{\frac{\sqrt{\left({p}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{q}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{3}\left({p}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)^{\mathrm{2}} }−\left({p}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{q}^{\mathrm{2}} \right)}{\mathrm{3}}} \\ $$$$\Rightarrow{c}=... \\ $$$$\Rightarrow{b}=\sqrt{{q}^{\mathrm{2}} −{a}^{\mathrm{2}} }=... \\ $$$$ \\ $$$${example}:\:{p}=\mathrm{7},\:{q}=\mathrm{5},\:{r}=\mathrm{4} \\ $$$$\Rightarrow{a}=\sqrt{\frac{\sqrt{\left(\mathrm{49}+\mathrm{16}−\mathrm{50}\right)^{\mathrm{2}} +\mathrm{3}\left(\mathrm{49}−\mathrm{16}\right)^{\mathrm{2}} }−\left(\mathrm{49}+\mathrm{16}−\mathrm{50}\right)}{\mathrm{3}}} \\ $$$$\Rightarrow{a}=\sqrt{\frac{\mathrm{6}\sqrt{\mathrm{97}}−\mathrm{15}}{\mathrm{3}}}\approx\mathrm{3}.\mathrm{834} \\ $$$$... \\ $$

Commented by ajfour last updated on 15/Jan/19

Thank you Sir, this was tough for me.

$${Thank}\:{you}\:{Sir},\:{this}\:{was}\:{tough}\:{for}\:{me}. \\ $$

Answered by MJS last updated on 16/Jan/19

a=(√(((−p^2 +2q^2 −r^2 )/3)+((2(√(p^4 +q^4 +r^4 −(p^2 q^2 +p^2 r^2 +q^2 r^2 ))))/3))) b=(√(((p^2 +q^2 +r^2 )/3)−((2(√(p^4 +q^4 +r^4 −(p^2 q^2 +p^2 r^2 +q^2 r^2 ))))/3))) c=(√(((2p^2 −q^2 −r^2 )/3)+((2(√(p^4 +q^4 +r^4 −(p^2 q^2 +p^2 r^2 +q^2 r^2 ))))/3))) with p=7 q=5 r=4 I get a=(√(−5+2(√(97))))≈3.83376 b=(√(30−2(√(97))))≈3.20972 c=(√(19+2(√(97))))≈6.22075

$${a}=\sqrt{\frac{−{p}^{\mathrm{2}} +\mathrm{2}{q}^{\mathrm{2}} −{r}^{\mathrm{2}} }{\mathrm{3}}+\frac{\mathrm{2}\sqrt{{p}^{\mathrm{4}} +{q}^{\mathrm{4}} +{r}^{\mathrm{4}} −\left({p}^{\mathrm{2}} {q}^{\mathrm{2}} +{p}^{\mathrm{2}} {r}^{\mathrm{2}} +{q}^{\mathrm{2}} {r}^{\mathrm{2}} \right)}}{\mathrm{3}}} \\ $$$${b}=\sqrt{\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} }{\mathrm{3}}−\frac{\mathrm{2}\sqrt{{p}^{\mathrm{4}} +{q}^{\mathrm{4}} +{r}^{\mathrm{4}} −\left({p}^{\mathrm{2}} {q}^{\mathrm{2}} +{p}^{\mathrm{2}} {r}^{\mathrm{2}} +{q}^{\mathrm{2}} {r}^{\mathrm{2}} \right)}}{\mathrm{3}}} \\ $$$${c}=\sqrt{\frac{\mathrm{2}{p}^{\mathrm{2}} −{q}^{\mathrm{2}} −{r}^{\mathrm{2}} }{\mathrm{3}}+\frac{\mathrm{2}\sqrt{{p}^{\mathrm{4}} +{q}^{\mathrm{4}} +{r}^{\mathrm{4}} −\left({p}^{\mathrm{2}} {q}^{\mathrm{2}} +{p}^{\mathrm{2}} {r}^{\mathrm{2}} +{q}^{\mathrm{2}} {r}^{\mathrm{2}} \right)}}{\mathrm{3}}} \\ $$$$\mathrm{with}\:{p}=\mathrm{7}\:{q}=\mathrm{5}\:{r}=\mathrm{4}\:\mathrm{I}\:\mathrm{get} \\ $$$${a}=\sqrt{−\mathrm{5}+\mathrm{2}\sqrt{\mathrm{97}}}\approx\mathrm{3}.\mathrm{83376} \\ $$$${b}=\sqrt{\mathrm{30}−\mathrm{2}\sqrt{\mathrm{97}}}\approx\mathrm{3}.\mathrm{20972} \\ $$$${c}=\sqrt{\mathrm{19}+\mathrm{2}\sqrt{\mathrm{97}}}\approx\mathrm{6}.\mathrm{22075} \\ $$

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