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Question Number 161015 by mr W last updated on 10/Dec/21

Commented by mr W last updated on 11/Dec/21

$T h e d i s t a n c e s f r o m t h e o r t h o c e n t e r$ $t o t h e v e r t e x e s o f a t r i a n g l e a r e$ $p, q, r r e s p e c t i v e l y . F i n d t h e s i d e$ $l e n g t h e s o f t h e t r i a n g l e .$
	Answered by mr W last updated on 11/Dec/21

	Commented by mr W last updated on 14/Dec/21
	$(p/v)=(q/u) ⇒pu=qv (p/w)=(r/u) ⇒pu=rw ⇒ determinant (((pu=qv=rw)))=(1/k), say ⇒u=(1/(pk)), v=(1/(qk)), w=(1/(rk)) say the area of triangle ABC is Δ and the side lengthes are a, b, c. Δ=(1/2)a(p+u)=(1/2)b(q+v)=(1/2)c(r+w) Δ=(1/2)(au+bv+cw) Δ=(1/2)(((2Δu)/(p+u))+((2Δv)/(q+v))+((2Δw)/(r+w))) ⇒ determinant ((((u/(p+u))+(v/(q+v))+(w/(r+w))=1))) (1/((p/u)+1))+(1/((q/v)+1))+(1/((r/w)+1))=1 (1/(p^2 k+1))+(1/(q^2 k+1))+(1/(r^2 k+1))=1 (p^2 k+1)(q^2 k+1)+(q^2 k+1)(r^2 k+1)+(r^2 k+1)(p^2 k+1)=(p^2 k+1)(q^2 k+1)(r^2 k+1) (p^2 q^2 +q^2 r^2 +r^2 p^2 )k^2 +2(p^2 +q^2 +r^2 )k+3=p^2 q^2 r^2 k^3 +(p^2 q^2 +q^2 r^2 +r^2 p^2 )k^2 +(p^2 +q^2 +r^2 )k+1 p^2 q^2 r^2 k^3 −(p^2 +q^2 +r^2 )k−2=0 determinant (((k^3 −((p^2 +q^2 +r^2 )/(p^2 q^2 r^2 )) k−(2/(p^2 q^2 r^2 ))=0))) (1/(p^4 q^4 r^4 ))−(1/(27))(((p^2 +q^2 +r^2 )/(p^2 q^2 r^2 )))^3 ≤0 ⇒3 real roots generally two of the three roots are suitable. determinant (((k=(2/(pqr))(√((p^2 +q^2 +r^2 )/3)) sin {((2𝛑)/3)−(1/3) sin^(−1) [pqr((3/(p^2 +q^2 +r^2 )))^(3/2) ]})),((k=−(2/(pqr))(√((p^2 +q^2 +r^2 )/3)) sin {(𝛑/3)−(1/3) sin^(−1) [pqr((3/(p^2 +q^2 +r^2 )))^(3/2) ]}))) k>0 for the case that the orthocenter lies inside the triangle and k<0 for the case that it lies outside. with k we can get: 𝚫=(1/( (√(((1/(p+(1/(pk))))+(1/(q+(1/(qk))))+(1/(r+(1/(rk)))))(−(1/(p+(1/(pk))))+(1/(q+(1/(qk))))+(1/(r+(1/(rk)))))((1/(p+(1/(pk))))−(1/(q+(1/(qk))))+(1/(r+(1/(rk)))))((1/(p+(1/(pk))))+(1/(q+(1/(qk))))−(1/(r+(1/(rk))))))))) a=((2𝚫)/(p+(1/(pk)))) b=((2𝚫)/(q+(1/(qk)))) c=((2𝚫)/(r+(1/(rk)))) example 1: p=7, q=6, r=5 ⇒a≈9.79245055 or 1.56227610 ⇒b≈10.43513717 or 3.92946646 ⇒c≈10.94952454 or 5.14205276 (see first diagram) example 2: p=q=r=2 ⇒a=b=c≈3.46410162 (=2(√3))$
	$\frac{p}{v} = \frac{q}{u} \Rightarrow p u = q v$ $\frac{p}{w} = \frac{r}{u} \Rightarrow p u = r w$ $\Rightarrow \begin{matrix} p u = q v = r w \end{matrix} = \frac{1}{k}, s a y$ $\Rightarrow u = \frac{1}{p k}, v = \frac{1}{q k}, w = \frac{1}{r k}$ $s a y t h e a r e a o f t r i a n g l e A B C i s Δ$ $a n d t h e s i d e l e n g t h e s a r e a, b, c .$ $Δ = \frac{1}{2} a (p + u) = \frac{1}{2} b (q + v) = \frac{1}{2} c (r + w)$ $Δ = \frac{1}{2} (a u + b v + c w)$ $Δ = \frac{1}{2} (\frac{2 Δ u}{p + u} + \frac{2 Δ v}{q + v} + \frac{2 Δ w}{r + w})$ $\Rightarrow \begin{matrix} \frac{u}{p + u} + \frac{v}{q + v} + \frac{w}{r + w} = 1 \end{matrix}$ $\frac{1}{\frac{p}{u} + 1} + \frac{1}{\frac{q}{v} + 1} + \frac{1}{\frac{r}{w} + 1} = 1$ $\frac{1}{p^{2} k + 1} + \frac{1}{q^{2} k + 1} + \frac{1}{r^{2} k + 1} = 1$ $(p^{2} k + 1) (q^{2} k + 1) + (q^{2} k + 1) (r^{2} k + 1) + (r^{2} k + 1) (p^{2} k + 1) = (p^{2} k + 1) (q^{2} k + 1) (r^{2} k + 1)$ $(p^{2} q^{2} + q^{2} r^{2} + r^{2} p^{2}) k^{2} + 2 (p^{2} + q^{2} + r^{2}) k + 3 = p^{2} q^{2} r^{2} k^{3} + (p^{2} q^{2} + q^{2} r^{2} + r^{2} p^{2}) k^{2} + (p^{2} + q^{2} + r^{2}) k + 1$ $p^{2} q^{2} r^{2} k^{3} - (p^{2} + q^{2} + r^{2}) k - 2 = 0$ $\begin{matrix} k^{3} - \frac{p^{2} + q^{2} + r^{2}}{p^{2} q^{2} r^{2}} k - \frac{2}{p^{2} q^{2} r^{2}} = 0 \end{matrix}$ $\frac{1}{p^{4} q^{4} r^{4}} - \frac{1}{27} {(\frac{p^{2} + q^{2} + r^{2}}{p^{2} q^{2} r^{2}})}^{3} ⩽ 0 \Rightarrow 3 r e a l r o o t s$ $g e n e r a l l y t w o o f t h e t h r e e r o o t s a r e$ $s u i t a b l e .$ $\begin{array}{cc} k = \frac{2}{p q r} \sqrt{\frac{p^{2} + q^{2} + r^{2}}{3}} \sin {\frac{2 π}{3} - \frac{1}{3} \sin^{- 1} [p q r {(\frac{3}{p^{2} + q^{2} + r^{2}})}^{\frac{3}{2}}]} \\ k = - \frac{2}{p q r} \sqrt{\frac{p^{2} + q^{2} + r^{2}}{3}} \sin {\frac{π}{3} - \frac{1}{3} \sin^{- 1} [p q r {(\frac{3}{p^{2} + q^{2} + r^{2}})}^{\frac{3}{2}}]} \end{array}$ $k > 0 f o r t h e c a s e t h a t t h e o r t h o c e n t e r$ $l i e s i n s i d e t h e t r i a n g l e a n d k < 0 f o r$ $t h e c a s e t h a t i t l i e s o u t s i d e .$ $w i t h k w e c a n g e t :$ $Δ = \frac{1}{\sqrt{(\frac{1}{p + \frac{1}{p k}} + \frac{1}{q + \frac{1}{q k}} + \frac{1}{r + \frac{1}{r k}}) (- \frac{1}{p + \frac{1}{p k}} + \frac{1}{q + \frac{1}{q k}} + \frac{1}{r + \frac{1}{r k}}) (\frac{1}{p + \frac{1}{p k}} - \frac{1}{q + \frac{1}{q k}} + \frac{1}{r + \frac{1}{r k}}) (\frac{1}{p + \frac{1}{p k}} + \frac{1}{q + \frac{1}{q k}} - \frac{1}{r + \frac{1}{r k}})}}$ $a = \frac{2 Δ}{p + \frac{1}{p k}}$ $b = \frac{2 Δ}{q + \frac{1}{q k}}$ $c = \frac{2 Δ}{r + \frac{1}{r k}}$ $\underset{―}{e x a m p l e 1 : p = 7, q = 6, r = 5}$ $\Rightarrow a \approx 9.79245055 o r 1.56227610$ $\Rightarrow b \approx 10.43513717 o r 3.92946646$ $\Rightarrow c \approx 10.94952454 o r 5.14205276$ $(s e e f i r s t d i a g r a m)$ $\underset{―}{e x a m p l e 2 : p = q = r = 2}$ $\Rightarrow a = b = c \approx 3.46410162 (= 2 \sqrt{3})$
	Commented by Tawa11 last updated on 11/Dec/21

	$Great sir$
	Commented by mr W last updated on 12/Dec/21

	Commented by mr W last updated on 14/Dec/21

	Commented by mr W last updated on 14/Dec/21

	Commented by mr W last updated on 14/Dec/21

	Commented by mr W last updated on 14/Dec/21

	Answered by mr W last updated on 27/Dec/21
	$according to Q162071 we can also calculate the side lengthes using following formulas: determinant ((((1/k)=(2/(pqr))(√((p^2 +q^2 +r^2 )/3)) sin {(π/3)−(1/3) sin^(−1) pqr((3/(p^2 +q^2 +r^2 )))^(3/2) })),(((1/k)=−(2/(pqr))(√((p^2 +q^2 +r^2 )/3)) sin {(π/3)+(1/3) sin^(−1) pqr((3/(p^2 +q^2 +r^2 )))^(3/2) }))) a=(√(q^2 +r^2 −2k)) b=(√(r^2 +p^2 −2k)) c=(√(p^2 +q^2 −2k))$
	$a c c o r d i n g t o Q 162071$ $w e c a n a l s o c a l c u l a t e t h e s i d e l e n g t h e s$ $u s i n g f o l l o w i n g f o r m u l a s :$ $\begin{array}{cc} \frac{1}{k} = \frac{2}{p q r} \sqrt{\frac{p^{2} + q^{2} + r^{2}}{3}} \sin {\frac{π}{3} - \frac{1}{3} \sin^{- 1} p q r {(\frac{3}{p^{2} + q^{2} + r^{2}})}^{\frac{3}{2}}} \\ \frac{1}{k} = - \frac{2}{p q r} \sqrt{\frac{p^{2} + q^{2} + r^{2}}{3}} \sin {\frac{π}{3} + \frac{1}{3} \sin^{- 1} p q r {(\frac{3}{p^{2} + q^{2} + r^{2}})}^{\frac{3}{2}}} \end{array}$ $a = \sqrt{q^{2} + r^{2} - 2 k}$ $b = \sqrt{r^{2} + p^{2} - 2 k}$ $c = \sqrt{p^{2} + q^{2} - 2 k}$
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