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Question Number 212822 by ajfour last updated on 24/Oct/24

Commented by ajfour last updated on 24/Oct/24

$C o n s i d e r t r i a n g l e i s o s c e l e s a b o u t$ $u p p e r v e r t e x .$
Commented by Ghisom last updated on 26/Oct/24
$an approach from the far side: (sorry I changed a⇄b) △_(ppq) ⇒ r=((q(√(2p−q)))/(2(√(2p+q)))) let q=1 ⇒ r=((√(2p−1))/( 2(√(2p+1)))) the vertices of the △ are P, Q, R P= ((((√(2p−1))/( 2(√(2p+1))))),(((√(2p−1))/( 2(√(2p+1))))) ) Q= (((((√(6p+5))+(√(2p−1)))/(2(√(2p+1))))),(0) ) R= ((((p/2)−(1/4)+(((√(6p+5))+(√(2p−1)))/( 4(√(2p+1)))))),((((√((6p+5)(2p−1)))/4)+((√(2p−1))/( 4(√(2p+1)))))) ) C_I = ((((1/4)−(1/(2(2p+1)))+(((√(6p+5))+2(√(2p−1)))/(4(√(2p+1)))))),((((√((6p+5)(2p−1)))/(4(2p+1)))+((√(2p−1))/( 4(√(2p+1)))))) ) ⇒ a=(((√(6p+5))+2(√(2p−1)))/(2(√(2p+1)))) b=((√((6p+5)(2p−1)))/4)+((√(2p−1))/( 4(√(2p+1)))) ∣PC_I ∣=((√p)/( (√(2p+1)))) if we want to match the picture ⇒ a≥b ⇒ (1/2)<p≤1.66853858571 { ((r=((√(2p−1))/( 2(√(2p+1)))))),((a=(((√(6p+5))+2(√(2p−1)))/(2(√(2p+1)))))),((b=((√((6p+5)(2p−1)))/4)+((√(2p−1))/( 4(√(2p+1)))))) :} we can find p for any ratio a:b further we can easily get { ((r=((√(2p−1))/( 2(√(2p+1)))))),((a=2r+(√(1−r^2 )))),((b=((1/2)+((2(√(1−r^2 )))/(1−4r^2 )))r)) :} (2) ⇔ (√(1−r^2 ))=a−2r inserting in (3) & transforming gives r^3 −2(b−1)r^2 −(a+(1/4))r+(b/2)=0 ⇒ r=((2b)/3)−(1/3)+((2(√(−3u)))/3)sin ((arcsin ((3(√3)v)/(2(√(−u^3 )))))/3) where u=−((4b^2 )/3)−a+((8b)/3)−((19)/(12)) v=−((16b^3 )/(27))+((16b^2 )/9)−((2ab)/3)+((2a)/3)−((13b)/9)+((41)/(54)) but a, b are not independent in this approach!$
$an approach from the far side :$ $(sorry I changed a ⇄ b)$ $△_{p p q} \Rightarrow r = \frac{q \sqrt{2 p - q}}{2 \sqrt{2 p + q}}$ $let q = 1 \Rightarrow r = \frac{\sqrt{2 p - 1}}{2 \sqrt{2 p + 1}}$ $the vertices of the △ are P, Q, R$ $P = (\begin{matrix} \frac{\sqrt{2 p - 1}}{2 \sqrt{2 p + 1}} \\ \frac{\sqrt{2 p - 1}}{2 \sqrt{2 p + 1}} \end{matrix})$ $Q = (\begin{matrix} \frac{\sqrt{6 p + 5} + \sqrt{2 p - 1}}{2 \sqrt{2 p + 1}} \\ 0 \end{matrix})$ $R = (\begin{matrix} \frac{p}{2} - \frac{1}{4} + \frac{\sqrt{6 p + 5} + \sqrt{2 p - 1}}{4 \sqrt{2 p + 1}} \\ \frac{\sqrt{(6 p + 5) (2 p - 1)}}{4} + \frac{\sqrt{2 p - 1}}{4 \sqrt{2 p + 1}} \end{matrix})$ $C_{I} = (\begin{matrix} \frac{1}{4} - \frac{1}{2 (2 p + 1)} + \frac{\sqrt{6 p + 5} + 2 \sqrt{2 p - 1}}{4 \sqrt{2 p + 1}} \\ \frac{\sqrt{(6 p + 5) (2 p - 1)}}{4 (2 p + 1)} + \frac{\sqrt{2 p - 1}}{4 \sqrt{2 p + 1}} \end{matrix})$ $\Rightarrow$ $a = \frac{\sqrt{6 p + 5} + 2 \sqrt{2 p - 1}}{2 \sqrt{2 p + 1}}$ $b = \frac{\sqrt{(6 p + 5) (2 p - 1)}}{4} + \frac{\sqrt{2 p - 1}}{4 \sqrt{2 p + 1}}$ $∣ {P C}_{I} ∣= \frac{\sqrt{p}}{\sqrt{2 p + 1}}$ $if we want to match the picture \Rightarrow$ $a ⩾ b \Rightarrow \frac{1}{2} < p ⩽ 1.66853858571$ ${\begin{cases} r = \frac{\sqrt{2 p - 1}}{2 \sqrt{2 p + 1}} \\ a = \frac{\sqrt{6 p + 5} + 2 \sqrt{2 p - 1}}{2 \sqrt{2 p + 1}} \\ b = \frac{\sqrt{(6 p + 5) (2 p - 1)}}{4} + \frac{\sqrt{2 p - 1}}{4 \sqrt{2 p + 1}} \end{cases}$ $we can find p for any ratio a : b$ $further we can easily get$ ${\begin{cases} r = \frac{\sqrt{2 p - 1}}{2 \sqrt{2 p + 1}} \\ a = 2 r + \sqrt{1 - r^{2}} \\ b = (\frac{1}{2} + \frac{2 \sqrt{1 - r^{2}}}{1 - 4 r^{2}}) r \end{cases}$ $(2) \Leftrightarrow \sqrt{1 - r^{2}} = a - 2 r$ $inserting in (3) & transforming gives$ $r^{3} - 2 (b - 1) r^{2} - (a + \frac{1}{4}) r + \frac{b}{2} = 0$ $\Rightarrow$ $r = \frac{2 b}{3} - \frac{1}{3} + \frac{2 \sqrt{- 3 u}}{3} \sin \frac{\arcsin \frac{3 \sqrt{3} v}{2 \sqrt{- u^{3}}}}{3}$ $where$ $u = - \frac{4 b^{2}}{3} - a + \frac{8 b}{3} - \frac{19}{12}$ $v = - \frac{16 b^{3}}{27} + \frac{16 b^{2}}{9} - \frac{2 a b}{3} + \frac{2 a}{3} - \frac{13 b}{9} + \frac{41}{54}$ $but a, b are not independent in this$ $approach!$
Commented by Frix last updated on 26/Oct/24

$Considering the originally posted picture :$ $We can solve for a (b, r) [degree 2] or for$ $b (a, r) [degree 2 for b^{2}] but not for r (a, b)$ $[degree 6] .$
Commented by Frix last updated on 27/Oct/24

$No problem, I will upload my workings$ $again later .$
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