Clawson Point

Triangle $ABC$ with vertex

$A=(-5,0)$

$B(t)=(4 \cos{t},4 \sin{t})$

$C(s)=(3 \cos{s},3 \sin{s})$

Solution of problem is triangle $ABC$ with Clawson point $(0,0)$ .

Examples of the locus of Clawson point for $t=const$ and $s=0..2 \pi$ - purple line

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Use Phyton.

Find $t,s$ and $B(t), C(s)$ - that

$\frac{h_a}{\tan {A}} = \frac{h_b}{\tan {B} }= \frac{h_c}{\tan {C} }$

$h_a$ - distance from point $(0,0)$ to line $BC$ , $h_b$ - distance from point $(0,0)$ to line $AC$ , $h_c$ - distance from point (0,0) to line $AB$ , $A,B,C$ - are angles of triangle $ABC$ .

Result is

$t =1.07831535$

$s=1.05850050$

$norma=1.0154e-11$

$S =1.9886140685$

My answer 19886 - is mistake. I change it to 19885 - it is true.

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variant t=1.07824935 s=1.0584355 norma=2.06515233480262e-10 S =1.988535489206
variant t=1.07825035 s=1.0584365 norma=1.82798782359453e-10 S =1.9885368450312
variant t=1.07825135 s=1.05843755 norma=1.70868599538155e-10 S =1.988538200853
variant t=1.07825235 s=1.0584385 norma=1.70724570969985e-10 S =1.9885395566739
variant t=1.07831035 s=1.0584955 norma=1.34638396283964e-10 S =1.988607289985
variant t=1.07831135 s=1.0584965 norma=8.617407644109922e-11 S =1.988608645690
variant t=1.07831235 s=1.0584975 norma=4.94936268490563e-11 S =1.9886100013925
variant t=1.07831335 s=1.0584985 norma=2.459693172609196e-11 S =1.9886113570929
variant t=1.07831435 s=1.0584995 norma=1.148388172697024e-11 S =1.98861271279128
variant t=1.07831535 s=1.0585005 norma=1.015435874097598e-11 S =1.9886140684875

P.S. Clawson Point max Area Problem