(
ukster: bedankt voor r2.)
Via de lengte van de zijden, exacte goniometrische gelijkheden en Heron's formule hebben we nu de exacte formule voor de 3 stralen r1, r2 en r3.
Dan resteert nog het aantonen van de gelijkheid; we beginnen met 1/r1:
\(\frac{1}{r_1} = \sqrt{\frac{1}{2-\frac{2}{5}\sqrt{5}}}= \sqrt{\frac{2+\frac{2}{5}\sqrt{5}}{4 - \frac{4}{5}}}= \frac{1}{4}\sqrt{10+2\sqrt{5}}\)
r2 in iets andere vorm:
\(r_2 = 1+\sqrt{5}-2\sqrt{1+\frac{2}{5}\sqrt{5}}\)
dus
\(\frac{1}{r2} = \frac{1+\sqrt{5}+2\sqrt{1+\frac{2}{5}\sqrt{5}}}{(1+\sqrt{5})^2-4\cdot (1+\frac{2}{5}\sqrt{5})} = \frac{1+\sqrt{5}+2\sqrt{1+\frac{2}{5}\sqrt{5}}} {2+\frac{2}{5}\sqrt{5}}\)
\(= \frac{(1+\sqrt{5}+2\sqrt{1+\frac{2}{5}\sqrt{5}})(2-\frac{2}{5}\sqrt{5})} {(2+\frac{2}{5}\sqrt{5})(2-\frac{2}{5}\sqrt{5}) }=\frac{1}{16}\left(1+\sqrt{5}+2\sqrt{1+\frac{2}{5}\sqrt{5}}\right)(10-2\sqrt{5})\)
\(=\frac{1}{4}\left( 2\sqrt{5} + 5\sqrt{1+\frac{2}{5}\sqrt{5}} - \sqrt{5+2\sqrt{5}} \right)\)
\(=\frac{1}{4}\left( 2\sqrt{5} + (5-\sqrt{5}) \sqrt{1+\frac{2}{5}\sqrt{5}}\right)\)
\(=\frac{1}{4}\left( 2\sqrt{5} + \sqrt{30-10\sqrt{5}} \sqrt{1+\frac{2}{5}\sqrt{5}}\right)\)
\(=\frac{1}{4}\left( 2\sqrt{5} + \sqrt{10+2\sqrt{5}}\right)\)
tenslotte 1/r3:
\(\frac{1}{r_3} = \frac{1}{4 - 2\sqrt{5}+2\sqrt{10-\frac{22}{5}\sqrt{5}}} = \frac{4 - 2\sqrt{5} - 2\sqrt{10-\frac{22}{5}\sqrt{5}}} {(4 - 2\sqrt{5})^2-\left(2\sqrt{10-\frac{22}{5}\sqrt{5}}\right)^2}\)
\(= \frac{4 - 2\sqrt{5} - 2\sqrt{10-\frac{22}{5}\sqrt{5}}} {(4 - 2\sqrt{5})^2-\left(2\sqrt{10-\frac{22}{5}\sqrt{5}}\right)^2} = \frac{2\sqrt{5} - 4 + 2\sqrt{10-\frac{22}{5}\sqrt{5}}} {4-\frac{8}{5}\sqrt{5}}\)
\(= \frac{1}{4} \left( \frac{(2\sqrt{5} - 4 + 2\sqrt{10-\frac{22}{5}\sqrt{5}})(1+\frac{2}{5}\sqrt{5})} {(1-\frac{2}{5}\sqrt{5})(1+\frac{2}{5}\sqrt{5})}\right)\)
\(= \frac{1}{4} \left( 2\sqrt{5} - 4 + 2\sqrt{10-\frac{22}{5}\sqrt{5}}\right)(5+2\sqrt{5})\)
\(= \frac{1}{4} \left( 2\sqrt{5} + 2(5+2\sqrt{5}) \sqrt{10-\frac{22}{5}\sqrt{5}}\right)\)
\(= \frac{1}{4} \left( 2\sqrt{5} + 2\sqrt{45+20\sqrt{5}} \sqrt{10-\frac{22}{5}\sqrt{5}}\right)\)
\(= \frac{1}{4} \left( 2\sqrt{5} + 2\sqrt{10+2\sqrt{5}}\right)\)
Uit bovenstaande volgt:
\(\frac{1}{r_3} = \frac{1}{r_1}+\frac{1}{r_2}\)
Het probleem komt over als een mooie sangaku
(
https://nl.wikipedia.org/wiki/Sangaku)
De vraag is dan: is er een eleganter bewijs dan bovenstaand brutekracht bewijs ?