Harvey Littleman

From Gogeometry Using : Additional points Similar triangles Equilateral triangle   Define 0 middle of BC. O is the center of semicircle OD diameter BC Define a=OB=OD=OC => AB=AC=BC=2a Arc BD=arc DE=arc EC => ∠BOD=∠DOE=∠EOC=Π/3 OB=OD=a and ∠BOD=Π/3 =>BD=a, ΔBOD is equilateral and BD//AC In the same way, DE=a, ΔDOE […]

Gogeometry Problem 326

From Gogeometry Using : Isometric transformation Additional points Pythagoras Similar triangles Tangents to a circle Define O middle of AB, r=ED AD^2+AO^2=DO^2 a^2+(a/2)^2=(a/2+r)^2 a^2+a^2/4=a^2/4+r^2+ar r^2+ar-a^2=0 ∆=a^2+4a^2=5a^2 r>0 => r=(a.sqr(5)-a )/2 Draw a square symetryc of ABCD by O =>D becomes D’, C becomes C’, E becomes E’ => C’D’BA is […]

Gogeometry Problem 458

From Gogeometry Using : Additional points Pythagoras Solution of a previous problem Tangents to a circle Proposition 13 Euclide Book II R radius of CircleO, and r of CircleF Define ro radius of CircleO , rd of CircleD, rc of CircleC and rf of CircleF Define G intersection of OA […]

Gogeometry Problem 285

From Gogeometry Using : Pythagoras   Define h=OF OA=OB=R D, E and C are aligned ΔCOD is right in O => DC^2=OC^2+OD^2 OD=r+h => (1) : (r+R/2)^2=(R/2)^2+(r+h)^2 => (1) : (r+R)^2=R^2+4(r+h)^2 OB=OA => R=2r+h => r+h=R-r (1) : (r+R)^2=R^2+4(R-r)^2 => r^2+2rR+R^2=R^2+4(R^2-2rR+r^2) => 6rR= 3r^2 Therefore 3R=r

Gogeometry Problem 284

From Gogeometry Using : Additional points Inscribed angles in a circle central and inscribed angle Tangents to a circle Inscribed angle and tangent to a circle Without using : Law of cosines Define : F the tangent intersection of CirleD and CircleO E intersection of CircleD and OA G intersection […]

Gogeometry Problem 283

AG and AH are tangent to a circle with center O =>AG=AH Proof AG and AH are tangent to a circle with center O <=>∠AHO=∠AGO= π/2 OH=OG=r => O is at the same distance of AH and AG => O in on the angle bissector of ΔGAH => ∠GAO=∠HAO =>ΔOAG […]

Tangents to a circle