Enigma

Enigmes & Problèmes scientifiques

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

dans Euclide Geometry Problem étiqueté Equal arcs par Harvey Littleman

From Gogeometry Using : Pythagoras Proposition 13 Euclide Book II In ΔDMF : DM^2=DF^2+FM^2 (1) DM^2= (r-x/2)^2+x^2 In ΔAMD : AM^2=AD^2+DM^2 -2xAD (Proposition 13 Euclide Book II) r^2= r^2+ DM^2 -2xr (2) DM^2=2xr (1) and (2) 2xr=(r-x/2)^2+x^2 2xr=r^2-xr+x^2/4+x^2 3xr=r^2+5x^2/4 Dividing by x^2 (x not equal to 0) =>3r/x=5/4+(r/x)^2 Let y=r/x […]

Gogeometry Problem 336

dans Euclide Geometry Problem étiqueté a Common Tangent and a Square / Two equal circles par Harvey Littleman

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

dans Euclide Geometry Problem étiqueté Circular Sector / Internal Common Tangent / Measurement / Semicircle / square par Harvey Littleman

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

dans Euclide Geometry Problem par Harvey Littleman

CircleO1 intersects CircleO2 in A and B By symmetry, CA=CB O2A=O2B => Δ AO2B is isosceles in O2 =>AC ⊥ CO2 In the same way with CircleO1: O1O2⊥ AB

Intersection of two circles

dans Euclide Geometry Principles par Harvey Littleman

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

dans Euclide Geometry Problem par Harvey Littleman

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

dans Euclide Geometry Problem par Harvey Littleman

Define h=HB Case 1) H in AC a^2=h^2+(b-m)^2 c^2=h^2+m^2 a^2-c^2=b^2-2mb Therefore a^2=b^2+c^2-2mb Case 2) A in HC a^2=h^2+(b+m)^2 c^2=h^2+m^2 a^2-c^2=b^2+2mb Therefore a^2=b^2+c^2+2mb

Proposition 13 Euclide Book II (Law of Cosines)

dans Euclide Geometry Principles par Harvey Littleman

AD tangent to circle with center O C on this circle => ∠BAD= ∠BCA Demonstration Define E in the circle such as E, O and A are colinear ∠0AD=90° and ∠ECA= 90° ∠BAE intercepts arc BE and ∠BCE intercepts arc BE => ∠BAE= ∠BCE ∠0AD= ∠0AB+∠BAD=90° ∠ECA= ∠ECB+∠BCA=90° […]

Inscribed angle and tangent to a circle

dans Euclide Geometry Principles par Harvey Littleman

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

dans Euclide Geometry Principles par Harvey Littleman