tag:blogger.com,1999:blog-6933544261975483399.post158846542264703970..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 480: Triangle, Circle, Center, Altitude, Angle, CircumcenterAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-6933544261975483399.post-60957976299926016942023-12-21T22:43:00.980-08:002023-12-21T22:43:00.980-08:00Simple Solution
< AOB = 180 - 2@
So Reflex <...Simple Solution<br /><br />< AOB = 180 - 2@<br />So Reflex < AOB = 180 + 2@<br /><br />Hence < ACB = Reflex <AOB /2 = 90 + @<br />So < BCD = 90 - @<br /><br />And Beta = 90- (90-@) = @<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-23234018376620464082023-12-20T20:42:26.761-08:002023-12-20T20:42:26.761-08:00[For easy typing, I use a for alpha & b for be...[For easy typing, I use a for alpha & b for beta]<br />Join OA, OA=OB (radii)<br /> a=bMarcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-85980476210934456372019-01-23T21:33:21.966-08:002019-01-23T21:33:21.966-08:00< AOC = 2B & < COB = 2A
So 2A + 2B + 2@ ...< AOC = 2B & < COB = 2A<br />So 2A + 2B + 2@ = 180<br />Hence A + B + @ = 90 ...(1)<br />Now < BCD = A + B<br />So A + B + Beta = 90......(2)<br /><br />Comparing (1) & (2) the result follows<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-50571273225146206212017-08-25T15:08:33.890-07:002017-08-25T15:08:33.890-07:00In general, ortocenter and circuncenter are isogon...In general, ortocenter and circuncenter are isogonal conjugates.Ignacio Larrosa Cañestrohttps://www.blogger.com/profile/03014700623279846626noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-42281595107659171552017-04-01T17:45:50.064-07:002017-04-01T17:45:50.064-07:001. Draw in AO.
2. Note triangles ABO ACO and BCO ...1. Draw in AO. <br />2. Note triangles ABO ACO and BCO are all isosceles.<br />3. Let angle OBA = x and angle ABC = y (greek letters are hard in comments)<br />4. Then angle OAB = OBA = x from the isosceles triangle.<br />5. Angle OCB = OBC must be x + y from its isosceles triangle<br />6. Angle AOC is 2y since its the central angle and ABC is the inscribed one.<br />6. Angle OAC = ACO = 90 - y since this is also isosceles.<br />7. Angle BCD = 180 - (angle ACO + angle OCB) = 90 - x<br />8. Therefore CBD = 180 - 90 - BCD = x. <br />Benjamin Leishttps://www.blogger.com/profile/10974191081762367425noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-91119164171537410772016-06-20T06:50:09.711-07:002016-06-20T06:50:09.711-07:00I take the E point of maximum arc AB (not arc AC...I take the E point of maximum arc AB (not arc ACB).Is <BCD=<AEB=γ,then β+γ=90.<br />But <AOB=2.<AEB=2γ,so 2α+2γ=180 or α+γ=90. Therefore α=β.<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-12956286997244870352010-07-26T12:15:55.367-07:002010-07-26T12:15:55.367-07:00join A to O, draw OT perpendicular to AB ( T on AB...join A to O, draw OT perpendicular to AB ( T on AB )<br />ang BCD = 90° - ß = 1/2 arc ACB (1)<br />ang AOB = arc ACB =><br />ang AOT = 1/2 arc ACB (2)<br />from (1) & (2)<br />ang BCD = ang AOT =90°-α =90°-ß (3)( AT=TB,OT perpen )<br />from (3) =><br /><br />α = ßc .t . e. onoreply@blogger.com