tag:blogger.com,1999:blog-6933544261975483399.post2549110956676272022..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1022: Circular Sector of 90 Degrees, Squares, Metric RelationsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-66783663678075156752018-10-30T03:03:59.337-07:002018-10-30T03:03:59.337-07:00Let AC = a, BC =b and OA = R
Let p be the height ...Let AC = a, BC =b and OA = R<br />Let p be the height of the altitude from O to AC, q from O to BC<br /> <br />OD2 + OG2<br />= {(p+a)2 + a2/4 } + {q+b)2 +b2/4}<br />= {R2 + 2ap + a2} + (R2 + 2bq + b2}<br />= 2R2 + a2 + b2 + 2(ap + bq)<br />= 2R2 + (2R2 – sqrt2.ab) +2{2S(OAC) + 2(OBC)}<br />= 4R2 – sqrt2.ab + 4{R2/2 + S(ABC)}<br />= 6R2 – sqrt2.ab + 4 X (1/2)/(sqrt2).ab<br />= 6R2<br />= 3AB2<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-72874133435456872382018-04-04T13:25:17.005-07:002018-04-04T13:25:17.005-07:00Que bueno! Sois unos geniosQue bueno! Sois unos geniosJ. Hilariohttps://www.blogger.com/profile/06366419615598878468noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-79772743819570260302015-02-18T00:46:29.744-08:002015-02-18T00:46:29.744-08:00We use complex numbers.
Let C = cos a + isin a be ...We use complex numbers.<br />Let C = cos a + isin a be a point on the unit circle. Then A = 0 + 1i and B = 1 + 0i.<br />Then AB² = 2, so we have to prove that OD² + OG² = 6.<br /><br />D = C -i(A - C) = cos a - sin a + 1 + i(coa s + sin a).<br />G = C + i(B - C) = cos a - sin a + i(sin a - cos a + 1).<br /><br />OD² + OG² = (cos a - sin a + 1)² + (cos a + sin a)² + (cos a - sin a)² + (sin a - cos a + 1)²<br /> = 4(cos²a + sin²a) + 2 <br /> = 6<br />Luc Gheysens, Belgiumhttp://www.gnomon.bloggen.benoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-17619992003222768412014-06-13T21:50:25.533-07:002014-06-13T21:50:25.533-07:00Let OA=R. Let S(ABC) be the area of ABC.
OD² + O...Let OA=R. Let S(ABC) be the area of ABC. <br /><br />OD² + OG²<br />= (AC² + R² − 2×AC×R cos∠OCD) + (BC² + R² − 2×BC×R cos∠OCG)<br />= (AC² + R² + 2×AC×R sin∠OCA) + (BC² + R² + 2×BC×R sin∠OCB)<br />= AC² + BC² + 2R² + 2R (AC sin∠OCA + BC sin∠OCB)<br />= AC² + BC² + 2R² + 4 [S(OAC) + S(OBC)]<br />= AC² + BC² + 2R² + 4 [S(OAB) + S(ABC)]<br />= AC² + BC² + 2R² + 2R² + 4×S(ABC)<br />= AC² + BC² + 4R² + 2×AC×BC sin∠ACB<br />= AC² + BC² + 4R² + 2×AC×BC sin135°<br />= AC² + BC² + 4R² − 2×AC×BC cos135°<br />= AC² + BC² + 4R² − 2×AC×BC cos∠ACB<br />= AB² + 4R²<br />= 3 AB²Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.com