tag:blogger.com,1999:blog-6933544261975483399.post8573201947735404813..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 101Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-3510552428450524972018-04-09T01:50:11.879-07:002018-04-09T01:50:11.879-07:00https://goo.gl/Hj5iWc
And if d is the side of ABC,...https://goo.gl/Hj5iWc<br />And if d is the side of ABC, d = √(c^2 + √3·a·b)Ignacio Larrosa Cañestrohttps://www.blogger.com/profile/03014700623279846626noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-41543765542661395272013-11-09T11:52:35.609-08:002013-11-09T11:52:35.609-08:00http://img199.imageshack.us/img199/3421/rte.bmp
Dr...http://img199.imageshack.us/img199/3421/rte.bmp<br />Draw equilateral triangle BDE (see sketch)<br />note that ∡ (EBC)= ∡ (DBA) and DE=DB=b and ∡ (EDB)=60<br />Triangle DBA congruent to EBC (case SAS) => DA=EC=a<br />In triangle CDE we have a^2=b^2+c^2 => ∡ (CDE)=90<br />So ∡ (CDB)=150<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-15838653935124809282013-11-08T09:17:56.690-08:002013-11-08T09:17:56.690-08:00Sea el punto O el punto simétrico de A respecto
a...Sea el punto O el punto simétrico de A respecto <br />a la recta AB, de tal forma que △CBO es equilátero<br />y congruente con △BCA. Defínase el punto E como<br />el punto interior a △CBO tal que CE=b, BE=c y OE=a.<br />Sea P el centro de simetría del rombo ABOC. <br />De este manera E es el punto simétrico a D respecto a P.<br />Dada estas condiciones es claro que ADOE,<br />DBEC y ABOC son todos paralelogramos.<br />Llamemos s la longitud del lado de los<br />triángulos equiláteros △ABC y △BOC. <br />Llamemos 'q' a la longitud de la diagonal ED,<br />'r' a la longitud del lado DO, y 't' a la longitud de la diagonal AO.<br />Por ley de paralelogramo sobre ABOC,<br />DBEC y ADOE, respectivamente:<br />4s²=s²+t²<br />2c²+2b²=2a²=q²+s²<br />2a²+2r²=q²+t²<br />En la segunda ecuación se ha usado la hipótesis<br />de la relación pitagórica a²=b²+c².<br />De estas relaciones anteriores se deduce que:<br />2r²=2a²-s²-2a²+3s²=2s².<br />Por lo tanto r=s. Esto implica que el cricuncentro<br />del triángulo CDB es precisamente el punto O.<br />Al ser el ángulo central ∡COB=300° el ángulo<br />∡CDB será su respectivo ángulo inscrito,<br />por lo que ∡CDB=∡COB/2=150°. Q.E.D.SagittaDeihttps://www.blogger.com/profile/11818515838560731969noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-56481563404688543882009-05-12T15:01:00.000-07:002009-05-12T15:01:00.000-07:00http://geometri-problemleri.blogspot.com/2009/05/p...http://geometri-problemleri.blogspot.com/2009/05/problem-18-ve-cozumu.htmlAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-81015583865913817772009-04-07T02:52:00.000-07:002009-04-07T02:52:00.000-07:00Here's a solution by analytical geometry for w...Here's a solution by analytical geometry for whatever it's worth: Let C be (-p,0), B:(p,0) and therefore A:(0,V3p)where V = square root. Let D be any point (x,y). Now slope CD = y/(p+x) and slope BD = y/(p-x)and thus, tan(BDC)=-tan(DCB+DBC)=[y/(p+x)+y/(p-x)]/[y/(p+x)*y/(p-x)-1]<BR/>or tan(BDC) = 2py/(x^2+y^2-p^2) ---------(1)<BR/>Now b^2 = (x-p)^2+y^2 & c^2 = (x+p)^2+y^2 while<BR/>a^2 =x^2+(y-V3p)^2. But a^2 =b^2+c^2. Therefore we can write:<BR/>(x-p)^2+y^2+(x+p)^2+y^2=x^2+(y-V3p)^2 which yields: x^2 + y^2 - p^2 + 2V3py = 0 or <BR/>x^2 + y^2 - p^2 = - 2V3py. Using this in equation (1) we've: tan(BDC) = 2py/- 2V3py <BR/> = -1/V3 <BR/>Hence angle BDC = 150 deg.<BR/>QED<BR/>Ajit: ajitathle@gmail.comAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com