tag:blogger.com,1999:blog-6933544261975483399.post84162539766547834..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1016. Triangle, 45 Degrees, Area, Quadrilateral, Altitude, BisectorAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-35462360815067414322017-03-27T02:20:33.314-07:002017-03-27T02:20:33.314-07:00Problem 1016
Let S2 = S(ABD) and let S3 = S(CBD)
...Problem 1016<br />Let S2 = S(ABD) and let S3 = S(CBD)<br /> <br />< CHE = 45, so AHEB is concyclic.<br />Hence < AEB = < AHB = 90<br />and so BE = c/√2<br />Similarly BF = a/√2<br /> <br />Now AD = bc/(a+c) so S2/S = {bc/(a+c)}/b,<br />Hence S2 = S{c/(a+c)}<br />Similarly S3 = S{a/(a+c)}<br /> <br />So S1 = S(BFD) + S(BDE) = {(a/ √2)/c}S2 + { (c/√2)/a}S3<br /> S1 = S{(a/√2/(a+c)} + S{ c/√2/(a+c)} = 1/√2<br />Therefore S^2 /S1^2 = 2.<br /> <br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-63530999827142517712014-05-23T13:48:17.674-07:002014-05-23T13:48:17.674-07:00Denote AB+BC=s and AB*BC=p. 2*[FBED]=2*[ABC]-2*[AF...Denote AB+BC=s and AB*BC=p. 2*[FBED]=2*[ABC]-2*[AFD+CED]=p/sqrt(2)-(AF+EC)h, where h is equal height from D to AF and EC. Note that BFHC and BEHA are concyclic, so E and F are feet of altitudes of AE=AB/sqrt(2) and CF=BC/sqrt(2) respectively. Therefore (AF+EC)=s-s/sqrt(2)=s(sqrt(2)-1)/sqrt(2), and h=p/(sqrt(2)*s). Plugging these into first equation yields<br /> 2*[FBED]=p/sqrt(2)-p(sqrt(2)-1)/2=sqrt(2)p/2-(sqrt(2)-1)p/2=p/2. 2*[ABC]=p/sqrt(2) <br />so [ABC]/[FBED]=sqrt(2)Ivan Bazarovnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-15943359698213164182014-05-18T21:00:04.071-07:002014-05-18T21:00:04.071-07:00Let Area(XYZ) denote area of triangle XYZ
Draw alt...Let Area(XYZ) denote area of triangle XYZ<br />Draw altitudes AE’ and CF’ of triangle ABC<br />Let N is orthocenter of triangle ABC<br />Triangles ABE’ and CBF’ are right isosceles triangles<br />Quadrilateral HNE’C is cyclic => ∠ (NHE)= ∠ (NCE)=45 <br />So E’ coincide to E<br />Similarly F’ coincide to F<br />Let BM and DL are altitudes of triangles BEF and DEF<br />Triangle EBF similar to ABC … ( case SAS)<br />So AB/BE=BC/BF = sqrt(2)= (BM+DL)/BM => DL/BM= sqrt(2)-1<br />We have S= Area(ABC)= 2.Area(BEF)…. (1)<br />And Area(DEF)/Area(BEF)=sqrt(2)-1.. ( both triangles have the same base)<br />So S1=Area(BEF)+Area(DEF)= sqrt(2). Area(BEF)…(2)<br />From (1) and (2) we have S/S1= sqrt(2) => S^2/S1^2= 2<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com