tag:blogger.com,1999:blog-6933544261975483399.post7507174748782833577..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1255: Right Triangle, Angle, 90 Degrees, Perpendicular, Collinear PointsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-66292100204862510072016-09-06T12:20:31.142-07:002016-09-06T12:20:31.142-07:00Let AB = BC = 1 => AC = Sqrt(2)
BD = 2 => DC...Let AB = BC = 1 => AC = Sqrt(2)<br />BD = 2 => DC = Sqrt(5)<br /><br />Join CE and form the Right triangle CDE<br />CD = Sqrt(5)<br />DE = 3*Sqrt(5)<br />CE = 5*Sqrt(2)<br /><br /><br />Drop a _|_ from D to EC that meets at F, DF = 3/Sqrt(2) <br /><br />Area of ADC = 0.5*AD*BC = 0.5*3*1 --------(1)<br />Area of DCE = 0.5*DC*DE = 0.5*Sqrt(5)*3*Sqrt(5) = 0.5*5*3 -------------(2)<br /><br />Adding (1) & (2) = 9 Sq.units ---------(3)<br /><br />Let us assume A,C & E are collinear<br /><br />Join AE and form the triangle ADE.<br /> <br />Area of ADE = 0.5*AE*DF = 0.5*6*Sqrt(2)*3/Sqrt(2) = 9 Sq.units ------------(4)<br /><br />Since (3) = (4), our assumption is true and A,C & E are collinearAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-39897700354554308552016-09-04T01:54:40.515-07:002016-09-04T01:54:40.515-07:00Trigonometry Solution
Let < BCD = m and let &l...Trigonometry Solution<br /><br />Let < BCD = m and let < DCE = n<br /><br />So tan(m+n) = (2+3)/(1-2X3) = -1<br />Hence m+n = 135 degrees and 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-59189994073998463582016-09-04T01:51:18.481-07:002016-09-04T01:51:18.481-07:00Geometry Solution
Drop a perpendicular EF from E ...Geometry Solution<br /><br />Drop a perpendicular EF from E to AF.<br />Let AB = BC = p and CD = q so that BD = 2p and DE = 3q<br /><br />Now triangles CBD and DFE are similar hence DE = 3p and FE = 6p since DE = 3DC. <br /><br />So AF = 6p = FE hence < FAE = 45. But < FAC = 45<br />It thus follows that A,C,E are collinear<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-30775725694829005982016-09-04T00:07:17.429-07:002016-09-04T00:07:17.429-07:00Geometry solution:
https://goo.gl/photos/sdw3M9H3K...Geometry solution:<br />https://goo.gl/photos/sdw3M9H3Ktv2F8PVA<br /><br />Let x=AB=BC and BD=2x<br />Let AC cut circumcircle of triangle BDC at F<br />Note that ADF is 45-45-90 triangle so<br />DF=AF=3x/√(2)<br />CF=AF-AC=3x/√(2) – x√2= x/√(2)<br />And CD/DE=CF/DF=1/3<br />Triangle DFC similar to EDC ( case SAS)<br />So angle (DCF)= angle (DCE)=> A, C and E are collinear.<br /><br />Trigonometry solution:<br />Let u= ∠ (BCD) and v= ∠ (DCE)<br />We have ∠ (BCE)=u+v<br />So tan(∠ (BCE))= (tan(u)+tan(v))/(1- tan(u).tan(v))= (2+3)/(1-2.3)= -1<br />So ∠ (BCE=135 degrees and ∠ (ACE)= 45+135= 180 degrees<br />So A,C and E are collinear<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-54384650026074709422016-09-03T23:12:09.333-07:002016-09-03T23:12:09.333-07:00Problem 1255
Solution 1
In accordance with probl...Problem 1255<br />Solution 1<br />In accordance with problem 848 in triangle BCD <BDC=26.5 so <BCD=90-26.5=63.5.<br />But accordance with problem 849 in triangle CDE <DEC=18.5 so <DCE=90-18.5=71.5.<br />Now <ACB+<BCD+<DCE=45+63.5+71.5=180.Therefore A,C, and E are collinear.<br />Solution 2<br />Let the points K,L on the side DE such that DK=KL=LE. If <DKC=α, <DLC=β and <DEC=θ<br />then in accordance with problem 363 in triangle DCE apply α+β+θ=90 or β+θ=45.<br />Is triangle BCD similar with triangle DCL(BC=2BD, DC=2DL) so <BCD=<DCL=90-β and<br /><DCE=90-θ. But <ACB+<BCD<DCE=45+(90-β)+(90-θ)=45+180-(β+θ)=45+180-45=180.<br />Therefore A,C, and E are collinear<br />APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.com