tag:blogger.com,1999:blog-6933544261975483399.comments2024-03-17T11:34:28.946-07:00Go Geometry (Problem Solutions)Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger6750125tag:blogger.com,1999:blog-6933544261975483399.post-30265973444283054962024-03-17T00:52:02.052-07:002024-03-17T00:52:02.052-07:00∆CFE ~ ∆DAE
CF/DA=EF/AE=1/3
CF=x/3
CE=√(100-x^2/9)...∆CFE ~ ∆DAE<br />CF/DA=EF/AE=1/3<br />CF=x/3<br />CE=√(100-x^2/9),DE=√(900-x^2 )<br />CE+DE=CD=x<br />√(100-x^2/9)+√(900-x^2 )=x<br />(4/3)* √(900-x^2 )=x<br />900-x^2=(9x^2)/16<br />x^2=576<br />x=24Marcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-37193601403893858672024-03-15T23:47:37.597-07:002024-03-15T23:47:37.597-07:00Triangle CKE ~ Triangle GKD
CK/KG=CE/GD=2/5
CK/CG=...Triangle CKE ~ Triangle GKD<br />CK/KG=CE/GD=2/5<br />CK/CG=CK/(CK+KG)=2/7<br /><br />Triangle CKF ~ Triangle CGD<br />KF/GD=CK/CG=2/7<br />x/5=2/7<br />x=10/7Marcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-60567450193779806392024-03-13T13:56:04.505-07:002024-03-13T13:56:04.505-07:00There are more general forms of this question.
The...There are more general forms of this question.<br />The first one is, BAD = 120-4t=90, BAC=30-t=22.5, BCA=30, ACD=2t=15 where t=7.5 and the result is 3t=22.5.<br /><br />The second one is, BAD = 60+4t=90, BAC=3t=22.5, BCA=60-4t, ACD=30-2t=15 where t=7.5 and the result is 30-t=22.5<br />I have not investigated yet that the above solutions are also the solutions of the general form. One can also try to give the solution for the general cases.<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-89522198401033178762024-03-12T12:40:47.786-07:002024-03-12T12:40:47.786-07:00t=20, 30+t=50, 2t=40, 90-3t=30, is the general for...t=20, 30+t=50, 2t=40, 90-3t=30, is the general form of the question. The answer is 150-2t=110. Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-31176858943071606592024-03-05T11:48:50.161-08:002024-03-05T11:48:50.161-08:00Let BO extended meet AC at X, BH extended meet AC ...Let BO extended meet AC at X, BH extended meet AC at U & OD extended at Y, Z be on BC extended, AH extended meet BC at V and let CH extended meet AB at W<br /><br />< VAB = 90 - 21 - 42 = 27 and < OAB = OBA = 21 & < OAV = 27 - 21 = 6<br />< AOC = 2 X < ABC = 126 & so < OCA = OAC = 27 & <VAC = 27 - 6 = 21<br />Now < BCO = < CBO = 42 & < BCW = 90 - 21 - 42 = 27 & so < OCW = 15<br />Hence < XOY = < CBO = 42 = < BCO = < COY & OY bisects < XOC<br />Also < CBU = 90 - 69 = 21 & so BY bisects < CBX<br />It follows that Y is the excentre of Triangle BOC & so YC bisects < OCZ<br />So < OCY = 69 & < DCY = 69 - 27 = 42 & since < CDY = < DCB = 69, < DYC = 69<br />But Triangles HCU & YCU are congruent SAS & So HC = YC = DC<br />Hence C is the Circumcentre pf Triangle HDY and so < DHY = 1/2 of < DCY = 21<br />Therefore < CHC = 21 + 48 = 69<br /><br />Sumith Peiris<br />Moratuwa<br />Sri Lanka Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-83956757463269617842024-03-03T03:27:47.225-08:002024-03-03T03:27:47.225-08:00See diagram here.
Extend BO to intersect AC in E,...<a href="https://drive.google.com/file/d/1koW872Jcj72ycisQDVOt_vhg8c0h3r9v/view?usp=sharing" rel="nofollow"> See diagram here.</a><br /><br />Extend BO to intersect AC in E, OD to intersect (O) in F and BH to intersect (O) in G.<br /><br />We have :<br />- ΔAOB and ΔBOC isosceles in O ⇒ ∠OAB = ∠ABO = 21° and ∠BCO = ∠CBO = 42°<br />- ΔAOC isosceles in O and ∠AOC = 2. ∠ABC = 126° ⇒ ∠OAC =∠OCA = (180-∠ABC)/2 = 27°<br />- ∠BAC = ∠OAB + ∠OAC = 48°<br />- CH being the altitude from C ⇒ ∠HCA = 90°- ∠BAC = 42°<br />- In ΔBCE, ∠BCE = ∠BCA = ∠BCO + ∠OAC = 69° and ∠BEC = 180°- ∠BCE - ∠CBO = 69° so ΔBCE is isosceles in B and therefore H, being the altitude from B, also lies on the bisector of ∠CBE<br />- Hence ∠OBG = ∠OBH = ∠EBH = 21°<br />- Also, since OD is // to BC, ∠ODA = ∠BCA = 69° = ∠OCA + ∠COD, therefore ∠COD = 42° and ∠BOD = 2.∠BAC + ∠COD = 138°<br /><br />By construction of F and G, ΔBOF and ΔBOG are both isosceles in O.<br />In ΔBOF, ∠OBF = (180-∠BOF )/2 = (180-∠BOD )/2 = 21° = ∠OBG = ∠OBH = means that H lies on BF, so that BH and OD actually intersect on (O) in F = G which are one and the same.<br /><br />Now :<br />- In ΔGCH, ∠CFH = ∠CFB = ∠CAB = 48° and ∠CHF = 90° - ∠HCA = 48° so ΔGCH is isosceles in C and CH = CF<br />- In ΔDCF, ∠CDF = ∠ODA = 69° and ∠CFD = 180° - ∠CDF - ∠DCF = 111 - ∠ABF = 69° so that ΔDCF is isosceles in C and CD = CF<br /><br />So: CH = CD, ΔDCH is isosceles in C and ∠DHC = (180-∠HCD)/2 = (180-∠HCA )/2 = 69°<br />Greghttps://www.blogger.com/profile/14941282981782772132noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-57393575466040163142024-02-22T02:31:45.557-08:002024-02-22T02:31:45.557-08:00Thak you for your warm wordsThak you for your warm wordsAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-12411551992526018962024-02-10T21:14:36.924-08:002024-02-10T21:14:36.924-08:00Let <CBD=x, <BCD=160-x
sin(160-x)/BD=sin20/B...Let <CBD=x, <BCD=160-x<br />sin(160-x)/BD=sin20/BC<br />BC/BD=sin20/sin(160-x)<br /><ABD=40<br />sin40/AD=sin70/BD<br />AD/BD=sin40/sin70<br />AD=BC<br />sin20/sin(160-x)=sin40/sin70<br />sin20sin70=sin(160-x)sin40<br />sin20cos20=sin(160-x)sin40<br />sin(160-x)=1/2<br />160-x=30 or 150<br />x=130 (rej.) or 10Marcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-6394338911479363792024-02-09T20:07:25.333-08:002024-02-09T20:07:25.333-08:00<BCA=40
<ABC=100
sin100/AC=sin40/BC
BC=ACsin...<BCA=40<br /><ABC=100<br />sin100/AC=sin40/BC<br />BC=ACsin40/sin100<br />CD=BC=ACsin40/sin100<br />sin30/CD=sinx/AC<br />CD=ACsin30/sinx<br />sin40/sin100=sin30/sinx<br />sin40sinx=sin100sin30<br />2sin40sinx=sin100<br />sinx=cos40<br />x=50Marcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-54590212271970822982024-02-05T21:37:20.567-08:002024-02-05T21:37:20.567-08:00. o 1 continue line segment df to point z on bc. o 1 continue line segment df to point z on bcAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-65820496180230878312024-02-01T04:11:33.642-08:002024-02-01T04:11:33.642-08:00very nice solution.thank you Sumithvery nice solution.thank you SumithAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-27196695141650246102024-01-31T14:29:20.361-08:002024-01-31T14:29:20.361-08:00thanks for you solution. How do you know that CD =...thanks for you solution. How do you know that CD = EF?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-30260273484778736392024-01-27T22:01:04.008-08:002024-01-27T22:01:04.008-08:00CD=sqrt20=AD
tan<EDC=2
<EDC=arctan(2)=63.434...CD=sqrt20=AD<br />tan<EDC=2<br /><EDC=arctan(2)=63.43494882<br /><EDA=116.5650512<br />AE^2=AD^2+2^2-2(2)(AD)cos<EDA<br />AE^2=32<br />cosx=(AE^2+ED^2-AD^2)/2(AE)(ED)<br />x=45Marcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-25057077664873860302024-01-26T05:43:14.400-08:002024-01-26T05:43:14.400-08:00simple solution.simple solution.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-13944764121684837102024-01-16T02:09:18.089-08:002024-01-16T02:09:18.089-08:00Let < ABD = p and let < ADC = q
So < DBC ...Let < ABD = p and let < ADC = q<br />So < DBC = q - C = B - p ==> q = B + C - p ......(1) <br />But 2q = B + C ......(2)<br />(2) - (1) gives q = p<br />Hence AB = AD<br /><br />Sumith Peiris<br />Moratuwa<br />Sri Lanka Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-22476094328592441062024-01-13T20:28:20.211-08:002024-01-13T20:28:20.211-08:00<ADB=(<ABC+<ACB)/2
<ADB=(180-<BAC)/...<ADB=(<ABC+<ACB)/2<br /><ADB=(180-<BAC)/2<br /><ADB=90-<BAC/2<br /><BAC=180-2<ADB<br /><ABD=180-<BAD-<ADB<br /><ABD=180-(180-2<ADB)-<ADB=<ADB<br />As <ABD=<ADB, AB=AD (sides opp eq. <s)Marcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-71085286902761731492024-01-12T22:26:51.505-08:002024-01-12T22:26:51.505-08:00Let the interecting point of CG & EF be Y
Let ...Let the interecting point of CG & EF be Y<br />Let <ECY=x<br /><EYC=90-x=<FYG<br /><CYF=180-<EYC=90+x<br /><CYF+<FYG=90+x+90-x=180<br />CHG is a st. lineMarconoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-11770216230490388362024-01-12T22:18:14.726-08:002024-01-12T22:18:14.726-08:00<HEF=<EAD (corr. <s EF//AG)
<EAD+<A...<HEF=<EAD (corr. <s EF//AG)<br /><EAD+<AED=90<br /><HEF+<AED=90<br /><HEF+<AED+<DEF=180<br />AEH is a st. lineMarconoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-70685634167431290382024-01-12T21:55:07.930-08:002024-01-12T21:55:07.930-08:00<AO'B=180-50=130
<ADB=180-<AO'B/2...<AO'B=180-50=130<br /><ADB=180-<AO'B/2=180-65=115Marconoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-38137815693234311912024-01-12T06:53:22.235-08:002024-01-12T06:53:22.235-08:00< ADC = 120
Since D is the mid point of arc AC,...< ADC = 120<br />Since D is the mid point of arc AC, < CAD = < ACD = 30<br /><br />Since < DEF = 60, AE = DE = x<br />Similarly CF = x<br /><br />So 3x = 12 and x = 4<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-12873623521064994022024-01-11T02:55:25.831-08:002024-01-11T02:55:25.831-08:00what is the step you cannot understand? do let me ...what is the step you cannot understand? do let me knowSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-54067513983466738422024-01-09T20:38:43.193-08:002024-01-09T20:38:43.193-08:00220 ?220 ?MateManiahttps://www.blogger.com/profile/02215817700048635017noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-33309075227694008662024-01-07T04:09:59.694-08:002024-01-07T04:09:59.694-08:002nd Geometry Solution
Extend DM to meet AB extende...2nd Geometry Solution<br />Extend DM to meet AB extended at N. Complete Rectangle EXBY, X on AB & Y on CM<br /><br />As before < AEM = 90<br /><br />So considering Triangle ADN,<br />AE = 20 X 40 / 20V5 = 8V5 and so <br />DE = 4V5 by using Pythagoras on Triangle ADE and EM = 10V5 - 4V5 = 6V5<br />Hence MY = 6 and EX = 16<br />So S(AEB) = 20 X 16 /2 = 160<br />S(ADE) = 8V5 X 4V5 /2 = 80<br />Adding S(ABED) = 160+80 = 240<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-45705666669518789682024-01-06T23:51:34.422-08:002024-01-06T23:51:34.422-08:00Let radius of the circle be r,
r^2+r^2-2r^2cos120=...Let radius of the circle be r,<br />r^2+r^2-2r^2cos120=144<br />r=4sqrt3<br />BD perpendicular to AC and intersects at H<br />BH=sqrt(12^2-6^2)=6sqrt3<br />DH=DB-BH-2r-6sqrt3=2sqrt3<br />x^2=EH^2+DH^2<br />x^2=4+12<br />x=4Marcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-12823714338530217152024-01-06T03:29:18.485-08:002024-01-06T03:29:18.485-08:00Trigonometry Solution
Extend BEM to meet AD exten...Trigonometry Solution<br /><br />Extend BEM to meet AD extended at N<br />< AEN = 90 as I showed before<br />If < MBC = @ then we can show that < EDM = 90 - @<br /><br />So S(DEM) = 1/2 X 20 X 10 X sin (90-2@) = 100.cos 2@<br />= (1-tan^2 @)/(1 + tan^2 @) = (1 - (1/2)^2) / (1 + (1/2)^2) = 60<br /><br />So S(ABED) = S(ABCD) - S(BCM) - S(DEM)<br /> = 400 - 100 - 60 = 240<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.com