See complete Problem 103
Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Monday, May 19, 2008
Geometry Problem 103
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See complete Problem 103
Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
HINT: Draw equilateral triangles CDE, ADF, and BDG. Apply SAS congruent triangles.
ReplyDeletewhere are points E,F,G?
ReplyDeletethanks
E, F, and G are exterior points in the plane of the triangle ABC so that CEAFBG is a hexagon.
ReplyDeleteSome errors have crept into my earlier proof. Please ignore the same. It must read as follows: ABC is the given equilateral triangle with a point D inside which is d, e & f units away from A, B & C resply. Rotate the triangle anti-clockwise thru' 60 deg. so that C goes to B, B to F and D to E. Now we've: AD=AE=d and BE=BD=e while FE=CD=f. It's easy to see that triangles DAC and EAB are congruent (corresponding sides equal) and, therefore, /_DAC = /_EAB which makes /_DAE=60 deg and since AD=AE, we conclude that triangle ADE is equilateral.
ReplyDeleteNow Tr. DAC + Tr. DAB = Tr. EAB + Tr. DAB = Tr. ADE + Tr. EDB and Tr. EDB has sides of d, e & f while Tr ADE is equilateral with side d.
Hence, Tr. DAC + Tr. DAB=(3)^(1/2)*(d^2)/4 +(s(s-a)(s-b)(s-c))^(1/2)/4 using Heron's Formula where 2s=d+e+f.
We can likewise prove that, Tr, DAB + Tr DBC=(3)^(1/2)*(e^2)/4 + (s(s-a)(s-b)(s-c))^(1/2)/4 and Tr. DBC + DCA =(3)^(1/2)*(f^2)/4+(s(s-a)(s-b)(s-c))^(1/2)/4
Add LHS and RHS of the three equations to get:
2*[Tr. ABC]=(3)^(1/2)*(d^2)/4+(3)^(1/2)*(e^2)/4 +(3)^(1/2)*(f^2)/4+3(s(s-a)(s-b)(s-c))^(1/2)/4 Or [Tr. ABC] = S =(3)^(1/2)(d^2)/8 +(3)^(1/2)*(e^2)/8+( 3)^(1/2)(f^2)/8+(3/8)( s(s-a)(s-b)(s-c))^(1/2) Or S=(1/2){(3)^(1/2)(d^2)/4 +(3)^(1/2)(e^2)/4+(3)^(1/2)(f^2)/4+3(s(s-a)(s-b)(s-c))^(1/2)} whics was to be proven.
Ajit
http://img198.imageshack.us/img198/538/problem103.png
ReplyDeleteDraw equilateral triangles CDE, DBG and ADF and we have a hexagon CEBGAF ( see picture)
We have triangles (CDB) congruence to (AGB) case SAS
similarly triangles (AFC) congruence to triangles (ADB) and triangles (CDA) congruence to triangles (CEB)
Area(CEBGAF)= 2.* Area(ABC)=2S
We also have congruence triangles EDB, GAD, CDF ( case SSS) with sides d, e, f .
Area(CEBGAF)= area(CDE)+Area(DBG)+Area(ADF)+3.Area(EDB)=2S
= f^2*SQRT(3)/4+ e^2*SQRT(3)/4+ d^2*SQRT(3)/4+3SQRT(s(s-e)(s-f)(s-d))
Divide both sides by 2 we will get the result
Peter Tran